Threshold dynamical analysis on a class of age-structured tuberculosis model with immigration of population

Abstract

Some studies show that latency and relapse, especially the age-dependent latency and relapse, may affect the transmission dynamics of tuberculosis model. Meanwhile, the immigration of infected individuals induces the loss of disease-free steady state and hence no basic reproduction number. In our work, a class of age-structured tuberculosis model with immigration is proposed, where the new individuals can immigrate into the susceptible, latent, infectious and removed compartments. We show that the endemic steady state is unique and globally asymptotically stable by using the Lyapunov functional. Numerical simulations are given to support our theoretical results.

1 Introduction

Tuberculosis (TB), mainly caused by Mycobacterium tuberculosis, is a widespread infectious disease and has become a global public health issue. Despite various treatment strategies and beneficial policies on TB patients, the current global TB remains a leading cause of death from an infectious disease. According to reports, there were one death in five in England in the 17th century [1]. About $8.7 \times10^{6}$ cases of TB globally is estimated in 2011, in which India has the largest total infected population, with an estimated 2.2 million new cases; China has the second largest TB epidemic, with more than $1.3 \times10^{6}$ new cases every year [2].

Mathematical models have been a useful tool to understand and analyze the transmission dynamics of TB and other infectious diseases. In [3], a SEI type of TB model with a general contact rate is considered, and the global stability of equilibria is derived. In [4], a TB model with early and late latent stages is introduced to discuss effectiveness of treating TB patients at different stages. The reader can refer to more related mathematical models for TB; see [5–7]. It is well known that TB experiences a latent phase as well as a relapse phase which are the removed individuals who have been previously infected, but revert back to the infectious compartments due to the reactivation (see [8, 9]). Hence, in this paper, we focus on an SEIR-type model.

The level of infectiousness and the likelihood of progression may depend on the types of infectious diseases and individuals’ status. Thus, the age-structured epidemic models are thought to be more practical to describe certain disease features. Taking into consideration the age-dependent latency and relapse, Liu [10] established and researched the following initial-boundary-value problem for a hybrid system of ordinary and partial differential equations:

$$ \textstyle\begin{cases} \frac{dS(t)}{dt}=\Lambda_{s}-\mu_{s} S(t)-\beta S(t)I(t), \\ (\frac{\partial}{\partial t}+\frac{\partial}{\partial a} ) e(t,a) =-\sigma(a)e(t,a)-\mu_{e} e(t,a), \\ \frac{dI(t)}{dt}= \int_{0}^{\infty}\sigma(a)e(t,a)\,da-(\mu_{i}+k)I(t)+\int_{0}^{\infty}\gamma (b)r(t,b)\,db, \\ (\frac{\partial}{\partial t}+\frac{\partial}{\partial b} ) r(t,b) =-\gamma(b)r(t,b)-\mu_{r} r(t,b), \end{cases} $$

(1.1)

with the boundary conditions

$$\begin{aligned} e(t,0)=\beta S(t)I(t),\qquad r(t,0)=kI(t) \end{aligned}$$

(1.2)

for $t\geq0$ and the initial conditions

$$\begin{aligned} S(0)=S_{0},\qquad e(0,a)=e_{0}(a),\qquad I(0)=I_{0},\qquad r(0,b)=r_{0}(b) \end{aligned}$$

(1.3)

for $a,b\geq0$. This model can be used to describe TB transmission dynamics. The initial conditions satisfy $S_{0},I_{0}\in\mathbb{R}^{+}$ and $e_{0}(a),r_{0}(b)\in L_{+}^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$, where $L_{+}^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$ is the space of functions on $[0,\infty)$ that are nonnegative and Lebesgue integrable. Here, $S(t)$ and $I(t)$ are the densities of susceptible and infectious individuals at time t, and $e(t,a)$ and $r(t,b)$ represent the densities of latent and removed individuals at time t, age a and b, respectively. a and b are referred to the latent age and the time spending at the removed compartment. $\Lambda _{s}$ denotes the recruitment of susceptible individuals, $\mu_{j}$ ($j=s,e,i,r$) denotes their per capita mortality rate and k denotes the recovery rate. $\sigma(a)$ and $\gamma(b)$ are the removal rate from latent and removed compartments, respectively. The authors showed the asymptotic smoothness of solutions and uniform persistence of system (1.1). Furthermore, they obtained the basic reproduction number $\Re_{0}$, which played as a threshold parameter and satisfied that if $\Re_{0}<1$, then the disease-free equilibrium of system (1.1) is locally and globally asymptotically stable, while if $\Re _{0}>1$, then the endemic equilibrium uniquely exists and it is locally and globally asymptotically stable. More related work is done to understand the dynamics of age-structured models; see [11–15]. Famous books on age-structured models are by Webb [16], Iannelli [17] and Smith et al. [18].

In the modern world, the worldwide transportation leads to tremendous movement of individuals and it is inevitable that infectious diseases may be introduced into a population from outside the population. Although most of the developed countries have screening policies for new immigrants, latent tuberculosis may take long time to become infectious and latent tuberculosis individuals may travel, thus, they are usually not detected by the TB screening. Furthermore, the removed tuberculosis patients who often have higher relapse rate may travel into another region; or the infectious individuals may also travel. In [19], McCluskey et al. proposed a TB model with immigration of both latent and infective compartments. Later, In [20], Guo et al. introduced a TB model with treatment and immigration into the latent compartment. It is well known that the model will no longer have the disease-free equilibrium and the basic reproduction number, and there will always have a unique endemic equilibrium which is globally asymptotically stable by using a global Lyapunov function. More related work can be found in [21–25] and references therein.

In this paper, we propose and investigate a TB model with immigration into the four compartments. We also incorporate into the continuous age-dependent in latent and removed compartments. It is a generalization of model proposed in [10]. According to mathematical analysis, we show that TB always exists in a region and the endemic equilibrium is unique and globally asymptotically stable. This paper is organized as follows. In the next section, we will formulate the model. In Section 3, we show the mathematical well-posedness of our model. In Section 4, we investigate the asymptotic smoothness of the semi-flow generated by our model and the existence of compact attractor. In Section 5, we show our dynamics results, including the existence and global stability of the unique endemic steady state. Some simulations and conclusion are provided in Section 6 and Section 7, respectively.

2 Model formulation

This section we devote to formulating our model.

Assume that there is constant recruitment into the susceptible and infectious compartments at rates $\Lambda_{s}$ and $\Lambda_{i}$, respectively. The recruitment into the latent and removed compartments with age-in-class a and b take place at rates denoted $\Lambda_{e}(a)$ and $\Lambda_{r}(b)$, respectively. The transfer diagram is shown in Figure 1. The corresponding hybrid system of ordinary and partial differential equations is the following form:

$$ \textstyle\begin{cases} \frac{dS(t)}{dt}=\Lambda_{s}-\mu_{s} S(t)-\beta S(t)I(t), \\ (\frac{\partial}{\partial t}+\frac{\partial}{\partial a} ) e(t,a) =\Lambda_{e}(a)-\sigma(a)e(t,a)-\mu_{e} e(t,a), \\ \frac{dI(t)}{dt}=\Lambda_{i}+ \int_{0}^{\infty}\sigma(a)e(t,a)\,da-(\mu_{i}+k)I(t)+\int_{0}^{\infty}\gamma (b)r(t,b)\,db, \\ (\frac{\partial}{\partial t}+\frac{\partial}{\partial b} ) r(t,b) =\Lambda_{r}(b)-\gamma(b)r(t,b)-\mu_{r}r(t,b), \end{cases} $$

(2.1)

with the boundary conditions

$$\begin{aligned} e(t,0)=\beta S(t)I(t),\qquad r(t,0)=kI(t),\quad t\geq0, \end{aligned}$$

and the initial conditions

$$\begin{aligned} S(0)=S_{0},\qquad e(0,a)=e_{0}(a),\qquad I(0)=I_{0},\qquad r(0,b)=r_{0}(b),\quad a,b\geq0. \end{aligned}$$

Note that system (2.1) considers the immigration of all compartments, which is different from system (1.1), where only considers the immigration of susceptible compartment. All parameters have the same biological meanings with system (1.1) and satisfy the following assumptions.

Figure 1

The schematic flow diagram of our model. Here, the terms $[\sigma E]$ and $[\gamma R]$ stand for $\int_{0}^{\infty}\sigma(a)e(t,a)\,da$ and $\int_{0}^{\infty}\gamma(b)r(t,b)\,db$.

Assumption 2.1

Consider system (2.1), we assume that:

(A₁):

All constant parameters $\Lambda_{s},\Lambda_{i},\mu_{s},\mu _{e},\mu_{i},\mu_{r},k>0$.

(A₂):

The functions $\sigma(a),\gamma(b)\in L^{\infty}(\mathbb {R}^{+},\mathbb{R}^{+})$, and denote $\sigma^{\mathrm{inf}}$, $\gamma ^{\mathrm{inf}}$ and $\sigma^{\mathrm{sup}}$, $\gamma^{\mathrm{sup}}$ as the essential infimums and the essential supremums of σ and γ, respectively.

(A₃):

$\sigma(a)$ and $\gamma(b)$ are Lipschitz continuous on $\mathbb{R}^{+}$ with Lipschitz coefficients $M_{\sigma}$ and $M_{\gamma}$, respectively.

(A₄):

$\int_{0}^{\infty}\sigma(a)\,da=0$ and $\int_{0}^{\infty}\gamma(b)\,db=0$.

(A₅):

$\Lambda_{e},\Lambda_{r}\in L^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$, and denote $\bar{\Lambda}_{e}=\int_{0}^{\infty}\Lambda_{e}(a)\,da$, $\bar{\Lambda}_{r}=\int_{0}^{\infty}\Lambda_{r}(b)\,db$.

(A₆):

$\bar{\Lambda}_{e},\bar{\Lambda}_{r}\in\mathbb{R}^{+}$, and $\bar{\Lambda}_{e}+\bar{\Lambda}_{r}>0$.

Define the space of functions X as $X=\mathbb{R}^{+}\times L_{+}^{1}(\mathbb{R}^{+},\mathbb{R}^{+})\times\mathbb{R}^{+}\times L_{+}^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$ with the norm

$$\bigl\Vert (x_{1},x_{2},x_{3},x_{4}) \bigr\Vert _{X}= \vert x_{1} \vert + \int_{0}^{\infty}\bigl\vert x_{2}(a) \bigr\vert \,da+ \vert x_{3} \vert + \int _{0}^{\infty}\bigl\vert x_{4}(b) \bigr\vert \,db. $$

Following the standard theory [16], it can be verified that system (2.1) with initial-boundary conditions has a unique nonnegative solution for all time. Thus, there exists a continuous semi-flow associated with system (2.1), that is, $\Phi:\mathbb{R}^{+}\times X\rightarrow X$ takes the following form:

$$\Phi(t,x_{0})=\bigl(S(t),e(t,\cdot),I(t),r(t,\cdot)\bigr),\quad t \geq0, x_{0}\in X, $$

with

$$\begin{aligned} \bigl\Vert \Phi(t,x_{0}) \bigr\Vert _{X}&= \bigl\Vert \bigl(S(t),e(t,\cdot),I(t),r(t,\cdot )\bigr) \bigr\Vert _{X} \\ &= \bigl\vert S(t) \bigr\vert + \int_{0}^{\infty}\bigl\vert e(t,a) \bigr\vert \,da+ \bigl\vert I(t) \bigr\vert + \int_{0}^{\infty}\bigl\vert r(t,b) \bigr\vert \,db. \end{aligned}$$

(2.2)

3 The well-posedness of system (2.1)

This section is devoted to the positivity and boundedness of solutions.

Assume that function $f(a,s)$ is continuous in $\mathbb{R}^{+}\times \mathbb{R}^{+}$, then one has

$$ \int_{0}^{t} \int_{0}^{a}f(a,s)\,ds\,da+ \int_{t}^{\infty}\int_{a-t}^{a}f(a,s)\,ds\,da= \int_{0}^{\infty}\int^{s+t}_{s}f(a,s)\,ds\,da, $$

(3.1)

which is achieved by changing the order of integration in two double integrals. This will be useful in the later proofs.

For simplification, we denote

$$\begin{aligned} &\varepsilon(a)=\sigma(a)+\mu_{e},\qquad \rho_{e}(a)=\exp \biggl(- \int _{0}^{a}\varepsilon(s)\,ds \biggr),\qquad \theta_{e}= \int_{0}^{\infty}\sigma(a)\rho _{e}(a)\,da, \\ &\eta(b)=\gamma(b)+\mu_{r},\qquad \rho_{r}(b)=\exp \biggl(- \int_{0}^{b}\eta (s)\,ds \biggr),\qquad \theta_{r}= \int_{0}^{\infty}\gamma(b)\rho_{r}(b)\,db. \end{aligned}$$

It follows from (A₃) of Assumption 2.1 that

$$\begin{aligned} 0\leq\rho_{e}(a)\leq e^{-\mu_{e} a}\leq1,\qquad 0\leq \rho_{r}(b)\leq e^{-\mu_{r}b}\leq1 \end{aligned}$$

(3.2)

and

$$\begin{aligned} \frac{d\rho_{e}(a)}{da}=-\varepsilon(a)\rho_{e}(a),\qquad \frac{d\rho_{r}(b)}{db}=-\eta(b)\rho_{r}(b). \end{aligned}$$

(3.3)

Similar to [16], solutions of the PDE parts are

$$\begin{aligned} e(t,a)= \textstyle\begin{cases} e(t-a,0)\rho_{e}(a)+\int_{0}^{a}\Lambda_{e}(s)\frac{\rho_{e}(a)}{\rho _{e}(s)}\,ds,&t> a\geq0; \\ e_{0}(a-t)\frac{\rho_{e}(a)}{\rho_{e}(a-t)}+ \int_{a-t}^{a}\Lambda_{e}(s)\frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds,&a\geq t\geq0, \end{cases}\displaystyle \end{aligned}$$

(3.4)

and

$$\begin{aligned} r(t,b)= \textstyle\begin{cases} r(t-b,0)\rho_{r}(b)+\int_{0}^{b}\Lambda_{r}(s)\frac{\rho_{r}(b)}{\rho _{r}(s)}\,ds,&t> b\geq0; \\ r_{0}(b-t)\frac{\rho_{r}(b)}{\rho_{r}(b-t)} +\int_{b-t}^{b}\Lambda_{r}(s)\frac{\rho_{r}(b)}{\rho_{r}(s)}\,ds,& b\geq t\geq0. \end{cases}\displaystyle \end{aligned}$$

(3.5)

Finally, we define the state space for system (2.1) as

$$\Omega= \biggl\{ \bigl(S(t),e(t,\cdot),I(t),r(t,\cdot)\bigr)\in X: S(t)+ \int_{0}^{\infty}e(t,a)\,da+I(t)+ \int_{0}^{\infty}r(t,b)\,db \leq\frac{\Lambda^{*}}{\mu^{*}} \biggr\} , $$

where $\Lambda^{*}=\Lambda_{s}+\bar{\Lambda}_{e}+{\Lambda}_{i}+\bar {\Lambda}_{r}$, $\mu^{*}=\min\{\mu_{j},j=s,e,i,r\}$. We can prove that Ω and Φ have the following properties.

Proposition 3.1

Consider system (2.1), we have

• (i) Ω is positively invariant for Φ, that is, $\Phi (t,x_{0})\in\Omega$, for $\forall t\geq0$ and $x_{0}\in\Omega$;

• (ii) Φ is point dissipative and Ω attracts all points in X.

Proof

From (2.2), we have

$$\begin{aligned} \frac{d}{dt} \bigl\Vert \Phi(t,x_{0}) \bigr\Vert _{X} =\frac{dS(t)}{dt}+\frac{d}{dt} \int_{0}^{\infty}e(t,a)\,da+\frac {dI(t)}{dt}+ \frac{d}{dt} \int_{0}^{\infty}r(t,b)\,db. \end{aligned}$$

(3.6)

Before calculating the derivative of semi-flow, we first show that

$$\begin{aligned} \int_{0}^{\infty}e(t,a)\,da =& \int_{0}^{t} e(t,a)\,da+ \int_{t}^{\infty}e(t,a)\,da \\ =& \int_{0}^{t}e(t-a,0)\rho_{e}(a)\,da + \int_{0}^{t} \int_{0}^{a}\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds\,da \\ &{} + \int_{t}^{\infty}e_{0}(a-t)\frac{\rho_{e}(a)}{\rho_{e}(a-t)} \,da + \int_{t}^{\infty}\int_{a-t}^{a}\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho _{e}(s)}\,ds\,da \\ =& \int_{0}^{t}e(t-a,0)\rho_{e}(a)\,da + \int_{t}^{\infty}e_{0}(a-t)\frac{\rho_{e}(a)}{\rho_{e}(a-t)} \,da \\ &{}+ \int_{0}^{\infty}\int_{s}^{s+t}\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho _{e}(s)}\,ds\,da. \end{aligned}$$

(3.7)

Note that the last term is obtained by using (3.1). Furthermore, let $\tau=t-a$ and $\tau=a-t$ in the first and second integrals on the right-hand side of (3.7), respectively, one has

$$\begin{aligned} \int_{0}^{\infty}e(t,a)\,da =& \int_{0}^{t}e(\tau,0)\rho_{e}(t-\tau)\,d \tau + \int_{t}^{\infty}e_{0}(\tau) \frac{\rho_{e}(t+\tau)}{\rho_{e}(\tau )}\,d\tau \\ &{}+ \int_{0}^{\infty}\int_{s}^{s+t}\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds\,da. \end{aligned}$$

Then we have

$$\begin{aligned} \frac{d}{dt} \int_{0}^{\infty}e(t,a)\,da =&\frac{d}{dt} \int_{0}^{t}e(\tau,0)\rho_{e}(t-\tau)\,d \tau +\frac{d}{dt} \int_{t}^{\infty}e_{0}(\tau) \frac{\rho_{e}(t+\tau)}{\rho _{e}(\tau)}\,d\tau \\ &{}+\frac{d}{dt} \int_{0}^{\infty}\int_{s}^{s+t}\Lambda_{e}(s) \frac{\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da \\ =&e(t,0)+ \int_{0}^{t}e(\tau,0)\frac{d}{dt} \rho_{e}(t-\tau)\,d\tau + \int_{t}^{\infty}e_{0}(\tau) \frac{\frac{d}{dt}\rho_{e}(t+\tau)}{\rho _{e}(\tau)}\,d\tau \\ &{}+ \int_{0}^{\infty}\Lambda_{e}(s) \frac{\rho_{e}(s+t)}{\rho_{e}(s)}\,ds \\ =&e(t,0)+ \int_{0}^{t}e(t-a,0)\frac{d}{da} \rho_{e}(a)\,da + \int_{t}^{\infty}e_{0}(a-t)\frac{\frac{d}{da}\rho_{e}(a)}{\rho_{e}(a-t)} \,da \\ &{}+ \int_{0}^{\infty}\Lambda_{e}(s) \frac{\rho_{e}(s+t)}{\rho_{e}(s)}\,ds. \end{aligned}$$

(3.8)

Here, we put our main attentions on dealing with the last term. Following (3.1), we have

$$\begin{aligned}& \int_{0}^{t} \int_{0}^{a}\Lambda_{e}(s) \frac{\frac{d}{da}\rho_{e}(a)}{\rho_{e}(s)}\,ds\,da + \int_{t}^{\infty}\int_{a-t}^{a}\Lambda_{e}(s) \frac{\frac{d}{da}\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da \\& \quad = \int_{0}^{\infty}\int_{s}^{s+t}\Lambda_{e}(s) \frac{\frac{d}{da}\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da. \end{aligned}$$

Furthermore, one has

$$\begin{aligned} \int_{0}^{\infty}\int_{s}^{s+t}\Lambda_{e}(s) \frac{\frac{d}{da}\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da &= \int_{0}^{\infty}\Lambda_{e}(s) \frac{\rho_{e}(s+t)-\rho_{e}(s)}{\rho_{e}(s)}\,ds \\ &= \int_{0}^{\infty}\Lambda_{e}(s) \frac{\rho_{e}(s+t)}{\rho_{e}(s)}\,ds - \int_{0}^{\infty}\Lambda_{e}(s) \,ds. \end{aligned}$$

(3.9)

Applying (3.9) to replace the last term of (3.8), one gets

$$\begin{aligned} \frac{d}{dt} \int_{0}^{\infty}e(t,a)\,da =&e(t,0)+ \int_{0}^{t}e(t-a,0)\frac{d}{da} \rho_{e}(a)\,da + \int_{t}^{\infty}e_{0}(a-t)\frac{\frac{d}{da}\rho_{e}(a)}{\rho _{e}(a-t)} \,da \\ &{}+ \int_{0}^{t} \int_{0}^{a}\Lambda_{e}(s) \frac{\frac{d}{da}\rho_{e}(a)}{\rho _{e}(s)}\,ds\,da \\ &{}+ \int_{t}^{\infty}\int_{a-t}^{a}\Lambda_{e}(s) \frac{\frac{d}{da}\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da+ \int_{0}^{\infty}\Lambda_{e}(s)\,da \\ =&e(t,0)- \int_{0}^{\infty}\varepsilon(a)e(t,a)\,da+\bar{ \Lambda}_{e}. \end{aligned}$$

(3.10)

Similarly, one can obtain

$$ \frac{d}{dt} \int_{0}^{\infty}r(t,b)\,db =r(t,0)- \int_{0}^{\infty}\eta(b)r(t,b)\,db+\bar{ \Lambda}_{r}. $$

(3.11)

Combining (3.10)-(3.11) and the first and third equations of (2.1) yields

$$\begin{aligned} & \frac{d}{dt} \biggl(S(t)+ \int_{0}^{\infty}e(t,a)\,da+I(t)+ \int_{0}^{\infty}r(t,b)\,db \biggr) \\ &\quad = (\Lambda_{s}+\bar{\Lambda}_{e}+ \Lambda_{i}+\bar{\Lambda}_{r}) - \biggl( \mu_{s}S(t)+\mu_{e} \int_{0}^{\infty}e(t,a)\,da+\mu_{i}I(t)+ \mu_{r} \int _{0}^{\infty}r(t,b)\,db \biggr) \\ &\quad \leq \Lambda^{*}-\mu^{*} \biggl(S(t)+ \int_{0}^{\infty}e(t,a)\,da+I(t)+ \int _{0}^{\infty}r(t,b)\,db \biggr), \end{aligned}$$

which implies

$$\begin{aligned} \bigl\Vert \Phi(t,x_{0}) \bigr\Vert _{X} \leq \frac{\Lambda^{*}}{\mu^{*}}-e^{-\mu^{*} t} \biggl(\frac{\Lambda^{*}}{\mu ^{*}}- \Vert x_{0} \Vert _{X} \biggr),\quad \mbox{for }t\geq0. \end{aligned}$$

(3.12)

Obviously, $\Phi(t,x_{0})\in\Omega$ holds true for any solution of (2.1) satisfying $x_{0}\in\Omega$ and $t\geq0$. Thus, Ω is positively invariant for semi-flow $\{\Phi(t)\}_{t\geq0}$.

Moreover, it follows from (3.12) that $\limsup_{t\rightarrow \infty}\|\Phi(t,x_{0})\|_{X}\leq{\Lambda^{*}}/{\mu^{*}}$ for any $x_{0}\in X$. Therefore, Φ is point dissipative and Ω attracts all points in X. This completes the proof. □

Combining Assumption 2.1 and Proposition 3.1, we have the following two propositions.

Proposition 3.2

If $x_{0}\in X$ and $\| x_{0}\|_{X}\leq M$ for some constant $M\geq {\Lambda^{*}}/{\mu^{*}}$, then the following statements hold true for $t\geq0$:

• (i) $0\leq S(t),\int_{0}^{\infty}e(t,a)\,da,I(t),\int_{0}^{\infty}r(t,b)\,db\leq M$;

• (ii) $e(t,0)\leq\beta M^{2}, r(t,0)\leq kM$.

Proposition 3.3

Let $B\subset X$ be bounded. Then

• (i) $\Phi(\mathbb{R}^{+},B)$ is bounded;

• (ii) Φ is eventually bounded on B;

• (iii) if $M\geq{\Lambda^{*}}/{\mu^{*}}$ is a bound for B, then M is also a bound for $\Phi(\mathbb{R}^{+},B)$;

• (iv) given any $L\geq{\Lambda^{*}}/{\mu^{*}}$, there exists $T=T(B,M)$ such that L is a bound for $\Phi(t,B)$ whenever $t\geq T$.

Now, we give the following proposition on the asymptotic lower bounds for system (2.1).

Proposition 3.4

If $x_{0}\in X$, then

$$\begin{aligned}& \liminf_{t\rightarrow\infty}S(t)\geq m_{s},\qquad \liminf _{t\rightarrow\infty}I(t)\geq m_{i}, \\& \liminf _{t\rightarrow\infty}e(t,0)\geq m_{e},\qquad \liminf _{t\rightarrow\infty}r(t,0)\geq m_{r}, \end{aligned}$$

where $m_{j}$, $j=s,i,e,r$ are defined below.

Proof

It follows from the first equation of system (2.1) that

$$S'=\Lambda_{s}-\mu_{s}S-\beta SI \geq \Lambda_{s}-(\mu_{s}+\beta M)S, $$

which implies that

$$\liminf_{t\rightarrow\infty}S(t)\geq\frac{\Lambda_{s}}{\mu _{s}+\beta M}\triangleq m_{s}. $$

Similarly, one has

$$\liminf_{t\rightarrow\infty}I(t)\geq\frac{\Lambda_{i}+(\sigma ^{\mathrm{inf}}+\gamma^{\mathrm{inf}})M}{\mu_{i}+k} \triangleq m_{i}. $$

Following (1.2), one can see that

$$\liminf_{t\rightarrow\infty}e(t,0)=\beta SI\geq\beta m_{s}m_{i} \triangleq m_{e},\qquad \liminf_{t\rightarrow\infty}r(t,0)=kI\geq km_{i} \triangleq m_{r}. $$

The proof is completed. □

4 Asymptotic smoothness

In order to obtain global properties of the semi-flow $\{\Phi(t)\} _{t\geq0}$, it is important to prove that the semi-flow is asymptotically smooth.

We introduce two definitions and a useful lemma.

Definition 4.1

A function $Y:\mathbb{R}\rightarrow X$ is called a total trajectory of Φ if Y satisfies $\Phi_{s}(Y(t))=Y(t+s)$ for all $t\in\mathbb {R}$ and $s\geq0$.

Definition 4.2

A non-empty invariant compact set A is called the compact attractor of a class $\mathcal{B}$ of sets if $\operatorname{dist}(\Phi (t,B),A)\rightarrow0$ for each $B\in\mathcal{B}$, where

$$\operatorname{dist}\bigl(\Phi(t,B),A\bigr)=\sup_{x\in\Phi_{t}(B)}\inf _{y\in A}\|x-y\|_{X}. $$

Remark 4.1

For any point $y_{0}\in A$, it follows from Definitions 4.1 and 4.2, and [18], Theorem 1.40, that there exists a total trajectory $Y(\cdot)$ with $y(0)=y_{0}$ and $y(t)\in A$ for all $t\in\mathbb{R}$.

Lemma 4.1

[13] Let $D\subseteq\mathbb{R}$. For $j=1,2$, suppose $f_{j}:D\rightarrow \mathbb{R}$ is a bounded Lipshitz continuous function with bound $K_{j}$ and Lipschitz coefficient $M_{j}$. Then the product function $f_{1}f_{2}$ is Lipschitz with coefficient $K_{1}M_{2}+K_{2}M_{1}$.

In order to prove the asymptotic smoothness of the semi-flow, we will apply the following result, which is a special case of [18], Theorem 2.46.

Lemma 4.2

If the following two conditions hold for any bounded closed set $B\subset X$, then the semi-flow $\Phi(t,x_{0})=\Phi_{1}(t,x_{0})+\Phi _{2}(t,x_{0}):\mathbb{R}^{+}\times X\rightarrow X$ is asymptotically smooth in X.

• (i) $\lim_{t\rightarrow\infty}\operatorname{diam}\Phi_{1}(t,B)=0$;

• (ii) for $t\geq0$, $\Phi_{2}(t,B)$ has compact closure.

The following result is used to verify (ii) of Lemma 4.2, which is based on Theorem B.2 in [18].

Lemma 4.3

A set $K\subset L^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$ has compact closure if and only if the following conditions hold:

• (i) $\sup_{f\in K}\int_{0}^{\infty}f(a)\,da<\infty$;

• (ii) $\lim_{h\rightarrow\infty}\int_{h}^{\infty}f(a)\,da\rightarrow0$ uniformly in $f\in K$;

• (iii) $\lim_{h\rightarrow{0^{+}}}\int_{0}^{\infty}|f(a+h)-f(a)|\,da\rightarrow0$ uniformly in $f\in K$;

• (iv) $\lim_{h\rightarrow{0^{+}}}\int_{0}^{h} f(a)\,da\rightarrow0$ uniformly in $f\in K$.

Based on Lemmas 4.2 and 4.3, we have the following theorem.

Theorem 4.1

The semi-flow $\{\Phi(t)\}_{t\geq0}$ is asymptotically smooth.

Proof

We first decompose $\Phi:\mathbb{R}^{+}\times X\rightarrow X$ into the following two operators $\Phi_{1}(t,X)$ and $\Phi_{2}(t,X)$. Let $\Phi_{1}(t,X):=(0,y_{2}(t,\cdot),0,y_{4}(t,\cdot))$ and $\Phi_{2}(t,X):=(S(t),\tilde{y}_{2}(t,\cdot),I(t),\tilde{y}_{4}(t,\cdot ))$, where

$$\begin{aligned}& \begin{aligned} &y_{2}(t,a):= \textstyle\begin{cases} \int_{0}^{a}\Lambda_{e}(s)\frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds,& t>a\geq0;\\ e(t,a),& a\geq t\geq0 \end{cases}\displaystyle \quad \text{and} \\ &y_{4}(t,b):= \textstyle\begin{cases} \int_{0}^{b}\Lambda_{r}(s)\frac{\rho_{r}(b)}{\rho_{r}(s)}\,ds,& t>b\geq0;\\ r(t,b),& b\geq t\geq0, \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(4.1)

$$\begin{aligned}& \begin{aligned} &\tilde{y}_{2}(t,a):= \textstyle\begin{cases} e(t-a,0)\rho_{e}(a),& t>a\geq0;\\ 0,&a\geq t\geq0 \end{cases}\displaystyle \quad \text{and} \\ &\tilde{y}_{4}(t,b):= \textstyle\begin{cases} r(t-b)\rho_{r}(b),& t>b\geq0;\\ 0,& b\geq t\geq0. \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(4.2)

Then we have $\Phi(t,X)=\Phi_{1}(t,X)+\Phi_{2}(t,X)$ for all $t\geq0$.

First, we show that $\lim_{t\rightarrow\infty}\operatorname{diam}\Phi_{1}(t,B)=0$. For $j=1,2$, let $x_{0}^{j}=(S_{0},e_{0}^{j}(\cdot),I_{0},r_{0}^{j}(\cdot))\in B$ be two initial conditions and $\Phi(t,x_{0}^{j})=(S^{j},e^{j}(t,\cdot),I^{j},r^{j}(t,\cdot))$ be their corresponding solutions. Thus, one has

$$ y_{2}^{1}(t,a)-y_{2}^{2}(t,a)= \textstyle\begin{cases} 0,& t>a\geq0;\\ (e_{0}^{1}(a-t)-e_{0}^{2}(a-t) )\frac{\rho_{e}(a)}{\rho _{e}(a-t)},&a\geq t\geq0. \end{cases} $$

Then

$$\begin{aligned} \begin{aligned} \bigl\Vert y_{2}^{1}(t,\cdot)-y_{2}^{2}(t, \cdot) \bigr\Vert _{1} &= \int_{t}^{\infty}\bigl\vert e_{0}^{1}(a-t)-e_{0}^{2}(a-t) \bigr\vert \frac{\rho_{e}(a)}{\rho _{e}(a-t)}\,da \\ &= \int_{0}^{\infty}\bigl\vert e_{0}^{1}(s)-e_{0}^{2}(s) \bigr\vert \frac{\rho_{e}(t+s)}{\rho _{e}(s)}\,ds \\ &= \int_{0}^{\infty}\bigl\vert e_{0}^{1}(s)-e_{0}^{2}(s) \bigr\vert e^{\int_{s}^{t+s}\varepsilon(\tau )\,d\tau}\,ds \\ &\leq e^{-\mu_{e}t} \bigl( \bigl\Vert e_{0}^{1} \bigr\Vert _{1}+ \bigl\Vert e_{0}^{2} \bigr\Vert _{1} \bigr) \\ &\leq2Me^{-\mu_{e}t}, \end{aligned} \end{aligned}$$

where $\|\cdot\|_{1}$ denotes the standard norm on $L^{1}$. Similarly, $\| y_{4}^{1}(t,\cdot)-y_{4}^{2}(t,\cdot)\|\leq2Me^{-\mu_{r}t}$. Consequently, the distance between $\Phi_{1}(t,x_{0}^{1})$ and $\Phi_{2}(t,x_{0}^{2})$ satisfies

$$\bigl\Vert \Phi_{1}\bigl(t,x_{0}^{1}\bigr)- \Phi_{2}\bigl(t,x_{0}^{2}\bigr) \bigr\Vert _{X} \leq2M \bigl(e^{-\mu_{e}t}+e^{-\mu_{r}t} \bigr), $$

which implies that $\Phi_{1}(t,B)\leq2M (e^{-\mu_{e}t}+e^{-\mu _{r}t} )$, due to the arbitrariness of $x_{0}^{j}\in B$, $j=1,2$. Thus, $\lim_{t\rightarrow\infty}\operatorname{diam}\Phi_{1}(t,B)=0$.

Now, we show that $\Phi_{2}(t,B)$ has compact closure. According to Proposition 3.3, $S(t)$ and $I(t)$ remain in the compact set $[0,\Lambda^{*}/\mu^{*}]\subset[0,M]$, where $M\geq\Lambda ^{*}/\mu^{*}$ is a bound for B. Thus, it is only to show that $\tilde {y}_{2}(t,a)$ and $\tilde{y}_{4}(t,b)$ satisfy conditions (i)-(iv) in Lemma 4.3.

Now, following (3.4) and (4.2), we have

$$\begin{aligned} 0&\leq\tilde{y}_{2}(t,a) = \textstyle\begin{cases} \beta e(t-a,0){\rho_{e}(a)},& t>a\geq0, \\ 0,& a\geq t\geq0 \end{cases}\displaystyle \\ &\leq\beta M^{2}e^{-\mu_{e}a}, \end{aligned}$$

which implies that conditions (i), (ii) and (iv) in Lemma 4.3 are satisfied. It suffices to verify that (iii) in Lemma 4.3 holds true. For sufficiently small $h\in(0,t)$, we have

$$\begin{aligned}& \int_{0}^{\infty}\bigl\vert \tilde{y}_{2}(t,a+h)- \tilde{y}_{2}(t,a) \bigr\vert \,da \\& \quad = \int_{0}^{t-h} \bigl\vert e(t-a-h,0) \rho_{e}(a+h)-e(t-a,0)\rho_{e}(a) \bigr\vert \,da + \int_{t-h}^{t} \bigl\vert 0-e(t-a,0) \rho_{e}(a) \bigr\vert \,da \\& \quad \leq H_{1}+H_{2}+ \int_{t-h}^{t} \bigl\vert e(t-a,0) \rho_{e}(a) \bigr\vert \,da, \end{aligned}$$

(4.3)

where

$$\begin{aligned}& H_{1}= \int_{0}^{t-h}e(t-a-h,0) \bigl\vert \rho_{e}(a+h)-\rho_{e}(a) \bigr\vert \,da, \\& H_{2}= \int_{0}^{t-h}\rho_{e}(a) \bigl\vert e(t-a-h,0)-e(t-a,0) \bigr\vert \,da. \end{aligned}$$

Recall that $0\leq\rho_{e}(a)\leq e^{-\mu_{e}a}\leq1$ and $\rho_{e}(a)$ is non-increasing function with respect to a, we have

$$\begin{aligned} \int_{0}^{t-h} \bigl\vert \rho_{e}(a+h)- \rho_{e}(a) \bigr\vert \,da &= \int_{0}^{t-h}\rho_{e}(a)\,da- \int_{0}^{t-h}\rho_{e}(a+h)\,da \\ & = \int_{0}^{t-h}\rho_{e}(a)\,da- \int_{h}^{t-h}\rho_{e}(a)\,da- \int_{t-h}^{t}\rho _{e}(a)\,da \\ & = \int_{0}^{h}\rho_{e}(a)\,da- \int_{t-h}^{t}\rho_{e}(a)\,da\leq h. \end{aligned}$$

(4.4)

Hence, from (ii) of Proposition 3.2, we yield $H_{1}\leq\beta M^{2}h$. It follows from (i) of Proposition 3.2 that $|dS(t)/dt|$ and $|dI(t)/dt|$ are bounded by $M_{S}=\Lambda_{s}+\mu_{s} M+\beta M^{2}$ and $M_{I}={\sigma}^{\mathrm{sup}}M+{\gamma}^{\mathrm{ sup}}M+(\mu_{i}+k)M$, respectively. Thus, $S(\cdot)$ and $I(\cdot)$ are Lipschitz continuous on $[0,\infty)$ with coefficients $M_{S}$ and $M_{I}$. Following Lemma 4.1, one sees that $S(\cdot)I(\cdot)$ is Lipschitz continuous on $[0,\infty)$ with coefficient $M_{SI}=M(M_{S}+M_{I})$. Thus,

$$H_{2}\leq \int_{0}^{t-h}\beta M_{SI}e^{-\mu_{e}a} \,da \leq\frac{\beta M_{SI}h}{\mu_{e}}. $$

Consequently, we have

$$\int_{0}^{\infty}\bigl\vert \tilde{y}_{2}(t,a+h)- \tilde{y}_{2}(t,a) \bigr\vert \,da\leq \biggl(2\beta M^{2}+ \frac{\beta M_{SI}}{\mu_{e}} \biggr)h. $$

This shows that $\int_{0}^{\infty}|\tilde{y}_{2}(t,a+h)-\tilde {y}_{2}(t,a)|\,da\rightarrow0$ as $h\rightarrow0$. Then $\tilde {y}_{2}(t,a)$ satisfies condition (iii) in Lemma 4.3, which implies that there exists a precompact subset $B_{e}\subset L_{+}^{1}(\mathbb {R}^{+},\mathbb{R}^{+})$ such that $\tilde{y}_{2}(t,a)$ remains in $B_{e}$. Similarly, $\tilde{y}_{4}(t,b)$ remains in a precompact subset $B_{r}\subset L_{+}^{1}(\mathbb{R}^{+},\mathbb{R}^{+})$. Thus, $\Phi _{2}(t,B)\subseteq[0,M]\times B_{e}\times[0,M]\times B_{r}$. Applying Lemma 4.3, we can conclude that $\Phi_{2}(t,B)$ has compact closure. Thus, two conditions of Lemma 4.2 are satisfied and $\{\Phi (t)\}_{t\geq0}$ is asymptotically smooth. This completes the proof. □

Propositions 3.1 and 3.3 and Theorem 4.1 show that Φ is point dissipative, eventually bounded on bounded sets, and asymptotically smooth. Thus, following [18], Theorem 2.33, we have the following proposition on the existence of a global attractor.

Theorem 4.2

The semigroup $\{\Phi(t)\}_{t\geq0}$ has a global attractor $\mathcal {A}$ contained in X, which attracts the bounded sets of X.

5 Dynamical results

This section is devoted to the existence and global stability of the steady state.

5.1 Existence of steady state

In this subsection, we consider the existence of steady state for system (2.1). Clearly, system (2.1) has no disease-free steady state. The steady state $(S^{*},e^{*}(\cdot),I^{*},r^{*}(\cdot))$ of system (2.1) satisfies the equalities

$$ \textstyle\begin{cases} 0=\Lambda_{s}-\mu_{s} S^{*}-\beta S^{*}I^{*}, \\ \frac{de^{*}(a)}{da} =\Lambda_{e}-\varepsilon(a)e^{*}(a), \\ 0=\Lambda_{i}+ \int_{0}^{\infty}\sigma(a)e^{*}(a)\,da-(\mu_{i}+k)I^{*}+\int_{0}^{\infty}\gamma (b)r^{*}(b)\,db, \\ \frac{dr^{*}(b)}{db} =\Lambda_{r}-\eta(b)r^{*}(b), \end{cases} $$

(5.1)

with the boundary conditions

$$ e^{*}(0)=\beta S^{*}I^{*},\qquad r^{*}(0)=kI^{*}. $$

(5.2)

It follows from the first equation of (5.1) that we have

$$ I^{*}=\frac{\Lambda_{s}-\mu_{s} S^{*}}{\beta S^{*}}. $$

(5.3)

From the second equation of (5.1) and boundary conditions (5.2), one has

$$\begin{aligned} e^{*}(a) &=e^{*}(0)\rho_{e}(a)+ \int_{0}^{\infty}\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho _{e}(s)}\,ds \\ &=\bigl(\Lambda_{s}-\mu_{s} S^{*}\bigr)\rho_{e}(a)+ \int_{0}^{\infty}\Lambda_{e}(s) \frac {\rho_{e}(a)}{\rho_{e}(s)}\,ds. \end{aligned}$$

(5.4)

Similarly, the fourth equation of (5.1) and the boundary condition $r^{*}(0)=kI^{*}$ lead to

$$\begin{aligned} r^{*}(b)=\frac{k(\Lambda_{s}-\mu_{s} S^{*})}{\beta S^{*}}\rho_{r}(b) + \int_{0}^{\infty}\Lambda_{r}(s) \frac{\rho_{r}(b)}{\rho_{r}(s)}\,ds. \end{aligned}$$

(5.5)

Inserting (5.4) and (5.5) into the third equation of (5.1), we have

$$f\bigl(S^{*}\bigr)=B_{1}\bigl(S^{*}\bigr)^{2}+B_{2}S^{*}+B_{3}=0, $$

where

$$\begin{aligned} &B_{1}=\beta\mu_{s}\theta_{e}>0, \\ &B_{2}=-\bigl[\mu_{s}(\mu_{i}+k-k \theta_{r})+\beta(\Lambda_{s}\theta_{e}+\Lambda _{i}+M_{1}+M_{2})\bigr]< 0, \\ &B_{3}=\Lambda_{s}(\mu_{i}+k-k \theta_{r})>0. \end{aligned}$$

Here, $\theta_{r}\in[0,1]$ denotes the probability of entering the removed compartment alive and $M_{i}>0$ ($i=1,2$) are defined as $M_{1}=\int_{0}^{\infty}\sigma(a)\int_{0}^{\infty}\Lambda_{e}(s){\rho _{e}(a)}/{\rho_{e}(s)}\,ds\,da$ and $M_{2}=\int_{0}^{\infty}\gamma(a)\int _{0}^{\infty}\Lambda_{r}(s){\rho_{r}(a)}/{\rho_{r}(s)}\,ds\,da$. Clearly, $f(0)>0$ and $f(\Lambda_{s}/\mu_{s})<0$. Thus, there exists a unique solution $S^{*}$ for $f(S^{*})=0$ in the interval $[0,\Lambda_{s}/\mu_{s}]$. It follows from equations (5.3)-(5.5) that $I^{*}$, and the functions $e^{*}(\cdot)$ and $r^{*}(\cdot)$ uniquely depend on $S^{*}$. Therefore, the unique steady state $T^{*}$ is determined. Obviously, $S^{*}\in(0,\Lambda_{s}/\mu_{s})$ implies that $I^{*}$, $e^{*}(\cdot)$ and $r^{*}(\cdot)$ are both positive. Thus, $T^{*}$ is strictly positive. Clearly, $\{T^{*}\}$ is an invariant bounded set.

From the above discussions, we have the following theorem.

Theorem 5.1

System (2.1) always has a unique endemic steady state $T^{*}=(S^{*}, e^{*}(\cdot), I^{*}, r^{*}(\cdot))$, and $T^{*}\in\mathcal{A}$.

5.2 Global stability

This subsection is devoted to showing the global stability of $T^{*}$, which implies that the attractor $\mathcal{A}$ only contains the unique endemic steady state.

The following lemma will be useful in the proof of our main result.

Lemma 5.1

Let $g(x)=x-1-\ln x$ for each $x\geq0$. Each solution of system (2.1) satisfies

$$\begin{aligned} &\frac{\partial}{\partial a}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr)= -\frac{\partial}{\partial t}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr), \\ &\frac{\partial}{\partial b}g \biggl(\frac{r(t,b)}{r^{*}(b)} \biggr)= -\frac{\partial}{\partial t}g \biggl(\frac{r(t,b)}{r^{*}(b)} \biggr). \end{aligned}$$

Proof

It follows from the fact that $g'(x)=1-{1}/{x}$ and the second equation of (5.1) that we have

$$\begin{aligned} \frac{\partial}{\partial a}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr) &= \biggl(1- \frac{e^{*}(a)}{e(t,a)} \biggr)\frac{\partial}{\partial a} \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr) \\ &= \biggl(1-\frac{e^{*}(a)}{e(t,a)} \biggr) \biggl[ \frac{1}{e^{*}(a)} \frac{\partial}{\partial a}e(t,a) +e(t,a)\frac{\partial}{\partial a}\frac{1}{e^{*}(a)} \biggr] \\ &= \biggl(\frac{e(t,a)}{e^{*}(a)}-1 \biggr) \biggl[ \frac{e_{a}(t,a)}{e(t,a)}+ \varepsilon(a)-\frac{\Lambda _{e}(a)}{e^{*}(a)} \biggr] \\ &=-\frac{\partial}{\partial t}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr), \end{aligned}$$

where ${e_{a}(t,a)}$ denotes ${\partial e(t,a)}/{\partial a}$. Similarly, we have

$$\begin{aligned} \frac{\partial}{\partial b}g \biggl(\frac{r(t,b)}{r^{*}(b)} \biggr) &= \biggl( \frac{r(t,b)}{r^{*}(b)}-1 \biggr) \biggl[ \frac{r_{b}(t,b)}{r(t,b)}+\eta(b)-\frac{\Lambda_{r}(b)}{r^{*}(b)} \biggr] \\ &=-\frac{\partial}{\partial t}g \biggl(\frac{r(t,b)}{r^{*}(b)} \biggr). \end{aligned}$$

 □

Based on the above preparations, we show our main result.

Theorem 5.2

The unique endemic steady state $T^{*}$ is globally asymptotically stable, and $\mathcal{A}=\{T^{*}\}$.

Proof

Define the Lyapunov functional as follows:

$$V=W_{s}+W_{e}+W_{i}+W_{r}, $$

where

$$\begin{aligned} &W_{s}=\theta_{e}S^{*}g \biggl(\frac{S}{S^{*}} \biggr), \qquad W_{e}= \int_{0}^{\infty}\omega_{e}(a)e^{*}(a)g \biggl( \frac {e(t,a)}{e^{*}(a)} \biggr)\,da, \\ &W_{i}=I^{*}g \biggl(\frac{I}{I^{*}} \biggr),\qquad W_{r}= \int_{0}^{\infty}\omega _{r}(b)r^{*}(b)g \biggl(\frac{r(t,b)}{r^{*}(b)} \biggr)\,db, \end{aligned}$$

where

$$\begin{aligned} &\omega_{e}(a)= \int_{a}^{\infty}\sigma(s)\exp \biggl(- \int_{a}^{s}\varepsilon (\tau)\,d\tau \biggr)\,ds, \\ &\omega_{r}(b)= \int_{b}^{\infty}\gamma(s)\exp \biggl(- \int_{b}^{s}\eta(\tau )\,d\tau \biggr)\,ds. \end{aligned}$$

Obviously, $\omega_{e}(a),\omega_{r}(b)>0$ for $a,b\geq0$ and $\omega _{j}(0)=\theta_{j}$ for $j=e,r$. The derivatives of $\omega_{e}(a)$ and $\omega_{r}(b)$ satisfy

$$ \frac{d\omega_{e}(a)}{da}=\omega_{e}(a)\varepsilon(a)- \sigma(a),\qquad \frac{d\omega_{r}(b)}{db}=\omega_{r}(b)\eta(b)-\gamma(b). $$

(5.6)

Since $\Lambda_{s}=\beta S^{*}I^{*}+\mu_{s} S^{*}$, the derivative of $W_{s}$ along with the solutions of (2.1) is

$$\begin{aligned} \frac{dW_{s}}{dt}&=\theta_{e}S^{*} \biggl(1-\frac{S^{*}}{S} \biggr) \frac{1}{S^{*}} \biggl[\mu_{s} S^{*} \biggl(1-\frac{S}{S^{*}} \biggr) +\beta S^{*}I^{*} \biggl(1-\frac{SI}{S^{*}I^{*}} \biggr) \biggr] \\ &=\theta_{e}\mu_{s} S^{*} \biggl(2-\frac{S^{*}}{S}- \frac{S}{S^{*}} \biggr) +\theta_{e}\beta S^{*}I^{*} \biggl(1- \frac{SI}{S^{*}I^{*}}-\frac {S^{*}}{S}+\frac{I}{I^{*}} \biggr). \end{aligned}$$

(5.7)

The derivative of $W_{e}$ along with the solutions of system (2.1) is calculated as

$$\begin{aligned} \frac{dW_{e}}{dt}&= \int_{0}^{\infty}\omega_{e}(a)e^{*}(a) \frac{\partial}{\partial t}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr)\,da \\ &=- \int_{0}^{\infty}\omega_{e}(a)e^{*}(a) \biggl(1-\frac{e^{*}(a)}{e(t,a)} \biggr)\frac{1}{e^{*}(a)} \biggl(\frac{\partial}{\partial a}e(t,a)+ \varepsilon(a)e(t,a) \biggr)\,da \\ &=- \int_{0}^{\infty}\omega_{e}(a)e^{*}(a) \biggl( \frac{e(t,a)}{e^{*}(a)}-1 \biggr) \biggl(\frac{e_{a}(t,a)}{e(t,a)}+\varepsilon(a)- \frac{\Lambda _{e}(a)}{e(t,a)} \biggr)\,da. \end{aligned}$$

It follows from (A₄) of Assumption 2.1 that we have $\lim_{a\rightarrow\infty}e(t,a)=0$, $\lim_{b\rightarrow\infty}r(t,b)=0$ for $t\geq0$. Combining the second equation of system (5.1), Lemma 5.1 and (5.6), using integration by parts, we have

$$\begin{aligned} \frac{dW_{e}}{dt} =&- \int_{0}^{\infty}\omega_{e}(a)e^{*}(a) \frac{\partial}{\partial a} g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr)\,da + \int_{0}^{\infty}\omega_{e}(a)e^{*}(a) \biggl( \frac{\Lambda_{e}(a)}{e^{*}(a)}-\frac{\Lambda _{e}(a)}{e(t,a)} \biggr)\,da \\ =&-\omega_{e}(a)e^{*}(a)g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr)\bigg|_{a=0}^{a=\infty} - \int_{0}^{\infty}g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr) \frac {d}{da} \bigl(\omega_{e}(a)e^{*}(a) \bigr)\,da \\ &{}+ \int_{0}^{\infty}\omega_{e}(a) \Lambda_{e}(a) \biggl(2-\frac {e(t,a)}{e^{*}(a)}-\frac{e^{*}(a)}{e(t,a)} \biggr) \,da \\ =&\omega_{e}(0)e^{*}(0)g \biggl(\frac{e(t,0)}{e^{*}(0)} \biggr) - \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac{e(t,a)}{e^{*}(a)} \biggr)\,da \\ &{}+ \int_{0}^{\infty}w_{e}(a) \Lambda_{e}(a)g \biggl(\frac{e(t,a)}{e^{*}(a)} \biggr)\,da \\ &{}+ \int_{0}^{\infty}\omega_{e}(a) \Lambda_{e}(a) \biggl(2-\frac {e(t,a)}{e^{*}(a)}-\frac{e^{*}(a)}{e(t,a)} \biggr) \,da \\ =&\omega_{e}(0)e^{*}(0)g \biggl(\frac{e(t,0)}{e^{*}(0)} \biggr) - \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac{e(t,a)}{e^{*}(a)} \biggr)\,da \\ &{}- \int_{0}^{\infty}w_{e}(a) \Lambda_{e}(a)g \biggl(\frac{e^{*}(a)}{e(t,a)} \biggr)\,da. \end{aligned}$$

Note that $\omega_{e}(0)=\theta_{e}$, $e^{*}(0)=\beta S^{*}I^{*}$ and $e(t,0)=\beta SI$, thus

$$\begin{aligned} \frac{dW_{e}}{dt} =& \theta_{e}\beta S^{*}I^{*}g \biggl( \frac{SI}{S^{*}I^{*}} \biggr) - \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac{S(t,a)}{S^{*}(a)} \biggr)\,da \\ &{}- \int_{0}^{\infty}w_{e}(a) \Lambda_{e}(a)g \biggl(\frac{e^{*}(a)}{e(t,a)} \biggr)\,da. \end{aligned}$$

(5.8)

Similarly, combining $\omega_{r}(0)=\theta_{r}$, $r^{*}(0)=kI^{*}$ and $r(t,0)=kI$ yields the derivative of $W_{r}$ along with the solutions of system (2.1) as follows:

$$\begin{aligned} \frac{dW_{r}}{dt} =& \theta_{r}kI^{*}g \biggl( \frac{I}{I^{*}} \biggr) - \int_{0}^{\infty}\gamma(b)r^{*}(b)g \biggl( \frac{r(t,b)}{r^{*}(b)} \biggr)\,db \\ &{}- \int_{0}^{\infty}w_{r}(b) \Lambda_{r}(b)g \biggl(\frac{r^{*}(b)}{r(t,b)} \biggr)\,db. \end{aligned}$$

(5.9)

Applying $\int_{0}^{\infty}\sigma(a)e^{*}(a)\,da+\int_{0}^{\infty}\gamma (b)r^{*}(b)\,db=(\mu_{i}+k)I^{*}$, we see that the derivative of $W_{i}$ along with the solutions of system (2.1) has the following form:

$$\begin{aligned} \frac{dW_{i}}{dt} =&I^{*} \biggl(1-\frac{I^{*}}{I} \biggr)\frac{1}{I^{*}} \biggl(\Lambda_{i}+ \int_{0}^{\infty}\sigma(a)e(t,a)\,da+ \int_{0}^{\infty}\gamma(b)r(t,b)\,db-( \mu_{i}+\delta_{i}+k)I \biggr) \\ =& \biggl(1-\frac{I}{I^{*}} \biggr) \biggl[\Lambda_{i} \biggl(1- \frac{I^{*}}{I} \biggr) + \int_{0}^{\infty}\sigma(a)e^{*}(a) \biggl( \frac{e(t,a)}{e^{*}(a)}-\frac {I}{I^{*}} \biggr)\,da \\ &{}+ \int_{0}^{\infty}\gamma(b)r^{*}(b) \biggl( \frac{r(t,b)}{r^{*}(b)}-\frac {I}{I^{*}} \biggr)\,db \biggr] \\ =&\Lambda_{i} \biggl(2-\frac{I^{*}}{I}-\frac{I}{I^{*}} \biggr) + \int_{0}^{\infty}\sigma(a)e^{*}(a) \biggl(1- \frac{I}{I^{*}}-\frac {I^{*}e(t,a)}{Ie^{*}(a)}+\frac{e(t,a)}{e^{*}(a)} \biggr)\,da \\ &{}+ \int_{0}^{\infty}\gamma(b)r^{*}(b) \biggl(1- \frac{I}{I^{*}}-\frac {I^{*}r(t,b)}{Ir^{*}(b)}+\frac{r(t,b)}{r^{*}(b)} \biggr) \,db. \end{aligned}$$

(5.10)

Summarizing (5.7)-(5.10), we have

$$\begin{aligned} \frac{dV}{dt} =&\theta_{e}\mu_{s} S^{*} \biggl(2- \frac{S^{*}}{S}-\frac{S}{S^{*}} \biggr) +\Lambda_{i} \biggl(2- \frac{I^{*}}{I}-\frac{I}{I^{*}} \biggr) - \int_{0}^{\infty}w_{e}(a) \Lambda_{e}(a)g \biggl(\frac{e^{*}(a)}{e(t,a)} \biggr)\,da \\ &{}+\theta_{r}kI^{*}g \biggl(\frac{I}{I^{*}} \biggr) - \int_{0}^{\infty}w_{r}(b) \Lambda_{r}(b)g \biggl(\frac{r^{*}(b)}{r(t,b)} \biggr)\,db +H_{si}+H_{e}+H_{r}, \end{aligned}$$

(5.11)

where

$$\begin{aligned} &H_{si}:=\theta_{e}\beta S^{*}I^{*} \biggl(1- \frac{SI}{S^{*}I^{*}}-\frac {S^{*}}{S}+\frac{I}{I^{*}} +\frac{SI}{S^{*}I^{*}}-1-\ln \frac{SI}{S^{*}I^{*}} \biggr), \\ &H_{e}:= \int_{0}^{\infty}\sigma(a)e^{*}(a) \biggl(1- \frac{I}{I^{*}}-\frac {I^{*}e(t,a)}{Ie^{*}(a)}+\frac{e(t,a)}{e^{*}(a)} +1-\frac{e(t,a)}{e^{*}(a)}+\ln \frac{e(t,a)}{e^{*}(a)} \biggr)\,da, \\ &H_{r}:= \int_{0}^{\infty}\gamma(b)r^{*}(b) \biggl(1- \frac{I}{I^{*}}-\frac {I^{*}r(t,b)}{Ir^{*}(b)}+\frac{r(t,b)}{r^{*}(b)} +1-\frac{r(t,b)}{r^{*}(b)}+\ln \frac{r(t,b)}{r^{*}(b)} \biggr)\,db. \end{aligned}$$

Note that

$$\begin{aligned}& \ln\frac{SI}{S^{*}I^{*}}=\ln\frac{I}{I^{*}}-\ln\frac{S^{*}}{S}, \\& \ln \frac{e(t,a)}{e^{*}(a)}=\ln\frac{I^{*}e(t,a)}{Ie^{*}(a)}+\ln\frac{I}{I^{*}}, \\& \ln \frac{r(t,b)}{r^{*}(b)}=\ln\frac{I^{*}r(t,b)}{Ir^{*}(b)}+\ln\frac{I}{I^{*}}, \end{aligned}$$

and

$$\begin{aligned} & \int_{0}^{\infty}\sigma(a)e^{*}(a)\,da = \theta_{e}\beta S^{*}I^{*} + \int_{0}^{\infty}\int_{0}^{a}\sigma(a)\Lambda_{e}(s) \frac{\rho _{e}(a)}{\rho_{e}(s)}\,ds\,da, \\ & \int_{0}^{\infty}\gamma(b)r^{*}(b)\,db = \theta_{r}kI^{*} + \int_{0}^{\infty}\int_{0}^{b}\gamma(b)\Lambda_{r}(s) \frac{\rho _{r}(b)}{\rho_{r}(s)}\,ds\,db. \end{aligned}$$

Thus, we have

$$\begin{aligned}& H_{si} =\theta_{e}\beta S^{*}I^{*} \biggl[ \biggl(1- \frac{S^{*}}{S}+\ln\frac {S^{*}}{S} \biggr) + \biggl(\frac{I}{I^{*}}-1-\ln \frac{I}{I^{*}} \biggr) \biggr] \\& \hphantom{H_{si}}=-\theta_{e}\beta S^{*}I^{*}g \biggl(\frac{S^{*}}{S} \biggr) +\theta_{e}\beta S^{*}I^{*}g \biggl(\frac{I}{I^{*}} \biggr), \end{aligned}$$

(5.12)

$$\begin{aligned}& H_{e} = \int_{0}^{\infty}\sigma(a)e^{*}(a) \biggl[ \biggl(1- \frac{I}{I^{*}}+\ln\frac{I}{I^{*}} \biggr) + \biggl(1-\frac{I^{*}e(t,a)}{Ie^{*}(a)}+\ln \frac {I^{*}e(t,a)}{Ie^{*}(a)} \biggr) \biggr]\,da \\& \hphantom{H_{e}}=- \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac{I}{I^{*}} \biggr)\,da - \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac {I^{*}e(t,a)}{Ie^{*}(a)} \biggr)\,da \\& \hphantom{H_{e}}=-\theta_{e}\beta S^{*}I^{*}g \biggl(\frac{I}{I^{*}} \biggr) -g \biggl(\frac{I}{I^{*}} \biggr) \int_{0}^{\infty}\int_{0}^{a}\sigma (a)\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds\,da \\& \hphantom{H_{e}={}}{}- \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac {I^{*}e(t,a)}{Ie^{*}(a)} \biggr)\,da, \end{aligned}$$

(5.13)

$$\begin{aligned}& H_{r} = \int_{0}^{\infty}\gamma(b)r^{*}(b) \biggl[ \biggl(1- \frac{I}{I^{*}}+\ln\frac{I}{I^{*}} \biggr) + \biggl(1-\frac{I^{*}r(t,b)}{Ir^{*}(b)}+\ln \frac {I^{*}r(t,b)}{Ir^{*}(b)} \biggr) \biggr]\,db \\& \hphantom{H_{r}}=- \int_{0}^{\infty}\gamma(b)r^{*}(b)g \biggl( \frac{I}{I^{*}} \biggr)\,db - \int_{0}^{\infty}\gamma(b)r^{*}(b)g \biggl( \frac {I^{*}r(t,b)}{Ir^{*}(b)} \biggr)\,db \\& \hphantom{H_{r}}=-\theta_{r}kI^{*}g \biggl(\frac{I}{I^{*}} \biggr) -g \biggl(\frac{I}{I^{*}} \biggr) \int_{0}^{\infty}\int_{0}^{b}\gamma (b)\Lambda_{r}(s) \frac{\rho_{r}(b)}{\rho_{r}(s)}\,ds\,db \\& \hphantom{H_{r}={}}{}- \int_{0}^{\infty}\gamma(b)r^{*}(b)g \biggl( \frac {I^{*}r(t,b)}{Ir^{*}(b)} \biggr)\,db. \end{aligned}$$

(5.14)

Consequently, we derive

$$\begin{aligned} \frac{dV}{dt} =&\theta_{e}\mu_{s} S^{*} \biggl(2- \frac{S^{*}}{S}-\frac {S}{S^{*}} \biggr) +\Lambda_{i} \biggl(2- \frac{I^{*}}{I}-\frac{I}{I^{*}} \biggr) \\ &{}- \int_{0}^{\infty}w_{e}(a) \Lambda_{e}(a)g \biggl(\frac{e^{*}(a)}{e(t,a)} \biggr)\,da - \int_{0}^{\infty}\sigma(a)e^{*}(a)g \biggl( \frac {I^{*}e(t,a)}{Ie^{*}(a)} \biggr)\,da \\ &{}- \int_{0}^{\infty}w_{r}(b) \Lambda_{r}(b)g \biggl(\frac{r^{*}(b)}{r(t,b)} \biggr)\,db - \int_{0}^{\infty}\gamma(b)r^{*}(b)g \biggl( \frac {I^{*}r(t,b)}{Ir^{*}(b)} \biggr)\,db \\ &{}-g \biggl(\frac{I}{I^{*}} \biggr) \int_{0}^{\infty}\int_{0}^{a}\sigma (a)\Lambda_{e}(s) \frac{\rho_{e}(a)}{\rho_{e}(s)}\,ds\,da \\ &{}-g \biggl(\frac{I}{I^{*}} \biggr) \int_{0}^{\infty}\int_{0}^{b}\gamma (b)\Lambda_{r}(s) \frac{\rho_{r}(b)}{\rho_{r}(s)}\,ds\,db, \end{aligned}$$

which implies that ${dV}/{dt}\leq0$ holds true. Furthermore, the strict equality holds only if ${(S,e(\cdot),I,r(\cdot))}={T^{*}}$. It is easy to obtain that the largest invariant subset of $\{(S,e(\cdot ),I,r(\cdot)):{dV}/{dt}=0\}$ is $\{T^{*}\}\subset\Omega$, and the Lyapunov-LaSalle invariance principle implies that the unique endemic steady state $T^{*}$ is globally asymptotically stable. □

6 Numerical simulations

In this section, we give some numerical simulations to show the epidemiological insights. In our application, we provide simulations of system (2.1) by using tuberculosis data from [4, 20, 26–28] to investigate the effects of immigration level on disease transmission. Here, we take

$$\mu_{s}=\mu_{e}=\mu_{r}=0.02, \qquad \mu_{i}=0.159,\qquad k=0.058, \qquad \beta=0.7\times10^{-4}. $$

Furthermore, we set the maximum age for the upper bound of latent and relapse age as 10 years. Then

$$\begin{aligned}& \sigma(a)=0.003 \biggl(1+\sin\frac{(a-5)\pi}{10} \biggr), \\& \gamma(b)=0.01 \biggl(1+\sin\frac{(b-5)\pi}{10} \biggr),\quad \mbox{for } 0\leq a,b\leq10. \end{aligned}$$

Thus, the averages of $\sigma(a)$ and $\gamma(b)$ are 0.003 and 0.01, which are the same as in [28] and [27], respectively. Set $\Lambda_{s}=7.5$, $\Lambda_{i}=0.2$ and

$$\begin{aligned}& \Lambda_{e}(a)=0.8 \biggl(1+\sin\frac{(a-5)\pi}{10} \biggr), \\& \Lambda_{r}(b)=1.5 \biggl(1+\sin\frac{(b-5)\pi}{10} \biggr),\quad \mbox{for } 0\leq a,b\leq10. \end{aligned}$$

Hence, $\bar{\Lambda}_{e}=0.8$ and $\bar{\Lambda}_{r}=1.5$. We have $\Lambda^{*}=10$.

6.1 The long time behaviors of system (2.1)

First, Theorem 5.2 asserts that the unique endemic steady state $T^{*}$ is globally asymptotically stable. This fact is revealed by Figure 2. From Figure 2, one can observe that the levels of all compartmental individuals tend to stable values, where $S(t)$, $e(t,a)$, $I(t)$ and $r(t,b)$ converge to the positive steady states $S^{*}$, $e^{*}(a)$, $I^{*}$ and $r^{*}(b)$. Easily, we can get $S^{*}=1.4\times10^{4}$ and $I^{*}=1.9\times10^{4}$. Obviously, it follows from Figure 2 that the numbers of the exposed and removed compartments are the distribution functions of a and b, respectively, particularly, $e^{*}(0)=1.96\times10^{4}$, $e^{*}(10)=1.26\times10^{4}$ and $r^{*}(0)=1\text{,}116$, $r^{*}(10)=693$. Furthermore, we show that the distribution of $e(t,a)$ and $r(t,a)$ for all age and at $a=5$. Figure 3 shows that this is a stationary distribution along all time.

Figure 2

The long time dynamical behaviors of system ( 2.1 ).

Figure 3

Distribution of the latent and removed populations prevalence for all age and at $\pmb{a=5}$ .

6.2 The age distribution of the latent and removed populations

Then we observe how the age distribution change the prevalence of the latent and removed populations. It follows from Figure 4 that with the increasing of age, the latent and removed populations prevalence decrease.

Figure 4

Distribution of the latent and removed populations prevalence for all time and at $\pmb{t=40}$ .

6.3 The stationary distribution

Finally, we are interested to observe that whether the stationary distribution exists or not. The numerical simulation is performed over a range of age, and then the stationary distribution of the latent and removed populations is obtained in Figure 5 for $t\in[0,60]$. From the color bar, we can observe that the distribution patterns of the latent and removed populations remain unchanged. Those distributions which do not change with time are called stationary patterns.

Figure 5

Distribution of the latent and removed populations prevalence for $\pmb{t\in[0,60]}$ and a range of age. Results are obtained for $\pmb{e(t,a)}$ (left) and $\pmb{r(t,b)}$ (right).

7 Conclusion and discussion

In this paper, we proposed and investigated a class of age-structured SEIR epidemic model with immigration. We show that, for all parameter values, the endemic steady state is unique and globally asymptotically stable by using the Lyapunov functional.

All simulation results show that the immigration of individuals leads to the insight that the TB cannot be fully eliminated from the population and will eventually reach a steady endemic level. The age of latent and removed leads to stationary patterns. This implies that TB will exist if either the immigration of population is allowed or there are TB infections in the region. Thus, in order to eradicate the TB, there are two choices: one is to prohibit the immigration of infected individuals, which is difficult to achieve; the other is to clear away the TB in any one region.