#  Asymptotic limits of the principal spectrum point of a nonlocal dispersal
cooperative system and application to a two-stage structured population model

Maria A. Onyido111Department of Mathematical Sciences, Northern Illinois
University, DeKalb, IL 60115, USA (monyido@niu.edu). , Rachidi B.
Salako222Department of Mathematical Sciences, University of Nevada Las Vegas,
NV 89154, USA (rachidi.salako@unlv.edu)., Markjoe O. Uba333Department of
Mathematical Sciences, Northern Illinois University, DeKalb, IL 60115, USA
(markjoeuba@gmail.com)., and Cyril I. Udeani444Department of Applied
Mathematics and Statistics, Comenius University in Bratislava, Mlynská dolina,
84248 Bratislava, Slovakia (cyril.izuchukwu@fmph.uniba.sk).

###### Abstract

This work examines the limits of the principal spectrum point, $\lambda_{p}$,
of a nonlocal dispersal cooperative system with respect to the dispersal
rates. In particular, we provide precise information on the sign of
$\lambda_{p}$ as one of the dispersal rates is : (i) small while the other
dispersal rate is arbitrary, and (ii) large while the other is either also
large or fixed. We then apply our results to study the effects of dispersal
rates on a two-stage structured nonlocal dispersal population model whose
linearized system at the trivial solution results in a nonlocal dispersal
cooperative system. The asymptotic profiles of the steady-state solutions with
respect to the dispersal rates of the two-stage nonlocal dispersal population
model are also obtained. Some biological interpretations of our results are
discussed.

Key words. Nonlocal-dispersal, stage-structured model, principal spectrum
point, steady state, asymptotic limit

AMS subject classifications. 92D40, 92D50, 35P15, 35K57

##  1 Introduction

Most population models typically investigate species under the assumption that
they share similar characteristics among its kind. However, biological species
could exhibit non-negligible stage-specific variations. Therefore, a stage-
structured model would help in incorporating stage-dependent physiological
parameters, thereby yielding better biological predictions of survival or
extinction. For example, many insect species pass from an egg stage into
several instar stages and then into the adult stage. For some studies on stage
structured populations, we refer interested readers to [5, 20] and the
references therein.

In this study, we consider the stage-structured population model with nonlocal
dispersal, introduced in [22], describing the dynamics of a species with
density function ${\bf u}(t,x)=(u_{1}(t,x),u_{2}(t,x))$ living in a bounded
habitat $\Omega\subset\mathbb{R}^{n}$ and structured in two stages: adults and
juveniles. Here, $u_{1}(t,x)$ denotes the density function of the juveniles
and $u_{2}(t,x)$ that of the adults who have attained reproductive maturity.
The adults have local reproductive rate $r(x)$, and the juveniles attain
reproductive maturity at the rate $s(x)$. The juveniles have local death rate
$a(x)$ and self-limitation rate $b(x)$ due to their size in space. The adults
have a local death rate $e(x)$ and local self-limitation rate $f(x)$ induced
by their size. Interaction between the adults and the juveniles may generate
an interspecific competition for local resources. We denote by $c(x)$ and
$g(x)$ the interspecific local competition rates of the adults and juveniles,
respectively. The following nonlocal system of PDE can be used to study the
dynamics of the species

$\begin{cases}\partial_{t}u_{1}=\mu_{1}\int_{\Omega}\kappa(x,y)(u_{1}(t,y)-u_{1}(t,x))dy+r(x)u_{2}-s(x)u_{1}-(a(x)+b(x)u_{1}+c(x)u_{2})u_{1}&x\in\Omega,\
t>0,\cr\partial_{t}u_{2}=\mu_{2}\int_{\Omega}\kappa(x,y)(u_{2}(t,y)-u_{2}(t,x))dy+s(x)u_{1}-(e(x)+f(x)u_{2}+g(x)u_{1})u_{2}&x\in\Omega,\
t>0,\end{cases}$ (1.1)

where $\mu_{1}$ and $\mu_{2}$ represent the dispersal rates. The nonlocal
dispersal operator in (1.1) is said to have Neumann-type boundary condition
(see [11] for more details). Nonlocal dispersal, modeled by an integral
operator like that appearing in system (1.1), is employed in modeling the
dynamics of species that exhibit long-range movement or position jump (see [2,
3, 10, 11, 12, 22, 23, 28, 29, 27] and the references therein for more
discussion on population models with nonlocal dispersal). Throughout this
study, we shall suppose that the following standing hypotheses hold for the
parameters in system (1.1):

(H1) $\kappa\in C(\bar{D}\times\bar{D})$, positive and symmetric, i.e.,
$\kappa(x,y)=\kappa(y,x)>0$ for every $x,y\in\bar{D}$.

(H2) The functions $a,b,c,e,f,g,r,$ and $s$ are Hölder continuous on
$\overline{\Omega}$ and nonnegative, $r$ and $s$ are non-identically zero, and
$b$ and $f$ are strictly positive.

Observe that ${\bf u}(t,x)=(0,0)$ is a soluton of (1.1), known as the trivial
solution. Linearizing (1.1) at this trivial solution yields the following
linear cooperative nonlocal system:

$\begin{cases}\partial_{t}U_{1}=\mu_{1}\int_{\Omega}\kappa(x,y)[U_{1}(t,y)-U_{1}(t,x)]dy+r(x)U_{2}-(a(x)+s(x))U_{1}&x\in\Omega,\
t>0,\cr\partial_{t}U_{2}=\mu_{2}\int_{\Omega}\kappa(x,y)[U_{2}(t,y)-U_{2}(t,x)]dy+s(x)U_{1}-e(x)U_{2}&x\in\Omega,\
t>0.\end{cases}$ (1.2)

Note that using equations (2.1), (2.4) and (2.5) below, (1.2) can be rewritten
as

$\frac{d{\bf U}}{dt}=\mu\circ\mathcal{K}({\bf U})+\mathcal{A}{\bf U}\quad
t>0.$ (1.3)

Denoting by $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ the principal
spectrum point of the linear operator $\mu\circ\mathcal{K}+\mathcal{A}$ (see
Definition (2.1)). It was established in [22] that the persistence and
extinction of the species modeled by (1.1) depend on the sign of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$. Hence, to understand the
effect of the dispersal rates on the species’ persistence, it is pertinent to
investigate the asymptotic dynamics of the principal spectrum point with
respect to the dispersal rates. Our first goal in this study is to examine
some characterizations of the principal spectrum point and determine its
behavior as the dispersal rates approach some critical values.

In general, cooperative nonlocal dispersal models appear in different
scenarios (see[16, 17] and the references therein), like the general two-
species Lotka-Volterra competition system (see [3] and the references
therein). We indicate that two-species competition systems can be transformed
into cooperative ones using standard arguments. Hence, our results in the
current work on $\lambda_{p}$ apply to that setting. For instance, [3]
established some criteria for the existence of principal eigenvalues of time-
periodic cooperative linear systems with nonlocal dispersal. Since the
principal eigenvalues (when it exists) coincide with the principal spectrum
point, our results in this study also apply to their study when the
coefficients are time-homogeneous.

For unstructured species modeled with nonlocal dispersal under a temporally
homogeneous but spatially heterogeneous environment with Neumann/Dirichlet-
type boundary conditions, Shen and Xie [28] examined the effects of the
dispersal rates on the asymptotic behavior of the principal spectrum point of
its linearization at the trivial solution. They obtained the limits of the
principal spectrum point as the dispersal rates go to zero or infinity. Hence,
our results on the limits of $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$
(see Theorem 2.2) extend their results to two species cooperative case.
Additionally, Shen and Vo [27] studied the unstructured model with Dirichlet-
type boundary conditions and extended some of the results of Shen and Xie to a
time-periodic environment.

It is pertinent to mention at this point that (1.1) is the nonlocal analog of
the following two-stage structured population system with random (local)
dispersal and Neumann boundary conditions

$\begin{cases}\partial_{t}u_{1}=\mu_{1}\Delta
u_{1}+r(x)u_{2}-s(x)u_{1}-(a(x)+b(x)u_{1}+c(x)u_{2})u_{1}&x\in\Omega,\
t>0,\cr\partial_{t}u_{2}=\mu_{2}\Delta
u_{2}+s(x)u_{1}-(e(x)+f(x)u_{2}+g(x)u_{1})u_{2}&x\in\Omega,\ t>0,\cr\nabla
u_{1}\cdot\nu=\nabla u_{2}\cdot\nu=0\;&x\in\partial\Omega,\;t>0,\end{cases}$
(1.4)

where $\nu$ is the outward unit normal to $\partial\Omega$ and the parameters
in (1.4) have the same meaning as those of (1.1). Cantrell, Cosner and Salako
[7] studied the effects of dispersal rates on the species modeled by (1.4) and
consequently obtained some results on the asymptotic behavior of solutions as
the dispersal rates go to zero or infinity. Our results extend theirs to the
nonlocal case. It is noteworthy to indicate that the semiflow generated by the
classical solution of system (1.1) is not compact. This noncompactness
introduces some difficulties in the analysis of the nonlocal system and most
of the techniques developed for (1.4) do not generally extend to (1.1).

When species persist and eventually stabilize, from an ecological viewpoint,
it is important to examine their spatial distributions for adequate
distribution of scarce resources. In this regard, the second goal of this
study is to apply the results obtained in the first part to establish the
asymptotic profiles of steady states of system (1.1). In particular, our
results in Theorems 2.5 provide complete information on the spatial
distribution of steady states of (1.1) when at least one of the dispersal
rates go to zero or infinity. It is appropriate to note here that [22] gave
some criteria for the existence, uniqueness and stability of steady states of
(1.1). Hence, our results in this regard complement those results.

The rest of the paper is organized as follows: Section 2 contains some
notations, definitions and our main results. We collect a few preliminary
results in section 3. The proofs of our main results are presented in sections
4–8.

##  2 Notations, Definitions and Main Results

###  2.1 Notations and Definitions

Let $X:=C(\overline{\Omega})$ denote the Banach space of uniformly continuous
and bounded functions on $\overline{\Omega}$ endowed with the sup-norm,
$\|u\|_{\infty}=\underset{x\in\Omega}{\sup}|u(x)|$ for every $u\in X$. Since
the density functions are nonnegative, we will be interested in the following
subsets of $X$:

$X^{+}=\\{u\in X\,|\,u(x)\geq 0,\quad x\in\bar{\Omega}\\},$

and

$X^{++}=\\{u\in X^{+}\,|\,\inf_{x\in\bar{\Omega}}u(x)>0\\}.$

We shall use bold-face letters to represent vectors. In particular, we use
${\bf u}=(u_{1},u_{2})$ to denote vectors in $\mathbb{R}^{2}$. For covenience,
we also use ${\bf u}=(u_{1},u_{2})$ for elements of $X\times X$. Hence, any
vector in $\mathbb{R}^{2}$ can be understood as a constant vector valued
function on $\overline{\Omega}$; this would not cause any confusion in the
paper. We endow $X\times X$ with the norm

$\|{\bf u}\|:=\max\\{\|u_{1}\|_{\infty},\|u_{2}\|_{\infty}\\}\quad\forall\
{\bf u}\in X\times X.$

Hence, $X\times X$ is also a Banach space. Given a function $h\in X$, define

$\hat{h}:=\frac{1}{|\Omega|}\int_{\Omega}h(x)dx,\quad
h_{\min}:=\min_{x\in\overline{\Omega}}h(x),\quad\text{and}\quad
h_{\max}=\max_{x\in\overline{\Omega}}h(x).$

Given a complex number $\lambda\in\mathbb{C}$, we denote by
$\mathcal{R}e(\lambda)$ its real part.

######  Definition 2.1 (Principal Spectrum Point).

Let $E$ be a Banach space and $B:dom(B)\to E$ be a linear map where $dom(B)$
is a linear subspace of $E$. Let $\sigma(B)$ denote the spectrum of the linear
map $B$. The principal spectrum point of $B$, denoted $\lambda_{p}(B)$, is
defined by

$\lambda_{p}(B):=\begin{cases}\sup\\{\mathcal{R}e(\lambda):\lambda\in\sigma(B)\\}&\text{if}\
\sigma(B)\neq\emptyset,\cr-\infty&\text{if}\ \sigma(B)=\emptyset.\end{cases}$

For our purpose in the current work, the Banach space in Definition 2.1 will
be either $E=X$ or $E=X\times X$, while the linear operator $B$ will always be
a bounded linear map on $E$. Hence, $\sigma(B)$ will always be nonempty and
bounded; so that the principal spectrum point $\lambda_{p}(B)$ is a real
number.

Given $h\in X$, consider the bounded linear operators $\mathcal{K},\
\mathcal{I},\ h\mathcal{I}\ :\ X\to X$ defined by

$\mathcal{K}u(x)=\int_{\Omega}\kappa(x,y)(u(y)-u(x))dy\quad\forall\ u\in\ X,\
x\in\overline{\Omega},$ (2.1) $\mathcal{I}u(x)=u(x)\quad\forall\ u\in X,\
x\in\overline{\Omega},$ (2.2) $(h\mathcal{I})u(x)=h(x)u(x)\qquad\forall\ u\in
X,\ x\in\overline{\Omega}.$ (2.3)

Next, given a bounded linear operator $\mathcal{B}$ on $X$ and a real vector
${\bf z}=(z_{1},z_{2})\in\mathbb{R}^{2}$, we denote by ${\bf
z}\circ\mathcal{B}$ the bounded linear operator on $X\times X$ given by

${\bf z}\circ\mathcal{B}{\bf
u}=\left(\begin{array}[]{c}z_{1}\mathcal{B}u_{1}\\\
z_{2}\mathcal{B}u_{2}\end{array}\right)\quad\forall\ {\bf u}\in X\times X.$
(2.4)

Define the operator $\mathcal{A}\ :\ X\times X\to X\times X$ by

$\mathcal{A}({\bf u})=A(x){\bf
u}=\left(\begin{array}[]{c}r(x)u_{2}-(a(x)+s(x))u_{1}\\\
su_{1}(x)-e(x)u_{2}(x)\end{array}\right)\quad\forall\ {\bf u}\in X\times X.$
(2.5)

where

$A(x)=\left(\begin{array}[]{cc}-(a(x)+s(x))&r(x)\\\
s(x)&-e(x)\end{array}\right)\quad\forall x\in\overline{\Omega}.$ (2.6)

For convenience, we introduce the matrix

$\hat{A}=\left(\begin{array}[]{cc}-(\hat{a}+\hat{s})&\hat{r}\\\
\hat{s}&-\hat{e}\end{array}\right),$ (2.7)

and let $\tilde{\Lambda}$ denote its maximal eigenvalue.

###  2.2 Main Results

We state our main results in this section. For every positive vector
${\bf\mu}=(\mu_{1},\,\mu_{2})\in\mathbb{R}^{+}\times\mathbb{R}^{+}$, let
$\sigma(\mu\circ\mathcal{K}+\mathcal{A})$ denote the spectrum of the bounded
linear operator $\mu\circ\mathcal{K}+\mathcal{A}$ and
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ its principal spectrum point.
If $\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ is an isolated
eigenvalue with a strictly positive eigenfunction, then
$\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ is called the
principal eigenvalue of $\mathcal{\mu\circ\mathcal{K}+\mathcal{A}}$. However,
in general, it is well-known that
$\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ may not be the
principal eigenvalue (see [12, 29] and the references therein). Thus, it is
essential to find some other useful ways to characterize
$\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ by comparing it to
some other relevant spectral quantities. This is accomplished in the next
subsection.

####  2.2.1 Characterization of $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$
in terms of generalized principal eigenvalues

Our first result in this subsection provides some useful characterizations of
the principal spectrum point
$\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ for two-species
nonlocal equations. Consider the following ordering in $X\times X$

${\bf u}\leq_{1}{\bf v}\;\text{if}\;u_{1}\leq v_{1}\ \text{and}\ u_{2}\leq
v_{2}.$ (2.8)

Note that for the positive constant vector ${\bf 1}:=(1,1)$, we have
$(\mu\circ\mathcal{K}+\mathcal{A})({\bf 1})=A(x){\bf 1}$ and

$-\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\}{\bf 1}\leq_{1}A(x){\bf
1}\leq_{1}\color[rgb]{0,0,0}\max\\{\|s\|_{\infty},\|r\|_{\infty}\\}{\bf
1}\quad\forall\ x\in\overline{\Omega}.$

Therefore, the quantities

$\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A}):=\sup\\{\lambda\in\mathbb{R}\ :\
\exists\varphi\in X^{++}\times X^{++}\ \text{satisfying}\
\lambda\varphi\leq_{1}(\mu\circ\mathcal{K}+\mathcal{A})(\varphi)\\}$ (2.9)

and

$\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A}):=\inf\\{\lambda\in\mathbb{R}\ :\
\exists\varphi\in X^{++}\times X^{++}\ \text{satisfying}\
(\mu\circ\mathcal{K}+\mathcal{A})(\varphi)\leq_{1}\lambda\varphi\\}$ (2.10)

are well-defined real numbers and satisfy

$-\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\}\leq\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})\leq\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})\leq\max\\{\|r\|_{\infty},\|s\|_{\infty}\\}.$
(2.11)

The fact that
$\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})\leq\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})$
is a consequence of the positivity of the uniformly continuous semigroup
$\\{{\bf U}^{\mu}(t)\\}_{t\geq 0}$ generated by the bounded linear operator
$\mu\circ\mathcal{K}+\mathcal{A}$. The quantities
$\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})$ and
$\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})$ are the generalized principal
eigenvalues of the linear operator $\mu\circ\mathcal{K}+\mathcal{A}$. The
following result holds.

######  Theorem 2.1.

Let $\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})$ and
$\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})$ be defined by (2.9) and (2.10),
respectively. Then

$\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})=\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A}).$

Theorem 2.1 shows that
$\lambda_{p}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ equals the
generalized principal eigenvalues
$\lambda_{*}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$ and
$\lambda^{*}(\mathcal{\mu\circ\mathcal{K}+\mathcal{A}})$; hence, it extends
the known result for the single species nonlocal equations (see [23, 4]) to
two species nonlocal equations. It would be of great interest to know whether
the conclusions of Theorem 2.1 hold in time-periodic and bounded spatially
heterogeneous environments or time-space periodic environments. We hope to
address these questions in our future works. The characterization of the
principal spectrum point as in Theorem 2.1 turns out to be an important tool
in the proofs of some of our main results on the limit of principal spectrum
point with respect to the dispersal rates. The next subsection discusses our
main results in this regard.

####  2.2.2  Asymptotic limits of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ with respect to $\mu$.

As established in [22, Theorem 1], the sign of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ completely determines the
persistence of the species. Hence, it is of great biological interest to
understand how the dispersal rate $\mu$ affect
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ since the influence of the
dispersal rates on the principal spectrum point translates to the effects of
dispersal rates on the species’ persistence and extinction. In this section,
we present several results on the limit of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ as $\mu$ approaches some
critical values. These results will help to precisely determine the behavior
of $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ in several instances.

Our first result in this section concerns the limit of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ for small $\mu$ and large
$\mu$.

######  Theorem 2.2.

  * (i)

For each $x\in\overline{\Omega}$, let $\Lambda(x)$ denote the maximal
eigenvalue of the cooperative matrix $A(x)$. Then

$\lim_{\max\\{\mu_{1},\mu_{2}\\}\to{0}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\Lambda_{\max}.$
(2.12)

  * (ii)

Let $\tilde{\Lambda}$ denote the maximal eigenvalue of the cooperative matrix
$\hat{A}$. Then

$\lim_{\min\\{\mu_{1},\mu_{2}\\}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\tilde{\Lambda}.$
(2.13)

Theorem 2.2 provides the exact information on the limits of the principal
spectrum point for large and small dispersal rates of the population. It is
important to indicate that $\tilde{\Lambda}$ is the maximum eigenvalue of the
cooperative matrix resulting from taking the spatial averages of the parameter
functions in the matrix $A(x)$, which is different from the average of the
function $\Lambda(x)$. This should not cause any confusion in the
presentation. We provide some remarks on the implications of the results.

######  Remark 2.1.

(i)

Thanks to Theorem 2.2-(i), the species persists for small dispersal rates if
and only if $\Lambda(x)>0$ for some $x\in\Omega$. Observe that $\Lambda(x)>0$
if and only if $r(x)s(x)>(s(x)+a(x))e(x)$. Note that $(a(x)+s(x))e(x)$ (resp.
$r(x)s(x)$ ) is the product of the loss (resp. growth) in the rates of change
of the density functions of the juveniles and adults. Hence, $\Lambda(x)>0$
indicates that, at such a location, the species are able to reproduce and grow
while having a small death rate. If the habitat has such locations, then small
dispersal rates will benefit species’ survival.

(ii)

Theorem 2.2-(ii) indicates that for large dispersal rates of the species,
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ approximate the maximal
eigenvalue of the matrix $\hat{A}$ defined by the spatial averages of the
parameters as defined by (2.7). Observe that $\tilde{\Lambda}>0$ if and only
if $\hat{r}\cdot\hat{s}-(\hat{a}+\hat{s})\hat{e}>0$. This shows that for large
dispersal rates, the persistence of the species depends on the spatial
averages of the functions $a,s,r$, and $e$ and not on their local spatial
distributions. It is important to note from Theorem 2.2-(i), that for small
dispersal rates, it is the local distributions of these functions that
determine the persistence of the species. Hence, if for example the functions
$r$ and $s$ have disjoint support with the product of their averages higher
than the product of the averages of $(a+s)$ and $e$, then the species will go
extinct if dispersing very slowly whereas they will persist if diffusing very
fast. In such scenario, small dispersal rates lead to extinction while large
dispersal rates leads to persistence. This is in contrast with the prediction
for persistence of species modeled with unstructured population models.

Now, we discuss the scenarios where one of the dispersal rate is small and the
other is fixed. For clarity in the statement of our results in these
scenarios, we first state the following proposition.

######  Proposition 2.1.

Let $h,l\in X^{+}$ and $\xi>0$ be given. The algebraic equation in $\nu$,

$0=\lambda_{p}\Big{(}\xi\mathcal{K}+\big{(}\frac{rs}{\nu+h}-l\big{)}\mathcal{I}\Big{)}-\nu,\quad\quad\
\nu>-h_{\min},$ (2.14)

has a (unique) zero, denoted by $\lambda_{\xi}^{h,l}$, if and only if

$\sup_{\nu>-h_{\min}}\Big{(}\lambda_{p}\Big{(}\xi\mathcal{K}+\big{(}\frac{rs}{\nu+h}-l\big{)}\mathcal{I}\Big{)}-\nu\Big{)}>0.$
(2.15)

Furthermore, $\lambda_{\xi}^{h,l}$, whenever it exists, has the same sign as
$\lambda_{p}\Big{(}\xi\mathcal{K}+\big{(}\frac{rs}{h}-l\big{)}\mathcal{I}\Big{)}$
if $h_{\min}>0$.

Thanks to Proposition (2.1), we can now state our result on the limit of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ as either $\mu_{1}\to 0$ or
$\mu_{2}\to 0$.

######  Theorem 2.3.

  * (i)

Let $\xi=\mu_{2}$, $h=a+s$ and $l=e$ in Proposition 2.1. Let
$\lambda_{\mu_{2}}^{a+s,e}$ denote the unique zero of the algebraic equation
(2.14) when inequality (2.15) holds, and
$\lambda_{\mu_{2}}^{a+s,e}:=-(a+s)_{\min}$ when (2.15) doesn’t hold. Then

$\lim_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})={\lambda}_{\mu_{2}}^{a+s,e}.$
(2.16)

  * (ii)

Let $\xi=\mu_{1}$, $h=e$ and $l=a+s$ in Proposition 2.1 . Let
$\lambda_{\mu_{1}}^{e,a+s}$ denote the unique solution of the algebraic
equation (2.14) when inequality (2.15) holds, and
$\lambda_{\mu_{1}}^{e,a+s}:=-e_{\min}$ when (2.15) doesn’t hold. Then

$\lim_{\mu_{2}\to
0^{+}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})={\lambda}_{\mu_{1}}^{e,a+s}.$
(2.17)

######  Remark 2.2.

Let us assume that $a+s$ is strictly positive on $\overline{\Omega}$ and fix
the dispersal rate $\mu_{2}$ of the adult.

  * (i)

Suppose that
$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}>0$.
It follows from Proposition 2.1 and Theorem 2.3-(i) that the species persists
for small dispersal rate of the juvenile. Precisely, there is
$\mu_{1,\mu_{2}}>0$ such that the species persists whenever
$\mu_{1}<\mu_{1,\mu_{2}}$. Observing that
$\lambda_{p}\Big{(}\tilde{\mu}_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}$
is non-increasing in $\tilde{\mu}_{2}$, (see [28, Theorem 2.2(1)]), then
$\lambda_{p}\Big{(}\tilde{\mu}_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}>0$
for every $\tilde{\mu}_{2}<\mu_{2}$. Therefore, it becomes an interesting
quest to know whether the positive number $\mu_{1,\mu_{2}}$ can be chosen such
that the species persists whenever the dispersal rates
$(\mu_{1},\tilde{\mu}_{2})$ satisfies $\mu_{1}<\mu_{1,\mu_{2}}$ and
$\tilde{\mu}_{2}<\mu_{2}$. This question will be completely settled by Theorem
2.6 below, which provides precise information on the sign of the principal
spectrum point when the dispersal rates are near the $\mu_{2}$-axis.

  * (ii)

If
$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}<0$,
it follows from Proposition 2.1 and Theorem 2.3-(i) that the species will go
extinct if the juveniles disperse very slowly. In this case, it is unclear
whether fast movement of the juvenile would be beneficial for the species
survival. So, a good understanding of the limit of the principal spectrum
point for large dispersal of only the juvenile is essential. This will be
investigated in the next result.

  * (iii)

Observe that
$\lambda_{p}\Big{(}\tilde{\mu}_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}\to\Big{(}\frac{rs}{a+s}-e\Big{)}_{\max}$
as $\tilde{\mu}_{2}\to 0$. Note also that
$\Big{(}\frac{rs}{a+s}-e\Big{)}_{\max}$ has the same sign as $\Lambda_{\max}$.
Hence the results of Theorems 2.2 and 2.3 on the persistence of the species
for small dispersal rates are consistent.

To discuss the limit of $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ as
either $\mu_{1}\to\infty$ or $\mu_{2}\to\infty$, we will need the following
intermediate result.

######  Proposition 2.2.

Fix $\xi>0$, $q,z\in X^{+}\setminus\\{0\\}$, and $l,h\in X^{+}$. The algebraic
equation in $\nu$,

$0=\int_{\Omega}q(x)(\nu\mathcal{I}-(\xi\mathcal{K}-l\mathcal{I}))^{-1}(z)(x)dx-\int_{\Omega}h(x)dx-\nu|\Omega|,\quad\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}),$
(2.18)

has a (unique) zero, denoted as $\tilde{\lambda}_{q,\xi,z}^{h,l}$, if and only
if

$\sup_{\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})}\Big{(}\int_{\Omega}q(x)(\nu\mathcal{I}-(\xi\mathcal{K}-l\mathcal{I}))^{-1}(z)(x)dx-\int_{\Omega}h(x)dx-\nu|\Omega|\Big{)}>0.$
(2.19)

If (2.19) holds, then

$\begin{cases}\tilde{\lambda}_{q,\xi,z}^{h,l}|\Omega|=\int_{\Omega}q\psi-\int_{\Omega}h\cr\tilde{\lambda}_{q,\xi,z}^{h,l}\psi=\xi\mathcal{K}(\psi)-l\psi+z\end{cases}$
(2.20)

where
$\psi=(\tilde{\lambda}_{q,\xi,z}^{h,l}\mathcal{I}-(\xi\mathcal{K}-l\mathcal{I}))^{-1}$.
Furthermore, the following holds.

  * (i)

If $\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$ is an eigenvalue of
$\xi\mathcal{K}-l\mathcal{I}$ with a positive eigenfunction, then (2.19)
holds.

  * (ii)

If $\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})<0$ and (2.19) holds, the unique
zero $\tilde{\lambda}_{q,\xi,p}^{h,l}$ has the same sign as the quantity
$\sigma_{q,\xi,z}^{h,l}$ defined by

$\sigma_{q,\xi,z}^{h,l}:=\int_{\Omega}q(x)(l\mathcal{I}-\xi\mathcal{K})^{-1}(z)(x)dx-\int_{\Omega}h(x)dx.$
(2.21)

The following result is concerned with the limit of
$\lambda_{p}(\mu\mathcal{K}+\mathcal{A})$ as either $\mu_{1}\to\infty$ or
$\mu_{2}\to\infty$.

######  Theorem 2.4.

  * (i)

Let $\xi=\mu_{2}$, $q=r$, $z=s$, $l=e$ and $h=a+s$ in Proposition 2.2. Let
$\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}$ denote the unique root of (2.18) when
(2.19) holds, and
$\tilde{\lambda}_{r,\mu_{2},s}^{a+s,e}=\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$
when (2.19) doesn’t hold. Then

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}.$
(2.22)

  * (ii)

Let $\xi=\mu_{1}$, $q=s$, $z=r$, $l=a+s$ and $h=e$ in Proposition 2.2. Let
$\tilde{\lambda}^{e,a+s}_{s,\mu_{1},r}$ denote the unique root of (2.18) when
(2.19) holds, and
$\tilde{\lambda}_{s,\mu_{1},r}^{e,a+s}=\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})$
when (2.19) doesn’t hold. Then

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\tilde{\lambda}^{e,a+s}_{s,\mu_{1},r}.$
(2.23)

The following result concerns the scenario where the members of one subgroup
of the population disperse very slowly while the members of the other subgroup
disperse very fast.

######  Theorem 2.5.

  * (i)

Let $\eta^{*}_{1}$ be given by

$\eta_{1}^{*}=\inf\left\\{\eta\in(-(a+s)_{\min},\infty)\ :\
\int_{\Omega}\left(\frac{rs}{\eta+a+s}-(e+\eta)\right)<0\right\\}.$ (2.24)

Then

$\lim_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\eta_{1}^{*}.$
(2.25)

Moreover $\eta_{1}^{*}$ and $\lim_{\eta\to
0^{+}}\int_{\Omega}\left(\frac{rs}{\eta+a+s}-(\eta+e)\right)$ have the same
sign.

  * (ii)

Let $\eta^{*}_{2}$ be given by

$\eta_{2}^{*}=\inf\left\\{\eta\in(-e_{\min},\infty)\ :\
\int_{\Omega}\left(\frac{rs}{\eta+e}-(a+s+\eta)\right)<0\right\\}.$ (2.26)

Then

$\lim_{\mu_{1}\to\infty,\mu_{2}\to
0}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\eta_{2}^{*}.$ (2.27)

Moreover $\eta_{2}^{*}$ and $\lim_{\eta\to
0^{+}}\int_{\Omega}\left(\frac{rs}{\eta+e}-(\eta+a+s)\right)$ have the same
sign.

Thanks to the above results on the limit of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ with respect to the dispersal
rates $\mu$, we can now provide a complete information on the sign of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ when $\mu$ is near either the
$\mu_{1}$-axis or $\mu_{2}$-axis in the $\mu_{1}\times\mu_{2}$-plane.

######  Theorem 2.6.

Let $\eta_{1}^{*}$ be given by Theorem 2.5-(i).

  * (i)

If $\eta_{1}^{*}>0$, then there exist $\delta_{1}^{*}>0$ and $m^{*}>0$ such
that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq m^{*}$ for every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1}^{*})\times\mathbb{R}_{+}$.

  * (ii)

If $\Lambda_{\max}<0$, then there exist $\delta_{1}^{*}>0$ and $m^{*}<0$ such
that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})<m^{*}$ for every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1}^{*})\times\mathbb{R}_{+}$.

  * (iii)

If $(a+s)_{\min}>0$ and $\eta_{1}^{*}<0<\Lambda_{\max}$, then for every
$0<\varepsilon\ll 1$ there exist $\delta_{1}^{*}>0$ and $m^{*}>0$ such that
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})>m^{*}$ for every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1}^{*})\times(0,\mu_{2}^{*}-\varepsilon)$
and $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})<-m^{*}$ for every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1}^{*})\times(\mu_{2}^{*}+\varepsilon,\infty),$
where $\mu_{2}^{*}>0$ is uniquely determined by
$\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=0$.

######  Remark 2.3.

Theorem 2.6 gives the sign of the principal spectrum point when the juvenile’s
dispersal rate is small. Similar result hold when the adult’s dispersal rate
is small. For simplicity, suppose that $e_{\min}>0$ and $(a+s)_{\min}>0$.
Following the theory of the basic reproduction number for population models or
infectious disease models [13, 14, 30], it can be shown using the next
generation matrix techniques that the quantity

$\mathcal{R}_{0}(x)=\frac{r(x)s(x)}{(a(x)+s(x))e(x)}\quad\forall\
x\in\overline{\Omega},$ (2.28)

is the local basic reproduction function for system (1.1). It tells us about
the local persistence of the population in the sense that if
$\mathcal{R}_{0}(x)>1$ at some location $x\in\Omega$, then the species can
persist in such location in the absence of dispersal. However, the species
dies out if not dispersing in the locations where $\mathcal{R}_{0}(x)<1$. It
is clear from the $\mathcal{R}_{0}$ expression given in (2.28) that when
$\min_{x\in\overline{\Omega}}\mathcal{R}_{0}(x)>1$, then
$\int_{\Omega}\Big{(}\frac{rs}{a+s}-e\Big{)}>0$ and
$\int_{\Omega}\Big{(}\frac{rs}{e}-(a+s)\Big{)}>0$. In this case, it follows
from Theorem 2.6 that there is $\eta^{*}>0$ such for any choice of dispersal
rate $\mu=(\mu_{1},\mu_{2})$ satisfying $\min\\{\mu_{1},\mu_{2}\\}<\eta^{*}$,
the species always persists. As a result, if the species’ local reproduction
function is bigger than one, the population persists as long as one of the
subgroups moves slowly.

####  2.2.3 Asymptotic profiles of steady states

For every $x\in\overline{\Omega}$, let ${\bf V}(x)=(V_{1}(x),V_{2}(x))$ denote
the unique nonnegative stable solution of the system of algebraic equations

$\begin{cases}0=r(x)V_{1}(x)-(a(x)+s(x)+b(x)V_{1}(x)+c(x)V_{2}(x))V_{1}(x)\cr
0=s(x)V_{1}(x)-(e(x)+f(x)V_{2}(x)+g(x)V_{1}(x))V_{2}(x).\end{cases}$ (2.29)

It follows from [22, Theorem 2] that ${\bf V}(x)$ is positive if and only if
$\Lambda(x)>0$. Thanks to [22, Theorem 1], system (1.1) has no positive
steady-state solution if $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq 0$.
Thus, by Theorem 2.2, to ensure that (1.1) has a positive steady state
solution for small dispersal rates $\mu$, it is necessary to assume that
$\Lambda_{\max}>0$. The following result demonstrates the asymptotic profiles
of positive steady-state solutions of (1.1) for small dispersal rates $\mu$.

######  Theorem 2.7.

Suppose that $\Lambda_{\max}>0$. Then every positive steady state solution
${\bf u}^{\mu}$ of (1.1) for small dispersal rate $\mu$ satisfies ${\bf
u}^{\mu}\to{\bf V}$ as $\max\\{\mu_{1},\mu_{2}\\}\to 0$, for $x$ uniformly in
$\Omega$.

Theorem 2.7 shows that spatial distribution of the steady-state solutions of
(1.1) for small dispersal rates of the population is completely determined by
the kinetic system (2.29). Our next result concerns the asymptotic profiles of
positive steady-state solutions of (1.1) for large dispersal rates, $\mu$.

######  Theorem 2.8.

Suppose that $\tilde{\Lambda}>0$. Then every positive steady state solution
${\bf u}^{\mu}$ of (1.1) for large dispersal rates $\mu$ satisfies ${\bf
u}^{\mu}\to\tilde{\bf V}$ as $\min\\{\mu_{1},\mu_{2}\\}\to{\infty}$, for $x$
uniformly in $\Omega$, where $\tilde{\bf V}$ is the unique positive solution
of

$\begin{cases}0=\hat{r}\cdot\tilde{V}_{1}-(\hat{a}+\hat{s}+\hat{b}\cdot\tilde{V}_{1}+\hat{c}\cdot\tilde{V}_{2})\tilde{V}_{1}\cr
0=\hat{s}\cdot\tilde{V}_{1}-(\hat{e}+\hat{f}\cdot\tilde{V}_{2}+\hat{g}\cdot\tilde{V}_{1})\tilde{V}_{2}.\end{cases}$
(2.30)

Theorem 2.8 shows that the steady-state solutions of (1.1) are uniformly
distributed on the whole habitat when the dispersal rates of the population
are significantly large. Moreover, their spatial distribution is determined by
that of the kinetic model for which the parameters are replaced by their
averages. The next result discusses the asymptotic profiles of positive steady
states when $\mu_{1}$ approaches zero for every fixed $\mu_{2}$. For
convenience, we introduce the function

$H(x,\tau)=\frac{1}{2b(x)}\Big{(}\sqrt{(a(x)+s(x)+c(x)\tau)^{2}+4b(x)r(x)\tau}-(a(x)+s(x)+c(x)\tau)\Big{)}\quad\forall\
\tau\geq 0,\ x\in\overline{\Omega}.$ (2.31)

The following result holds.

######  Theorem 2.9.

Fix $\mu_{2}>0$ and suppose that $(a+s)_{\min}>0$. Suppose also that
$\lambda_{\mu_{2}}^{a+s,e}>0$, where $\lambda_{\mu_{2}}^{a+s,e}$ is given by
Theorem 2.3-(i). Then any positive steady-state solution ${\bf u}^{\mu}$ of
(1.1) for small values of $\mu_{1}$ satisfies ${\bf
u}^{\mu}\to(H(\cdot,w^{*}),w^{*})$ as $\mu_{1}\to 0$ uniformly in $\Omega$
where $H$ is defined by (2.31) and $w^{*}\in X^{++}$ is the unique positive
solution of

$0=\mu_{2}\mathcal{K}w^{*}+sH(\cdot,w^{*})-(e+fw^{*}+gH(\cdot,w^{*}))w^{*}\quad
x\in\Omega.$ (2.32)

Note that similar result will hold if we fix $\mu_{1}$ and let $\mu_{2}\to 0$.
We complete our results with the asymptotic profiles of positive steady-state
solutions of (1.1) as one of the diffusion get arbitrarily large. In this
direction, the following result holds.

######  Theorem 2.10.

Fix $\mu_{2}>0$ and suppose that $e\in X^{+}\setminus\\{0\\}$. Suppose also
that $\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}>0$, where
$\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}$ is given by Theorem 2.4-(i). Then any
positive steady-state solution ${\bf u}^{\mu}$ of (1.1) for large values of
$\mu_{1}$ satisfies, up to a subsequence, ${\bf
u}^{\mu}\to(l^{*},\tilde{w}^{*})$ as $\mu_{1}\to\infty$ uniformly in $\Omega$
where $l^{*}>0$, $\tilde{w}^{*}\in X^{++}$ and satisfy

$\begin{cases}0=\int_{\Omega}r\tilde{w}^{*}-\Big{(}\int_{\Omega}(a+s)+l^{*}\int_{\Omega}b+\int_{\Omega}c\tilde{w}^{*}\Big{)}l^{*}\cr
0=\mu_{2}\mathcal{K}\tilde{w}^{*}+sl^{*}-(e+f\tilde{w}^{*}+gl^{*})\tilde{w}^{*}&x\in\Omega.\end{cases}$
(2.33)

Note that similar result will hold if we fix $\mu_{1}$ and let
$\mu_{2}\to\infty$. Theorems 2.9 and 2.10 provide information on the spatial
distribution of the steady state solution of system 1.1 when one of the
dispersal rate is fixed while the other is either very small or very large.
The limiting equation (2.32) satisfied by $w^{*}$, is uniquely solvable for a
positive solution. We expect that such information could be used to obtain
some partial results on the uniqueness and (linear) stability of steady-state
solutions of system (1.1) when one dispersal rate is sufficiently small.
However, due to the nonlocal terms in (2.33), we could not establish the
uniqueness of its solution. This highlights some of the difficulties
associated with the questions of uniqueness and stability of steady-state
solutions of (1.1). We hope that the results established in the current study
will be of great help in tackling these open questions.

##  3 Preliminaries

In the current section, we collect some preliminary results needed for
establishing our main results in the subsequent sections. Recall that the
linearization of (1.1) at the trivial solution was given by (1.2). Let
$\\{{\bf U}^{\mu}(t)\\}_{t\geq 0}$ denote the uniformly continuous semigroup
generated by the bounded linear operator $\mu\circ\mathcal{K}+\mathcal{A}$ on
${X}\times{X}$. Hence, for every ${\bf U}_{0}\in X\times X$, ${\bf U}(t,{\bf
U}_{0}):={\bf U}^{\mu}(t){\bf U}_{0}$ is the unique solution of (1.3),
equivalently (1.2), with initial data ${\bf U}_{0}$. Since $A(x)$ is
cooperative for each $x\in\overline{\Omega}$, the uniformly continuous
semigroup $\\{{\bf U}^{\mu}(t)\\}_{t\geq 0}$ is strongly positive, in the
sense that

${\bf U}^{\mu}(t)(X^{+}\times X^{+})\subset X^{+}\times
X^{+}\quad\text{and}\quad{\bf U}^{\mu}(t)((X^{+}\times X^{+})\setminus\\{{\bf
0}\\})\subset X^{++}\times X^{++}\quad\forall\ t>0.$ (3.1)

The following result shows that the principal spectrum point
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ equals the exponential growth
bound of the uniformly continuous semigroup $\\{{\bf U}^{\mu}(t)\\}_{t\geq
0}$. This result turns out to be of particular importance in the arguments
used to prove some of our main results.

######  Proposition 3.1.

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\in\sigma(\mu\circ\mathcal{K}+\mathcal{A})$
and

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\lim_{t\to\infty}\frac{\ln(\|{\bf
U}^{\mu}(t)\|)}{t}.$ (3.2)

###### Proof.

See [22, Proposition A1]. ∎

Let $\\{e^{t\mathcal{K}}\\}_{t\geq 0}$ denote the uniformly continuous
semigroup of bounded linear operators generated by the bounded linear operator
$\mathcal{K}$ on $X$. This means that given any $u_{0}\in X$,
$u(t,x;u_{0}):=(e^{t\mathcal{K}}u_{0})(x)$ is the unique solution of

$\begin{cases}\partial_{t}u=\mathcal{K}u(t,x)&x\in\bar{\Omega},\cr
u(0,\cdot)=u_{0}(x)&x\in\bar{\Omega}.\end{cases}$ (3.3)

By the comparison principle for linear nonlocal operators,
$\\{e^{t\mathcal{K}}\\}_{t\geq 0}$ is strongly monotone, in the sense that

$e^{t\mathcal{K}}u<e^{t\mathcal{K}}v\quad\text{for all}\ t>0\ \text{whenever
}\ u\leq\not\equiv v,\ u,v\in X.$ (3.4)

Let

$\beta_{*}:=\inf_{u\in L^{2}(\Omega)\setminus\\{0\\},\
\int_{\Omega}u=0}\frac{\frac{1}{2}\int_{\Omega}\int_{\Omega}\kappa(x,y)(u(x)-u(y))^{2}dydx}{\int_{\Omega}u^{2}(x)dx}.$
(3.5)

It follows from [10] that $\beta_{*}$ is a positive number. With respect to
the asymptotic behavior of solutions of (3.3), the following result holds.

######  Lemma 3.1 ([10]).

For every $u_{0}\in L^{2}(\Omega)$, it holds that

$\int_{\Omega}e^{t\mathcal{K}}u_{0}=\int_{\Omega}u_{0}\quad\forall\ t>0,$
(3.6)

hence $\\{e^{t\mathcal{K}}\\}_{t\geq 0}$ leaves invariant the Banach space
$\\{u\in L^{2}(\Omega):\int_{\Omega}u=0\\}$. Furthermore,

$\Big{\|}e^{t\mathcal{K}}u-\frac{1}{|\Omega|}\int_{\Omega}u\Big{\|}_{L^{2}(\Omega)}\leq
e^{-t\beta_{*}}\Big{\|}u-\frac{1}{|\Omega|}\int_{\Omega}u\Big{\|}_{L^{2}(\Omega)}\quad\forall\
t>0,\ u\in L^{2}(\Omega),$ (3.7)

and

$\Big{\|}u-\frac{1}{|\Omega|}\int_{\Omega}u\Big{\|}_{L^{2}(\Omega)}^{2}\leq\frac{1}{2\beta_{*}}\int_{\Omega}\int_{\Omega}\kappa(x,y)(u(x)-u(y))^{2}dydx\quad
u\in L^{2}(\Omega).$ (3.8)

For convenience, we define

$K(x)=\int_{\Omega}\kappa(x,y)dy\quad x\in\overline{\Omega}.$ (3.9)

By hypothesis (H1), $K$ is Hölder continuous on $X$ and $K\in X^{++}$; hence,
$K_{\min}>0.$ Next, let $\xi>0$ and $h\in X$ be fixed. Let
$\lambda_{p}(\xi\mathcal{K}-h\mathcal{I})$ denote the principal spectrum point
of the bounded linear operator $\xi\mathcal{K}-h\mathcal{I}$ on $X$. Thus, as
in Proposition 3.1,
$\lambda_{p}(\xi\mathcal{K}-h\mathcal{I})\in\sigma(\xi\mathcal{K}-h\mathcal{I})$
and

$\lambda_{p}(\xi\mathcal{K}-h\mathcal{I})=\lim_{t\to\infty}\frac{\ln\big{(}\big{\|}e^{t(\xi\mathcal{K}-h\mathcal{I})}\big{\|}\big{)}}{t},$
(3.10)

where $\\{e^{t(\xi\mathcal{K}-h\mathcal{I})}\\}_{t\geq 0}$ denotes the
strongly positive uniformly continuous semigroup generated by the bounded
linear operator $\xi\mathcal{K}-h\mathcal{I}$ on $X$. Note that for every
$\nu>\lambda_{p}(\xi\mathcal{K}-h\mathcal{I})$, the bounded linear operator
$\nu\mathcal{I}-(\xi\mathcal{K}-h\mathcal{I})$ is invertible. Denote by
$\Psi_{\xi,h}(\cdot,\nu,\cdot)=(\nu\mathcal{I}-(\xi\mathcal{K}-h\mathcal{I}))^{-1}$
the inverse operator of $\nu\mathcal{I}-(\xi\mathcal{K}-h\mathcal{I})$, that
is for every $w\in X$, $\Psi_{\xi,h}(x,\nu,w)$ is the unique solution of the
equation

$\nu\Psi=\xi\mathcal{K}\Psi-h\Psi+w.$ (3.11)

Note that since (3.10) holds (see also [18]), then

$\Psi_{\xi,h}(\cdot,\nu,w)=\int_{0}^{\infty}e^{-\nu
t}e^{t(\xi\mathcal{K}-h\mathcal{I})}wdt\quad\forall\ w\in X,$ (3.12)

where the above integral is defined with respect to the uniform topology.

##  4  Proofs of Theorems 2.1 and 2.2

This section is devoted for the proofs of Theorems 2.1 and 2.2.

###### Proof of Theorem 2.1.

The proof is divided into two steps.

Step 1. In this step, we show that

$\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A}).$
(4.1)

It is clear from the positivity of the uniformly continuous semigroup $\\{{\bf
U}^{\mu}(t)\\}_{t\geq 0}$ that
$\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})\geq\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$.
Now, let $\varepsilon>0$ be fixed. Note from the definition of
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ that
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon$ belongs to the
resolvent set of the bounded linear operator
$\mu\circ\mathcal{K}+\mathcal{A}$. Furthermore, by (3.2), it follows from the
Hille–Yosida theorem that
$((\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\mathcal{I}-(\mu\circ\mathcal{K}+\mathcal{A}))^{-1}$
is the Laplace transform of the positive uniformly continuous semigroup
$\\{{\bf U}^{\mu}(t)\\}_{t\geq 0}$, where
$((\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\mathcal{I}-(\mu\circ\mathcal{K}+\mathcal{A}))^{-1}$
denotes the inverse of the bounded linear operator
$(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\mathcal{I}-(\mu\circ\mathcal{K}+\mathcal{A})$.
Hence, for every $\psi\in X^{++}\times X^{++}$, we have that
$((\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\mathcal{I}-(\mu\circ\mathcal{K}+\mathcal{A}))^{-1}\psi\in
X^{++}\times X^{++}$. In particular for $\psi={\bf 1}$,

$\varphi:=((\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\mathcal{I}-(\mu\circ\mathcal{K}+\mathcal{A}))^{-1}{\bf
1}\in X^{++}\times X^{++}$ (4.2)

and

$(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\varphi=(\mu\circ\mathcal{K}+\mathcal{A})\varphi+{\bf
1}>(\mu\circ\mathcal{K}+\mathcal{A})\varphi.$ (4.3)

Therefore,
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon\geq\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})$.
Since $\varepsilon>0$ was arbitrarily chosen, then
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})$,
which completes the proof of Step 1.

Step 2. In the current step, we show that

$\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A}).$
(4.4)

To this end, let
$\sigma:=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})-\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})$.
It is clear from (2.11) and (4.1) that $\sigma\geq 0$. We shall show that
$\sigma=0$. Suppose on the contrary that $\sigma>0$. Hence,

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A}-\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A}){\bf
1}\circ\mathcal{I})=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})-\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})=\sigma>0.$
(4.5)

Therefore, it follows from [22, Theorem 4-(i)] that when $\mathcal{A}$ is
replaced with $\mathcal{A}-\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A}){\bf
1}\circ\mathcal{I}$ in system (1.1) with $c=g\equiv 0$, there is a positive
steady-state solution ${\bf u}^{**}$ of the system

$\begin{cases}0=\mu_{1}\mathcal{K}u_{1}^{**}+ru_{2}^{**}-(s+a+\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})+bu_{1}^{**})u_{1}^{**}&x\in\overline{\Omega},\cr
0=\mu_{2}\mathcal{K}u_{2}^{**}+su_{1}^{**}-(e+\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})+fu_{2}^{**})u_{2}^{**}&x\in\overline{\Omega}.\end{cases}$
(4.6)

Hence, with
$\varepsilon^{*}:=\min_{x\in\overline{\Omega}}\min\\{b(x)u_{1}^{**}(x),f(x)u_{2}^{**}(x)\\}>0$,
it follows from (4.6) that

$(\varepsilon^{*}+\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})){\bf
u}^{**}\leq(\mu\circ\mathcal{K}+\mathcal{A}){\bf u}^{**},$

which implies that
$\varepsilon^{*}+\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})\leq\lambda_{*}(\mu\circ\mathcal{K}+\mathcal{A})$.
This is impossible since $\varepsilon^{*}>0$. Therefore, $\sigma=0$, hence
(4.4) holds. This completes the proof of the Theorem.

∎

Now, for every $x\in\overline{\Omega}$, let $\Lambda(x)$ denote the maximal
eigenvalue of the matrix $A(x)$ and ${\bf Q}(x)$ be an associated nonnegative
eigenvector satisfying $Q_{1}(x)+Q_{2}(x)=1$. By direct computations,

$\Lambda(x)=\frac{1}{2}\Big{(}\sqrt{(a(x)+s(x)-e(x))^{2}+4r(x)s(x))}-(a(x)+s(x)+e(x))\Big{)}\quad\forall\
x\in\overline{\Omega},$ (4.7)

and $Q_{1}(x)$ and $Q_{2}(x)$ satisfy

$\displaystyle(\Lambda(x)+a(x)+s(x))Q_{1}(x)=r(x)Q_{2}(x)\quad\forall\
x\in\overline{\Omega}.$ (4.8)

Hence, since $Q_{1}(x)=1-Q_{2}(x)$,

$\Lambda(x)+a(x)+s(x)=(r(x)+\Lambda(x)+a(x)+s(x))Q_{2}(x).$ (4.9)

However,

$\displaystyle\Lambda(x)+a(x)+s(x)=\frac{1}{2}\Big{(}\sqrt{(a(x)+s(x)-e(x))^{2}+4r(x)s(x)}+(a(x)+s(x)-e(x))\Big{)}.$
(4.10)

Hence, if $r(x)s(x)>0$, we have that $r(x)+a(x)+s(x)>0$, in which case it
follows from (4.9) that

$Q_{2}(x)=\frac{\Lambda(x)+a(x)+s(x)}{r(x)+\Lambda(x)+a(x)+s(x)}>0$ (4.11)

and

$Q_{1}(x)=\frac{r(x)}{r(x)+\Lambda(x)+a(x)+s(x)}>0.$ (4.12)

In view of (4.11) and (4.12), if $r,s\in X^{++}$, then ${\bf Q}\in
X^{++}\times X^{++}$.

######  Lemma 4.1.

Let $\Lambda(x)$ be the maximal eigenvalue of the cooperative matrix $A(x)$,
then,

$\limsup_{\max\\{\mu_{1},\mu_{2}\\}\to{0}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\Lambda_{\max}.$
(4.13)

###### Proof.

Let $\varepsilon>0$ be fixed and set $r^{\varepsilon}=r+\varepsilon$ and
$s^{\varepsilon}=s+\varepsilon$. Next, let ${\bf Q}^{\varepsilon}(x)$ be given
by (4.11) and $\eqref{HK13}$ where $r$ and $s$ are replaced with
$r^{\varepsilon}$ and $s^{\varepsilon}$. Then, there is $\sigma_{0}>0$ such
that

$(\Lambda_{\max}^{\varepsilon}+2\varepsilon){\bf
Q}^{\varepsilon}\geq(\mu\circ\mathcal{K}+\mathcal{A}){\bf
Q}^{\varepsilon}\quad\forall\ 0<\mu_{1},\mu_{2}\leq\sigma_{0},$ (4.14)

where $\Lambda^{\varepsilon}(x)$ is given by (4.7) with $r$ and $s$ replaced
with $r^{\varepsilon}$ and $s^{\varepsilon}$. To see this, observe that from
(4.11) and (4.12), we know that ${\bf Q}^{\varepsilon}\in X^{++}\times
X^{++}.$ Hence, by computations

$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)(Q_{1}^{\varepsilon}(y)-Q_{1}^{\varepsilon}(x))dy+r(x)Q_{2}^{\varepsilon}(x)-(a(x)+s(x))Q_{1}^{\varepsilon}(x)$
$\displaystyle=$
$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)(Q_{1}^{\varepsilon}(y)-Q_{1}^{\varepsilon}(x))dy+\Big{(}r^{\varepsilon}(x)Q_{2}^{\varepsilon}(x)-(a(x)+s^{\varepsilon}(x))Q_{1}^{\varepsilon}(x)\Big{)}+\varepsilon(Q_{1}^{\varepsilon}(x)-Q_{2}^{\varepsilon}(x))$
$\displaystyle=$
$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)(Q_{1}^{\varepsilon}(y)-Q_{1}^{\varepsilon}(x))dy+\Lambda^{\varepsilon}(x)Q_{1}^{\varepsilon}(x)+\varepsilon(Q_{1}^{\varepsilon}(x)-Q_{2}^{\varepsilon}(x))$
$\displaystyle\leq$
$\displaystyle\mu_{1}\|K\|_{\infty}\|Q_{1}^{\varepsilon}\|_{\infty}|\Omega|+(\Lambda_{\max}^{\varepsilon}+\varepsilon)Q_{1}^{\varepsilon}(x)$
$\displaystyle\leq$
$\displaystyle(\Lambda_{\max}^{\varepsilon}+2\varepsilon)Q_{1}^{\varepsilon}(x)$

whenever $\mu_{1}\leq\frac{\varepsilon
Q_{1,\min}^{\varepsilon}}{\|K\|_{\infty}\|Q_{1}^{\varepsilon}\|_{\infty}|\Omega|}$.
Similar computations show that

$\mu_{2}\int_{\Omega}\kappa(x,y)(Q_{2}^{\varepsilon}(y)-Q_{2}^{\varepsilon}(x))dy+s(x)Q_{1}^{\varepsilon}(x)-e(x)Q_{2}^{\varepsilon}(x)\leq(\Lambda_{\max}^{\varepsilon}+2\varepsilon)Q_{2}^{\varepsilon}(x)$

whenever $\mu_{2}\leq\frac{\varepsilon
Q_{2,\min}^{\varepsilon}}{\|K\|_{\infty}\|Q_{2}^{\varepsilon}\|_{\infty}|\Omega|}$.
Whence, taking
$\sigma_{0}:=\frac{\varepsilon}{\|K\|_{\infty}|\Omega|}\min\Big{\\{}\frac{Q_{1,\min}^{\varepsilon}}{\|Q_{1}^{\varepsilon}\|_{\infty}},\frac{Q_{2,\min}^{\varepsilon}}{\|Q_{2}^{\varepsilon}\|_{\infty}}\Big{\\}}$,
then (4.14) follows. Now, from (4.14), since ${\bf Q}^{\varepsilon}\in
X^{++}\times X^{++}$, then by Theorem 2.1,

$\Lambda_{\max}^{\varepsilon}+2\varepsilon\geq\lambda^{*}(\mu\circ\mathcal{K}+\mathcal{A})=\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\quad\forall\
0<\mu_{1},\mu_{2}<\sigma_{0}.$

Since $\varepsilon$ is arbitrarily chosen and $\lim_{\varepsilon\to
0^{+}}(\Lambda_{\max}^{\varepsilon}+2\varepsilon)=\Lambda_{\max}$, (4.13) then
follows. ∎

######  Lemma 4.2.

Let $\Lambda(x)$ be the maximal eigenvalue of the cooperative matrix $A(x)$,
then,

$\Lambda_{\max}\leq\liminf_{\max\\{\mu_{1},\mu_{2}\\}\to{0}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A}).$
(4.15)

###### Proof.

Let $\varepsilon_{0}>0$ be fixed. Then there is $x_{0}\in\Omega$ and $r_{0}>0$
such that

$B(x_{0},4r_{0})\subset\Omega\quad\text{and}\quad\Lambda(x)\geq\Lambda_{\max}-\varepsilon_{0}\quad\forall\
x\in B(x_{0},4r_{0}).$ (4.16)

Now, choose $\psi\in C^{\infty}(\mathbb{R}^{n})$ satisfying

$0\leq\psi\leq 1,\quad\psi\equiv 1\ \text{on}\
B(x_{0}.r_{0})\quad\text{and}\quad\psi\equiv 0\ \text{on}\
\mathbb{R}^{n}\setminus B(x_{0},2r_{0}).$ (4.17)

We claim that

$(\Lambda_{\max}-\varepsilon_{0}-(\mu_{1}+\mu_{2})\|K\|_{\infty})\psi{\bf
Q}\leq(\mu\circ\mathcal{K}+\mathcal{A})\psi{\bf Q}\quad\forall\ \mu=(\mu_{1},\
\mu_{2}):\mu_{1},\mu_{2}>0.$ (4.18)

Indeed, by computations,

$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)(\psi(y)Q_{1}(y)-Q_{1}(x)\psi(x))dy+r(x)Q_{2}(x)\psi(x)-(a(x)+s(x))Q_{1}(x)\psi(x)$
$\displaystyle\geq$
$\displaystyle-\mu_{1}Q_{1}(x)\psi(x)\|K\|_{\infty}+(r(x)Q_{2}(x)-(a(x)+s(x))Q_{1}(x))\psi(x)$
$\displaystyle=$
$\displaystyle-\mu_{1}Q_{1}(x)\psi(x)\|K\|_{\infty}+\Lambda(x)Q_{1}(x)\psi(x)$
$\displaystyle\geq$
$\displaystyle-\mu_{1}Q_{1}(x)\psi(x)\|K\|_{\infty}+(\Lambda_{\max}-\varepsilon_{0})Q_{1}(x)\psi(x)$

since (4.16) and (4.17) hold. Hence

$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)(\psi(y)Q_{1}(y)-Q_{1}(x)\psi(x))dy+r(x)Q_{2}(x)\psi(x)-(a(x)+s(x))Q_{1}(x)\psi(x)$
$\displaystyle\geq$
$\displaystyle(\Lambda_{\max}-\varepsilon_{0}-(\mu_{1}+\mu_{2})\|K\|_{\infty})Q_{1}(x)\psi(x)\quad\forall\
x\in\overline{\Omega}.$

Similar arguments show that

$\displaystyle\mu_{2}\int_{\Omega}\kappa(x,y)(\psi(y)Q_{2}(y)-Q_{2}(x)\psi(x))dy+s(x)Q_{1}(x)\psi(x)-e(x)Q_{2}(x)\psi(x)$
$\displaystyle\geq$
$\displaystyle(\Lambda_{\max}-\varepsilon_{0}-(\mu_{1}+\mu_{2})\|K\|_{\infty})Q_{1}(x)\psi(x)\quad\forall\
x\in\overline{\Omega},$

hence (4.18) holds. Now, since $\\{{\bf U}^{\mu}(t)\\}_{t\geq 0}$ is positive
and $\frac{\ln(\|{\bf
U}^{\mu}(t)\|)}{t}\to\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ as
$t\to\infty$ (Proposition 3.1), it follows from (4.18) that

$\Lambda_{\max}-\varepsilon_{0}-(\mu_{1}+\mu_{2})\|K\|_{\infty}\leq\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\quad\forall\
\varepsilon_{0},\ \mu_{1},\ \mu_{2}>0.$

This implies that (4.15) holds since $\varepsilon_{0}$ is arbitrarily chosen.
∎

For every $\mu=(\mu_{1},\mu_{2})\in\mathbb{R}^{+}\times\mathbb{R}^{+}$, let
$\Lambda_{\mu}(x)$ denote the maximal eigenvalue of the cooperative matrix

$A_{\mu}(x):=A(x)-K(x)\text{Diag}(\mu)$ (4.19)

where $\text{Diag}(\mu)$ is the $2\times 2$ diagonal matrix with diagonal
entries $\mu_{1}$ and $\mu_{2}$. Hence,

$\Lambda_{\mu}(x)\leq\Lambda(x)-\min\\{\mu_{1},\mu_{2}\\}K_{\min}\leq\Lambda_{\max}-\min\\{\mu_{1},\mu_{2}\\}K_{\min}.$
(4.20)

######  Lemma 4.3.

For every $\mu$ satisfying
$\min\\{\mu_{1},\mu_{2}\\}>\frac{\Lambda_{\max}+\max\\{\|e\|_{\infty},\|a+s\|_{\infty}\\}}{K_{\min}}$,
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ is the principal eigenvalue of
the bounded linear operator $\mu\circ\mathcal{K}+\mathcal{A}$.

###### Proof.

For every $\mu$ satisfying
$\frac{\Lambda_{\max}+\max\\{\|e\|_{\infty},\|a+s\|_{\infty}\\}}{K_{\min}}<\min\\{\mu_{1},\mu_{2}\\}$,
it follows from inequalities (2.11) and (4.20) and Theorem 2.1 that
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})>\max_{x\in\overline{\Omega}}\Lambda_{\mu}(x)$,
hence the result follows from [3, Theorem 2.1]. ∎

###### Proof of Theorem 2.2 .

(i) It follows from Lemmas 4.1 and 4.2.

(ii)

For every $\mu=(\mu_{1},\mu_{2})$ satisfying
$\min\\{\mu_{1},\mu_{2}\\}>\frac{\Lambda_{\max}+\max\\{\|e\|_{\infty},\|a+s\|_{\infty}\\}}{K_{\min}}$,
by Lemma 4.3, $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ is the principal
eigenvalue of the bounded linear operator $\mu\circ\mathcal{K}+\mathcal{A}$;
so we can choose an associated strictly positive eigenfunction $\varphi^{\mu}$
satisfying

$\|\varphi_{1}^{\mu}\|_{L^{2}(\Omega)}+\|\varphi_{2}^{\mu}\|_{L^{2}(\Omega)}=1.$
(4.21)

From this point, we set
$w_{i}^{\mu}=\varphi_{i}^{\mu}-\frac{1}{|\Omega|}\int_{\Omega}\varphi_{i}^{\mu}$,
$i=1,2$. The rest of the proof is divided into two steps.

Step 1. In the current step, we shall show that

$\lim_{\mu\to\infty}\sum_{i=1}^{2}\|w_{i}^{\mu}\color[rgb]{0,0,0}\|_{L^{2}(\Omega)}=0.$
(4.22)

To this end, we first note from Lemma 3.1 that

$2\beta_{*}\|w_{i}^{\mu}\|_{L^{2}(\Omega)}^{2}\leq\int_{\Omega}\int_{\Omega}\kappa(x,y)(\varphi_{i}^{\mu}(y)-\varphi_{i}^{\mu}(x))^{2}dydx,\quad
i=1,2.$

Therefore,

$\displaystyle\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\int_{\Omega}(\varphi_{1}^{\mu})^{2}=$
$\displaystyle\mu_{1}\int_{\Omega}\int_{\Omega}\kappa(x,y)(\varphi_{1}^{\mu}(y)-\varphi_{1}^{\mu}(x))\varphi_{1}^{\mu}(x)dydx+\int_{\Omega}r\varphi_{1}^{\mu}\varphi_{2}^{\mu}-\int_{\Omega}(a+s)(\varphi_{1}^{\mu})^{2}$
$\displaystyle=$
$\displaystyle-\frac{\mu_{1}}{2}\int_{\Omega}\int_{\Omega}\kappa(x,y)(\varphi_{1}^{\mu}(y)-\varphi_{1}^{\mu}(x))^{2}dydx+\int_{\Omega}r\varphi_{1}^{\mu}\varphi_{2}^{\mu}-\int_{\Omega}(a+s)(\varphi_{1}^{\mu})^{2}$
$\displaystyle\leq$
$\displaystyle-\mu_{1}\beta_{*}\|w_{1}^{\mu}\|_{L^{2}(\Omega)}^{2}+\int_{\Omega}r\varphi_{1}^{\mu}\varphi_{2}^{\mu}-\int_{\Omega}(a+s)(\varphi_{1}^{\mu})^{2}.$

Hence, in view of (2.11) and (4.21) and the fact that $\varphi_{i}^{\mu}>0$
for each $i=1,2$, we obtain that

$\displaystyle\Big{\|}w_{1}^{\mu}\|_{L^{2}(\Omega)}^{2}\leq$
$\displaystyle\frac{1}{\mu_{1}\beta_{*}}\left(\int_{\Omega}r\varphi_{1}^{\mu}\varphi_{2}^{\mu}-\int_{\Omega}(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+(a+s))(\varphi_{1}^{\mu})^{2}\right)$
(4.23) $\displaystyle\leq$
$\displaystyle\frac{1}{\mu_{1}\beta_{*}}\left(\frac{\|r\|_{\infty}}{2}\|\varphi_{2}^{\mu}\|_{L^{2}(\Omega)}^{2}+\frac{\|r\|_{\infty}-2\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})}{2}\|\varphi_{1}^{\mu}\|_{L^{2}(\Omega)}^{2}\right)$
(4.24) $\displaystyle\leq$
$\displaystyle\frac{1}{\mu_{1}\beta_{*}}\left(\frac{\|r\|_{\infty}}{2}\|\varphi_{2}^{\mu}\|_{L^{2}(\Omega)}^{2}+\frac{\|r\|_{\infty}+2\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\}}{2}\|\varphi_{1}^{\mu}\|_{L^{2}(\Omega)}^{2}\right)$
(4.25) $\displaystyle\leq$
$\displaystyle\frac{1}{\mu_{1}\beta_{*}}\left(\frac{\|r\|_{\infty}}{2}+\frac{\|r\|_{\infty}+2\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\}}{2}\right)$
(4.26) $\displaystyle=$
$\displaystyle\frac{(\|r\|_{\infty}+\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\})}{\mu_{1}\beta_{*}}.$
(4.27)

Similar arguments show that

$\Big{\|}w_{2}^{\mu}\|_{L^{2}(\Omega)}^{2}\leq\frac{(\|s\|_{\infty}+\max\\{\|a+s\|_{\infty},\|e\|_{\infty}\\})}{\mu_{2}\beta_{*}}.$
(4.28)

Therefore (4.22) holds from the last two inequalities.

Step 2. In the current step, we complete the proof of the theorem. Observe
that

$\begin{cases}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\left[\frac{1}{|\Omega|}\int_{\Omega}\varphi_{1}^{\mu}\right]=\frac{1}{|\Omega|}\int_{\Omega}rw_{2}^{\mu}-\frac{1}{|\Omega|}\int_{\Omega}(a+s)w_{1}^{\mu}+\frac{\hat{r}}{|\Omega|}\int_{\Omega}\varphi^{\mu}_{2}-\frac{(\hat{a}+\hat{s})}{|\Omega|}\int_{\Omega}\varphi^{\mu}_{1}\cr\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\left[\frac{1}{|\Omega|}\int_{\Omega}\varphi_{2}^{\mu}\right]=\frac{1}{|\Omega|}\int_{\Omega}sw_{1}^{\mu}-\frac{1}{|\Omega|}\int_{\Omega}ew_{2}^{\mu}+\frac{\hat{s}}{|\Omega|}\int_{\Omega}\varphi^{\mu}_{1}-\frac{\hat{e}}{|\Omega|}\int_{\Omega}\varphi^{\mu}_{2}.\end{cases}$
(4.29)

Since $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})$ is bounded (see (2.11)),
without loss of generality after passing to a subsequence, we may suppose that
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\to\tilde{\lambda}$ as
$\mu\to\infty$ for some real mumber $\tilde{\lambda}$. Since
$\\{\Big{(}\frac{1}{|\Omega|}\int_{\Omega}\varphi_{1}^{\mu},\frac{1}{|\Omega|}\int_{\Omega}\varphi_{2}^{\mu}\Big{)}\\}_{\mu}$
is also nonnegative and bounded (see (4.21)), we may also suppose that there
is some nonnegative real vector ${\bf Q}^{*}$ such that
$\Big{(}\frac{1}{|\Omega|}\int_{\Omega}\varphi_{1}^{\mu},\frac{1}{|\Omega|}\int_{\Omega}\varphi_{2}^{\mu}\Big{)}\to{\bf
Q}^{*}$ as $\mu\to\infty$. Hence, by (4.21) and (4.22), we have that
$Q_{1}^{*}+Q_{2}^{*}=\frac{1}{\sqrt{|\Omega|}}$ and $\varphi^{\mu}\to{\bf
Q}^{*}$ as $\mu\to\infty$, in $L^{2}(\Omega)$. As a result, by letting
$\mu\to\infty$ in (4.29) and recalling (4.22), we obtain

$\tilde{\lambda}{\bf Q}^{*}=\hat{A}{\bf Q}^{*},$

where the matrix $\hat{A}$ is given by (2.7). Finally, since ${\bf Q}^{*}$ is
nonnegative, $Q_{1}^{*}+Q_{2}^{*}=\frac{1}{\sqrt{|\Omega|}}>0$ and the matrix
$\hat{A}$ is cooperative and irreducible, it then follows from Perron-
Frobenius theorem that $\tilde{\lambda}=$ $\tilde{\Lambda}$ is the maximal
eigenvalue of the matrix $\hat{A}$. Finally, since $\tilde{\lambda}=$
$\tilde{\Lambda}$ is independent of the subsequence of $\mu$ that we chose, we
conclude that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\to$
$\tilde{\Lambda}$ as $\mu\to\infty$. ∎

##  5 Proofs of Proposition 2.1 and Theorem 2.3

Next, we discuss the proofs of Proposition 2.1 and Theorem 2.3. To this end,
we first present the proof of Propostion 2.1.

###### Proof of Proposition 2.1.

Let $\xi>0$ and $h,l\in X^{+}$ be fixed. Note that the function

$L(\nu):=\lambda_{p}\Big{(}\xi\mathcal{K}+\big{(}\frac{rs}{\nu+h}-l\big{)}\mathcal{I}\Big{)}-\nu\quad\quad\
\nu>-h_{\min},$ (5.1)

is continuous, strictly decreasing and

$\lim_{\nu\to\infty}L(\nu)=-\infty.$

Therefore, the algebraic equation $L(\nu)=0$ has a unique solution, say
$\lambda^{h,l}_{\xi}$ if and only if

$\lim_{\nu\to-h_{\min}^{+}}L(\nu)=\sup_{\nu>-h_{\min}}L(\nu)>0,$

which proves the first part of the proposition. Moreover, when $h_{\min}>0$,
if $L(0)<0$ then $\lambda_{\xi}^{h,l}<0$; if $L(0)>0$ then
$\lambda_{\xi}^{h,l}>0$; and if $L(0)=0$ then $\lambda_{\xi}^{h,l}=0$.
Therefore, since $L(0)=\lambda(\xi\mathcal{K}+\frac{rs}{h}-l)$, then
$\lambda_{\xi}^{h,l}$ and $\lambda(\xi\mathcal{K}+\frac{rs}{h}-l)$ have the
same sign. ∎

For every $\eta\in X$ and $\xi>0$, it follows as in Theorem (2.1) (see also
[4]) that the following sup-inf characterization of the principal spectrum
point, $\lambda_{p}(\xi\mathcal{K}+\eta\mathcal{I})$,

$\lambda_{p}\Big{(}\xi\mathcal{K}+\eta\mathcal{I}\Big{)}=\lambda_{*}\Big{(}\xi\mathcal{K}+\eta\mathcal{I}\Big{)}=\lambda^{*}\Big{(}\xi\mathcal{K}+\eta\mathcal{I}\Big{)},$
(5.2)

holds, where

$\lambda_{*}\Big{(}\xi\mathcal{K}+\eta\mathcal{I}\Big{)}:=\sup\Big{\\{}\lambda\in\mathbb{R}:\exists\psi\in
X^{++}\ \text{satisfying}\
\lambda\psi\leq\xi\mathcal{K}\psi+\eta\psi\Big{\\}}$

and

$\lambda^{*}\Big{(}\xi\mathcal{K}+\eta\mathcal{I}\Big{)}:=\inf\Big{\\{}\lambda\in\mathbb{R}:\exists\psi\in
X^{++}\ \text{satisfying}\
\lambda\psi\geq\xi\mathcal{K}\psi+\eta\psi\Big{\\}}.$

In particular, for every $\nu>-(a+s)_{\min}$, it follows as in Theorem (2.1)
that

$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+(\frac{rs}{\nu+a+s}-e)\mathcal{I}\Big{)}=\lambda_{*}\Big{(}\mu_{2}\mathcal{K}+(\frac{rs}{\nu+a+s}-e)\mathcal{I}\Big{)}=\lambda^{*}\Big{(}\mu_{2}\mathcal{K}+(\frac{rs}{\nu+a+s}-e)\mathcal{I}\Big{)},$
(5.3)

###### Proof of Theorem 2.3 .

(i) Let $\mu_{2}>0$ be fixed. First, suppose that
$\lambda_{\mu_{2}}^{a+s,e}>-(a+s)_{\min}$. Therefore,

$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{\lambda_{\mu_{2}}^{a+s,e}+a+s}-e\Big{)}\mathcal{I}\Big{)}=\lambda_{\mu_{2}}^{a+s,e}.$

Hence, by (5.3) for every $\varepsilon>0$, there exist
$\varphi_{2,\varepsilon}^{+}\in X^{++}$ and $\varphi_{2,\varepsilon}^{-}\in
X^{++}$ such that

$(\lambda_{\mu_{2}}^{a+s,e}+\varepsilon)\varphi_{2,\varepsilon}^{+}\geq\mu_{2}\mathcal{K}\varphi_{2,\varepsilon}^{+}+\frac{rs}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\varphi^{+}_{2,\varepsilon}-e\varphi_{2,\varepsilon}^{+}$
(5.4)

and

$(\lambda_{\mu_{2}}^{a+s,e}-\varepsilon)\varphi_{2,\varepsilon}^{-}\leq\mu_{2}\mathcal{K}\varphi_{2,\varepsilon}^{-}+\frac{rs}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\varphi^{-}_{2,\varepsilon}-e\varphi_{2,\varepsilon}^{-}.$
(5.5)

First, let
$\varphi_{1,\varepsilon}^{+}:=\frac{r+\varepsilon}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\varphi_{2,\varepsilon}^{+}$.
Then, it follows from (5.4) that

$\Big{(}\lambda_{\mu_{2}}^{a+s,e}+\varepsilon\Big{(}1+\frac{s}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\Big{)}\Big{)}\varphi_{2,\varepsilon}^{+}\geq\mu_{2}\mathcal{K}\varphi_{2,\varepsilon}^{+}+s\varphi^{+}_{1,\varepsilon}-e\varphi_{2,\varepsilon}^{+}$

and

$\displaystyle\lambda_{\mu_{2}}^{a+s,e}\varphi_{1,\varepsilon}^{+}=$
$\displaystyle(r+\varepsilon)\varphi_{2,\varepsilon}^{+}-(a+s)\varphi_{1,\varepsilon}^{+}$
$\displaystyle=$
$\displaystyle\mu_{1}\mathcal{K}\varphi_{1,\varepsilon}^{+}+r\varphi_{2,\varepsilon}^{+}-(a+s)\varphi_{1,\varepsilon}^{+}+\Big{(}\varepsilon-\mu_{1}\frac{\mathcal{K}\varphi_{1,\varepsilon}^{+}}{\varphi_{2,\varepsilon}^{+}}\Big{)}\varphi_{2,\varepsilon}^{+}$
$\displaystyle\geq$
$\displaystyle\mu_{1}\mathcal{K}\varphi_{1,\varepsilon}^{+}+r\varphi_{2,\varepsilon}^{+}-(a+s)\varphi_{1,\varepsilon}^{+}$

whenever
$\mu_{1}<\frac{\varepsilon}{1+\Big{\|}\frac{\mathcal{K}\varphi_{1,\varepsilon}^{+}}{\varphi_{2,\varepsilon}^{+}}\Big{\|}_{\infty}}$.
Therefore, by Theorem 2.1, we have that

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\lambda_{\mu_{2}}^{a+s,e}+\varepsilon\Big{\|}1+\frac{s}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\Big{\|}_{\infty}\quad\text{whenever}\quad\mu_{1}<\frac{\varepsilon}{1+\Big{\|}\frac{\mathcal{K}\varphi_{1,\varepsilon}^{+}}{\varphi_{2,\varepsilon}^{+}}\Big{\|}_{\infty}}.$
(5.6)

Next, let
$\varphi_{1,\varepsilon}^{-}:=\frac{r}{\lambda_{\mu_{2}}^{a+s,e}+a+s}\varphi_{2,\varepsilon}^{-}$.
Then, by (5.5),

$(\lambda_{\mu_{2}}^{a+s,e}-\varepsilon)\varphi_{2,\varepsilon}^{-}\leq\mu_{2}\mathcal{K}\varphi_{2,\varepsilon}^{-}+s\varphi^{-}_{1,\varepsilon}-e\varphi_{2,\varepsilon}^{-}$

and

$\displaystyle(\lambda_{\mu_{2}}^{a+s,e}-\varepsilon)\varphi_{1,\varepsilon}^{-}=$
$\displaystyle
r\varphi_{2,\varepsilon}^{-}-(a+s)\varphi_{1,\varepsilon}^{-}-\varepsilon\varphi_{1,\varepsilon}^{-}$
$\displaystyle=$
$\displaystyle\mu_{1}\mathcal{K}\varphi_{1,\varepsilon}^{-}+r\varphi_{2,\varepsilon}^{-}-(a+s)\varphi_{1,\varepsilon}^{-}-(\varepsilon+\mu_{1}\mathcal{K})\varphi_{1,\varepsilon}^{-}$
$\displaystyle=$
$\displaystyle\mu_{1}\mathcal{K}\varphi_{1,\varepsilon}^{-}+r\varphi_{2,\varepsilon}^{-}-(a+s)\varphi_{1,\varepsilon}^{-}-(\varepsilon-\mu_{1}K(\cdot))\varphi_{1,\varepsilon}^{-}-\mu_{1}\int_{\Omega}\kappa(\cdot,y)\varphi_{1,\varepsilon}^{-}dy$
$\displaystyle\leq$
$\displaystyle\mu_{1}\mathcal{K}\varphi_{1,\varepsilon}^{-}+r\varphi_{2,\varepsilon}^{-}-(a+s)\varphi_{1,\varepsilon}^{-}$

whenever, $\mu_{1}<\frac{\varepsilon}{\|K\|_{\infty}}$. Therefore,

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\lambda_{\mu_{2}}^{a+s,e}-\varepsilon\quad\text{whenever}\
0<\mu_{1}<\frac{\varepsilon}{\|K\|_{\infty}}.$ (5.7)

By (5.6) and (5.7), we obtain that
$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\to\lambda_{\mu_{2}}^{a+s,e}$ as
$\mu_{1}\to 0^{+}$, which yields the desired result.

Next, we suppose that $\lambda_{\mu_{2}}^{a+s,e}=-(a+s)_{\min}$. Since the
function $L(\nu)$ defined by (5.1) is strictly decreasing,
$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{\varepsilon-(a+s)_{\min}+a+s}-e\Big{)}\Big{)}<\varepsilon-(a+s)_{\min}$
for every $\varepsilon>0$. Therefore, as in (5.6), we have that

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\varepsilon-(a+s)_{\min}+\varepsilon\Big{\|}1+\frac{s}{\varepsilon-(a+s)_{\min}+a+s}\Big{\|}_{\infty}\quad\text{whenever}\
0<\mu_{1}\ll 1,$

which yields that

$\limsup_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq-(a+s)_{\min}.$ (5.8)

On the other hand, for every $\mu_{1}>0$ and $\varepsilon>0$, it follows from
Theorem 2.1, that there is
$\varphi^{\varepsilon}=(\varphi^{\varepsilon}_{1},\varphi_{2}^{\varepsilon})\in
X^{++}\times X^{++}$, such that

$(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\varphi^{\varepsilon}\geq\mu\circ\mathcal{K}\varphi^{\varepsilon}+\mathcal{A}\varphi^{\varepsilon},$

which implies

$(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\varphi^{\varepsilon}_{1}\geq\mu_{1}\mathcal{K}\varphi^{\varepsilon}_{1}+r\varphi_{2}^{\varepsilon}-(a+s)\varphi_{1}^{\varepsilon}\geq\mu_{1}\mathcal{K}\varphi^{\varepsilon}_{1}-(a+s)\varphi_{1}^{\varepsilon},$

since $r\varphi_{2}^{\varepsilon}\geq 0$. Therefore,

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})$

since $\varepsilon>0$ is arbitrarily chosen and $\varphi_{1}^{\varepsilon}\in
X^{+}$. However, by [28, Theorem 2.2 (iv)],

$\lim_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})=-(a+s)_{\min}.$

As a result,

$\liminf_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\liminf_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})=-(a+s)_{\min},$

which combined with (5.8) yield

$\lim_{\mu_{1}\to
0^{+}}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=-(a+s)_{\min}.$

This completes the proof of the result.

(ii) It follows by proper modification of the proof of (i). ∎

##  6 Proofs of Proposition 2.2 and Theorem 2.4

We first give a proof of Proposition 2.2 and one more intermediate result (see
Lemma 6.1 below).

###### Proof of Proposition 2.2.

Fix $\xi>0$, $q,z\in X^{+}\setminus\\{0\\}$ and $l,h\in X^{+}$. Since the
resolvent function is locally analytic, then the function
$(\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}),\infty)\ni\nu\mapsto\Psi_{\xi,l}(\cdot,\nu,z)$
is smooth. Moreover, it follows from the resolvent identity that

$\frac{\partial\Psi_{\xi,l}(\cdot,\nu,z)}{\partial{\nu}}=-(\nu\mathcal{I}-(\xi\mathcal{K}-l\mathcal{I}))^{-2}(z)=-\Psi_{\xi,l}(\cdot,\nu,\Psi_{\xi,l}(\cdot,\nu,z)).$
(6.1)

Thus the function

$\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)=\int_{\Omega}q(x)\Psi_{\xi,l}(x,\nu,z)-\int_{\Omega}h-\nu|\Omega|,\quad\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$
(6.2)

is smooth with derivative function given by

$\frac{d\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)}{d\nu}=-\int_{\Omega}q(x)\Psi_{\xi,l}(x,\nu,\Psi_{\xi,l}(\cdot,\nu,z))dx-|\Omega|\quad\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}).$
(6.3)

But, since $\xi\mathcal{K}-l\mathcal{I}$ generates a strongly continuous and
strongly positive semigroup
$\\{e^{t(\xi\mathcal{K}-l(\cdot))\mathcal{I})}\\}_{t\geq 0}$, and $p\in
X^{+}\setminus\\{0\\}$, then $\Psi_{\xi,l}(\cdot,\nu,p)\in X^{++}$, hence
$\Psi_{\xi,l}(\cdot,\nu,\Psi_{\xi,l}(\cdot,\nu,z))\in X^{++}$. Thus, in view
of the fact that $q\in X^{+}\setminus\\{0\\}$, it follows from (6.3) that
$\frac{d\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)}{d\nu}<0$ for all
$\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$. Furthermore,

$\lim_{\nu\to\infty}\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)=-\infty.$ (6.4)

Indeed, an integration gives

$\nu\int_{\Omega}\Psi_{\xi,l}(x,\nu,z)dx=-\int_{\Omega}l(x)\Psi_{\xi,l}(x,\nu,z)dx+\int_{\Omega}z(x)dx\leq\int_{\Omega}z(x)dx\quad\forall\
\nu>\lambda_{p}(\xi\mathcal{K}-l(\cdot)\mathcal{I}).$

Observing that
$\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})\leq\lambda_{p}(\xi\mathcal{K})=0$,
then

$\displaystyle\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)\leq$
$\displaystyle\|q\|_{\infty}\int_{\Omega}{\Psi}_{\xi,l}(x,\nu,z)dx-\int_{\Omega}h(x)dx-\nu|\Omega|$
$\displaystyle\leq$
$\displaystyle\frac{\|q\|_{\infty}}{\nu}\int_{\Omega}z(x)dx-\int_{\Omega}h(x)dx-\nu|\Omega|,$

for every $\nu>0$. Hence (6.4) holds since the right hand side of the last
inequality converges to negative infinity as $\nu$ approaches infinity. As a
result, since $\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)$ is the expression at the
right side of the equation (2.18), therefore, the algebraic equation (2.18)
has a (unique) root if and only if (2.19) holds. (2.20) easily holds by
inspection. Next, we proceed to prove assertions (i) and (ii).

(i) Suppose that $\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$ is an eigenvalue
of $\xi\mathcal{K}-l\mathcal{I}$ with a positive function. We claim that

$\lim_{\nu\to\lambda_{p}^{+}(\xi\mathcal{K}-l\mathcal{I})}\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)=\infty.$
(6.5)

Indeed, if both (6.4) and (6.5) hold, then (2.19) holds, and by the
intermediate value theorem and the strict monotonicity of
$\tilde{\Psi}_{\xi,l}^{h,q,z}$ in $\nu$, there is a unique
$\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$ satisfying
$\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)=0$. Now, we proceed to show that (6.5)
hold. From (3.12), for every $\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$,
we have

$\displaystyle\Psi_{\xi,l}(\cdot,\nu,z)=$ $\displaystyle\int_{0}^{1}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}zdt+\int_{1}^{\infty}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}zdt$ $\displaystyle=$
$\displaystyle\int_{0}^{1}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}zdt+e^{-\nu}\int_{0}^{\infty}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}e^{\mu_{2}\mathcal{K}-e\mathcal{I}}zdt$
$\displaystyle=$ $\displaystyle\int_{0}^{1}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}zdt+e^{-\nu}\Psi_{\xi,l}(\cdot,\nu,\tilde{z}).$

where $\tilde{z}:=e^{(\xi\mathcal{K}-l\mathcal{I})}z$. Thus, since
$\\{e^{t(\xi\mathcal{K}-l\mathcal{I})}\\}_{t\geq 0}$ is strongly positive and
$z\in X^{+}\setminus\\{0\\}$, $\tilde{z}\in X^{++}$; that is
$\tilde{z}_{\min}>0$, and

$\Psi_{\xi,l}(\cdot,\nu,z)\geq
e^{-\nu}\Psi_{\xi,l}(\cdot,\nu,\tilde{z})\geq\tilde{z}_{\min}e^{-\nu}\Psi_{\xi,l}(\cdot,\nu,1).$

Hence,

$\tilde{\Psi}_{\xi,l}^{h,q,z}(\nu)=\int_{\Omega}q(x)\Psi_{\xi,l}(x,\nu,z)dx-\int_{\Omega}h-\nu|\Omega|\geq\tilde{z}_{\min}e^{-\nu}\int_{\Omega}q(x)\Psi_{\xi,l}(x,\nu,1)dx-\int_{\Omega}h-\nu|\Omega|.$
(6.6)

Now, let $\psi\in X^{++}$ be an eigenfunction of
$\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})$ satisfying $\psi_{\max}=1$. Then

$e^{t(\xi\mathcal{K}-l\mathcal{I})}\psi=e^{t\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})}\psi\quad\forall\
t>0.$

Hence, since $0<\psi\leq 1$ and
$\\{e^{t(\xi\mathcal{K}-l\mathcal{I})}\\}_{t\geq}$ is positive, we obtain that

$\Psi_{\xi,l}(\cdot,\nu,1)=\int_{0}^{\infty}e^{-\nu
t}e^{t(\xi\mathcal{K}-l\mathcal{I})}(t)dt\geq\int_{0}^{\infty}e^{-(\nu-\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}))}\psi=\frac{1}{\nu-\lambda_{p}(\xi\mathcal{K}-e\mathcal{I})}\psi(\cdot)\quad\forall\
\nu>\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}).$

This implies that

$\int_{\Omega}q(x)\Psi_{\xi,l}(x,\nu,1)dx\geq\frac{1}{\nu-\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})}\int_{\Omega}q(x)\psi(x)dx\to\infty\quad\text{as}\
\nu\to\lambda_{p}^{+}(\xi\mathcal{K}-l\mathcal{I}).$

(Note that we have used the fact that $\int_{\Omega}q(x)\psi(x)dx>0$ since
$\psi\in X^{++}$ and $q\in X^{+}\setminus\\{0\\}$.) This together with
inequality (6.6) implies that (6.5) holds. Clearly, by (6.5), we have that
(2.19) holds.

(ii) Suppose that $\lambda_{p}(\xi\mathcal{K}-l\mathcal{I})<0$ and (2.19)
holds. Then, $0\in(\lambda_{p}(\xi\mathcal{K}-l\mathcal{I}),\infty)$ and
$\tilde{\Psi}_{\xi,l}^{h,q,z}(\tilde{\lambda}_{q,\xi,z}^{h,l})=0$. Therefore,
since $\tilde{\Psi}_{\xi,l}^{h,q,z}$ is strictly decreasing, then
$\tilde{\lambda}_{q,\xi,z}^{h,l}$ and
$\tilde{\Psi}_{\xi,l}^{h,q,z}(0)=\sigma_{q,\xi,z}^{h,l}$ have the same sign.

∎

For future use, we set

$(\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r)w(x)=\Psi_{\mu_{2},e}(x,\nu,sw)r(x)\quad\forall\
w\in X,\ x\in\overline{\Omega},\quad\text{and}\
\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$ (6.7)

The following result will also be needed to complete the proof of Theorem 2.4.

######  Lemma 6.1.

Let $\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$ be fixed, so that the
bounded linear map $\nu\mathcal{I}-(\mu_{2}\mathcal{K}-e\mathcal{I})$ is
invertible and let $\Psi_{\mu_{2},e}(\cdot,\nu,\cdot)$ be given by (3.12). For
every $\mu_{1}>0$, let
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})$
denote the principal spectrum point of the bounded linear operator
$\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I}$
on $X$. Then there is $\mu_{1}^{*}\gg 0$, such that for every
$\mu_{1}>\mu_{1}^{*}$,
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})$
is a geometrically simple eigenvalue whose eigenspace is spanned by a strictly
positive eigenfunction. Furthermore,

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})=\frac{1}{|\Omega|}\left(\int_{\Omega}\Psi_{\mu_{2},e}(\cdot,\nu,s)rdx-\int_{\Omega}(a+s)dx\right).$
(6.8)

In particular, if (2.19) holds for the choices of $\xi=\mu_{2}$, $q=r$, $z=s$,
$l=e$ and $h=a+s$, then $\nu:=\tilde{\lambda}_{r,\mu_{2},s}^{a+s,e}$, the
unique zero of (2.18), satisfies

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}_{r,\mu_{2},s}^{a+s,e},s\mathcal{I})r-(a+s)\mathcal{I})=\tilde{\lambda}_{r,\mu_{2},s}^{a+s,e}.$
(6.9)

###### Proof.

Let $\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e(\cdot)\mathcal{I})$ be fixed. Taking
$w\equiv 1$, for every $\mu_{1}>0$, we have

$\displaystyle\mu_{1}\mathcal{K}(w)(x)+\Psi_{\mu_{2},e}(x,\nu,sw)r(x)-(a(x)+s(x))w(x)$
$\displaystyle=\Psi_{\mu_{2},e}(x,\nu,sw)r(x)-(a(x)+s(x))w(x)$
$\displaystyle\geq-\|a+s\|_{\infty}w(x)\quad\forall\ x\in\overline{\Omega},$

since $r\geq 0$, $s\geq 0$ and $\Psi_{\mu_{2},e}(\cdot,\nu,\cdot)$ is
positive. Thus

$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})\geq-\|a+s\|_{\infty}\quad\forall\
\mu_{1}>0.$ (6.10)

Recall that

$\displaystyle((\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})w)(x)$
(6.11) $\displaystyle=$
$\displaystyle\mu_{1}\int_{\Omega}\kappa(x,y)w(y)dy-\mu_{1}K(x)w(x)+\Psi_{\mu_{2},e}(x,\nu;sw)r(x)-(a(x)+s(x))w(x)\quad\forall\
w\in X,\ x\in\overline{\Omega}.$ (6.12)

Observe also that, with $w\equiv 1$,

$\displaystyle-\mu_{1}K(x)w(x)+\Psi_{\mu_{2},e}(x,\nu;sw)r(x)-(a(x)+s(x))w(x)$
$\displaystyle\leq$
$\displaystyle-\mu_{1}K_{\min}+\Psi_{\mu_{2},e}(x,\nu,s)r(x)-(a(x)+s(x))$
$\displaystyle\leq$
$\displaystyle(-\mu_{1}K_{\min}+\|\Psi_{\mu_{2},e}(\cdot,\nu,s)r\|_{\infty}-(a+s)_{\min})w(x)\quad\forall\
x\in\overline{\Omega},\ \mu_{1}>0.$

Hence, for every $\mu_{1}>0$, it holds that

$\lambda_{p}(-\mu_{1}K\mathcal{I}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})\leq-\mu_{1}K_{\min}+\|\Psi_{\mu_{2},e}(\cdot,\nu,s)r\|_{\infty}-(a+s)_{\min}.$

Therefore, in view of (6.10), we have that

$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})>\lambda_{p}(-\mu_{1}K\mathcal{I}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})$

whenever

$\mu_{1}>\mu_{1}^{*}:=\frac{\|\Psi_{\mu_{2},e}(\cdot,\nu,s)\|_{\infty}+\|a+s\|_{\infty}-(a+s)_{\min}}{K_{\min}}.$

Hence, since the mapping

$X\ni w\mapsto\mu_{1}\int_{\Omega}\kappa(\cdot,y)w(y)dy\in X,$

is strictly positive and compact, and
$\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I}$
is a compact perturbation of
$\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I}$, it follows by
similar arguments as in the proof [28, Proposition 3.8] that
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})$
is the principal eigenvalue of
$\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I}$
whose eigenspace is spanned by a strictly positive eigenfunction in $X^{++}$
for every $\mu_{1}>\mu_{1}^{*}$. Now, it remains to show that (6.8) holds.

For every $\mu_{1}>\mu_{1}^{*}$, let $\psi^{\mu_{1}}$ be the positive
principal eigenfunction of
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})$
satisfying $\|\psi^{\mu_{1}}\|_{L^{2}(\Omega)}=1$. It follows by computations
similar to that leading to inequality (4.23) that

$\lim_{\mu_{1}\to\infty}\Big{\|}\psi^{\mu_{1}}-\frac{1}{|\Omega|}\int_{\Omega}\psi^{\mu_{1}}\Big{\|}_{L^{2}(\Omega)}=0.$

which together with the fact that $\|\psi^{\mu_{1}}\|_{L^{2}(\Omega)}=1$ for
every $\mu_{1}>\mu_{1}^{*}$ yield

$\lim_{\mu_{1}\to\infty}\frac{1}{|\Omega|}\int_{\Omega}\psi^{\mu_{1}}dx=\frac{1}{\sqrt{|\Omega|}}\quad\text{and}\quad\lim_{\mu_{1}\to\infty}\Big{\|}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{\|}_{L^{2}(\Omega)}=0.$
(6.14)

Now, integrating the equation satisfied by $\psi^{\mu_{1}}$ on $\Omega$ yield

$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})=\frac{\int_{\Omega}\Psi_{\mu_{2},e}(x,\nu,s\psi^{\mu_{1}})r(x)dx-\int_{\Omega}(a(x)+s(x))\psi^{\mu_{1}}dx}{\int_{\Omega}\psi^{\mu_{1}}(x)dx}.$
(6.15)

for every $\mu_{1}>\mu_{1}^{*}$. Note that, since $\kappa$ is symmetric, it
follows from [28, Proposition 3.9] that

$\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})=\sup_{w\in
L^{2}(\Omega)\setminus\\{0\\}}\frac{\int_{\Omega}\int_{\Omega}\mu_{2}\kappa(x,y)[w(y)-w(x)]w(x)dydx-\int_{\Omega}e(x)w^{2}(x)dx}{\int_{\Omega}w^{2}dx}.$
(6.16)

Observe that, for every $\mu_{1}>\mu_{1}^{*},$ the function

$W_{\mu_{1}}(\cdot):=\Psi_{\mu_{2},e}(\cdot,\nu,s\psi^{\mu_{1}})-\frac{1}{\sqrt{|\Omega|}}\Psi_{\mu_{2},e}(\cdot,\nu,s)=\Psi_{\mu_{2},e}\Big{(}\cdot,\nu,\Big{(}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{)}s\Big{)}$

satisfies,

$\nu
W_{\mu_{1}}=\mu_{2}\mathcal{K}W_{\mu_{1}}-eW_{\mu_{1}}+\Big{(}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{)}s$

Multiplying this equation by $W_{\mu_{1}}$ and integrating the resulting
equation on $\Omega$ and recalling (6.16), we obtain

$\nu\int_{\Omega}W^{2}_{\mu_{1}}(x)dx\leq\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})\int_{\Omega}W_{\mu_{1}}^{2}(x)dx+\int_{\Omega}\Big{(}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{)}s(x)W_{\mu_{1}}(x)dx,$

equivalently,

$\Big{(}\nu-\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})\Big{)}\|W_{\mu_{1}}\|_{L^{2}(\Omega)}^{2}\leq\int_{\Omega}\Big{(}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{)}s(x)W_{\mu_{1}}(x)dx.$

Therefore, since $\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$, applying
Hölder’s inequality to the right hand side of the last inequality, we obtain

$\displaystyle\left\|\Psi_{\mu_{2},e}(\cdot,\nu,s\psi^{\mu_{1}})-\frac{1}{\sqrt{|\Omega|}}\Psi_{\mu_{2},e}(\cdot,\nu,s)\right\|_{L^{2}(\Omega)}\leq$
$\displaystyle\frac{\Big{\|}\Big{(}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{)}s\Big{\|}_{L^{2}(\Omega)}}{\nu-\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})}$
$\displaystyle\leq$
$\displaystyle\frac{\|s\|_{\infty}\Big{\|}\psi^{\mu_{1}}-\frac{1}{\sqrt{|\Omega|}}\Big{\|}_{L^{2}(\Omega)}}{\nu-\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})}.$

As a consequence, we deduce from (6.14) that

$\left\|\Psi_{\mu_{2},e}(\cdot,\nu,s\psi^{\mu_{1}})-\frac{1}{\sqrt{|\Omega|}}\Psi_{\mu_{2},e}(\cdot,\nu,s)\right\|_{L^{2}(\Omega)}\to
0\ \text{as}\ \mu_{1}\to\infty.$

Hence

$\lim_{\mu_{1}\to\infty}\int_{\Omega}\Psi_{\mu_{2},e}(x,\nu,s\psi^{\mu_{1}})r(x)dx=\frac{1}{\sqrt{|\Omega|}}\int_{\Omega}\Psi_{\mu_{2},e}(x,\nu,s)r(x)dx.$

Note also from (6.14) that

$\lim_{\mu_{1}\to\infty}\int_{\Omega}(a(x)+s(x))\psi^{\mu_{1}}(x)dx=\frac{1}{\sqrt{|\Omega|}}\int_{\Omega}(a(x)+s(x))dx\quad\text{and}\quad\lim_{\mu_{1}\to\infty}\int_{\Omega}\psi^{\mu_{1}}dx=\frac{|\Omega|}{\sqrt{|\Omega|}}.$

Therefore, letting $\mu_{1}\to\infty$ in (6.15) we obtain (6.8). Finally, if
(2.19) holds for $\xi=\mu_{2}$, $q=r$, $z=s$, $l=e$, and $h=a+s$, and
$\nu=\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}$, then (6.9) holds by (2.20) and
(6.8). This completes the proof of the lemma. ∎

Thanks to Lemma 6.1 and Proposition 2.2, we can now present the proof of
Theorem 2.4.

###### Proof of Theorem 2.4.

We only present the proof of (i) since (ii) can be proved by similar
arguments. Let $\xi=\mu_{2}$, $q=r$, $z=s$, $l=e$, and $h=a+s$. We distinguish
two cases.

Case 1. In this case, we suppose that (2.19) holds. Let
$\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$
be the unique zero of (2.18). By Lemma 6.1, there is $\mu_{1}^{*}\gg 1$ such
that
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\mathcal{I})r-(a+s)\mathcal{I})$
is the principal eigenvalue of
$\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\mathcal{I})r-(a+s)\mathcal{I}$
with a strictly positive eigenfunction $\varphi_{1}^{\mu_{1}}$. Set
$\varphi_{2}^{\mu_{2}}=\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\varphi_{1}^{\mu_{1}})$.
Then, for every $\mu_{1}>\mu_{1}^{*}$,
$(\varphi_{1}^{\mu_{1}},\varphi_{2}^{\mu_{2}})\in X^{++}\times X^{++}$ and
satisfies

$\begin{cases}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\mathcal{I})r-(a+s)\mathcal{I})\varphi_{1}^{\mu_{1}}=\mu_{1}\mathcal{K}\varphi_{1}^{\mu_{1}}+r\varphi_{2}^{\mu_{2}}-(a+s)\varphi_{1}^{\mu_{1}}\cr\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}\varphi_{2}^{\mu_{2}}=\mu_{2}\mathcal{K}\varphi_{2}^{\mu_{2}}+s\varphi_{1}^{\mu_{1}}-e\varphi_{2}^{\mu_{2}}.\end{cases}$

Hence, it follows from Theorem 2.1 that

$\displaystyle\min\big{\\{}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\mathcal{I})r-(a+s)\mathcal{I}),\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}\big{\\}}$
$\displaystyle\leq$
$\displaystyle\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\max\big{\\{}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},s\mathcal{I})r-(a+s)\mathcal{I}),\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s}\big{\\}}$

for every $\mu_{1}>\mu_{1}^{*}$. Therefore, letting $\mu_{1}\to\infty$ in the
last inequalities and recalling (6.9), we obtain that

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\tilde{\lambda}^{a+s,e}_{r,\mu_{2},s},$

which yields the desired result.

Case 2. Here we suppose that (2.19). Hence,
$\tilde{\Psi}_{\mu_{2},e}^{a+e,r,s}(\nu)<0$ for every
$\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$, where
$\tilde{\Psi}_{\mu_{2},e}^{a+e,r,s}$ is defined by (6.2). This together with
(6.8) implies that for every
$\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$,

$\lim_{\mu_{1}\to\infty}\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})=\frac{1}{|\Omega|}\left(\int_{\Omega}\Psi_{\mu_{2},e}(\cdot,\nu,s)rdx-\int_{\Omega}(a+s)dx\right)<\nu$
(6.17)

Therefore, for every $\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$, there
is $\mu_{1}^{1,\nu}>0$ such that

$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu,s\mathcal{I})r-(a+s)\mathcal{I})<\nu\quad\mu_{1}>\mu_{1}^{1,\nu},\
\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$ (6.18)

Furthermore, by Lemma 6.1, for every
$\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$, there is
$\mu_{1}^{2,\nu}\geq\mu_{1}^{1,\nu}$, such that
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})$
is a geometrically simple eigenvalue whose eigenspace is spanned by a strictly
positive eigenfunction. Now, fix
$\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$. For every
$\mu_{1}>\mu_{1}^{2,\nu}$, with $\mu=(\mu_{1},\mu_{2})$ let
$\varphi_{1}^{\mu}$ be the strictly positive eigenfunction of
$\lambda_{p}(\mu_{1}\mathcal{K}+\Psi_{\mu_{2},e}(\cdot,\nu;s\mathcal{I})r-(a+s)\mathcal{I})$
satisfying $\max_{x\in\overline{\Omega}}\varphi_{1}^{\mu}(x)=1$ and set
$\varphi_{2}^{\mu}=\Psi_{\mu_{2},e}(\cdot,\nu,s\varphi_{1}^{\mu})$. Then
$(\varphi_{1}^{\mu},\varphi_{2}^{\mu})\in X^{++}\times X^{++}$ and by (6.18),

$\begin{cases}\nu\varphi_{1}^{\mu}>\mu_{1}\mathcal{K}\varphi_{1}+r\varphi_{2}^{\mu}-(a+s)\varphi_{1}^{\mu_{1}}\cr\nu\varphi_{2}^{\mu}=\mu_{2}\mathcal{K}\varphi_{2}^{\mu}+s\varphi_{1}^{\mu}-e\varphi_{2}^{\mu}.\end{cases}$

It then follows from Theorem 2.1 that

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\nu\quad\forall\
\mu_{1}>\mu_{1}^{2,\nu}.$

Therefore, since $\nu>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$ is
arbitrary,

$\limsup_{\mu_{1}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$
(6.19)

On the other hand, for every $\varepsilon>0$ and dispersal rate
$\mu=(\mu_{1},\mu_{2})$, with $\varphi=(\varphi_{1},\varphi_{2})\in
X^{++}\times X^{++}$ given by (4.2), it follows from (4.3) that

$(\lambda_{p}(\mu\circ\mathcal{K}-\mathcal{A})+\varepsilon)\varphi_{2}>\mu_{2}\mathcal{K}\varphi_{2}+s\varphi_{1}-e\varphi_{2}\geq\mu_{2}\mathcal{K}\varphi_{2}-e\varphi_{2}.$

It then follows from (5.2) that
$\lambda_{p}(\mu\circ\mathcal{K}-\mathcal{A})+\varepsilon\geq\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$
Since $\varepsilon$ is arbitrary, then for any dispersal rate
$\mu=(\mu_{1},\mu_{2})$, it always holds that
$\lambda_{p}(\mu\circ\mathcal{K}-\mathcal{A})\geq\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$
Therefore,

$\liminf_{\mu_{1}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I}).$
(6.20)

We deduce the desired result from (6.19) and (6.20).

∎

##  7 Proofs of Theorems 2.5 and 2.6

###### Proof of Theorem 2.5.

(i) Suppose that $r_{\min}>0$. Let $\eta_{1}^{*}$ be defined as in (2.24). We
distinguish two cases.

Case 1. $\eta_{1}^{*}>-(a+s)_{\min}$. Consider the function

$\mathcal{L}\ :\
(-(a+s)_{\min},\infty)\ni\eta\mapsto\int_{\Omega}\left(\frac{rs}{\eta+a+s}-(e+\eta)\right).$

Clearly, $\mathcal{L}$ is strictly decreasing in $\eta$ and
$\eta_{1}^{*}=\inf\\{\eta\in(-(a+s)_{\min},\infty)\ :\
\mathcal{L}(\eta)<0\\}$. Note that $\mathcal{L}(\eta)\to-\infty$ as
$\eta\to\infty$. Observe that, since $\eta_{1}^{*}>-(a+s)_{\min}$,
$\mathcal{L}(\eta^{*}_{1})=0$, which implies that

$\eta_{1}^{*}=\frac{1}{|\Omega|}\int_{\Omega}\frac{rs}{\eta_{1}^{*}+a+s}dx-\hat{e}>-\hat{e}.$

Thus, $\eta_{1}^{*}>-\min\\{(a+s)_{\min},\hat{e}\\}$. Fix
$\eta>-\min\\{(a+s)_{\min},\hat{e}\\}$. Since

$\lim_{\mu_{1}\to
0}\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})=\max_{x\in\overline{\Omega}}\\{-(a(x)+s(x))\\}=-(a+s)_{\min},$

then there exists $\mu_{1}^{\eta}>0$ such that
$\eta>\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})$ for every
$0<\mu_{1}<\mu_{1}^{\eta}$. Hence
$\eta\mathcal{I}-(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})$ is invertible for
every $0<\mu_{1}<\mu_{1}^{\eta}$. For every $0<\mu_{1}<\mu_{1}^{\eta}$, let

$\varphi_{1}:=\left(\eta-(\mu_{1}\mathcal{K}-(a+s))\right)^{-1}(r).$ (7.1)

Hence, $\varphi_{1}\in X^{++}$ (since $r\in X^{+}\setminus\\{0\\}$) and
satisfies

$\eta\varphi_{1}=\mu_{1}\mathcal{K}(\varphi_{1})+r-(a+s)\varphi_{1}$ (7.2)

Note from the comparison principle for nonlocal operators that
$\|\varphi_{1}\|_{\infty}\leq\frac{\|r\|_{\infty}}{\eta+(a+s)_{\min}}$.
Moreover, it follows from (7.2) that

$\varphi_{1}-\frac{r}{\eta+a+s}=\frac{\mu_{1}\mathcal{K}(\varphi_{1})}{\eta+a+s}$

from which we deduce that

$\left\|\varphi_{1}-\frac{r}{\eta+a+s}\right\|_{\infty}\leq\mu_{1}\left\|\frac{\mathcal{K}(\varphi_{1})}{\eta+a+s}\right\|_{\infty}\leq\mu_{1}\frac{\|\mathcal{K}\|\|\varphi_{1}\|_{\infty}}{\eta+(a+s)_{\min}}\leq\mu_{1}\frac{\|\mathcal{K}\|\|r\|_{\infty}}{(\eta+(a+s)_{\min})^{2}}\to
0\ \text{as}\ \mu_{1}\to 0.$ (7.3)

Next, since

$\lim_{\mu_{2}\to\infty}\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})=-\hat{e},$

then there is $\mu_{2}^{\eta}>0$ such that
$\eta>\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})$ for every
$\mu_{2}>\mu_{2}^{\eta}$. For every $0<\mu_{1}<\mu_{1}^{\eta}$ and
$\mu_{2}>\mu_{2}^{\eta}$, let

$\varphi_{2}:=(\eta\mathcal{I}-(\mu_{2}\mathcal{K}-e\mathcal{I}))^{-1}(s\varphi_{1})$

where $\varphi_{1}$ is given by (7.1). Since $s\varphi_{1}\in
X^{+}\setminus\\{0\\}$, then $\varphi_{2}\in X^{++}$. Now, observe that
$(\varphi_{1},\varphi_{2})$ satisfies

$\begin{cases}\eta\varphi_{1}=\mu_{1}\mathcal{K}(\varphi_{1})+r\varphi_{2}-(a+s)\varphi_{1}&x\in\Omega,\cr\eta\varphi_{2}=\mu_{2}\mathcal{K}(\varphi_{2})+s\varphi_{1}-e\varphi_{2}&x\in\Omega.\end{cases}$
(7.4)

Observe that from the equation

$\eta\varphi_{2}=\mu_{2}\mathcal{K}(\varphi_{2})+s\varphi_{1}-e\varphi_{2}$
(7.5)

that

$\varphi_{2}=\hat{\varphi_{2}}+\frac{1}{K(x)}\left[\int_{\Omega}\kappa(x,y)(\varphi_{2}(y)-\hat{\varphi_{2}})dy-\frac{1}{\mu_{2}}\left((\eta+e)\varphi_{2}-s\varphi_{1}\right)\right].$
(7.6)

Thus, multiplying (7.5) by $\varphi_{2}$, integrating the resulting equation
and employing arguments similar to that leading to (4.23), we get that

$\displaystyle\left\|\varphi_{2}-\hat{\varphi_{2}}\right\|_{L^{2}(\Omega)}^{2}\leq$
$\displaystyle\frac{1}{\mu_{2}\beta_{*}}\left(\int_{\Omega}s\varphi_{1}\varphi_{2}-\eta\int_{\Omega}\varphi_{2}^{2}\right)\leq\frac{(\|s\|_{\infty}+|\eta|)}{\mu_{2}\beta_{*}}\sum_{i=1}^{2}\int_{\Omega}\varphi_{i}^{2}.$

Therefore, it follows from (7.6) that

$\displaystyle\|\varphi_{2}-\hat{\varphi_{2}}\|_{\infty}\leq$
$\displaystyle\frac{1}{K_{\min}}\left(\kappa_{\max}\int_{\Omega}|\varphi_{2}-\hat{\varphi_{2}}|dy+\frac{(\|\eta+e\|_{\infty}+\|s\|_{\infty})}{\mu_{2}}\sum_{i=1}^{2}\|\varphi_{i}\|_{\infty}\right)$
(7.7) $\displaystyle\leq$
$\displaystyle\frac{1}{K_{\min}}\left(\kappa_{\max}\sqrt{|\Omega|}\|\varphi_{2}-\hat{\varphi_{2}}\|_{L^{2}(\Omega)}+\frac{(\|\eta+e\|_{\infty}+\|s\|_{\infty})}{\mu_{2}}\sum_{i=1}^{2}\|\varphi_{i}\|_{\infty}\right)$
(7.8) $\displaystyle\leq$
$\displaystyle\frac{1}{K_{\min}}\left(\frac{\kappa_{\max}\sqrt{|\Omega|(\|s\|_{\infty}+|\eta|)}}{\sqrt{\mu_{2}\beta_{*}}}\sqrt{\sum_{i=1}^{2}|\Omega|\|\varphi_{i}\|_{\infty}^{2}}+\frac{(\|\eta+e\|_{\infty}+\|s\|_{\infty})}{\mu_{2}}\sum_{i=1}^{2}\|\varphi_{i}\|_{\infty}\right)$
(7.9) $\displaystyle\leq$
$\displaystyle\frac{1}{K_{\min}}\left(\frac{\kappa_{\max}|\Omega|\sqrt{2(\|s\|_{\infty}+|\eta|)}}{\sqrt{\mu_{2}\beta_{*}}}+\frac{(\|\eta+e\|_{\infty}+\|s\|_{\infty})}{\mu_{2}}\right)\sum_{i=1}^{2}\|\varphi_{i}\|_{\infty}.$
(7.10)

Next, we claim that

$\limsup_{\mu_{2}\to\infty,\mu_{1}\to 0}\|\varphi_{2}\|_{\infty}<\infty.$
(7.11)

We proceed by contradiction to establish that (7.11) holds. Suppose that there
is a sequence $\mu_{2}^{n}\to\infty$ and $\mu_{1}^{n}\to 0$ such that

$\lim_{n\to\infty}\|\varphi_{2,n}\|_{\infty}=\infty,$ (7.12)

where $(\varphi_{1,n},\varphi_{2,n})$ solves (7.4) with
$\mu=(\mu_{1,n},\mu_{2,n})$ for every $n\geq 1$. Set
$\psi_{i,n}=\frac{\varphi_{i,n}}{\|\varphi_{2,n}\|_{\infty}},i=1,2$ for every
$n\geq 1$. Hence,

$\|\psi_{1,n}\|_{\infty}\leq\frac{\|\varphi_{1,n}\|_{\infty}}{\|\varphi_{2,n}\|_{\infty}}\leq\frac{\|r\|_{\infty}}{(\eta+(a+s)_{\min})\|\varphi_{2,n}\|_{\infty}}\to
0\ \text{as}\ n\to\infty$ (7.13)

and

$\|\psi_{2,n}\|_{\infty}=1\quad\forall\ n\geq 1.$ (7.14)

Thus, dividing both sides of (7.7) by $\|\varphi_{2,n}\|_{\infty}$, we obtain

$\left\|\psi_{2,n}-\frac{1}{|\Omega|}\int_{\Omega}{\psi_{2,n}}\right\|_{\infty}\to
0\quad\text{as}\ n\to\infty,$

which together with (7.14) yields

$\lim_{n\to\infty}\frac{1}{|\Omega|}\int_{\Omega}\psi_{2,n}=1\quad\text{and}\quad\lim_{n\to\infty}\|\psi_{2,n}-1\|_{\infty}=0.$
(7.15)

If we integrate both sides of (7.5), we obtain that

$\eta\frac{1}{|\Omega|}\int_{\Omega}\psi_{2,n}(x)dx=-\frac{1}{|\Omega|}\int_{\Omega}e(x)\psi_{2,n}(x)dx+\frac{1}{|\Omega|}\int_{\Omega}s(x)\psi_{1,n}(x)dx.$

Letting $n\to\infty$ in this equation and recalling (7.13) and (7.14), we
obtain that $\eta=-\hat{e}$, which contradicts our initial assumption that
$\eta>-\hat{e}$. Therefore, (7.11) holds. Now, thanks to (7.11) and the fact
that $\|\varphi_{1}\|_{\infty}\leq\frac{\|r\|_{\infty}}{\eta+(a+s)_{\min}}$,
we conclude from (7.7) that

$\lim_{\mu_{1}\to
0,\mu_{2}\to\infty}\|\varphi_{2}-\hat{\varphi_{2}}\|_{\infty}=0.$

As a result, if we integrate (7.5) and recall (7.3), we get

$\varphi_{2}\to\frac{\int_{\Omega}\frac{sr}{\eta+a+s}}{|\Omega|(\eta+\hat{e})}=1+\frac{\mathcal{L}(\eta)}{|\Omega|(\eta+\hat{e})}\quad\text{as}\
\mu_{1}\to 0\ \text{and}\ \mu_{2}\to\infty\ \text{uniformly in }\ \Omega.$
(7.16)

From this point, we distinguish two subcases.

Subcase 1. $\eta>\eta_{1}^{*}$. Then
$\mathcal{L}(\eta)<\mathcal{L}(\eta_{1}^{*})=0$. Then, it follows from (7.16)
that there is $M\gg 1$ such that $\varphi_{2}<1$ for every
$0<\mu_{1}<\frac{1}{M}$ and $\mu_{2}>M$. As a result, we conclude from (7.4)
that

$\begin{cases}\eta\varphi_{1}\geq\mu_{1}\mathcal{K}(\varphi_{1})+r\varphi_{2}-(a+s)\varphi_{1}&x\in\Omega,\cr\eta\varphi_{2}\geq\mu_{2}\mathcal{K}(\varphi_{2})+s\varphi_{1}-e\varphi_{2}&x\in\Omega,\end{cases}$

for every $0<\mu_{1}<\frac{1}{M}$ and $\mu_{2}>M$, which implies that

$\limsup_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\eta.$

Since $\eta>\eta_{1}^{*}$ is arbitrary, then $\limsup_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\eta_{1}^{*}.$

Subcase 2. $\eta<\eta_{1}^{*}$. Then
$\mathcal{L}(\eta)>\mathcal{L}(\eta_{1}^{*})=0$. Then, it follows from (7.16)
that there is $M\gg 1$ such that $\varphi_{2}>1$ for every
$0<\mu_{1}<\frac{1}{M}$ and $\mu_{2}>M$. As a result, we conclude from (7.4)
that

$\begin{cases}\eta\varphi_{1}\leq\mu_{1}\mathcal{K}(\varphi_{1})+r\varphi_{2}-(a+s)\varphi_{1}&x\in\Omega,\cr\eta\varphi_{2}\leq\mu_{2}\mathcal{K}(\varphi_{2})+s\varphi_{1}-e\varphi_{2}&x\in\Omega,\end{cases}$

for every $0<\mu_{1}<\frac{1}{M}$ and $\mu_{2}>M$, which implies that

$\liminf_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\eta.$

Since $\eta<\eta_{1}^{*}$ is arbitrary, then $\liminf_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\eta_{1}^{*}.$

From Subcases 1 and 2, we deduce that $\lim_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})=\eta_{1}^{*}.$

Case 2. $\eta_{1}^{*}=-(a+s)_{\min}$. In this case, we must have that
$-(a+s)_{\min}\geq-\hat{e}$. Indeed, if this was false, that is
$-\hat{e}>-(a+s)_{\min}$, we would have

$\mathcal{L}(-\hat{e})=\int_{\Omega}\left(\frac{rs}{a+s-\hat{e}}-(e-\hat{e})\right)=\int_{\Omega}\frac{rs}{a+s-\hat{e}}>0,$

from which we deduce that $\eta_{1}^{*}>-\hat{e}$. So, we obtain a
contradiction to the fact that $\eta_{1}^{*}=-(a+s)_{\min}<-\hat{e}$. Next,
since $\eta_{1}^{*}=-(a+s)_{\min}=-\min\\{(a+s)_{\min},\hat{e}\\}$, then
$\mathcal{L}(\eta)<0$ for every $\eta>-(a+s)_{\min}$. It then follows from
(7.16) and subcase 1 above that

$\limsup_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\eta_{1}^{*}.$

Now, it remains to show that

$\liminf_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq\eta_{1}^{*}.$
(7.17)

Let $\varepsilon>0$. By Theorem 2.1, for every $\mu_{1}>0$ and $\mu_{2}>0$,
there is $\varphi\in X^{++}\times X^{++}$ such that

$(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\varphi_{1}\geq\mu_{1}\mathcal{K}(\varphi_{1})+r\varphi_{2}-(a+s)\varphi_{1}\geq\mu_{1}\mathcal{K}(\varphi_{1})-(a+s)\varphi_{1}$

and

$(\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon)\varphi_{2}\geq\mu_{2}\mathcal{K}(\varphi_{2})+s\varphi_{1}-e\varphi_{2}\geq\mu_{2}\mathcal{K}(\varphi_{2})-e\varphi_{2}.$

Hence,

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})+\varepsilon\geq\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})\quad\forall\
\mu_{1}>0,\ \mu_{2}>0.$

Therefore, since $\varepsilon$ is arbitrary,

$\displaystyle\liminf_{\mu_{1}\to
0,\mu_{2}\to\infty}\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\geq$
$\displaystyle\lim_{\mu_{1}\to
0}\lambda_{p}(\mu_{1}\mathcal{K}-(a+s)\mathcal{I})=-(a+s)_{\min}=\eta_{1}^{*},$

which yields (7.17). Finally, it follows from the monotonicity of the function
$\mathcal{L}$ that $\eta^{*}_{1}$ and $\lim_{\eta\to 0^{+}}\mathcal{L}(\eta)$
have the same sign.

(ii) It follows by a proper modification of the proof of (i).

∎

We complete this section with a proof of Theorem 2.6.

###### Proof of Theorem 2.6.

Suppose that $r_{\min}>0$.
(i) Suppose on the contrary that there exist a sequence
$\mu^{n}:=(\mu_{1}^{n},\mu_{2}^{n})$ of dispersal rates with $\mu_{1}^{n}\to
0^{+}$ as $n\to\infty$ such that

$\limsup_{n\to\infty}\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\leq 0.$
(7.18)

We now distinguish three cases.

Case 1. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to\infty$ as $n\to\infty$. Therefore, it follows from Theorem
2.5-(i) that $\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\eta_{1}^{*}$
as $n\to\infty$. This contradicts with (7.18) since $\eta_{1}^{*}>0$.

Case 2. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to 0$ as $n\to\infty$. Therefore, it follows from Theorem 2.2-(i)
that $\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\Lambda_{\max}$ as
$n\to\infty$. Hence $\Lambda_{\max}\leq 0$ by (7.18). As a result, we obtain
from (4.7) that

$\sqrt{(a(x)+s(x)-e(x))^{2}+4r(x)s(x)}\leq(a(x)+s(x)+e(x))\quad\forall\
x\in\Omega,$

which is equivalent to

$r(x)s(x)\leq(a(x)+s(x))e(x)\quad\forall\ x\in\Omega.$

Thus for every $\eta>-(a+s)_{\min}$,

$\int_{\Omega}\left(\frac{rs}{\eta+a+s}-(\eta+e)\right)\leq\int_{\Omega}\left(\frac{(a+s)e}{\eta+a+s}-(e+\eta)\right)=-\eta\int_{\Omega}\left(\frac{e}{\eta+a+s}+1\right).$

In particular, since $\eta_{1}^{*}>0\geq-(a+s)_{\min}$, then

$0=\int_{\Omega}\left(\frac{rs}{\eta_{1}^{*}+a+s}-(\eta_{1}^{*}+e)\right)\leq-\eta_{1}^{*}\int_{\Omega}\left(\frac{e}{\eta_{1}^{*}+a+s}+1\right)$

This is clearly impossible since $\eta_{1}^{*}>0$.

Case 3. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to\mu_{2}$ as $n\to\infty$ for some positive number $\mu_{2}$.
Therefore, it follows from Theorem 2.3-(i) that
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\lambda_{\mu_{2}}^{a+s,e}$
as $n\to\infty$, which in view of (7.18) implies that

$\lambda_{\mu_{2}}^{a+s,e}\leq 0.$ (7.19)

However,

$0=\frac{1}{|\Omega|}\int_{\Omega}\left(\frac{rs}{\eta_{1}^{*}+a+s}-(\eta_{1}^{*}+e)\right)\leq\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{\eta_{1}^{*}+a+s}-(e+\eta_{1}^{*})\Big{)}\mathcal{I}\Big{)}.$

Hence, it follows from Proposition 2.1 and the fact that the mapping
$(-(a+s)_{\min},\infty)\ni\eta\mapsto\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{\eta+a+s}-(e+\eta)\Big{)}\mathcal{I}\Big{)}$
is strictly decreasing that $\lambda_{\mu_{2}}^{a+s,e}\geq\eta_{1}^{*}$. This
clearly contradicts with (7.19) since $\eta_{1}^{*}>0$.

From cases 1, 2 and 3, we deduce that there is $\delta_{1}^{*}>0$ such the
statement of Theorem 2.6-(i) holds.

(ii) Suppose that $\Lambda_{\max}<0$. Then $\eta_{1}^{*}<0$. Indeed, if it was
the case that $\eta_{1}^{*}\geq 0$, then

$\int_{\Omega}\left(\frac{rs}{\eta_{1}^{*}+a+s}-(\eta_{1}^{*}+e)\right)=0,$

which would imply that there is some $x_{0}\in\overline{\Omega}$ such that

$r(x_{0})s(x_{0})\geq(\eta_{1}^{*}+e(x_{0}))(\eta_{1}^{*}+(a(x_{0})+s(x_{0})))\geq
e(x_{0})(a(x_{0})+s(x_{0})).$

This implies that $\Lambda_{\max}\geq 0$, which is not true. Hence, we must
have that $\eta_{1}^{*}<0$. Now, we proceed by contradiction to establish the
desired result. Suppose to the contrary that there exist a sequence
$\mu^{n}:=(\mu_{1}^{n},\mu_{2}^{n})$ of dispersal rates with $\mu_{1}^{n}\to
0^{+}$ as $n\to\infty$ such that

$\liminf_{n\to\infty}\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\geq 0.$
(7.20)

We distinguish three cases.

Case 1. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to\infty$ as $n\to\infty$. Therefore, it follows from Theorem
2.5-(i) that $\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\eta_{1}^{*}$
as $n\to\infty$. This contradicts with (7.20) since $\eta_{1}^{*}<0$.

Case 2. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to 0$ as $n\to\infty$. Therefore, it follows from Theorem 2.2-(i)
that $\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\Lambda_{\max}$ as
$n\to\infty$. This contradicts with (7.20) since $\Lambda_{\max}<0$.

Case 3. In this case, we suppose without loss of generality that
$\mu_{2}^{n}\to\mu_{2}$ as $n\to\infty$ for some positive number $\mu_{2}$.
Therefore, it follows from Theorem 2.3-(i) that
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\lambda_{\mu_{2}}^{a+s,e}$
as $n\to\infty$, which in view of (7.20) implies that

$\lambda_{\mu_{2}}^{a+s,e}\geq 0.$ (7.21)

However, since $\eta_{1}^{*}<0$, then
$-(a+s)_{\min}<0\leq\lambda_{\mu_{2}}^{a+s,e}$. Therefore, by Proposition 2.1,

$0=\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{\lambda_{\mu_{2}}^{a+s,e}+a+s}-(e+\lambda_{\mu_{2}}^{a+s,e})\Big{)}\mathcal{I}\Big{)},$

which implies that there is some $x_{1}\in\overline{\Omega}$ such that

$r(x_{1})s(x_{1})\geq(\lambda_{\mu_{2}}^{a+s,e}+e(x_{1}))(\lambda_{\mu_{2}}^{a+s,e}+(a(x_{1})+s(x_{1})))>e(x_{1})(a(x_{1})+s(x_{1})).$

This implies that $\Lambda_{\max}>0$, which is not true.

From cases 1, 2 and 3, we deduce that there is $\delta_{1}^{*}>0$ such the
statement of Theorem 2.6-(ii) holds.

(iii) Suppose that $(a+s)_{\min}>0$ and $\eta_{1}^{*}<0<\Lambda_{\max}$. Then

$\int_{\Omega}\left(\frac{rs}{a+s}-e\right)<0<\left(\frac{rs}{a+s}-e\right)_{\max}.$

This implies that the function $\frac{rs}{a+s}-e$ is not constant. Hence, the
function
$(0,\infty)\ni\mu_{2}\mapsto\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}$
is strictly decreasing ([28, Theorem 2.2(i)]). Moreover, since

$\lim_{\mu_{2}\to\infty}\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=\frac{1}{|\Omega|}\int_{\Omega}\Big{(}\frac{rs}{a+s}-e\Big{)}$

and

$\lim_{\mu_{2}\to
0}\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=\Big{(}\frac{rs}{a+s}-e\Big{)}_{\max},$

there is a unique $\mu_{2}^{*}>0$ such that
$\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=0$.
Next, fix $0<\varepsilon<\mu_{2}^{*}$ and we proceed in two steps to complete
the proof of the theorem.

Step 1. In this step, we show that there exist $\delta_{1,\varepsilon}>0$ and
$m^{*}>0$ such that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})>m^{*}$ for
every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1,\varepsilon})\times(0,\mu_{2}^{*}-\varepsilon)$.
If this was false, there would exist a sequence
$\mu^{n}=(\mu_{1}^{n},\mu_{2}^{n})$ of dispersal rates with
$0<\mu_{2}^{n}\leq\mu_{2}^{*}-\varepsilon$ for every $n\geq 1$ and
$\mu_{1}^{n}\to 0^{+}$ as $n\to\infty$ such that (7.18) holds. If, up to a
subsequence, $\mu_{2}^{n}\to 0$ as $n\to\infty$, then by Theorem 2.2,
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\Lambda_{\max}>0$ as
$n\to\infty$. This contradicts with (7.18). On the other hand, if
$\mu_{2}^{n}\to\mu_{2}\in(0,\mu_{2}^{*}-\varepsilon]$ (up to a subsequence) as
$n\to\infty$, then by Theorem 2.3-(i),
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\lambda_{\mu_{2}}^{a+s,e}$
as $n\to\infty$. Hence, in view of (7.18), we must have that
$\lambda_{\mu_{2}}^{a+s,e}\leq 0$. However, since $\mu_{2}<\mu_{2}^{*}$, then
$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}>\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=0$,
which implies that $\lambda_{\mu_{2}}^{a+s,e}>0$, so we get a contradiction.
Therefore, we conclude that there exist $\delta_{1,\varepsilon}>0$ and
$m^{*}>0$ such that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})>m^{*}$ for
every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1,\varepsilon})\times(0,\mu_{2}^{*}-\varepsilon)$.

Step 2. In this step, we show that there exist $\delta_{1,\varepsilon}>0$ and
$m^{*}>0$ such that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})<-m^{*}$ for
every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1,\varepsilon})\times(\mu_{2}^{*}+\varepsilon,\infty)$.
If this was false, there would exist a sequence
$\mu^{n}=(\mu_{1}^{n},\mu_{2}^{n})$ of dispersal rates with
$\mu_{2}^{n}\geq\mu_{2}^{*}-\varepsilon$ for every $n\geq 1$ and
$\mu_{1}^{n}\to 0^{+}$ as $n\to\infty$ such that (7.20) holds.

If, up to a subsequence, $\mu_{2}^{n}\to\infty$ as $n\to\infty$, then by
Theorem 2.5-(i),
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\eta_{1}^{*}<0$ as
$n\to\infty$. This contradicts with (7.20). On the other hand, if
$\mu_{2}^{n}\to\mu_{2}\in[\mu_{2}^{*}+\varepsilon,\infty)$ (up to a
subsequence) as $n\to\infty$, then by Theorem 2.3-(i),
$\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\to\lambda_{\mu_{2}}^{a+s,e}$
as $n\to\infty$. Hence, in view of (7.20), we must have that
$\lambda_{\mu_{2}}^{a+s,e}\geq 0$. However, since $\mu_{2}>\mu_{2}^{*}$, then
$\lambda_{p}\Big{(}\mu_{2}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}<\lambda_{p}\Big{(}\mu_{2}^{*}\mathcal{K}+\Big{(}\frac{rs}{a+s}-e\Big{)}\mathcal{I}\Big{)}=0$,
which implies that $\lambda_{\mu_{2}}^{a+s,e}<0$, so we get a contradiction.
Therefore, we conclude that there exist $\delta_{1,\varepsilon}>0$ and
$m^{*}>0$ such that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})<-m^{*}$ for
every
$\mu=(\mu_{1},\mu_{2})\in(0,\delta_{1,\varepsilon})\times(\mu_{2}^{*}+\varepsilon,\infty)$.
This completes the proof of the theorem.

∎

##  8 Proofs of Theorems 2.7, 2.8, 2.9, and 2.10

###### Proof of Theorems 2.7.

We proceed by contradiction. To this end, suppose that there exists a sequence
of dispersal rate $\mu^{n}=(\mu_{1}^{n},\mu_{2}^{n})$ converging to zero, a
sequence of positive steady states solutions ${\bf u}^{n}$ of (1.1) associated
with the dispersal rates $\mu^{n}$, and a sequence $\\{x_{n}\\}_{n\geq 1}$ of
elements of $\overline{\Omega}$ such that

$\inf_{n\geq 1}\|{\bf u}^{n}(x^{n})-{\bf V}(x^{n})\|>0.$ (8.1)

Step 1. Uniform bound on $\\{{\bf u}^{n}\\}_{n\geq 1}$. It is clear that any
positive steady state solution ${\bf u}$ of (1.1) is a subsolution of the
cooperative and subhomogeneous system

$\begin{cases}0=\mu_{1}\mathcal{K}u_{1}+r(x)u_{2}-(a(x)+s(x)+b(x)u_{1}(x))u_{1}(x)&x\in\Omega,\cr
0=\mu_{2}\mathcal{K}u_{2}+s(x)u_{1}-(e(x)+f(x)u_{2})u_{2}&x\in\Omega.\end{cases}$
(8.2)

Now, since $\Lambda_{\max}>0$, it follows from Theorem 2.2 that there is
$\eta_{0}>0$, such that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})>0$ for
every $\mu=(\mu_{1},\mu_{2})$ with $0<\mu_{i}<\eta_{0}$, $i=1,2$. Hence, by
[22, Theorem 4-(i)], system (8.2) has a unique globally stable positive steady
state solution ${\bf u}^{\mu}$ for every $\mu=(\mu_{1},\mu_{2})$ with
$0<\mu_{i}<\eta_{0}$, $i=1,2$. However, an easy computation shows that the
constant vector function ${\bf M}^{*}:=(M_{1}^{*},M_{2}^{*})$ where
$M_{1}^{*}=M_{2}^{*}=\max\\{\frac{r_{\max}}{b_{\min}},\frac{s_{\max}}{f_{\min}}\\}$
is a supersolution of (8.2). Hence, by stability of ${\bf u}^{\mu}$, we deduce
that ${\bf u}^{\mu}\leq_{1}M^{*}$ for every $\mu=(\mu_{1},\mu_{2})$ with
$0<\mu_{i}<\eta_{0}$, $i=1,2$. Since ${\bf u}^{n}$ is a subsolution of (8.2)
with $\mu^{n}=\mu$ and $\mu^{n}\to 0$ as $n\to\infty$, then, without loss of
generality, we have that ${\bf u}^{n}\leq_{1}{\bf M}^{*}$ for every $n\geq 1$.

Step 2. For every $x\in\overline{\Omega}$ and $n\geq 1$, it holds that

$\Lambda(x)\leq\max\\{u_{1}^{n}(x),u_{2}^{n}(x)\\}(b(x)+f(x)+c(x)+g(x)+\mu_{1}^{n}K(x)+\mu_{2}^{n}K(x)).$
(8.3)

Indeed, observe that ${\bf u}^{n}$ satisfies

$\begin{cases}(b(x)u_{1}^{n}(x)+c(x)u_{2}^{n}(x)+\mu_{1}K(x))u_{1}^{n}(x)=r(x)u_{2}^{n}(x)-(a(x)+s(x))u_{1}^{n}(x)+\mu_{1}\int_{\Omega}\kappa(x,y)u_{1}^{n}(y)dy\cr(f(x)u_{2}^{n}(x)+g(x)u_{1}^{n}(x)+\mu_{2}K(x))u_{2}^{n}(x)=s(x)u_{1}^{n}(x)-e(x)u_{1}^{n}(x)+\mu_{2}\int_{\Omega}\kappa(x,y)u_{2}^{n}(y)dy.\end{cases}$

Hence,

$\max\\{u_{1}^{n}(x),u_{2}^{n}(x)\\}(b(x)+f(x)+c(x)+g(x)+\mu_{1}^{n}K(x)+\mu_{2}^{n}K(x)){\bf
u}^{n}(x)\geq\mathcal{A}(x){\bf u}^{n}(x).$

As a result, we deduce from Theorem 2.1 that (8.3) holds since ${\bf
u}^{n}(x)>0$.

Step 3. In this step, we complete the proof of the theorem by deriving
contradiction to (8.1). By Step 1 and the Bolzano-Weierstrass theorem, if
possible after passing to a subsequence, we may suppose that ${\bf
u}^{n}(x^{n})\to{\bf u}^{\infty}$ as $n\to\infty$ for some ${\bf
u}^{\infty}\in\mathbb{R}^{+}\times\mathbb{R}^{+}$. Furthermore, since
$\overline{\Omega}$ is compact, again after passing to a further subsequence,
we may suppose that there is some $x_{\infty}\in\overline{\Omega}$ such that
$x_{n}\to x_{\infty}$ as $n\to\infty$. Therefore, observing from Step 1 that
$\mu_{i}^{n}\|\mathcal{K}u_{i}^{n}\|_{\infty}\to 0$ as $n\to\infty$, then
${\bf u}^{\infty}$ satisfies

$\begin{cases}0=r(x_{\infty})u_{2}^{\infty}-(a(x_{\infty})+s(x_{\infty})+b(x_{\infty})u_{1}^{\infty}+c(x_{\infty})u_{2}^{\infty})u_{1}^{\infty}\cr
0=s(x_{\infty})u_{1}^{\infty}-(e(x_{\infty})u_{2}^{\infty}+g(x_{\infty})u_{1}^{\infty})u_{2}^{\infty}.\end{cases}$
(8.4)

Now, we distinguish two cases.

Case 1. $\Lambda(x_{\infty})\leq 0$. In this case, it follows from (8.4) and
[22, Theorem 2] that ${\bf u}^{\infty}={\bf 0}$. Since ${\bf V}(x_{\infty})$
is also a nonnegative solution of (8.4), we also have by [22, Theorem 2] that
${\bf V}(x_{\infty})={\bf 0}$. This clearly contradicts with (8.1) since ${\bf
V}(x_{n})\to{\bf V}(x_{\infty})$ and ${\bf u}^{n}(x_{n})\to{\bf u}^{\infty}$
as $n\to\infty$.

Case 2. $\Lambda(x_{\infty})>0$. From Step 2, we have that

$0<\Lambda(x_{\infty})\leq\max\\{u_{1}^{\infty},u_{2}^{\infty}\\}(b(x_{\infty})+f(x_{\infty})+c(x_{\infty})+g(x_{\infty})),$

which implies that ${\bf u}^{\infty}\neq{\bf 0}$. However, by [22, Theorem 2],
${\bf V}(x_{\infty})$ is the unique nonnegative solution of (8.4) different
from ${\bf 0}$. Therefore, ${\bf V}(x_{\infty})={\bf u}^{\infty}$, which
contradicts with (8.1).

Thanks to cases 1 and 2, the desired result holds.

∎

###### Proof of Theorem 2.8.

Suppose that $\tilde{\Lambda}>0$ and let ${\bf u}^{\mu}$ be a positive steady-
state solution of (1.1) for ${\mu}>>1$. This gives

$\begin{cases}0=\mu_{1}\int_{\Omega}\kappa(x,y)(u_{1}^{\mu}(y)-u_{1}^{\mu}(x))dy+ru_{2}^{\mu}-su_{1}^{\mu}-(a+bu_{1}^{\mu}+\tau
cu_{2}^{\mu})u_{1}^{\mu}&x\in\Omega,\cr
0=\mu_{2}\int_{\Omega}\kappa(x,y)(u_{2}^{\mu}(y)-u_{2}^{\mu}(x))dy+su_{1}^{\mu}-(e+fu_{2}^{\mu}+\tau
gu_{1}^{\mu})u_{2}^{\mu}&x\in\Omega.\end{cases}$ (8.5)

Set $v_{i}^{\mu}=u_{i}^{\mu}-\frac{1}{|\Omega|}\int_{\Omega}u_{i}^{\mu}$,
$i=1,2$.

Step 1. In this step, we show that for each $i=1,2$,

$\lim_{\mu\to\infty}\|v_{i}^{\mu}\|_{\infty}=0.$ (8.6)

Multiplying the first equation of (8.5) by $u_{1}^{\mu}$ and integrating the
resulting equation, it follows by similar arguments leading to (4.23) that

$\|v_{1}^{\mu}\|_{L^{2}(\Omega)}^{2}\leq\frac{1}{\mu_{1}\beta_{*}}\int_{\Omega}ru_{1}^{\mu}u_{2}^{\mu}\leq\frac{\|u_{1}^{\mu}\|_{\infty}\|u_{2}^{\mu}\|_{\infty}}{\mu_{1}\beta_{*}}\int_{\Omega}r.$

Similarly,

$\|v_{2}^{\mu}\|_{L^{2}(\Omega)}^{2}\leq\frac{1}{\mu_{2}\beta_{*}}\int_{\Omega}su_{1}^{\mu}u_{2}^{\mu}\leq\frac{\|u_{1}^{\mu}\|_{\infty}\|u_{2}^{\mu}\|_{\infty}}{\mu_{2}\beta_{*}}\int_{\Omega}s.$

Recalling from Step 1 of the proof of Theorem 2.7 that
$\|u_{i}^{\mu}\|_{\infty}\leq\max\\{\frac{r_{\max}}{b_{\min}},\frac{s_{\max}}{f_{\min}}\\}$,
we deduce from the last two inequalities that

$\lim_{\mu\to\infty}\|v_{i}^{\mu}\|_{L^{2}(\Omega)}=0.$ (8.7)

Next, using (8.7), we show that (8.6) holds. To this end, observe that

$v_{1}^{\mu}=\frac{1}{K(x)}\int_{\Omega}\kappa(x,y)v_{1}^{\mu}(y)dy+\frac{1}{\mu_{1}K(x)}\left(ru_{2}^{\mu}-(a+s+bu_{1}^{\mu}+cu_{2}^{\mu})u_{1}^{\mu}\right).$

Hence, by the Hölder’s inequality,

$\displaystyle\|v_{1}^{\mu}\|_{\infty}\leq\frac{\|\kappa\|_{\infty}|\Omega|^{\frac{1}{2}}\|v_{1}^{\mu}\|_{L^{2}(\Omega)}}{K_{\min}}+\frac{\left(\|r\|_{\infty}\|u_{2}^{\mu}\|_{\infty}+(\|a+s\|_{\infty}+\|b\|_{\infty}\|u_{1}^{\mu}\|_{\infty}+\|c\|_{\infty}\|u_{2}^{\mu}\|)\|u_{1}^{\mu}\|_{\infty}\right)}{\mu_{1}K_{\min}}$

Similarly,

$\displaystyle\|v_{2}^{\mu}\|_{\infty}\leq\frac{\|\kappa\|_{\infty}|\Omega|^{\frac{1}{2}}\|v_{2}^{\mu}\|_{L^{2}(\Omega)}}{K_{\min}}+\frac{\left(\|s\|_{\infty}\|u_{1}^{\mu}\|_{\infty}+(\|e\|_{\infty}+\|f\|_{\infty}\|u_{2}^{\mu}\|_{\infty}+\|g\|_{\infty}\|u_{1}^{\mu}\|)\|u_{2}^{\mu}\|_{\infty}\right)}{\mu_{1}K_{\min}}.$

Therefore, since
$\|u_{i}^{\mu}\|_{\infty}\leq\max\\{\frac{r_{\max}}{b_{\min}},\frac{s_{\max}}{f_{\min}}\\}$,
(8.6) follows from the last two inequalities.

Step 2. We show that

$\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\leq\max\\{\|u_{1}^{\mu}\|_{\infty},\|u_{2}^{\mu}\|_{\infty}\\}(\|b\|_{\infty}+\|e\|_{\infty}+\|c\|_{\infty}+\|g\|_{\infty}).$
(8.8)

Indeed, observe that

$\begin{cases}\max\\{\|u_{1}^{\mu}\|_{\infty},\|u_{2}^{\mu}\|_{\infty}\\}(\|b\|_{\infty}+\|e\|_{\infty}+\|c\|_{\infty}+\|g\|_{\infty})u_{1}^{\mu}\geq(bu_{1}^{\mu}+cu_{2}^{\mu})u_{1}=\mu_{1}\mathcal{K}u_{1}^{\mu}+ru_{2}^{\mu}-(a+s)u_{1}^{\mu}\cr\max\\{\|u_{1}^{\mu}\|_{\infty},\|u_{2}^{\mu}\|_{\infty}\\}(\|b\|_{\infty}+\|e\|_{\infty}+\|c\|_{\infty}+\|g\|_{\infty})u_{2}^{\mu}\geq(fu_{2}^{\mu}+gu_{1}^{\mu})u_{2}^{\mu}=\mu_{2}\mathcal{K}u_{2}^{\mu}+su_{1}^{\mu}-eu_{2}^{\mu}.\end{cases}$

Therefore, (8.8) follows from Theorem 2.1 since ${\bf u}^{\mu}\in X^{++}\times
X^{++}$.

Step 3. In this step, we complete the proof of the theorem. By integrating
both equations in (1.1), and using the fact that $\kappa$ is symmetric, we
obtain

$\begin{cases}0=\int_{\Omega}ru_{2}^{\mu}-\int_{\Omega}(a+s+bu_{1}^{\mu}+cu_{2}^{\mu})u_{1}^{\mu}\cr
0=\int_{\Omega}su_{1}^{\mu}-\int_{\Omega}(e+fu_{2}^{\mu}+gu_{1}^{\mu})u_{2}^{\mu}\end{cases}$
(8.9)

Now, since $\|u^{\mu}\|_{\infty}$ is uniformly bounded in $\mu$, then after
passing to a subsequence, we may suppose, there is ${\bf
Q}=(Q_{1},Q_{2})\in\mathbb{R}^{+}\times\mathbb{R}^{+}$ such that
$(\frac{1}{|\Omega|}\int_{\Omega}u_{1}^{\mu},\frac{1}{|\Omega|}\int_{\Omega}u_{2}^{\mu})\to{\bf
Q}$ as $\mu\to\infty$. It then follows from Step 1 that ${\bf u}^{\mu}\to{\bf
Q}$ as $\mu\to\infty$ uniformly on $\overline{\Omega}$. Recalling from Theorem
2.4 that $\lambda_{p}(\mu\circ\mathcal{K}+\mathcal{A})\to$ $\tilde{\Lambda}$
as $\mu\to\infty$ and $\tilde{\Lambda}>0$, we deduce from Step 2 that ${\bf
Q}\neq{\bf 0}$. Now, letting $\mu\to\infty$ in (8.9), we obtain that

$\begin{cases}0=\hat{r}Q_{2}-(\hat{a}+\hat{s}+\hat{b}Q_{1}+\hat{c}Q_{2})Q_{1}\cr
0=\hat{s}Q_{1}-(\hat{e}+\hat{f}Q_{2}+\hat{g}Q_{1})Q_{2}.\end{cases}$ (8.10)

This shows that ${\bf Q}$ is a positive solution of (2.30). Since by [22,
Theorem 2], $\hat{V}$ is the unique positive solution of (2.30), then ${\bf
Q}=\hat{\bf V}$, and the result follows.

∎

###### Proof of Theorem 2.9.

Suppose that $(a+s)_{\min}>0$ and $r_{\min}>0$. Let $\mu_{2}>0$ be fixed and
$\lambda_{\mu_{2}}^{a+s,e}$ be given by Theorem 2.3-(i). Suppose that
$\lambda_{\mu_{2}}^{a+s,e}>0$. Let ${\bf u}^{\mu}$ be a positive steady-state
solution of (1.1) for small values of $\mu_{1}$. Recall that

$\begin{cases}0=\mu_{1}\mathcal{K}u_{1}^{\mu}+ru_{2}^{\mu}-(a+s+bu_{1}^{\mu}+cu_{2}^{\mu})u_{1}^{\mu}&x\in\Omega,\cr
0=\mu_{2}\mathcal{K}u_{2}^{\mu}+su_{1}^{\mu}-(e+fu_{2}^{\mu}+gu_{1}^{\mu})u_{2}^{\mu}&x\in\Omega.\end{cases}$
(8.11)

Solving for $u_{1}^{\mu}$ in the first equation of (8.11), we get

$u_{1}^{\mu}=\frac{1}{2b}\left(\sqrt{(a+s+cu_{2}^{\mu})^{2}+4b(ru_{2}^{\mu}+\mu_{1}\mathcal{K}u_{1}^{\mu})}-(a+s+cu_{2}^{\mu})\right).$
(8.12)

Hence, introducing the functions

$G(x,\tau):=a(x)+s(x)+c(x)\tau$ (8.13)

and

$\displaystyle F(x,\tau,\nu):=$
$\displaystyle\frac{1}{2b(x)}\left(\sqrt{G^{2}(x,\tau)+4b(x)(r(x)\tau+\nu)}-G(x,\tau)\right)$
(8.14) $\displaystyle=$ $\displaystyle\frac{2(\tau
r(x)+\nu)}{\sqrt{G^{2}(x,\tau)+4b(x)(\tau r(x)+\nu)}+G(x,\tau)}$ (8.15)

for every $x\in\overline{\Omega}$, $\tau\geq 0$ and $\nu\geq 0$, we have that
$u_{1}^{\mu}=F(\cdot,u_{2}^{\mu},\mu_{1}\mathcal{K}u_{1}^{\mu})$. Observe that
$F(x,\tau,0)=H(x,\tau)$ for every $\tau\geq 0$ and $x\in\overline{\Omega}$,
where $H$ is defined in (2.31). Hence, it follows from (8.11) and (8.12) that

$0=\mu_{2}\mathcal{K}u_{2}^{\mu}+sF(x,u_{2}^{\mu},\mu_{1}u_{1}^{\mu}(x))-(e(x)+f(x)u_{2}^{\mu}(x)+g(x)F(x,u_{2}^{\mu},\mu_{1}\mathcal{K}u_{1}^{\mu}(x)))u_{2}^{\mu}(x)\quad\forall\
x\in\overline{\Omega}.$ (8.16)

Note that for every $x\in\overline{\Omega}$ and $\tau\geq 0$, the function
$[0,\infty)\ni\nu\mapsto F(x,\tau,\nu)$ is nondecresing. Recall also from Step
1 of the proof of Theorem 2.7 that

$\|u_{i}^{\mu}\|_{\infty}\leq
M^{*}:=\frac{r_{\max}+s_{\max}}{\min\\{b_{\min},f_{\min}\\}},$

hence, it follows from (8.16) that

$0\leq\mu_{2}\mathcal{K}u_{2}^{\mu}+s(x)F(x,u_{2}^{\mu},\mu_{1}M^{*})-(e(x)+f(x)u_{2}^{\mu}(x)+g(x)H(x,u_{2}^{\mu}))u_{2}^{\mu}(x)\quad\forall\
x\in\overline{\Omega}$ (8.17)

and

$0\geq\mu_{2}\mathcal{K}u_{2}^{\mu}+s(x)H(x,u_{2}^{\mu})-(e(x)+f(x)u_{2}^{\mu}(x)+g(x)F(x,u_{2}^{\mu},\mu_{1}M^{*}))u_{2}^{\mu}(x)\quad\forall\
x\in\overline{\Omega}.$ (8.18)

Form this point, the proof is divided into three parts.

Step 1. In this step, we show that that there is $\mu_{1}^{*}>0$ such for
every $0<\mu<\mu_{1}^{*}$, there is a unique globally stable solution
$\underline{u}_{2}^{\mu}$ of

$0=\mu_{2}\mathcal{K}\underline{u}_{2}^{\mu}+\Big{(}s(x)\frac{H(x,\underline{u}_{2}^{\mu})}{\underline{u}_{2}^{\mu}}-e(x)-f(x)\underline{u}_{2}^{\mu}(x)-g(x)F(x,\underline{u}_{2}^{\mu},\mu_{1}M^{*})\Big{)}\underline{u}_{2}^{\mu}(x)\quad\forall\
x\in\overline{\Omega}$ (8.19)

Now, observe from (8.14) that, for every $x\in\overline{\Omega}$, the mapping

$(0,\infty)\ni\tau\mapsto\frac{H(x,\tau)}{\tau}=\frac{F(x,\tau,0)}{\tau}=\frac{2r(x)}{\sqrt{G^{2}(x,\tau)+4b(x)\tau
r(x)}+G(x,\tau)\color[rgb]{1,0,0})\color[rgb]{0,0,0}}$

is strictly decreasing and

$\lim_{\tau\to
0}\frac{H(x,\tau)}{\tau}=\frac{r(x)}{(a(x)+s(x))}\quad\text{for}\ x\
\text{uniformly on }\ \Omega.$

Note that $[0,\infty)\ni\tau\mapsto F(x,\tau,\nu)$ is of class $C^{1}$ and

$\displaystyle f(x)+\partial_{\tau}F(x,\tau,\nu)$ $\displaystyle=$
$\displaystyle
f(x)+\frac{\left(\Big{(}2b(x)r(x)+c(x)G(x,\tau)\Big{)}-c(x)\sqrt{G^{2}(x,\tau)+4b(x)(\tau
r(x)+\nu)}\right)}{2b(x)\sqrt{G^{2}(x,\tau)+4b(x)(\tau
r(x)+\nu)}}=\mathbb{I}_{1}(x)+\mathbb{I}_{2}(x)$

where

$\displaystyle\mathbb{I}_{1}(x,\tau,\nu):=$
$\displaystyle\frac{\left(\Big{(}2b(x)r(x)+c(x)G(x,\tau)\Big{)}-c(x)\sqrt{G^{2}(x,\tau)+4b(x)\tau
r(x)}\right)}{2b(x)\sqrt{G^{2}(x,\tau)+4b(x)(\tau r(x)+\nu)}}$ (8.20)

and

$\displaystyle\mathbb{I}_{2}(x,\tau,\nu):=f(x)+\frac{c(x)\left(\sqrt{G^{2}(x,\tau)+4b(x)\tau
r(x)}-\sqrt{G^{2}(x,\tau)+4b(x)(\tau
r(x)+\nu)}\right)}{2b(x)\sqrt{G^{2}(x,\tau)+4b(x)(\tau r(x)+\nu)}}=$
$\displaystyle f(x)-\frac{4b(x)c(x)\nu}{2b(x)\sqrt{G^{2}(x,\tau)+4b(x)(\tau
r(x)+\nu)}\left(\sqrt{G^{2}(x,\tau)+4b(x)\tau
r(x)}+\sqrt{G^{2}(x,\tau)+4b(x)(\tau r(x)+\nu)}\right)}$

An easy computation shows that $\mathbb{I}_{1}(x,\tau,\nu)\geq 0$ for every
$x\in\overline{\Omega}$, $\tau\geq 0$, and $\nu\geq 0$, and

$\lim_{\nu\to
0}\max_{x\in\overline{\Omega},\tau\in[0,M^{*}]}\left|\mathbb{I}_{2}(x,\tau,\nu)-f(x)\right|=0.$

Therefore, since $f_{\min}>0$, there is $\tilde{\mu}_{1}^{*}>0$ such that for
every $0<\mu_{1}\leq\tilde{\mu}_{1}^{*}$ and $x\in\overline{\Omega}$, the
mapping

$(0,M^{*}]\ni\tau\mapsto\tilde{F}(x,\tau,\mu_{1}):=s(x)\frac{H(x,\tau)}{\tau}-e(x)-f(x)\tau-g(x)F(x,\tau,\mu_{1}M^{*})$

is strictly decreasing. Observe that
$0\leq\frac{H(x,\tau)}{\tau}\leq\frac{r(x)s(x)}{(a(x)+s(x))}$ for all
$x\in\overline{\Omega}$ and $\tau>0$, hence

$\tilde{F}(x,M^{*},\mu_{1})\leq\frac{r(x)s(x)}{a(x)+s(x)}-f(x)M^{*}\leq
r(x)-f(x)M^{*}<0\quad\forall\ x\in\overline{\Omega},\
0<\mu_{1}<\tilde{\mu}^{*}_{1}.$

Finally, it is easy to see that

$\lim_{\mu_{1}\to 0,\tau\to
0}\max_{x\in\overline{\Omega}}\left|\tilde{F}(x,\tau,\mu_{1})-\Big{(}\frac{r(x)s(x)}{a(x)+s(x)}-e(x)\Big{)}\right|=0.$

This shows that

$\lim_{\mu_{1}\to
0}\lambda_{p}(\mu_{2}\mathcal{K}+\tilde{F}(x,0,\mu_{1}))=\lambda_{p}(\mu_{2}\mathcal{K}+(\frac{rs}{a+s}-e)\mathcal{I}).$

As a result, since
$\lambda_{p}(\mu_{2}\mathcal{K}+(\frac{rs}{a+s}-e)\mathcal{I})$ and
$\lambda_{\mu_{2}}^{a+s,e}$ have the same sign by Proposition 2.1, and
$\lambda_{\mu_{2}}^{a+s,e}>0$, there is $0<\mu_{1}^{*}<\tilde{\mu}_{1}^{*}$
such that $\lambda_{p}(\mu_{2}\mathcal{K}+\tilde{F}(x,0,\mu_{1}))>0$ for every
$0<\mu_{1}<\mu_{1}^{*}$. Hence, it follows from standard arguments on the
nonlocal-dispersal equations of Fisher-KPP type that for every
$0<\mu_{1}<\mu_{1}^{*}$, there is a unique stable solution
$\underline{u}_{2}^{\mu}\in X^{++}$ which solves (8.19). Moreover,
$\|\underline{u}_{2}^{\mu}\|_{\infty}\leq M^{*}$ for every
$0<\mu_{1}<\mu_{1}^{*}$. Furthermore, since $\mu_{1}\mapsto
F(x,\tau,\mu_{1}M^{*})$ is strictly decreasing on $(0,\mu_{1}^{*})$ for every
$x\in\overline{\Omega}$ and $\tau\in[0,M^{*}]$, then by the comparison
principle for cooperative nonlocal equations, we have that
$\underline{u}_{2}^{\mu}$ is decreasing in $\mu_{1}$. And there is
$\underline{u}_{2}^{*}\in X^{++}$ such that
$\underline{u}_{2}^{\mu}\to\underline{u}_{2}^{*}$. Note that
$\underline{u}_{2}^{*}$ also solves (2.32). Since (2.32) has a unique
solution, then $\underline{u}_{2}^{*}=w^{*}$.

Step 2. In this step, we show that that there is
$0<\hat{\mu}_{1}^{*}<\mu_{1}^{*}$ such for every $0<\mu<\hat{\mu}_{1}^{*}$,
there is a unique stable solution
$\overline{u}_{2}^{\mu}\geq\underline{u}_{2}^{\mu}$ of

$0=\mu_{2}\mathcal{K}\overline{u}_{2}^{\mu}+s(x)F(x,\overline{u}_{2}^{\mu},\mu_{1}M^{*})-(e(x)+f(x)\overline{u}_{2}^{\mu}(x)+g(x)H(x,\overline{u}_{2}^{\mu}))\overline{u}_{2}^{\mu}(x)\quad\forall\
x\in\overline{\Omega}.$ (8.21)

The proof follows by a proper modification of the arguments of Step 1. Note
also here that $\overline{u}_{2}^{\mu}$ is decreasing in $\mu_{1}$ and
converges uniformly to $w^{*}$.

Step 3. In this step, we complete the proof of the theorem. Indeed, it follows
from (8.17) and (8.18) that
$\underline{u}_{2}^{\mu}<u_{2}^{\mu}<\overline{u}_{2}^{\mu}$ for every
$0<\mu_{1}<\hat{\mu}_{1}^{*}$, which implies that
$\|u_{2}^{\mu}-w^{*}\|_{\infty}\to 0$ as $\mu_{1}\to 0$. This in turn together
with (8.12) implies that $u_{1}^{\mu}\to H(\cdot,w^{*}(\cdot))$ as $\mu_{1}\to
0$ uniformly on $\Omega$.

∎

We complete this section with a proof of Theorem 2.10.

###### Proof of Theorem 2.10.

Fix $\mu_{2}>0$ and suppose that $e\in X^{+}\setminus\\{0\\}$. Suppose also
that $\lambda_{\mu_{2}}^{\infty}>0$, where $\lambda_{\mu_{2}}^{\infty}$ is
given by Theorem 2.4-(i). Let ${\bf u}^{\mu}$, $\mu=(\mu_{1},\mu_{2})$, be a
collection of steady-state solutions of (1.1) for sufficiently large values of
$\mu_{1}$. Hence, as in Step 1 of proof of Theorem 2.8, we have

$\lim_{\mu_{1}\to\infty}\left\|u_{1}^{\mu}-\frac{1}{|\Omega|}\int_{\Omega}u_{1}^{\mu}\right\|_{\infty}=0.$

Hence, there is a nonnegative number $l^{*}$ and a sequence
$\mu_{1,n}\to\infty$ with $\mu^{n}=(\mu_{1,n},\mu_{2})$ such that

$\lim_{n\to\infty}\|u_{1}^{\mu^{n}}-l^{*}\|_{\infty}=0.$ (8.22)

Now, since $u_{2}^{\mu^{n}}$ satisfies

$0=\mu_{2}\mathcal{K}(u_{2}^{\mu^{n}})+su_{1}^{\mu^{n}}-(e+fu_{2}^{\mu^{n}}+gu_{1}^{\mu^{n}})u_{2}^{\mu^{n}}$
(8.23)

and $\lambda_{p}(\mu_{2}\mathcal{K}-e\mathcal{I})<0$ since $e\in
X^{+}\setminus\\{0\\}$, and $\Psi_{\mu_{2},e}(\cdot,0,\cdot)$ is strongly
positive, then

$0\leq
u_{2}^{\mu^{n}}=\Psi_{\mu_{2},e}\big{(}\cdot,0,su_{1}^{\mu^{n}}-(fu_{2}^{\mu^{n}}+gu_{1}^{\mu^{n}})u_{2}^{\mu^{n}}\big{)}\leq\Psi_{\mu_{2},e}(su_{1}^{\mu^{n}})\leq\|\Psi_{\mu_{2},e}(\cdot,0,\cdot)\|s_{\max}\|u_{1}^{\mu^{n}}\|_{\infty}\quad\forall\
n\geq 1.$ (8.24)

It then follows from (8.8) and (8.22) that

$\displaystyle
0<\lambda_{\mu_{2}}^{\infty}=\lim_{n\to\infty}\lambda_{p}(\mu^{n}\circ\mathcal{K}+\mathcal{A})\leq
l^{*}\max\\{1,\|\Psi_{\mu_{2},e}(\cdot,0,\cdot)\|s_{\max}\\}(\|b\|_{\infty}+\|e\|_{\infty}+\|c\|_{\infty}+\|g\|_{\infty}),$

which implies that $l^{*}>0$. As a result, defining the function

$\tilde{F}^{*}(x,\tau)=l^{*}s(x)-(e(x)+f(x)\tau+l^{*}g(x))\tau\quad
x\in\overline{\Omega},\ \tau\geq 0,$

we have that $\tilde{F}^{*}(\cdot,0)=l^{*}s\in X^{+}\setminus\\{0\\}$. Hence,
$\lambda_{p}(\mu_{2}\mathcal{K}+\tilde{F}^{*}(\cdot,0))>0$. Moreover, for
every $x\in\overline{\Omega}$, the mapping $\tau\mapsto\tilde{F}^{*}(x,\tau)$
is strictly decreasing and $\tilde{F}^{*}(x,\tau)\color[rgb]{0,0,0}<0$ for
$\tau\geq\sqrt{\frac{2l^{*}s_{\max}}{f_{\min}}}$. It follows from standard
arguments that the nonlocal equation

$0=\mu_{2}\mathcal{K}\tilde{w}+\tilde{F}^{*}(x,\tilde{w})$ (8.25)

has a unique positive solution $\tilde{w}^{*}\in X^{++}$. Observing from
(8.23) that

$0=\mu_{2}\mathcal{K}(u_{2}^{\mu^{n}})+\tilde{F}^{*}(x,u_{2}^{\mu^{n}})+(s-u_{2}^{\mu^{n}}g)(u_{1}^{\mu^{n}}-l^{*}),$

and by (8.22) and (8.24)

$\lim_{n\to\infty}\|(s-u_{2}^{\mu^{n}}g)(u_{1}^{\mu^{n}}-l^{*})\|_{\infty}=0,$

we can employ a perturbation argument to conclude that
$u_{2}^{\mu^{n}}\to\tilde{w}^{*}$ as $n\to\infty$, uniformly in $\Omega$.
Finally, integrating over $\Omega$ the equation

$0=\mu_{1,n}\mathcal{K}(u_{1}^{\mu^{n}})+ru_{2}^{\mu^{n}}-(a+s+bu_{1}^{\mu^{n}}+cu_{2}^{\mu^{n}})u_{1}^{\mu^{n}}$

and letting $n\to\infty$ in the resulting equation yields

$0=\int_{\Omega}r\tilde{w}^{*}-l^{*}\int_{\Omega}((a+s)+bl^{*}+c\tilde{w}^{*}).$

Therefore, $(l^{*},\tilde{w}^{*})$ satisfies (2.33) and ${\bf
u}^{\mu^{n}}\to(l^{*},\tilde{w}^{*})$ as $n\to\infty$, uniformly in $\Omega$.

∎

### Acknowledgement.

We sincerely thank the anonymous reviewer for the helpful comments and
suggestions.

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