# Global asymptotic stability for Gurtin-MacCamy’s population dynamics model

Zhaohai Ma(a) and Pierre Magal(b)
(a)School of Science, China University of Geosciences,
Beijing 100083, People’s Republic of China
(b)Univ. Bordeaux, IMB, UMR 5251, F-33400 Talence, France.
CNRS, IMB, UMR 5251, F-33400 Talence, France. Corresponding author. e-mail:
<EMAIL_ADDRESS>
###### Abstract

In this paper, we investigate the global asymptotic stability of an age-
structured population dynamics model with a Ricker’s type of birth function.
This model is a hyperbolic partial differential equation with a nonlinear and
nonlocal boundary condition. We prove a uniform persistence result for the
semi-flow generated by this model. We obtain the existence of global
attractors and we prove the global asymptotic stability of the positive
equilibrium by using a suitable Lyapunov functional. Furthermore, we prove
that our global asymptotic stability result is sharp, in the sense that Hopf
bifurcation may occur as close as we want from the region global stability in
the space of parameter.

##  1 Introduction

The age-structured population dynamics models have a long history starting
from the articles of Lotka [5] in (1907) and the article of Sharpe and Lotka
[12] (1911). The first nonlinear age-structured model was introduced by Gurtin
and MacCamy [2] in the 70th. They proved a result of the existence and
uniqueness of the solutions and a local stability result for an equilibrium.
More recently, age-structured models received a lot of interest in the
literature and we refer to the books [3, 9, 14, 15] for more results about
this topic.

The article is devoted to the global asymptotic stability of the following
Gurtin-MacCamy’s age-structured model,

$\left\\{\begin{array}[]{ll}(\partial_{t}+\partial_{a})u(t,a)=-\mu
u(t,a),&t>0,a>0,\\\ u(t,0)=\alpha
f\left(\displaystyle\int_{0}^{\infty}\beta(a)u(t,a)\mathrm{d}a\right),&t>0,\end{array}\right.$
(1.1)

with the initial distribution
$u(0,a)=u_{0}(a)\in{\mathrm{L}}^{1}_{+}\left(\left(0,+\infty\right),\mathbb{R}\right).$
Here the parameters $\alpha>0,\mu>0$ and $\beta$ and $f$ satisfy the following
Assumption 1.1.

###### Assumption 1.1

We assume that $f(u)=ue^{-u},u\geq 0$ and
$\beta\in{\mathrm{L}}^{\infty}_{+}(0,+\infty)$ is normalized by

$\int_{0}^{+\infty}\beta(a)e^{-\mu a}\mathrm{d}a=1.$

To explain our result, we first need to recall that the solution integrated
along the characteristic of the system (1.1) is

$u(t,a)=\left\\{\begin{array}[]{cc}e^{-\mu t}u_{0}(a-t),&\text{ if }a\geq
t,\\\ e^{-\mu a}b(t-a),&\text{ if }a\leq t,\end{array}\right.$

where $t\to b(t)$ is the unique continuous solution of the Volterra’s integral
equation

$b(t)=\alpha f\left(\int_{t}^{\infty}\beta(a)e^{-\mu
t}u_{0}(a-t)\mathrm{d}a+\int_{0}^{t}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a\right),\forall t\geq 0.$ (1.2)

For each $\alpha>1$, the system (1.1) has a unique positive equilibrium

$\overline{u}(a)=\overline{u}(0)e^{-\mu a},$ (1.3)

and $\bar{u}(0)=\ln\alpha>0$ is the unique strictly positive solution of the
scalar equation

$\overline{u}(0)=\alpha f(\overline{u}(0)).$

If we assume that
$\beta(.)\,e^{-\mu.}\overset{\ast}{\rightharpoonup}\delta_{\tau}(a)$ for the
weak star topology (in the dual of the space of some suitable exponential
bounded continuous functions) and where $\delta_{\tau}(a)$ is the Dirac mass
at $\tau>0$. This implies in particular

$\int_{0}^{\infty}\beta(a)\,e^{-\mu
a}\varphi(a)\mathrm{d}a\to\varphi(\tau),\forall\varphi\in C_{c}(0,+\infty),$

where $C_{c}(0,+\infty)$ is the space of continuous functions with compact
support on $(0,+\infty)$.

Then, by replacing $\beta(a)\,e^{-\mu a}$ by $\delta_{\tau}(a)$) in equation
(1.2), we obtain a continuous-time difference equation of the form

$\displaystyle b(t)=\alpha f(b(t-\tau)),\forall t\geq\tau.$ (1.4)

In Theorem 6.5 we will see that the positive equilibrium $\overline{u}$ is
global stable for the system (1.1) whenever

$1<\alpha\leq e^{2}.$ (1.5)

As far as we know, global asymptotic stability and Lyapunov functional for the
age-structured model was first considered by Magal, McCluskey, and Webb [7].
We also refer to Smith and Thieme [13] for another approach for the same
infection age-structured model. Since then, the Lyapunov method of [7] has
applied to many examples of age-structured models in the literature. In the
present article, we apply the recent results of Magal, Seydi, and Wang [10],
and we reconsider the decomposition of $L^{1}_{+}$ into two invariant
subregions which are used later on in the uniform persistence section. The
invariance of subregions is easy to explain. Indeed, our global stability
result should reflect the fact that if we start only with old individuals, who
can no longer reproduce, the population goes extinct. Therefore the state-
space decomposition of $L^{1}_{+}$ should reflect this simple fact (see
Section 3 for precisions). This idea was neglected in many articles and we
hope that this new method will help to understand this part of the problem.

One may also observe that condition (1.5) corresponds to the global asymptotic
stability of the positive equilibrium $\overline{b}=\ln(\alpha)$ for the
difference equation (1.4). Moreover, $\alpha=e^{2}$ corresponds to a Hopf’s
bifurcation point for the difference equation (1.4) (when it is regarded as a
discrete-time problem). Indeed,

$\alpha f^{\prime}(\overline{b})=\alpha[1-\ln(\alpha)]e^{-\ln(\alpha)}=-1.$

Hopf’s bifurcation for the model (1.1) was first considered by Magal and Ruan
[8, 9] and Liu, Magal and Ruan [4]. In Section 7, we prove that for each
$\alpha^{\star}>e^{2}$ (as close as we want from $e^{2}$) we can find a
function $\beta(a)$ such that Hopf bifurcation occurs at
$\alpha=\alpha^{\divideontimes}\in(e^{2},\alpha^{\star})$ for the system
(1.1). In this sense, our global stability result is sharp with respect to the
parameter $\alpha$.

The paper is organized as follows. In Section 2, we state a dissipativity
result. The invariant subregion under the semiflow generated by (1.1) is
studied in Section 3. In Section 4, we consider the asymptotic smoothness of
the semiflow generated by (1.1). In Section 5, we prove that the semiflow
generated by (1.1) is uniformly persistent in the invariant subregion
$\widehat{M}_{0}$ and obtain the existence of the global attractor. In section
6, we prove that the positive equilibrium is global asymptotically stable by
using the Lyapunov functional. Section 7 is devoted to the existence of Hopf’s
bifurcation whenever $\alpha>e^{2}$.

##  2  Dissipativity

Set $\Lambda=\displaystyle\frac{\alpha}{\mu}\sup\limits_{x\geq 0}f(x).$
Consider the total number of individuals at time $t$

$U(t)=\int_{0}^{\infty}u(t,a)\mathrm{d}a,\forall t\geq 0\text{ and
}U_{0}=\int_{0}^{\infty}u_{0}(a)\mathrm{d}a.$

Then we have the following result about the total number of individuals $U(t)$
.

###### Lemma 2.1

We have

$U(t)\leq e^{-\mu t}U_{0}+\left(1-e^{-\mu t}\right)\Lambda,\forall t\geq 0.$

##  3 Invariant subregion

In this section, we use the recent approach presented in Magal, Seydi and Wang
[10]. The following comparison result can be obtained by comparison of the
Volterra integral equations.

###### Theorem 3.1

Assume that $u_{\pm}(t,a)$ satisfies

$\left\\{\begin{array}[]{l}\partial_{t}u_{\pm}(t,a)+\partial_{a}u_{\pm}(t,a)=-\mu
u_{\pm}(t,a),\text{ for }a\geq 0,\ t\geq 0,\\\
u_{\pm}(t,0)=\alpha\delta_{\pm}\int_{0}^{+\infty}\beta(a)u_{\pm}(t,a)\mathrm{d}a,\\\
u_{\pm}(0,.)=u_{0}\in{\mathrm{L}}^{1}_{+}(\left(0,+\infty\right),\mathbb{R}).\end{array}\right.$
(3.1)

Then

$u_{-}(t,a)\leq u(t,a)\leq u_{+}(t,a),\ \mbox{for}\ t\geq 0,\ \mbox{for a.
e.}\ a\geq 0,$

where
$\delta_{-}=e^{-\|\beta\|_{{\mathrm{L}}^{\infty}}\max(U_{0},\Lambda)}<1=\delta_{+}.$

Proof. We just need to observe that

$b(t)=\alpha\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a\exp\left(-\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a\right){.}$

Therefore by using Lemma 2.1, we have

$\alpha\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}ae^{-\|\beta\|_{{\mathrm{L}}^{\infty}}\max(U_{0},\Lambda)}\leq
b(t)\leq\alpha\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a.$

Then the result follows by using the comparison result in [10].

Let

$\Gamma^{\pm}(a)=\alpha\delta_{\pm}\int_{a}^{\infty}e^{-\int_{a}^{\theta}[\mu+\lambda^{\pm}]\mathrm{d}\sigma}\beta(\theta)\mathrm{d}\theta,$

where $\lambda^{\pm}$ is chosen to satisfy that $\Gamma^{\pm}(0)=1$. That is
to say that $\lambda^{\pm}\in\mathbb{R}$ satisfies

$\alpha\delta_{\pm}\int_{0}^{\infty}e^{-\int_{0}^{\theta}[\mu+\lambda^{\pm}]\mathrm{d}\sigma}\beta(\theta)\mathrm{d}\theta=1.$

###### Remark 3.2

$\lambda^{\pm}\in\mathbb{R}$ is the dominant eigenvalue of the linear operator
associated to the system (3.1). The above integral equation corresponds to the
characteristic equation of this linear operator.

The following lemma is showing that $\Gamma^{\pm}(a)$ is an adjoint
eigenfunction associated the eigenvalues $\lambda^{\pm}$.

###### Lemma 3.3

For each $t\geq 0$,

$\int_{0}^{\infty}\Gamma^{\pm}(a)u_{\pm}(t,a)\mathrm{d}a=e^{\lambda^{\pm}t}\int_{0}^{\infty}\Gamma^{\pm}(a)u_{0}(a)\mathrm{d}a.$

Proof. The function $a\to\Gamma^{\pm}(a)$ satisfies

$\left\\{\begin{array}[]{ll}(\Gamma^{\pm})^{\prime}(a)=[\mu+\lambda^{\pm}]\Gamma^{\pm}(a)-\alpha\delta_{\pm}\beta(a),\
\mbox{for a. e.}\ a\geq 0,\\\ \Gamma^{\pm}(0)=1.\end{array}\right.$

By using classical solutions of (3.1), we deduce that

$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{\infty}\Gamma^{\pm}(a)u_{\pm}(t,a)\mathrm{d}a=\int_{0}^{\infty}\Gamma^{\pm}(a)[-\partial_{a}u_{\pm}(t,a)-\mu
u_{\pm}(t,a)]\mathrm{d}a.$

Using integration by parts, it follows that

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{\infty}\Gamma^{\pm}(a)u_{\pm}(t,a)\mathrm{d}a$
$\displaystyle=$
$\displaystyle\Gamma^{\pm}(0)u_{\pm}(t,0)+\int_{0}^{\infty}(\Gamma^{\pm})^{\prime}(a)u_{\pm}(t,a)\mathrm{d}a$
$\displaystyle-\int_{0}^{\infty}\Gamma^{\pm}(a)\mu u_{\pm}(t,a)\mathrm{d}a.$

Using the facts $\Gamma^{\pm}(0)=1$ and
$u_{\pm}(t,0)=\alpha\delta_{\pm}\int_{0}^{+\infty}\beta(a)u_{\pm}(t,a)\mathrm{d}a$,
we have

$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{\infty}\Gamma^{\pm}(a)u_{\pm}(t,a)\mathrm{d}a=\lambda^{\pm}\int_{0}^{\infty}\Gamma^{\pm}(a)u_{\pm}(t,a)\mathrm{d}a,$

and the result follows from the fact that the set of initial values giving a
classical solution is dense in $L^{1}$.

By Assumption 1.1 we have $\beta\neq 0$. Therefore we can define

$a^{\star}:=\sup\left\\{a>0:\int_{a}^{\infty}\beta(\sigma)e^{-\mu\sigma}\mathrm{d}\sigma>0\right\\}\in(0,\infty].$

###### Remark 3.4

In practice, the number $a^{\star}$ corresponds to the maximal value at which
individuals can reproduce.

Assume for simplicity that $a\to\beta(a)$ is a continuous function. Then,
$a^{\star}=\infty$ if and only if for each $a\geq 0$, there exists
$\widehat{a}>a$ such that $\beta(\widehat{a})>0.$ If $a^{\star}<\infty$, then

$\beta(a)=0,\forall a\geq a^{\star}.$

Define the interior sub-domain

$\widehat{M}_{0}=\left\\{u\in{\mathrm{L}}^{1}_{+}(0,+\infty):\int_{0}^{a^{\star}}u(a)\mathrm{d}a>0\right\\},$

and the boundary sub-domain

$\partial\widehat{M}_{0}=\left\\{u\in{\mathrm{L}}^{1}_{+}(0,+\infty):\int_{0}^{a^{\star}}u(a)\mathrm{d}a=0\right\\}.$

Actually, the boundary domain $\partial\widehat{M}_{0}$ corresponds to a case
where the distribution $u$ contains only individuals that will not reproduce.
However, some individuals in the interior region will produce newborn
individuals. Due to the irreducible structured of the semiflow generated by
(1.1), we can obtain the invariance of $\partial\widehat{M}_{0}$ and
$\widehat{M}_{0}$.

###### Theorem 3.5

Let Assumption 1.1 be satisfied. The domains
$\left[0,\infty\right)\times\widehat{M}_{0}$ and
$\left[0,\infty\right)\times\partial\widehat{M}_{0}$ are positively invariant
by the semiflow generated by (1.1). That is to say that

$\int_{0}^{a^{\star}}u_{0}(a)\mathrm{d}a>0\Rightarrow\int_{0}^{a^{\star}}u(t,a)\mathrm{d}a>0,\forall
t\geq 0,$

and

$\int_{0}^{a^{\star}}u_{0}(a)\mathrm{d}a=0\Rightarrow\int_{0}^{a^{\star}}u(t,a)\mathrm{d}a=0,\forall
t\geq 0.$

Moreover if $\int_{0}^{a^{\star}}u_{0}(a)\mathrm{d}a=0$, then

$\int_{0}^{\infty}\beta(a)u(t,a)\mathrm{d}a=0,\forall t\geq 0,$

and the solution is explicitly given by

$u(t,a)=\left\\{\begin{array}[]{l}e^{-\mu t}u_{0}(a-t),\text{{\rm if }}a-t\geq
0,\\\ 0,\text{{\rm if }}t-a\geq 0.\end{array}\right.$

Therefore, $\lim_{t\to\infty}\|u(t,.)\|_{{\mathrm{L}}^{1}}=0,$ and the
convergence is exponential.

Proof. Assume first $u_{0}\in\partial\widehat{M}_{0}$. Then,
$\int_{0}^{\infty}\Gamma^{+}(a)u_{0}(a)\mathrm{d}a=0.$ Moreover,

$\int_{0}^{\infty}\Gamma^{+}(a)u(t,a)\mathrm{d}a\leq\int_{0}^{\infty}\Gamma^{+}(a)u_{+}(t,a)\mathrm{d}a=e^{\lambda^{+}t}\int_{0}^{\infty}\Gamma^{+}(a)u_{0}(a)\mathrm{d}a=0{.}$

Therefore, $\int_{0}^{\infty}\Gamma^{+}(a)u(t,a)\mathrm{d}a=0,\forall t\geq
0.$ Hence,

$u(t,.)\in\partial\widehat{M}_{0},\forall t\geq 0.$

Assume next that $u_{0}\in\widehat{M}_{0}$. Then it follows that
$\int_{0}^{\infty}\Gamma^{-}(a)u_{0}(a)\mathrm{d}a>0,$ and

$\int_{0}^{\infty}\Gamma^{-}(a)u(t,a)\mathrm{d}a\geq\int_{0}^{\infty}\Gamma^{-}(a)u_{-}(t,a)\mathrm{d}a=e^{\lambda^{-}t}\int_{0}^{\infty}\Gamma^{-}(a)u_{0}(a)\mathrm{d}a>0.$

Thus,

$u(t,.)\in\widehat{M}_{0},\forall t\geq 0.$

Let $u_{0}\in\widehat{M}_{0}$. Since the function $a\to\beta(a)$ is not
assumed to be strictly positive on the interval $[0,a^{*})$, the previous
theorem does not imply that

$b(t)=\int_{0}^{\infty}\beta(a)u(t,a)\mathrm{d}a>0,\forall t\geq 0.$

By using the previous assumption, we should consider two different cases: 1)
$a^{*}=\infty$, then

$\int_{a}^{\infty}\beta(a)e^{-\mu a}\mathrm{d}a>0,\forall a>0;$

and 2) $a^{*}<\infty$, then

$\int_{a}^{a^{*}}\beta(a)e^{-\mu a}\mathrm{d}a>0,\forall a\in(0,a^{*}).$

###### Proposition 3.6

Let Assumption 1.1 be satisfied. For each $u_{0}\in\widehat{M}_{0}$, there
exists $t_{0}=t_{0}(u_{0})>0$ such that $b(t)>0,\forall t\geq t_{0}.$

Proof. Due to the fact that $u(t,.)\geq u_{-}(t,.),\forall t\geq 0$, it is
sufficient to consider that $t\to
B_{-}(t):=\int_{0}^{+\infty}\beta(a)u_{-}(t,a)\mathrm{d}a$ is the unique
solution of the Volterra integral equation

$B_{-}(t)=F_{-}(t)+\int_{0}^{t}\gamma(a)B_{-}(t-a)\mathrm{d}a,$

where $\gamma(a):=\beta(a)e^{-\mu a}$ and

$F_{-}(t)=e^{-\mu t}\int_{t}^{\infty}\beta(a)u_{0}(a-t)\mathrm{d}a=e^{-\mu
t}\int_{0}^{\infty}\beta(\sigma+t)u_{0}(\sigma)\mathrm{d}\sigma.$

By using the continuity of the shift operator in $L^{1}$, we deduce that the
map

$t\to\int_{t}^{\infty}\beta(a)u_{0}(a-t)\mathrm{d}a$

is continuous. Therefore, the map $t\to F_{-}(t)$ is continuous.

First part of the proof: Let $\Delta t>0$. By using Fubini’s theorem, we have

$\int_{t}^{t+\Delta
t}\int_{0}^{\infty}\beta(\sigma+\theta)u_{0}(\sigma)\mathrm{d}\sigma\mathrm{d}\theta=\int_{0}^{\infty}\int_{t}^{t+\Delta
t}\beta(\sigma+\theta)\mathrm{d}\theta u_{0}(\sigma)\mathrm{d}\sigma.$

Then, by using the definition of $a^{*}$, we deduce that there exists
$t^{*}\in(0,a^{*})$ such that

$\int_{0}^{\infty}\beta(\sigma+t^{*})u_{0}(\sigma)\mathrm{d}\sigma>0\Leftrightarrow
F_{-}(t^{*})>0.$

Second part of the proof:  In order to prove this theorem, we observe that for
all $t\geq 0$

$\int_{t}^{t+\Delta t}B_{-}(\theta)\mathrm{d}\theta=\int_{t}^{t+\Delta
t}F_{-}(\theta)\mathrm{d}\theta+\int_{t}^{t+\Delta
t}\int_{0}^{t}\gamma(\theta-a)B_{-}(a)\mathrm{d}a\mathrm{d}\theta.$

Then, by using Fubini’s theorem, we obtain for all $t\geq 0$

$\int_{t}^{t+\Delta t}B_{-}(l)\mathrm{d}l=\int_{t}^{t+\Delta
t}F_{-}(l)\mathrm{d}l+\int_{0}^{t}\int_{t-a}^{t-a+\Delta
t}\gamma(\theta)\mathrm{d}\theta B_{-}(a)\mathrm{d}a.$

Now, you can work like if $\gamma(q)$ were a continuous function, and we
deduce that the length of the support $t\to B_{-}(t)$ increases from
$(0,a^{*})$ to $(a^{*},2a^{*})$. By induction, we deduce that there exists an
integer $m\geq 1$ such that $t\to B_{-}(t)$ is strictly positive on
$(ma^{*},(m+1)a^{*})$. The result follows.

##  4 Asymptotic smoothness of the semiflow

In this section, we prove that the semiflow generated by (1.1) is
asymptotically smooth. Define

$u_{1}(t,a)=\left\\{\begin{array}[]{cc}e^{-\mu t}u_{0}(a-t),&\text{ if }a\geq
t,\\\ e^{-\mu a}b_{1}(t-a),&\text{ if }a\leq t,\end{array}\right.$

and

$u_{2}(t,a)=\left\\{\begin{array}[]{cc}0,&\text{ if }a\geq t,\\\ e^{-\mu
a}b_{2}(t-a),&\text{ if }a\leq t,\end{array}\right.$

where

$\begin{array}[]{l}b_{1}(t)=\alpha\left[f\left(B(t)\right)-f(B_{2}(t))\right],\quad
b_{2}(t)=\alpha f(B_{2}(t)),\end{array}$

and

$B(t)=\int_{0}^{\infty}u(t,a)\mathrm{d}a=\underset{B_{1}(t)}{\underbrace{e^{-\mu
t}\int_{t}^{\infty}\beta(a)u_{0}(a-t)\mathrm{d}a}}+\underset{B_{2}(t)}{\underbrace{\int_{0}^{t}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a}}.$

###### Theorem 4.1

The semiflow generated by (1.1) is asymptotically smooth. Moreover, precisely
the semiflow $U(t)$ generated by (1.1) on $L^{1}_{+}$ can be decomposed into

$U(t)u_{0}=C(t)u_{0}+V(t)u_{0},\forall t\geq 0,\forall u_{0}\in L^{1}_{+},$

where $C(t):L^{1}_{+}\to L^{1}$ and $V(t):L^{1}_{+}\to L^{1}$ are defined by

$C(t)u_{0}=u_{2}(t,.),\text{ and }V(t)u_{0}=u_{1}(t,.),\forall t\geq 0,\forall
u_{0}\in L^{1}_{+}.$

Then we have the following properties

  * (i)

The nonlinear operator $C(t):L^{1}_{+}\to L^{1}$ is completely continuous for
each $t\geq 0$;

  * (ii)

There exists a constant $\chi>0$ such that $\|V(t)u_{0}\|\leq\chi e^{-\mu
t},\forall t\geq 0.$

Proof. Proof of (i): It is sufficient to observe that $t\to B_{2}(t)$ is
uniformly continuous on $[0,\tau]$ uniformly with respect to $b(t)$. Indeed,
we have $\forall t,s\in[0,\tau]$ (with $t\geq s$)

$|B_{2}(t)-B_{2}(s)|\leq\sup_{\sigma\in[0,\tau]}|b(\sigma)|\left[\int_{0}^{t-s}g(l)\mathrm{d}l+\int_{0}^{s}|g(t-s+l)-g(l)|\mathrm{d}l\right],$

where $g(a)=\beta(a)e^{-\mu a}$. By applying Arzela-Ascoli’s theorem, it
follows that $C(t)$ maps the bounded sets into relatively compact sets.

Proof of (ii): Moreover, since $f(u)$ is Lipschitz continuous on $[0,\infty)$,
we deduce that

$\|u_{1}(t,.)\|_{L^{1}}\leq e^{-\mu
t}\|u_{0}\|_{L^{1}}+\|e^{-\mu.}\|_{L^{1}}\alpha\|f^{\prime}\mid_{[0,\infty)}\|_{\infty}B_{1}(t).$

Thus, there exists a constant
$\chi=1+\|e^{-\mu.}\|_{L^{1}}\alpha\|f^{\prime}\mid_{[0,\infty)}\|_{\infty}\|\beta\|_{L^{\infty}}$
such that

$\|u_{1}(t,.)\|_{L^{1}}\leq\chi e^{-\mu t}\|u_{0}\|_{L^{1}},\forall t\geq 0.$

The proof is completed.

###### Theorem 4.2

The semiflow generated by (1.1) is asymptotically smooth. That is to say that

$\lim_{t\to\infty}\kappa\left(U(t)B\right)=0,$

where $\kappa$ is the Kuratovsky’s measure of non-compactness defined by

$\kappa(B)=\inf\left\\{\varepsilon>0:B\text{ can be covered by a finite number
of balls of radius }\leq\varepsilon\right\\},$

for any bounded set $B$ of $L^{1}_{+}(0,\infty)$.

Proof. For each bounded set $B\subset L^{1}_{+}(0,\infty)$, we have the
Kuratovsky’s measure of non-compactness satisfies

$\kappa(U(t)B)\leq\kappa(V(t)B)+\kappa(C(t)B)=\kappa(V(t)B)\leq\chi e^{-\mu
t}\sup_{u\in B}\|u\|_{L^{1}}.$

The proof is completed.

##  5 Uniform persistence and global attractor

In this section, motivated by the Magal [6], Magal and Zhao [11], Zhao [16],
we obtain the existence of the global attractor.

###### Assumption 5.1

We assume that $\alpha>1$.

###### Theorem 5.2

Let Assumptions 1.1 and 5.1 be satisfied. Then the semiflow $U(t)$ is
uniformly persistent in $\widehat{M}_{0}$. That is, there exists
$\varepsilon>0$ such that for any $u_{0}\in\widehat{M}_{0}$,

$\liminf_{t\to+\infty}\|u(t,\cdot)\|_{L^{1}}\geq\varepsilon.$

Furthermore, the semiflow $U$ has a compact global attractor $\mathcal{A}_{0}$
in $\widehat{M}_{0}$.

Proof.

Assume that the semiflow $U(t)$ is not uniformly persistent in
$\widehat{M}_{0}$. Then for any $\varepsilon>0$, there exists some
$u_{0}\in\widehat{M}_{0}$ such that

$\|u(t,\cdot)\|_{L^{1}}\leq\varepsilon,\quad\forall t\geq 0.$ (5.1)

Thus, we have

$\exp\left({-\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a}\right)\geq\exp\left({-\|\beta\|_{L^{\infty}}\|u(t,\cdot)\|_{L^{1}}}\right)\geq\delta_{*},$

where $\delta_{*}=e^{-\varepsilon\|\beta\|_{L^{\infty}}}$. Thanks to the
comparison principle of Theorem 3.1, we have

$u(t,a)\geq u_{*}(t,a),\quad t\geq 0,\mbox{for a. e.}\ a\geq 0,$

where $u_{*}(t,a)$ satisfies

$\left\\{\begin{array}[]{l}\partial_{t}u_{*}(t,a)+\partial_{a}u_{*}(t,a)=-\mu
u_{*}(t,a),\text{ for }a\geq 0,\ t\geq 0,\\\
u_{*}(t,0)=\alpha\delta_{*}\int_{0}^{+\infty}\beta(a)u_{*}(t,a)\mathrm{d}a,\\\
u_{*}(0,.)=u_{0}\in\widehat{M}_{0}.\end{array}\right.$

For this $u_{0}\in\widehat{M}_{0}$, we have
$\int_{0}^{\infty}\Gamma^{-}(a)u_{0}(a)\mathrm{d}a>0$ and

$\int_{0}^{\infty}\Gamma^{-}(a)u(t,a)\mathrm{d}a\geq\int_{0}^{\infty}\Gamma^{-}(a)u_{*}(t,a)\mathrm{d}a=e^{\lambda_{*}t}\int_{0}^{\infty}\Gamma^{-}(a)u_{0}(a)\mathrm{d}a>0,$

where $\lambda_{*}\in\mathbb{R}$ satisfies

$\alpha\delta_{*}\int_{0}^{+\infty}\beta(a)e^{-(\mu+\lambda_{*})a}\mathrm{d}a=1.$

By Assumption 5.1, we can choose $\varepsilon>0$ small enough to ensure that
$\alpha\delta_{*}>1$. Therefore, it follows from the definition of the
$\lambda_{*}$ and Assumption 1.1 that $\lambda_{*}>0$. Thus,

$\lim_{t\to+\infty}\|u(t,\cdot)\|_{L^{1}}\geq\lim_{t\to\infty}\int_{0}^{\infty}\Gamma^{-}(a)u(t,a)\mathrm{d}a=+\infty,$

which contradicts to (5.1).

Since the semiflow $U$ is asymptotically smooth, point dissipative and
positive orbits of compact subsets of $\widehat{M}_{0}$ are bounded, we deduce
that $U$ has a compact global attractor $\mathcal{A}_{0}$ in
$\widehat{M}_{0}$.

###  5.1 Complete orbits

###### Proposition 5.3

Let Assumption 1.1 be satisfied. There exists $\delta>0$ such that

$\int_{0}^{+\infty}\beta(a)u_{0}(a)\mathrm{d}a\geq\delta,\quad\forall
u_{0}\in\mathcal{A}_{0}.$

Proof. Let $u(t,\cdot)$ be the solution to (1.1) with the initial condition
$u_{0}\in\mathcal{A}_{0}$. By Proposition 3.6, there exists
$t_{0}=t_{0}(u_{0})>0$ such that

$\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a>0,\quad\forall t\geq t_{0}.$

Recall that the mapping $(t,u_{0})\to U(t)u_{0}$ is continuous. Also, the
mapping $u_{0}\to\int_{0}^{+\infty}\beta(a)u_{0}(a)\mathrm{d}a$ is continuous.
Thus, there exists $\varepsilon=\varepsilon(u_{0})>0$ such that if
$\tilde{u}_{0}\in\mathcal{A}_{0}$ with
$\|\tilde{u}_{0}-u_{0}\|\leq\varepsilon,$ and the solution to (1.1) with the
initial condition $\tilde{u}_{0}$ is denoted by $\tilde{u}(t,\cdot)$, then

$\int_{0}^{+\infty}\beta(a)\tilde{u}(t,a)\mathrm{d}a>0,\quad\forall
t\in[t_{0},t_{0}+a_{0}],$

for some $a_{0}>0$. Thus, it follows from the proof of the Proposition 3.6
that

$\int_{0}^{+\infty}\beta(a)\tilde{u}(t,a)\mathrm{d}a>0,\quad\forall t\geq
t_{0}.$

Therefore, by using the compactness of $\mathcal{A}_{0}$, we deduce that there
exists $\hat{t}>0$ (independent of $\mathcal{A}_{0}$) such that for each
$u_{0}\in\mathcal{A}_{0}$,

$\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a>0,\quad\forall t\geq\hat{t}.$

By using the fact that $\mathcal{A}_{0}$ is invariant under $U$, it follows
that for each $u_{0}\in\mathcal{A}_{0}$,

$\int_{0}^{+\infty}\beta(a)u_{0}(a)\mathrm{d}a>0.$

Now, by using the continuity of
$u_{0}\to\int_{0}^{+\infty}\beta(a)u_{0}(a)\mathrm{d}a$ again and the
compactness of $\mathcal{A}_{0}$, we obtain the results and complete the
proof.

###### Theorem 5.4

Assume that $t\to u(t,a)$ is a complete orbit of the system (1.1). Then

$u(t,a)=e^{-\mu a}b(t-a),\text{ for almost every }a\geq 0,\forall
t\in\mathbb{R},$ (5.2)

and the map $t\to b(t)$ satisfies the renewal equation

$b(t)=\alpha f\left(\int_{0}^{\infty}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a\right),\forall t\in\mathbb{R}.$ (5.3)

Moreover, if we assume in addition that

$\gamma_{-}\leq\int_{0}^{a_{*}}u(t,a)\mathrm{d}a\leq\gamma_{+},\forall
t\in\mathbb{R},$ (5.4)

for some $\gamma_{+}>\gamma_{-}>0$, then

$\delta^{-}\leq\int_{0}^{\infty}\beta(a)u(t,a)\mathrm{d}a\leq\delta^{+},\forall
t\in\mathbb{R},$

for some $\delta^{+}\geq\delta^{-}>0$.

Proof. Let $t_{0}\leq 0$ and $t\geq t_{0}$. Due to $t\to u(t,a)$ is the
complete orbit of the system (1.1), the solution integrated along the
characteristic of (1.1) with the initial value $u(t_{0},a)$ is

$u(l,a)=\left\\{\begin{array}[]{ll}e^{-\mu(l-t_{0})}u(t_{0},a-(l-t_{0})),&\text{
if }a\geq l-t_{0},\\\ e^{-\mu a}b(l-t_{0}-a),&\text{ if }a\leq
l-t_{0},\end{array}\right.$ (5.5)

where $l\to b(l)$ is the unique continuous solution of the Volterra’s integral
equation

$b(l-t_{0})=\alpha
f\left(\int_{l-t_{0}}^{\infty}\beta(a)e^{-\mu(l-t_{0})}u(t_{0},a-(l-t_{0}))\mathrm{d}a+\int_{0}^{l-t_{0}}\beta(a)e^{-\mu
a}b(l-t_{0}-a)\mathrm{d}a\right).$ (5.6)

Noticing (5.5), (5.6) and (5.3) and setting $t=l-t_{0}$, the result (5.2)
follows.

It follows from (5.2) that

$\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a=\int_{0}^{+\infty}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a\leq\delta^{+},$

where $\delta^{+}=\alpha\max\limits_{u\geq 0}f(u)$.

Since $\mathcal{A}_{0}$ is the global attractor, which implies that
$U(t)\mathcal{A}_{0}=\mathcal{A}_{0}$ for all $t\geq 0$, we can find a
complete orbit $u(t,\cdot)$ through $u(0,\cdot)$. Due to (5.4), it follows
that for all $t\in\mathbb{R}$, there exists a $\hat{t}\leq t$ such that
$\int_{0}^{+\infty}\beta(a)u(\hat{t},a)\mathrm{d}a>0.$ Then, by the method of
Proposition 3.6, we have

$\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a>0,\quad t\in\mathbb{R}.$

Since $\mathcal{A}_{0}$ is compact, it follows that there exists some
$\delta^{-}>0$ such that

$\int_{0}^{+\infty}\beta(a)u(t,a)\mathrm{d}a\geq\delta^{-},$

which completes the proof.

##  6 Global stability result

In this section, we use a Lyapunov functional to show that all solutions of
(1.1) for which the population is initially present tend to the positive
equilibrium under the Assumption 6.2.

Define

$V(u(t,\cdot))=\int_{0}^{+\infty}\gamma(a)g\left(\frac{u(t,a)}{\bar{u}(a)}\right)\mathrm{d}a,$
(6.1)

where

$\gamma(a)=\int_{a}^{+\infty}\beta(\sigma)e^{-\mu\sigma}\mathrm{d}\sigma,\quad
g(u)=(u-1)^{2}.$

###### Remark 6.1

We observe that $V$ is well-defined on the attractor $\mathcal{A}_{0}$ for
every $t\geq 0$, while $V$ is not well-defined on $L^{1}(0,+\infty)$ because
of the function $g$ under the integral.

###### Assumption 6.2

We assume that $1<\alpha\leq e^{2}$.

The following Lemma is due to Fisher and Goh [1].

###### Lemma 6.3

Let Assumption 6.2 be satisfied. Then for each $u\geq 0$,

$\left|{\alpha f(u)-\overline{u}(0)}\right|=\left|{\alpha f(u)-\alpha
f(\overline{u}(0))}\right|\leq\left|{u-\overline{u}(0)}\right|.$ (6.2)

Moreover the above inequality is strict whenever $u\neq 0$ and
$u\neq\overline{u}(0)$.

######  Proposition 6.4 (Lyapunov function)

Let Assumptions 1.1 and 6.2 be satisfied. Assume that $t\to u(t,a)$ is a
complete orbit of the system (1.1) satisfying

$\gamma_{-}\leq\int_{0}^{a_{*}}u(t,a)\mathrm{d}a\leq\gamma_{+},\forall
t\in\mathbb{R},$

for some $\gamma_{+}>\gamma_{-}>0$. Then the function $t\to V(u(t,\cdot))$ is
decreasing. Moreover, $t\to V(u(t,\cdot))$ is constant if
$u(t,a)=\overline{u}(a),\forall t\in\mathbb{R}.$

Proof. We have

$\displaystyle\frac{\mathrm{d}V(u(t,\cdot))}{\mathrm{d}t}=$
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{+\infty}\gamma(a)g\left(\frac{u(t,a)}{\bar{u}(a)}\right)\mathrm{d}a=\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{+\infty}\gamma(a)g\left(\frac{b(t-a)}{\bar{u}(0)}\right)\mathrm{d}a$
$\displaystyle=$
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{t}\gamma(t-s)g\left(\frac{b(s)}{\bar{u}(0)}\right)\mathrm{d}s$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{t}\gamma^{\prime}(t-s)g\left(\frac{b(s)}{\bar{u}(0)}\right)\mathrm{d}s+\gamma(0)g\left(\frac{b(t)}{\bar{u}(0)}\right).$

It follows from the definition of $\gamma(a)$ that
$\gamma^{\prime}(a)=-\beta(a)e^{-\mu a}$ and $\gamma(0)=1$. Then,

$\frac{\mathrm{d}V(u(t,\cdot))}{\mathrm{d}t}=g\left(\frac{b(t)}{\bar{u}(0)}\right)-\int_{0}^{+\infty}\beta(a)e^{-\mu
a}g\left(\frac{b(t-a)}{\bar{u}(0)}\right)\mathrm{d}a.$ (6.3)

By using Lemma 6.3, we have

$\begin{array}[]{ll}g\left(\dfrac{b(t)}{\bar{u}(0)}\right)=&\dfrac{1}{\bar{u}^{2}(0)}\left[\alpha
f\left(\displaystyle\int_{0}^{+\infty}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a\right)-\bar{u}(0)\right]^{2}\\\
&\leq\dfrac{1}{\bar{u}^{2}(0)}\left[{\displaystyle\int_{0}^{+\infty}\beta(a)e^{-\mu
a}b(t-a)\mathrm{d}a-\bar{u}(0)}\right]^{2}.\end{array}$ (6.4)

Thus,

$g\left(\frac{b(t)}{\bar{u}(0)}\right)\leq
g\left(\int_{0}^{+\infty}\beta(a)e^{-\mu
a}\frac{b(t-a)}{\bar{u}(0)}\mathrm{d}a\right).$ (6.5)

Substituting the inequality (6.5) to (6.3), we obtain

$\displaystyle\frac{\mathrm{d}V(u(t,\cdot))}{\mathrm{d}t}\leq$ $\displaystyle
g\left(\int_{0}^{+\infty}\beta(a)e^{-\mu
a}\frac{b(t-a)}{\bar{u}(0)}\mathrm{d}a\right)-\int_{0}^{+\infty}\beta(a)e^{-\mu
a}g\left(\frac{b(t-a)}{\bar{u}(0)}\right)\mathrm{d}a,$

and we deduce that $\frac{\mathrm{d}V(u(t,\cdot))}{\mathrm{d}t}\leq 0$ by
using Jensen’s inequality and the fact that $g$ is a convex function.

To prove the last part of the proposition, it is sufficient to observe that by
Lemma 6.3, the inequality (6.4) is strict whenever
$\int_{0}^{+\infty}\beta(a)e^{-\mu a}b(t-a)\mathrm{d}a\neq\overline{u}(0)$.
Therefore, the function $t\to V(u(t,\cdot))$ is constant if and only if

$\int_{0}^{+\infty}\beta(a)e^{-\mu a}b(t-a)\mathrm{d}a=\overline{u}(0),\forall
t\in\mathbb{R},$

and the result follows by using (5.2) and (5.3).

###### Theorem 6.5

Let Assumptions 1.1 and 6.2 be satisfied. Then, the positive equilibrium
$\bar{u}(a)$ of the system (1.1) is global asymptotically stable.

Proof. Let $u(t,\cdot)$ be a complete solution to (1.1) that lies in the
attractor $\mathcal{A}_{0}\subset\widehat{M}_{0}$. It follows from Proposition
5.3 that there exist $\delta_{1},\delta_{2}>0$ such that

$\delta_{1}\leq\frac{u(t,a)}{\bar{u}(a)}\leq\delta_{2},\quad\forall a\geq
0,t\in\mathbb{R}.$ (6.6)

By the definition of the function $g$ in (6.1), we observe that
$g:(0,+\infty)\to\mathbb{R}_{+}$ is decreasing for $u\in(0,1)$ and is
increasing for $u>1$ and has global minimum $g(1)=0$. It follows from (6.6)
that $g\left(\frac{u(t,a)}{\bar{u}(a)}\right)$ is bounded. Thus, the Lyapunov
functional $V(t)$ in (6.1) is bounded for all $t\in\mathbb{R}$. Furthermore,
by Proposition 6.4, the function $t\to V(u(t,\cdot))$ is decreasing and is
constant if $u(t,a)=\overline{u}(a),\quad\forall t\in\mathbb{R}.$ Thus, we
have now shown that the arbitrary complete solution $u(t,\cdot)$ in the
attractor $\mathcal{A}_{0}$ must be the positive equilibrium $\bar{u}(a)$,
namely $\mathcal{A}_{0}=\\{\bar{u}(a)\\}$.

##  7 Hopf’s bifurcation

This section is based on Magal and Ruan [8]. In order to study Hopf
bifurcation, we should consider the parameter $\alpha$ in the region
$\\{\alpha\in\mathbb{R}:\alpha>e^{2}\\}$. The linearized system (1.1) around
$\bar{u}(a)$ is

$\left\\{\begin{array}[]{ll}(\partial_{t}+\partial_{a})u(t,a)=-\mu
u(t,a),&t>0,a>0,\\\
u(t,0)=(1-\ln\alpha)\displaystyle\int_{0}^{\infty}\beta(a)u(t,a)\mathrm{d}a,&t>0.\end{array}\right.$
(7.1)

Then, the characteristic equation of the linear system (7.1) can be expressed
by

$\Delta(\lambda,\alpha)=1-(1-\ln\alpha)\int_{0}^{+\infty}\beta(a)e^{-(\lambda+\mu)}\mathrm{d}a=0.$
(7.2)

###### Assumption 7.1

Assume the birth function

$\beta(a)=C_{0}(a-\tau)^{n}e^{-\kappa(a-\tau)}1_{[\tau,+\infty)}(a),$

with $\tau>0,\kappa>0,n\in\mathbb{N}$ or $\tau>0,\kappa=0,n=0$, where the
constant $C_{0}$ is dependent on the parameters $\tau,\kappa$ and $n$ to
ensure that $\int_{0}^{+\infty}\beta(a)e^{-\mu a}\mathrm{d}a=1$.

Under Assumption 7.1, the characteristic equation (7.2) can be rewritten as

$\Delta(\lambda,\alpha)=1-(1-\ln\alpha)e^{-\lambda\tau}\left(1+\frac{\lambda}{\mu+\kappa}\right)^{-n-1}=0.$
(7.3)

Now we are position to look for purely imaginary roots
$\lambda=\pm\mathrm{i}\omega$ with $\omega>0$ of the characteristic equation
(7.3).

###### Proposition 7.2

Let Assumption 7.1 be satisfied and the parameters
$\mu>0,\tau>0,\kappa>0,n\in\mathbb{N}$ be fixed. Then the characteristic
equation (7.3) has a pair of purely imaginary solution $\pm\mathrm{i}\omega$
with $\omega>0$ if and only if $\omega>0$ is a solution of

$\omega\tau+(n+1)\arctan\frac{\omega}{\mu+\kappa}=\pi+2k\pi,\quad
k\in\mathbb{N},$ (7.4)

and

$\alpha=\exp\left(1+\left(1+\frac{\omega^{2}}{(\mu+\kappa)^{2}}\right)^{\frac{n+1}{2}}\right).$
(7.5)

Furthermore, for each $k\in\mathbb{N}$, there exists exactly one solution
$\omega_{k}$ for equation (7.4), i.e., the characteristic equation (7.3) has
exactly one pair of purely imaginary eigenvalues $\pm\mathrm{i}\omega_{k}$
with $\omega_{k}>0$ for each

$\alpha_{k}=\exp\left(1+\left(1+\frac{\omega_{k}^{2}}{(\mu+\kappa)^{2}}\right)^{\frac{n+1}{2}}\right).$
(7.6)

Moreover, $\omega_{k}\to+\infty\text{ and }\alpha_{k}\to+\infty,\text{ as
}k\to+\infty.$ If $\omega_{k}(n)>0$ is the solution of equation (7.4) with
fixed $k\in\mathbb{N}$ for any $n\in\mathbb{N}$, then

$\omega_{k}(n)\to 0\text{ and }\alpha_{k}(n)\to e^{2},\text{ as }n\to+\infty.$
(7.7)

Proof. Under Assumption 7.1, if the characteristic equation (7.3) admits a
pair of purely imaginary solution $\lambda=\pm\mathrm{i}\omega$ with
$\omega>0$, then (7.3) can be expressed

$(1-\ln\alpha)(r(\omega))^{-n-1}e^{-\mathrm{i}[\omega\tau+(n+1)\theta(\omega)]}=1,$
(7.8)

where

$r(\omega)=\sqrt{1+\frac{\omega^{2}}{(\mu+\kappa)^{2}}},\quad\theta(\omega)=\arctan\frac{\omega}{\mu+\kappa}.$

Therefore, by separating the real and imaginary parts of (7.8), we can obtain
that $\omega$ satisfies (7.4)–(7.5). Thus, the characteristic equation (7.3)
has a pair of purely imaginary solution $\lambda=\pm\mathrm{i}\omega$ with
$\omega>0$ if and only if $\omega$ is a solution of (7.4) and (7.5).

Let $h(\omega)=\omega\tau+(n+1)\arctan\frac{\omega}{\mu+\kappa}$ for
$\omega\geq 0$. Then, $h(w)$ is strictly increasing function for $\omega>0$.
Noticing $h(0)=0$ and $h(\omega)\to+\infty$ as $\omega\to+\infty$, it follows
from the continuity of $h(\omega)$ that the equation (7.5) has exactly one
solution $\omega_{k}$ for each $k\in\mathbb{N}$ with the parameter
$\alpha_{k}$ satisfying (7.6). Moreover, let $\omega_{k}>0(k\in\mathbb{N})$
satisfy

$\omega_{k}\tau+(n+1)\arctan\frac{\omega_{k}}{\mu+\kappa}=\pi+2k\pi.$

It follows from that $\omega_{k}\to+\infty$ by letting $k\to+\infty$ on both
sides of the above equality. The result $\alpha_{k}\to+\infty\text{ as
}k\to+\infty$ can be obtained from (7.6) and $\omega_{k}\to+\infty$ as
$k\to+\infty$.

Let $\omega_{k}(n)>0$ be the solution of equation (7.4) with fixed
$k\in\mathbb{N}$ for any $n\in\mathbb{N}$, i.e.,

$\omega_{k}(n)\tau+(n+1)\arctan\frac{\omega_{k}(n)}{\mu+\kappa}=\pi+2k\pi.$
(7.9)

Thus, we can obtain from (7.9) that

$\omega_{k}(n)\tau+(n+1)\arctan\frac{\omega_{k}(n)}{\mu+\kappa}=\pi+2k\pi=\omega_{k}(n+1)\tau+(n+2)\arctan\frac{\omega_{k}(n+1)}{\mu+\kappa}.$
(7.10)

It follows from mean value theorem that there exists some $\xi$ between
$\frac{\omega_{k}(n)}{\mu+\kappa}$ and $\frac{\omega_{k}(n+1)}{\mu+\kappa}$
such that

$\arctan\frac{\omega_{k}(n)}{\mu+\kappa}-\arctan\frac{\omega_{k}(n+1)}{\mu+\kappa}=\frac{1}{1+\xi^{2}}\frac{1}{\mu+\kappa}(\omega_{k}(n)-\omega_{k}(n+1)).$
(7.11)

Therefore, by replacing (7.11) to (7.10), we can rewrite (7.10) as

$\left(\tau+\frac{n+1}{(\mu+\kappa)(1+\xi^{2})}\right)(\omega_{k}(n)-\omega_{k}(n+1))=\arctan\frac{\omega_{k}(n+1)}{\mu+\kappa}.$
(7.12)

Thanks to the positivity of $\omega_{k}(n)$, we obtain from (7.12) that

$\omega_{k}(n)>\omega_{k}(n+1)>0,\text{ for all }n\in\mathbb{N},$

which implies that $\omega_{k}(n)$ is strictly decreasing for
$n\in\mathbb{N}$. Thus, the monotone bounded convergence theorem for the
sequence $\\{\omega_{k}(n)\\}$ implies that the limit
$\lim\limits_{n\to+\infty}\omega_{k}(n)$ exists. Let
$\omega^{*}_{k}=\lim\limits_{n\to+\infty}\omega_{k}(n)$. Then we have
$\omega_{k}^{*}\geq 0$ by the positivity of the sequence
$\\{\omega_{k}(n)\\}$. We claim that $\omega_{k}^{*}=0$. In fact, if
$\omega_{k}^{*}>0$, we will obtain a contradiction by letting $n\to+\infty$ on
both sides of (7.9) Therefore,
$\omega_{k}^{*}=\lim\limits_{n\to+\infty}\omega_{k}(n)=0$. By letting
$n\to+\infty$ on both sides of equality (7.9), we have

$\displaystyle\pi+2k\pi=\lim_{n\to+\infty}\left(\omega_{k}(n)\tau+(n+1)\arctan\frac{\omega_{k}(n)}{\mu+\kappa}\right)=\lim_{n\to+\infty}(n+1)\arctan\frac{\omega_{k}(n)}{\mu+\kappa},$

which implies that

$\omega_{k}(n)\sim\frac{(\pi+2k\pi)(\mu+\kappa)}{n+1}\text{ as }n\to+\infty.$
(7.13)

Therefore, the result $\alpha_{k}(n)\to e^{2}\text{ as }n\to+\infty$ in (7.7)
can be obtained from (7.5) and (7.13).

The following theorem provides the transversality condition for the model
(1.1).

###### Theorem 7.3

Let Assumption 7.1 be satisfied and the parameters
$\mu>0,\tau>0,\kappa>0,n\in\mathbb{N}$ be fixed. For each $k\geq 0$, if
$\pm\mathrm{i}\omega_{k}$ with $\omega_{k}>0$ is the purely imaginary root of
the characteristic equation associated to $\alpha_{k}$ defined in Proposition
7.2, then there exists $\rho_{k}>0$ and a $C^{1}$-map
$\hat{\lambda}_{k}:(\alpha_{k}-\rho_{k},\alpha_{k}+\rho_{k})\to\mathbb{C}$
such that

$\hat{\lambda}_{k}(\alpha_{k})=\mathrm{i}\omega_{k},\quad\Delta(\hat{\lambda}_{k}(\alpha),\alpha)=0,\quad\forall\alpha\in(\alpha_{k}-\rho_{k},\alpha_{k}+\rho_{k})$

satsfying the transversality condition

$\mathrm{Re}\frac{\mathrm{d}\hat{\lambda}_{k}(\alpha_{k})}{\mathrm{d}\alpha}>0.$
(7.14)

Proof. By differentiating the characteristic function $\Delta(\lambda,\alpha)$
with $\lambda$ and $\alpha$ and noticing (7.3), we have

$\frac{\partial\Delta(\lambda,\alpha)}{\partial\lambda}=\tau+\frac{n+1}{\lambda+\mu+\kappa},\quad\frac{\partial\Delta(\lambda,\alpha)}{\partial\alpha}=\frac{1}{\alpha(1-\ln\alpha)}.$

Therefore, by setting $(\lambda,\alpha)=(\mathrm{i}\omega_{k},\alpha_{k})$ in
above equations, we have

$\mathrm{Re}\frac{\partial\Delta(\mathrm{i}\omega_{k},\alpha_{k})}{\partial\lambda}=\tau+\frac{(n+1)(\mu+\kappa)}{(\mu+\kappa)^{2}+\omega_{k}^{2}}>0,\quad\frac{\partial\Delta(\mathrm{i}\omega_{k},\alpha_{k})}{\partial\alpha}=\frac{1}{\alpha_{k}(1-\ln\alpha_{k})}<0.$
(7.15)

It follows from the implicit function theorem around each
$(\mathrm{i}\omega_{k},\alpha_{k})$ that for each $k\geq 0$, there exists
$\rho_{k}>0$ and a $C^{1}$-map
$\hat{\lambda}_{k}:(\alpha_{k}-\rho_{k},\alpha_{k}+\rho_{k})\to\mathbb{C}$
such that

$\hat{\lambda}_{k}(\alpha_{k})=\mathrm{i}\omega_{k},\quad\Delta(\hat{\lambda}_{k}(\alpha),\alpha)=0,\quad\forall\alpha\in(\alpha_{k}-\rho_{k},\alpha_{k}+\rho_{k}).$

Furthermore, by differentiating $\Delta(\hat{\lambda}(\alpha),\alpha)=0$ with
$\alpha$, we can obtain

$\frac{\partial\Delta(\hat{\lambda}_{k}(\alpha),\alpha)}{\partial\alpha}+\frac{\partial\Delta(\hat{\lambda}_{k}(\alpha),\alpha)}{\partial\lambda}\frac{\mathrm{d}\hat{\lambda}_{k}(\alpha)}{\mathrm{d}\alpha}=0,\quad\forall\alpha\in(\alpha_{k}-\rho_{k},\alpha_{k}+\rho_{k}).$
(7.16)

The result (7.14) can be obtained by substituting (7.15) into (7.16) with
$\alpha=\alpha_{k}$.

By Proposition 7.2 and Theorem 7.3 and using the Hopf bifurcation theorem, we
can immediately obtain the following Hopf bifurcation result.

###### Theorem 7.4

Let Assumption 7.1 be satisfied. Then, for each $k\geq 0$, the number
$\alpha_{k}$ defined in Proposition 7.2 is a Hopf bifurcation point for the
system (1.1) parameterized by $\alpha$ around the positive equilibrium
$\bar{u}$.

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