Theory and Modern Applications

# The stability analysis of an epidemic model with saturating incidence and age-structure in the exposed and infectious classes

## Abstract

In this paper, an HBV epidemic model that incorporates saturating incidence and age-structure in the exposed and infectious classes is proposed. We study the asymptotic smoothness of semi-flow generated by the model. By calculating the basic reproduction number and analyzing the characteristic equations, the local stability of disease-free and endemic steady states is studied. We investigate the global dynamics of this model by using Lyapunov functionals and LaSalle’s invariance principle and prove that, if the basic reproduction number is less than unity, the disease-free steady state is globally asymptotically stable; if the basic reproduction number is greater than unity, the endemic steady state is globally asymptotically stable.

## 1 Introduction

Hepatitis B is very contagious and it is hard to control its transmission. Thus studying the law of its infection has become the focus of attention. Researchers have presented an HBV epidemic model to study hepatitis B transmission and some results of the investigation have been given. The model’s dynamics is determined by the basic reproduction number (the average number of secondary infections caused by one infectious individual in its duration of infection). When the basic reproduction number is less than unity, the disease-free steady state is globally asymptotically stable and all hepatitis B patients will recover; when the basic reproduction number is larger than unity, there exists a unique endemic steady state and it is globally asymptotically stable.

Although there exist many epidemic models, some are not appropriate to show the hepatitis B transmission rules. In [1], Liu divided hepatitis B patients into acute and chronic patients and presented an ordinary differential model. However, the scaled probability of hepatitis B virus infection is connected with the age of infection and the risk per unit time of activation appears to be higher in the early stages of infection than in later stages. Therefore, in [2], Li presented the following model:

\begin{aligned} &\dot{S}(t) =\Lambda- (\mu+ p) S(t)- S(t) \int_{0}^{\infty}\beta_{1}(a) i(t,a) \, \mathrm{d}a-S(t) \int_{0}^{\infty}\beta_{3}(a) j(t,a) \, \mathrm {d}a , \\ &\dot{V}(t) = p S(t)- (\mu+ \rho) V(t)-V(t) \int_{0}^{\infty}\beta_{2}(a) i(t,a) \, \mathrm{d}a \\ &\hphantom{\dot{V}(t) ={}}{}-V(t) \int_{0}^{\infty}\beta_{4}(a) j(t,a) \, \mathrm {d}a, \\ &\frac{\partial e(t,a)}{\partial a}+\frac{\partial e(t,a)}{\partial t}=-\theta_{1}(a) e(t,a), \\ &\frac{\partial i(t,a)}{\partial a}+\frac{\partial i(t,a)}{\partial t}=-\theta_{2}(a) i(t,a), \\ &\frac{\partial j(t,a)}{\partial a}+\frac{\partial j(t,a)}{\partial t}=-\theta_{3}(a) j(t,a), \\ &\dot{R}(t) =\rho V(t)+ \int_{0}^{\infty}\delta_{1}(a) i(t,a) \, \mathrm {d}a+ \int_{0}^{\infty}\delta_{2}(a) j(t,a) \, \mathrm{d}a -\mu R(t), \end{aligned}
(1.1)

with boundary conditions

\begin{aligned}& \begin{aligned} e(t,0)&=S(t) \int_{0}^{\infty}\bigl(\beta_{1}(a) i(t,a) + \beta_{3}(a) j(t,a) \bigr)\,\mathrm{d}a \\ &\quad {}+V(t) \int_{0}^{\infty}\bigl(\beta_{2}(a) i(t,a)+ \beta_{4}(a) j(t,a) \bigr) \,\mathrm{d}a, \end{aligned} \\& i(t,0)= \int_{0}^{\infty}\gamma_{1}(a)e(t,a)\, \mathrm{d}a, \\& j(t,0)= \int_{0}^{\infty}\gamma_{2}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}\xi(a) i(t,a)\,\mathrm{d}a. \end{aligned}

It seems that vaccination has been the most effective prevention measure against hepatitis B. However, clinic evidence shows that some vaccines gradually lost immunity soon after vaccination. In other words, even though all individuals have been vaccinated, many of them may still be infected. Therefore, it is necessary to consider that a small part of vaccines will be susceptible individuals after a period of time. Furthermore, acute hepatitis B patients are highly infectious and they have apparent symptoms, the government and hospital have to take strategies as soon as possible, such like quarantine measures, therapeutic measures and so on. Considering the real transmission rules of hepatitis B, it is necessary to introduce saturating incidence to describe the law of hepatitis B’s infection. Therefore, we propose an epidemic model as follows:

\begin{aligned} &\dot{S}(t) =\Lambda- (\mu+ p) S(t)+\eta V(t)- S(t) \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a) i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a) j(t,a) \biggr)\,\mathrm{d}a, \\ &\dot{V}(t) = p S(t)- (\mu+ \rho+\eta) V(t)-V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a) i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a) j(t,a) \biggr) \,\mathrm{d}a, \\ &\frac{\partial e(t,a)}{\partial a}+\frac{\partial e(t,a)}{\partial t}=-\theta_{1}(a)e(t,a), \\ &\frac{\partial i(t,a)}{\partial a}+\frac{\partial i(t,a)}{\partial t}=-\theta_{2}(a)i(t,a), \\ &\frac{\partial j(t,a)}{\partial a}+\frac{\partial j(t,a)}{\partial t}=-\theta_{3}(a)j(t,a), \\ &{\dot{R}(t) =\rho V(t)+ \int_{0}^{\infty}\delta_{1}(a) i(t,a) \, \mathrm {d}a+ \int_{0}^{\infty}\delta_{2}(a) j(t,a) \, \mathrm{d}a -\mu R(t)}, \end{aligned}
(1.2)

where

\begin{aligned}& {\theta_{1}(a)=\mu+\gamma_{1}(a)+\gamma_{2}(a)}, \qquad {\theta_{2}(a)=\mu +\delta_{1}(a)+ \varepsilon_{1}(a)+\xi(a)}, \\& {\theta_{3}(a)=\mu+\delta_{2}(a)+\varepsilon_{2}(a)}, \end{aligned}

with boundary conditions

\begin{aligned} &e(t,0)= S(t) \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a) i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a) j(t,a) \biggr)\,\mathrm{d}a \\ &\hphantom{e(t,0)= {}}{}+V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a) i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a) j(t,a) \biggr) \,\mathrm{d}a, \\ &i(t,0)= \int_{0}^{\infty}\gamma_{1}(a)e(t,a)\, \mathrm{d}a, \\ &j(t,0)= \int_{0}^{\infty}\gamma_{2}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}\xi(a) i(t,a)\,\mathrm{d}a, \end{aligned}
(1.3)

and initial conditions

\begin{aligned} &S(0)=\varphi_{S}\geq0, \qquad V(0)= \varphi_{V}\geq0,\qquad e(0,a)=\varphi_{e}(a)\in L_{+}^{1}(0,\infty), \\ &i(0,a)=\varphi_{i}(a)\in L_{+}^{1}(0,\infty),\qquad j(0,a)= \varphi_{j}(a)\in L_{+}^{1}(0,\infty),\qquad R(0)=R_{0} \geq0, \end{aligned}
(1.4)

where $$e(t,a)$$ represents the density of exposed individuals with age of latency a at time t. $$i(t,a)$$, $$j(t,a)$$ represent the density of patients with acute hepatitis B and chronic hepatitis B with age of infection a at time t, respectively. α means the saturating incidence coefficient. The parameters of model (1.2) are biologically explained in Table 1.

Since the variable $$R(t)$$ does not appear in the first five equations of (1.2), in this paper, we consider the following subsystem:

\begin{aligned}& \dot{S}(t) =\Lambda- (\mu+ p) S(t)+\eta V(t)- S(t) \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a) i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a) j(t,a) \biggr)\,\mathrm{d}a, \\& \dot{V}(t) = p S(t)- (\mu+ \rho+\eta) V(t)-V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a) i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a) j(t,a) \biggr) \,\mathrm{d}a, \\& \frac{\partial e(t,a)}{\partial a}+\frac{\partial e(t,a)}{\partial t}=-\theta_{1}(a)e(t,a), \\& \frac{\partial i(t,a)}{\partial a}+\frac{\partial i(t,a)}{\partial t}=-\theta_{2}(a)i(t,a), \\& \frac{\partial j(t,a)}{\partial a}+\frac{\partial j(t,a)}{\partial t}=-\theta_{3}(a)j(t,a), \\& e(t,0)=S(t) \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a) i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a) j(t,a) \biggr)\,\mathrm{d}a \\& \hphantom{e(t,0)={}}{}+V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a) i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a) j(t,a) \biggr) \,\mathrm{d}a, \\& i(t,0)= \int_{0}^{\infty}\gamma_{1}(a)e(t,a)\, \mathrm{d}a, \\& j(t,0)= \int_{0}^{\infty}\gamma_{2}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}\xi(a) i(t,a)\,\mathrm{d}a. \end{aligned}
(1.5)

This paper is organized as follows. In Sect. 2, we introduce some basic results of system (1.5), including state space, assumptions and boundedness of the solutions. Asymptotic smoothness of the semi-flow is analyzed in Sect. 3, which is generated by the system (1.5). Then we study the existence of equilibria and obtain the expression of the basic reproduction number $$R_{0}$$ in Sect. 4. The local stability of equilibria is proved in Sect. 5, while the uniform persistence of the system (1.5) is verified in Sect. 6. In Sect. 7, we give a proof of the global stability of equilibria. A brief remark is given in Sect. 8 to conclude this work.

More details concerning the global stability analysis of epidemic model, we refer the reader to [323].

## 2 Preliminaries

To make the model be biologically significant, we list the assumption as follows:

### Assumption 1

We assume that

1. (i)

$$\beta_{1}(a)$$, $$\beta_{2}(a)$$, $$\beta_{3}(a)$$, $$\beta_{4}(a)$$, $$\theta_{1}(a)$$, $$\theta_{2}(a)$$, $$\xi(a)$$ are non-negative and belong to $$L_{+}^{\infty}(0,\infty)$$ with respective essential upper bound $$\bar{\beta}_{1}$$, $$\bar{\beta }_{2}$$, $$\bar{\beta}_{3}$$, $$\bar{\beta}_{4}$$, $$\bar{\theta}_{1}$$, $$\bar {\theta}_{2}$$, $$\bar{\xi}\in(0,\infty)$$;

2. (ii)

$$\beta_{1}(a)$$, $$\beta_{2}(a)$$, $$\beta_{3}(a)$$, $$\beta_{4}(a)$$, $$\xi(a)$$ are Lipschitz continuous on $${R}_{+}$$ with coefficients $$M_{\beta_{1}}$$, $$M_{\beta_{2}}$$, $$M_{\beta_{3}}$$, $$M_{\beta_{4}}$$, $$M_{\xi}$$, respectively;

3. (iii)

there exists a positive constant $$\mu_{0}\in(0,\mu]$$ such that $$|\theta_{1}(a)-\gamma_{1}(a)-\gamma_{2}(a)|\geq\mu_{0}$$, $$|\theta _{2}(a)-\xi(a)|\geq\mu_{0}$$, $$\theta_{3}(a)\geq\mu_{0}$$, for all $$a>0$$.

### 2.1 State space

Define the space of functions X as

$${X}={R}_{+}\times{R}_{+} \times L_{+}^{1}(0,\infty) \times L_{+}^{1}(0,\infty )\times L_{+}^{1}(0,\infty),$$

equipped with the norm

$$\bigl\Vert ( x_{1},x_{2},x_{3},x_{4},x_{5} ) \bigr\Vert _{\mathcal {X}}= \vert x_{1} \vert + \vert x_{2} \vert + \int_{0}^{\infty}\bigl\vert x_{3}(a) \bigr\vert \,\mathrm{d}a+ \int_{0}^{\infty}\bigl\vert x_{4}(a) \bigr\vert \,\mathrm{d}a+ \int_{0}^{\infty}\bigl\vert x_{5}(a) \bigr\vert \,\mathrm{d}a.$$

Then the initial values (1.4) of system (1.5) are taken to be included in X:

$$\bigl(S(0),V(0),e(0,a),i(0,a),j(0,a) \bigr)= \bigl(S_{0},V_{0}, \varphi _{e}(a),\varphi_{i}(a),\varphi_{j}(a) \bigr)\in{X}.$$

By the standard theory of functional differential equations [24], it can be verified that system (1.5) with initial conditions (1.4) has a unique non-negative solution. Thus, we have a continuous semi-flow associated with system (1.5), that is,

$$\Phi_{t}(X_{0}):=X(t)= \bigl(S(t),V(t),e(t,\cdot),i(t, \cdot),j(t,\cdot) \bigr)\in \mathcal{X},\quad t\geq0,$$
(2.1)

with

\begin{aligned} \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}}&= \bigl\Vert S(t),V(t),e(t,\cdot),i(t,\cdot ),j(t,\cdot) \bigr\Vert _{\mathcal{X}} \\ &= \bigl\vert S(t) \bigr\vert + \bigl\vert V(t) \bigr\vert + \int_{0}^{\infty}\bigl\vert e(t,a) \bigr\vert \, \mathrm{d}a + \int_{0}^{\infty}\bigl\vert i(t,a) \bigr\vert \, \mathrm{d}a + \int_{0}^{\infty}\bigl\vert j(t,a) \bigr\vert \, \mathrm{d}a. \end{aligned}

Finally, we define the state space for system (1.5) as

\begin{aligned} \Upsilon :=& \biggl\{ \bigl(S(t),V(t),e(t,\cdot),i(t,\cdot),j(t,\cdot) \bigr)\in \mathcal{X}:0\leq S(t)+V(t)+ \int_{0}^{\infty}e(t,a)\,\mathrm{d}a \\ &{}+ \int _{0}^{\infty}i(t,a)\,\mathrm{d}a+ \int_{0}^{\infty}j(t,a) \,\mathrm{d}a\leq \frac{\Lambda}{\mu_{0}} \biggr\} , \end{aligned}

which can be proved to be positive invariant by the following proposition.

### 2.2 Boundedness

The last three equations of system (1.5) can be reformulated as Volterra equations by use of Volterra formulation. In order to be convenient for computation, we denote

\begin{aligned}& {B_{1}(a)=\exp \biggl(- \int_{0}^{a} \theta_{1}(\tau)\,\mathrm{d} \tau \biggr)}, \qquad {B_{2}(a)=\exp \biggl(- \int_{0}^{a} \theta_{2}(\tau)\,\mathrm{d} \tau \biggr)}, \\& {B_{3}(a)=\exp \biggl(- \int_{0}^{a} \theta_{3}(\tau)\,\mathrm{d} \tau \biggr)}. \end{aligned}

From the expressions of $$B_{1}(a)$$, $$B_{2}(a)$$ and $$B_{3}(a)$$, according to Assumption 1, it is easy to see that, for all $$a\geq0$$,

\begin{aligned}& 0\leq B_{1}(a),B_{2}(a),B_{3}(a)\leq e^{-\mu_{0}a}, \\& B'_{1}(a)=-\theta_{1}(a)B_{1}(a), \qquad B'_{2}(a)=-\theta _{2}(a)B_{2}(a), \qquad B'_{3}(a)=-\theta_{3}(a)B_{3}(a). \end{aligned}

By integrating the terms $$e(t,a)$$, $$i(t,a)$$ and $$j(t,a)$$ along the characteristic line $$t-a=\mathrm{const.}$$, respectively, we get the following expressions:

\begin{aligned}& e(t,a)= \textstyle\begin{cases} e(t-a,0)B_{1}(a) &\mbox{for }0\leq a\leq t,\\ \varphi_{e}(a-t) \frac{B_{1}(a)}{B_{1}(a-t)} &\mbox{for }0\leq t\leq a, \end{cases}\displaystyle \end{aligned}
(2.2)
\begin{aligned}& i(t,a)= \textstyle\begin{cases} i(t-a,0)B_{2}(a) &\mbox{for }0\leq a\leq t, \\ \varphi_{i}(a-t) \frac{B_{2}(a)}{B_{2}(a-t)}& \mbox{for }0\leq t\leq a, \end{cases}\displaystyle \end{aligned}
(2.3)
\begin{aligned}& j(t,a)= \textstyle\begin{cases} j(t-a,0)B_{3}(a) &\mbox{for }0\leq a\leq t, \\ \varphi_{j}(a-t) \frac{B_{3}(a)}{B_{3}(a-t)}& \mbox{for }0\leq t\leq a. \end{cases}\displaystyle \end{aligned}
(2.4)

Before analyzing the boundedness of system (1.5), we first show the non-negative of the solution. From (2.2)–(2.4), it is not difficult to verify that $$e(t,a)$$, $$i(t,a)$$ and $$j(t,a)$$ are non-negative, due to the non-negativity of $$B_{1}(a)$$, $$B_{2}(a)$$, $$B_{3}(a)$$, and the non-negative initial conditions (1.4). From the first equation of (1.3), we see that $$S(t)>0$$ holds or $$V(t)>0$$ holds. For the sake of contradiction, if $$V(t)>0$$ and $$S(t) \le0$$, from the second equation of (1.5), we know that $$\dot{V}(t)<0$$. Then $$V(t)$$ is a monotone decreasing function with respect to t and there exists at least one zero solution, which contradicts $$V(t)>0$$. If $$S(t)>0$$ and $$V(t) \le 0$$, $$V(t)$$ is a monotone increasing function with respect to t, which contradicts the later boundedness analysis. Thus, all the solutions of system (1.5) remain non-negative.

In order to imply the boundedness of system (1.5), we have the following proposition.

### Proposition 2.1

Consider system (1.5) and Eq. (2.1), we have

1. (i)

ϒ is positively invariant for $$\Phi_{t}$$, that is, $$\Phi _{t}(X_{0})\in\Upsilon$$, for $$\forall t\geq0$$, $$X_{0}\in\Upsilon$$;

2. (ii)

$$\Phi_{t}$$ is point dissipative: there is a bounded set that attracts all points in $$\mathcal{X}$$.

### Proof

Note that

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}} =&\frac {\mathrm{d}S(t)}{\mathrm{d}t}+\frac{\mathrm{d}V(t)}{\mathrm{d}t}+\frac {\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}e(t,a){\, \mathrm{d}}a \\ &{}+\frac {\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}i(t,a){\,\mathrm{d}}a+ \frac {{\mathrm{d}}}{\mathrm{d}t} \int_{0}^{\infty}j(t,a)\,\mathrm{d}a. \end{aligned}
(2.5)

By Eq. (2.2), we get

$$\int_{0}^{\infty}e(t,a)\,\mathrm{d}a= \int_{0}^{t} e(t-a,0)B_{1}(a)\,\mathrm {d}a + \int_{t}^{\infty}\varphi_{e}(a-t) \frac{B_{1}(a)}{B_{1}(a-t)}\, \mathrm{d}a.$$

Taking the substitution $$\sigma=t-a$$ and $$\tau=a-t$$ in the first and second integral, respectively, and differentiating by t, we get

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}e(t,a)\,\mathrm{d}a =& \frac {\mathrm{d}}{\mathrm{d}t} \int_{0}^{t} e(\sigma,0)B_{1}(t-\sigma)\, \mathrm {d}\sigma+ \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}\varphi_{e}(\tau ) \frac{B_{1}(t+\tau)}{B_{1}(\tau)}\,\mathrm{d}\tau \\ =&e(t,0)B_{1}(0)+ \int_{0}^{\infty}\varphi_{e}(\tau) \frac{B'_{1}(t+\tau )}{B_{1}(\tau)}\,\mathrm{d}\tau+ \int_{0}^{t} e(\sigma-a)B'_{1}(t- \sigma)\, \mathrm{d}\sigma. \end{aligned}

Noting that $$B_{1}(0)=1$$ and $$B'_{1}(a)=-\theta_{1}(a)B_{1}(a)$$, we obtain

$$\frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}e(t,a)\,\mathrm {d}a=e(t,0)- \int_{0}^{\infty}\theta_{1}(a)e(t,a)\, \mathrm{d}a.$$
(2.6)

Similarly, we have

\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}i(t,a)\,\mathrm {d}a=i(t,0)- \int_{0}^{\infty}\theta_{2}(a)i(t,a)\, \mathrm{d}a, \end{aligned}
(2.7)
\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{\infty}j(t,a)\,\mathrm {d}a=j(t,0)- \int_{0}^{\infty}\theta_{3}(a)j(t,a)\, \mathrm{d}a. \end{aligned}
(2.8)

By (2.6), (2.7) and (2.8), Eq. (2.5) becomes

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}} =&\Lambda -\mu S(t)-(\mu+\rho)V(t)-S(t) \int_{0}^{\infty}\biggl(\frac{\beta _{1}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}-V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}+S(t) \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}+V(t) \int_{0}^{\infty}\biggl(\frac{\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\theta_{1}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}\gamma _{1}(a)e(t,a)\, \mathrm{d}a- \int_{0}^{\infty}\theta_{2}(a)i(t,a)\,\mathrm {d}a \\ &{}+ \int_{0}^{\infty}\gamma_{2}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}\xi (a)i(t,a)\,\mathrm{d}a- \int_{0}^{\infty}\theta_{3}(a)j(t,a)\, \mathrm{d}a \\ =&\Lambda-\mu S(t)-(\mu+\rho)V(t)- \int_{0}^{\infty}\bigl(\theta _{1}(a)- \gamma_{1}(a)-\gamma_{2}(a) \bigr)e(t,a)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\bigl(\theta_{2}(a)-\xi(a) \bigr)i(t,a)\,\mathrm{d}a- \int _{0}^{\infty}\theta_{3}(a)j(t,a)\, \mathrm{d}a. \end{aligned}

Thus, from (iii) of Assumption 1, we can get

\begin{aligned} \frac{d}{dt} \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}} \leq&\Lambda-\mu S(t)-(\mu +\rho)V(t) \\ &{}-\mu_{0} \biggl( \int_{0}^{\infty}e(t,a)\,\mathrm{d}a+ \int_{0}^{\infty}i(t,a)\, \mathrm{d}a+ \int_{0}^{\infty}j(t,a)\,\mathrm{d}a \biggr) \\ \leq&\Lambda-\mu_{0} \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}}. \end{aligned}

Hence, it follows from the variation of constants formula that, for $$t\geq0$$,

$$\bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}} \leq\frac{\Lambda}{\mu_{0}}-e^{-\mu _{0}t} \biggl(\frac{\Lambda}{\mu_{0}}- \bigl\Vert \Phi_{t}(X_{0}) \bigr\Vert _{\mathcal{X}} \biggr),$$
(2.9)

which implies that $$\Phi_{t}(X_{0})\in\Upsilon$$ for any solution of (1.5) satisfying $$X_{0}\in\Upsilon$$ and all $$t\geq0$$. Thus, the positive invariance of set ϒ for semi-flow Φ can be verified.

Moreover, by (2.9) we can make inferences that $$\lim\sup_{t\rightarrow \infty}\|\Phi_{t}(X_{0})\|_{\mathcal{X}}\leq\Lambda/\mu_{0}$$ for any $$X_{0}\in\mathcal{X}$$. Therefore, Φ is point dissipative and ϒ attracts all points in $$\mathcal{X}$$. This completes the proof. □

### Proposition 2.2

If $$X_{0}\in\mathcal{X}$$ and $$\|X_{0}\| _{\mathcal{X}}\leq M$$ for some constant $$M\geq\Lambda/\mu_{0}$$, then the following statements hold for $$t\geq0$$:

1. (i)

$$0\leq S(t)$$, $$V(t)$$, $$\int_{0}^{\infty}e(t,a)\,\mathrm{d}a$$, $$\int _{0}^{\infty}\frac{i(t,a)}{1+\alpha i(t,a)}\,\mathrm{d}a$$, $$\int _{0}^{\infty}j(t,a)\,\mathrm{d}a\leq M$$;

2. (ii)

$$e(t,0)\leq(\bar{\beta}_{1}+\bar{\beta}_{2}+\bar{\beta}_{3}+\bar {\beta}_{4})M^{2}$$, $$i(t,0)\leq\bar{\gamma}_{1}M$$, $$j(t,0)\leq(\bar {\gamma}_{2}+\bar{\xi})M$$.

### Proposition 2.3

Let $$C\in\mathcal{X}$$ be bounded, then:

1. (i)

$$\Phi_{t}(C)$$ is bounded;

2. (ii)

$$\Phi_{t}$$ is eventually bounded on C.

## 3 Asymptotic smoothness

In order to obtain global properties of the semi-flow $${\Phi(t)}_{t\geq 0}$$, it is necessary to prove that the semi-flow is asymptotically smooth. Before giving the results, we first introduce some lemmas for later use.

### Lemma 3.1

([8])

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

The definition of asymptotic smoothness is as follows:

### Definition 3.1

([25])

A semi-flow $$\Phi (t,X_{0}):={R}^{+}\times\mathcal{X}\rightarrow\mathcal{X}$$ is said to be asymptotically smooth, if, for any nonempty, closed bounded set $$B\subset\mathcal{X}$$ for which $$\Phi(t,B)\subset B$$, there is a compact set $$B_{0}\subset B$$ such that $$B_{0}$$ attracts B.

In order to prove the asymptotic smoothness of the semi-flow, we will apply the following results, which is based on Lemma 3.2.3 in [25].

### Lemma 3.2

([21, 22])

If the following two conditions hold then the semi-flow $$\Phi(t,X_{0})=\phi(t,X_{0})+\varphi(t,X_{0}): {R^{+}}\times\mathcal{X}\rightarrow\mathcal{X}$$ is asymptotically smooth in $$\mathcal{X}$$.

1. (i)

There exists a continuous function $$w: {R}^{+}\times {R}^{+}\rightarrow{R}^{+}$$ such that $$w(t,h)\rightarrow0$$ as $$t\rightarrow\infty$$ and $$\|\varphi(t,X_{0})\|_{\mathcal{X}}\leq w(t,h)$$ if $$\|X_{0}\|_{\mathcal{X}}\leq h$$;

2. (ii)

For $$t\geq0$$, $$\phi(t,X_{0})$$ is completely continuous.

To verify that the two conditions are fulfilled for system (1.5), we decompose $$\Phi:{R}^{+}\times\mathcal{X}\rightarrow\mathcal{X}$$ into the following two operators $$\phi(t,X_{0}), \varphi (t,X_{0}):{R}^{+}\times{X}\rightarrow{X}$$. Let $$\phi(t,X_{0}):=(S(t),V(t),\tilde{e}(t,\cdot),\tilde{i}(t,\cdot ),\tilde{j}(t,\cdot))$$ and $$\varphi(t,X_{0}):=(0,0,\tilde{\varphi}_{e}(t,\cdot),\tilde{\varphi }_{i}(t,\cdot), \tilde{\varphi}_{j}(t,\cdot))$$ where

\begin{aligned}& \tilde{\varphi}_{e}(t,a):= \textstyle\begin{cases} 0 &\mbox{for } 0\leq a\leq t,\\ e(t,a) &\mbox{for } 0\leq t\leq a \end{cases}\displaystyle \quad \mbox{and}\quad \tilde{e}(t,a):= \textstyle\begin{cases} e(t,a) &\mbox{for } 0\leq a\leq t,\\ 0 &\mbox{for } 0\leq t\leq a, \end{cases}\displaystyle \end{aligned}
(3.1)
\begin{aligned}& \tilde{\varphi}_{i}(t,a):= \textstyle\begin{cases} 0 &\mbox{for } 0\leq a\leq t,\\ i(t,a) &\mbox{for } 0\leq t\leq a \end{cases}\displaystyle \quad \mbox{and}\quad \tilde{i}(t,a):= \textstyle\begin{cases} i(t,a) &\mbox{for } 0\leq a\leq t,\\ 0 &\mbox{for } 0\leq t\leq a, \end{cases}\displaystyle \end{aligned}
(3.2)
\begin{aligned}& \tilde{\varphi}_{j}(t,a):= \textstyle\begin{cases} 0 &\mbox{for } 0\leq a\leq t,\\ j(t,a) &\mbox{for } 0\leq t\leq a \end{cases}\displaystyle \quad \mbox{and}\quad \tilde{j}(t,a):= \textstyle\begin{cases} j(t,a) &\mbox{for } 0\leq a\leq t,\\ 0 &\mbox{for } 0\leq t\leq a. \end{cases}\displaystyle \end{aligned}
(3.3)

Then we have $$\Phi(t,X_{0})=\varphi(t,X_{0})+\phi(t,X_{0})$$ for all $$t\geq0$$. In order to verify that condition (i) of Lemma 3.2 holds true, we turn to a proof of the following proposition.

### Proposition 3.1

([26])

For $$h>0$$, let $$w(t,h)=he^{-\mu _{0}t}$$. Then $$\lim_{t\rightarrow\infty}w(t,h)=0$$ and $$\|\varphi (t,X_{0})\|_{{X}}\leq w(t,h)$$ if $$\| X_{0} \|_{{X}}\leq h$$.

### Proof

Obviously, $$\lim_{t\rightarrow\infty}w(t,h)=0$$. For $$X_{0}\in\Upsilon$$ and $$\| X_{0}\|_{{X}}\leq h$$, we have

\begin{aligned} \bigl\Vert \varphi(t,X_{0}) \bigr\Vert _{{X}} =&|0|+|0|+ \int_{0}^{\infty} \bigl\vert \tilde{\varphi }_{e}(t,a) \bigr\vert \,\mathrm{d}a+ \int_{0}^{\infty} \bigl\vert \tilde{ \varphi}_{i}(t,a) \bigr\vert \, \mathrm{d}a + \int_{0}^{\infty} \bigl\vert \tilde{ \varphi}_{j}(t,a) \bigr\vert \,\mathrm{d}a \\ =& \int_{t}^{\infty} \biggl\vert \varphi_{e}(a-t) \frac{B_{1}(a)}{B_{1}(a-t)} \biggr\vert \,\mathrm{d}a+ \int_{t}^{\infty} \biggl\vert \varphi_{i}(a-t) \frac {B_{2}(a)}{B_{2}(a-t)} \biggr\vert \,\mathrm{d}a \\ &{}+ \int_{t}^{\infty} \biggl\vert \varphi_{j}(a-t) \frac{B_{3}(a)}{B_{3}(a-t)} \biggr\vert \,\mathrm{d}a \\ =& \int_{0}^{\infty} \biggl\vert \varphi_{e}( \tau) \frac{B_{1}(t+\tau )}{B_{1}(\tau)} \biggr\vert \,\mathrm{d}\tau+ \int_{0}^{\infty} \biggl\vert \varphi _{i}( \tau) \frac{B_{2}(t+\tau)}{B_{2}(\tau)} \biggr\vert \,\mathrm{d}\tau \\ &{}+ \int_{0}^{\infty} \biggl\vert \varphi_{j}( \tau)\frac{B_{3}(t+\tau)}{B_{3}(\tau )} \biggr\vert \,\mathrm{d}\tau \\ =& \int_{0}^{\infty} \biggl\vert \varphi_{e}( \tau) \exp \biggl(- \int_{\tau }^{t+\tau}\theta_{1}(\upsilon)\, \mathrm{d}\upsilon \biggr) \biggr\vert \, \mathrm{d}\tau \\ &{}+ \int_{0}^{\infty} \biggl\vert \varphi_{i}( \tau) \exp \biggl(- \int _{\tau}^{t+\tau}\theta_{2}(\upsilon)\, \mathrm{d}\upsilon \biggr) \biggr\vert \, \mathrm{d}\tau \\ &{}+ \int_{0}^{\infty} \biggl\vert \varphi_{j}( \tau)\exp \biggl(- \int_{\tau}^{t+\tau }\theta_{3}(\upsilon)\, \mathrm{d}\upsilon \biggr) \biggr\vert \,\mathrm{d}\tau. \end{aligned}

By (iii) of Assumption 1, $$\theta_{1}(a)$$, $$\theta_{2}(a)$$, $$\theta _{2}(a)\geq\mu_{0}$$ for $$a\geq0$$, we have

\begin{aligned} \bigl\Vert \varphi(t,X_{0}) \bigr\Vert _{{X}} \leq& e^{-\mu_{0}t} \biggl(|0|+|0|+ \int _{0}^{\infty} \bigl\vert \varphi_{e}( \tau) \bigr\vert \,\mathrm{d}\tau+ \int_{0}^{\infty } \bigl\vert \varphi_{i}( \tau) \bigr\vert \,\mathrm{d}\tau + \int_{0}^{\infty} \bigl\vert \varphi_{j}( \tau) \bigr\vert \,\mathrm{d}\tau \biggr) \\ =&e^{-\mu_{0}t} \Vert X_{0} \Vert _{{X}}\leq he^{-\mu_{0}t}\equiv w(t,h). \end{aligned}

This completes the proof. □

To verify (ii) of Lemma 3.2, we need to give the following lemma.

### Lemma 3.3

([27])

Let $$K\subset L^{p}(0,\infty)$$ be closed and bounded where $$p\geq1$$. Then K is compact if the following conditions hold true.

1. (1)

$${\sup_{f \in K}}\int_{0}^{\infty}{f(a)\,\mathrm{d}a} < \infty$$.

2. (2)

$${\lim_{r \to\infty}}\int_{r}^{\infty}{f(a)\,\mathrm{d}a} \to0$$ uniformly in $$f \in K$$.

3. (3)

$${\lim_{h \to{0^{+} }}}\int_{0}^{\infty}{ \vert {f(a + h) - f(a)} \vert \,\mathrm{d}a} \to0$$ uniformly in $$f \in K$$.

4. (4)

$${\lim_{h \to{0^{+} }}}\int_{0}^{h} {f(a)\,\mathrm{d}a} \to0$$ uniformly in $$f \in K$$.

### Proposition 3.2

([27])

For $$t\geq0$$, $$\phi(t,X_{0})$$ is completely continuous.

### Proof

From Lemma 3.3, for any closed and bounded set $$B \subset {X}$$, we see that $$\phi(t,B)$$ is compact. According to Proposition 2.2, $$S(t)$$ and $$V(t)$$ remain in the compact set $$[0,\Lambda/\mu_{0}]\subset[0,M]$$, where $$M\geq\Lambda/\mu_{0}$$ is the bound for B. Thus, it is only to show that $$\tilde{e}(t,a)$$, $$\tilde {i}(t,a)$$ and $$\tilde{j}(t,a)$$ remain in a precompact subset of $$L_{+}^{1}(0,\infty)$$, which is independent of $$X_{0}\in\Upsilon$$.

Now, from (2.2) and (3.1) we have

$$0\leq\tilde{e}(t,a)= \textstyle\begin{cases} e(t-a,0)B_{1}(a) &\mbox{for }0\leq a < t, \\ 0 &\mbox{for }0\leq t\leq a. \end{cases}$$

Then combining (i) of Proposition 2.2, we have

$$\tilde{e}(t,a)\leq(\bar{\beta}_{1}+\bar{\beta}_{2}+\bar{ \beta}_{3}+\bar {\beta}_{4})M^{2}e^{-\mu_{0}a},$$

which implies that (1), (2) and (4) in Lemma 3.3 are satisfied. To check condition (3), for sufficiently small $$h\in(0,t)$$, we have

\begin{aligned} \int_{0}^{\infty} \bigl\vert \tilde{e}(t,a+h)- \tilde{e}(t,a) \bigr\vert \,\mathrm{d}a =& \int _{0}^{t} \bigl\vert e(t,a+h)-e(t,a) \bigr\vert \,\mathrm{d}a \\ =& \int_{0}^{t-h} \bigl\vert e(t-a-h,0)B_{1}(a+h)-e(t-a,0)B_{1}(a) \bigr\vert \,\mathrm{d}a \\ &{}+ \int_{t-h}^{t} \bigl\vert 0-e(t-a,0)B_{1}(a) \bigr\vert \,\mathrm{d}a \\ \leq& \int_{0}^{t-h}e(t-a-h,0) \bigl\vert B_{1}(a+h)-B_{1}(a) \bigr\vert \,\mathrm{d}a \\ &{}+ \int_{0}^{t-h}B_{1}(a) \bigl\vert e(t-a-h,0)-e(t-a,0) \bigr\vert \,\mathrm{d}a \\ &{}+ \int_{t-h}^{t} \bigl\vert e(t-a,0)B_{1}(a) \bigr\vert \,\mathrm{d}a. \end{aligned}

Recall that $$0\leq B_{1}(a)\leq e^{-\mu_{0}a}\leq1$$ and $$B_{1}(a)$$ is non-increasing function with respect to a, it follows that

\begin{aligned} \int_{0}^{t-h} \bigl\vert B_{1}(a+h)-B_{1}(a) \bigr\vert \,\mathrm{d}a =& \int _{0}^{t-h}B_{1}(a)\,\mathrm{d}a- \int_{0}^{t-h}B_{1}(a+h)\,\mathrm{d}a \\ =& \int_{0}^{t-h}B_{1}(a)\,\mathrm{d}a- \int_{h}^{t}B_{1}(a)\,\mathrm {d}a \\ =& \int_{0}^{t-h}B_{1}(a)\,\mathrm{d}a- \int_{h}^{t-h}B_{1}(a)\,\mathrm {d}a- \int_{t-h}^{t}B_{1}(a)\,\mathrm{d}a \\ =& \int_{0}^{h}B_{1}(a)\,\mathrm{d}a- \int_{t-h}^{t}B_{1}(a)\,\mathrm {d}a\leq h. \end{aligned}

Hence, from (ii) of Proposition 2.2, we have

$$\int_{0}^{\infty} \bigl\vert \tilde{e}(t,a+h)- \tilde{e}(t,a) \bigr\vert \,\mathrm{d}a\leq 2(\bar{\beta}_{1}+\bar{ \beta}_{2}+\bar{\beta}_{3}+\bar{\beta }_{4})M^{2}h+ \Delta,$$
(3.4)

where

$$\Delta= \int_{0}^{t-h}B_{1}(a) \bigl\vert e(t-a-h,0)-e(t-a,0) \bigr\vert \,\mathrm{d}a.$$

From (i) of Proposition 2.2, we find that $$|\mathrm{d}S(t)/\mathrm{d}t|$$ is bounded by $$M_{S}=\Lambda+(\mu+p)M+\bar{\beta}_{1}M^{2}+\bar{\beta}_{3}M^{2}$$ and $$|\mathrm{d}V(t)/\mathrm{d}t|$$ is bounded by $$M_{V}=(\mu+p)M+\bar{\beta}_{2}M^{2}+\bar {\beta}_{4}M^{2}$$. Therefore, $$S(\cdot)$$ and $$V(\cdot)$$ are Lipschitz on $$[0,\infty)$$ with coefficients $$M_{S}$$ and $$M_{V}$$. By Lemma 3.1 of [28], there exist two Lipschitz coefficients $$M_{I_{1}}$$, $$M_{I_{2}}$$, $$M_{J_{1}}$$, $$M_{J_{2}}$$ for $$\int_{0}^{\infty}\beta_{1}(a)i(\cdot,a)\, \mathrm{d}a$$, $$\int_{0}^{\infty}\beta_{2}(a)i(\cdot,a)\,\mathrm{d}a$$, $$\int_{0}^{\infty}\beta_{3}(a)j(\cdot,a)\,\mathrm{d}a$$, $$\int _{0}^{\infty}\beta_{4}(a)j(\cdot,a)\,\mathrm{d}a$$, respectively. Thus, $$\int_{0}^{\infty}\beta_{1}(a)i(\cdot,a)S(\cdot)\,\mathrm{d}a$$, $$\int _{0}^{\infty}\beta_{2}(a)i(\cdot, a)$$ $$V(\cdot)\,\mathrm{d}a$$, $$\int_{0}^{\infty}\beta_{3}(a)j(\cdot,a)S(\cdot )\,\mathrm{d}a$$, $$\int_{0}^{\infty}\beta_{4}(a)j(\cdot,a)V(\cdot)\, \mathrm{d}a$$ are Lipschitz continuous on $$[0,\infty)$$ with coefficients $$M_{SI}= KM_{I_{1}}+KM_{S}$$, $$M_{SJ}=KM_{J_{1}}+KM_{S}$$, $$M_{VI}= KM_{I_{2}}+KM_{V}$$, $$M_{VJ}=KM_{J_{2}}+ KM_{V}$$, respectively. Denote $$M=M_{SI}+M_{SJ}+M_{VI}+M_{VJ}$$. Then

$$\Delta\leq Mh \int_{0}^{t-h}e^{-\mu_{0}a}\,\mathrm{d}a\leq \frac{Mh}{\mu _{0}}.$$
(3.5)

Finally, by (3.4) and (3.5), we have

$$\int_{0}^{\infty} \bigl\vert \tilde{e}(t,a+h)- \tilde{e}(t,a) \bigr\vert \,\mathrm{d}a\leq \biggl(2\bar{\beta}M^{2}+ \frac{M}{\mu_{0}} \biggr)h,$$
(3.6)

where $$\bar{\beta}=\bar{\beta}_{1}+\bar{\beta}_{2}+\bar{\beta}_{3}+\bar {\beta}_{4}$$. The right hand of (3.6) converges uniformly to 0 as $$h\rightarrow0$$ and condition (3) is proved for $$\tilde{e}(t,a)$$. Noting that (3.4) holds for any $$X_{0}\in B$$, thus, $$\tilde{e}(t,a)$$ remains in a precompact subset $$B_{\tilde{e}}$$ of $$L_{+}^{1}(0,\infty)$$. Similarly, $$\tilde{i}(t,a)$$ and $$\tilde{j}(t,a)$$ remain in a precompact subset $$B_{\tilde{i}}$$, $$B_{\tilde{j}}$$ of $$L_{+}^{1}(0,\infty)$$, respectively. Thus, the proof is completed. □

From Propositions 3.1 and 3.2, we apply Lemma 3.2 and conclude that the following theorem holds.

### Theorem 3.1

The semi-flow $${\Phi(t)}_{t\geq0}$$ generated by system (1.5) is asymptotically smooth.

## 4 The existence of steady states

System (1.5) always has the steady state $$E_{0}=(S_{0},V_{0},0,0,0)$$, where

$$S_{0}=\frac{\Lambda(\mu+\rho+\eta)}{(\mu+p)(\mu+\rho+\eta)-p\eta},\qquad V_{0}=\frac{p\Lambda}{(\mu+p)(\mu+\rho+\eta)-p\eta}.$$

Define the basic reproduction number as follows

\begin{aligned} R_{0} :=&S_{0} \biggl[ \int_{0}^{\infty}\gamma_{1}(a)B_{1}(a) \,\mathrm {d}a \int_{0}^{\infty}\beta_{1}(a)B_{2}(a) \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{3}(a)B_{3}(a) \,\mathrm{d}a \biggl( \int _{0}^{\infty}\gamma_{2}(a)B_{1}(a) \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \gamma_{1}(a)B_{1}(a) \,\mathrm{d}a \int_{0}^{\infty}\xi(a)B_{2}(a)\, \mathrm{d}a \biggr) \biggr] \\ &{}+V_{0} \biggl[ \int_{0}^{\infty}\gamma _{1}(a)B_{1}(a) \,\mathrm{d}a \int_{0}^{\infty}\beta_{2}(a)B_{2}(a) \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{4}(a)B_{3}(a) \,\mathrm{d}a \biggl( \int _{0}^{\infty}\gamma_{2}(a)B_{1}(a) \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \gamma_{1}(a)B_{1}(a) \,\mathrm{d}a \int_{0}^{\infty}\xi(a)B_{2}(a)\, \mathrm{d}a \biggr) \biggr]. \end{aligned}

Now we consider the existence of positive steady state of system (1.5). The steady state $$(S^{*},V^{*},e^{*}(\cdot), i^{*}(\cdot),j^{*}(\cdot) )$$ of system (1.5) satisfies the following equalities:

\begin{aligned}& \Lambda-(\mu+p)S^{*}+\eta V^{*}-S^{*} \int_{0}^{\infty}\biggl(\frac{\beta _{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)j^{*}(a) \biggr)\,\mathrm {d}a=0, \\& pS^{*}-(\mu+\rho+\eta)V^{*}-V^{*} \int_{0}^{\infty}\biggl( \frac{\beta _{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)j^{*}(a) \biggr)\,\mathrm {d}a=0, \\& \frac{de^{*}(a)}{\mathrm{d}a}=-\theta_{1}(a)e^{*}(a), \\& \frac{di^{*}(a)}{\mathrm{d}a}=-\theta_{2}(a)i^{*}(a), \\& \frac{dj^{*}(a)}{\mathrm{d}a}=-\theta_{3}(a)j^{*}(a), \\& e^{*}(0)=S^{*} \int_{0}^{\infty}\biggl(\frac{\beta_{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)j^{*}(a) \biggr)\,\mathrm{d}a \\& \hphantom{e^{*}(0)={}}{}+V^{*} \int_{0}^{\infty}\biggl( \frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+\beta _{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a, \\& i^{*}(0)= \int_{0}^{\infty}\gamma_{1}(a)e^{*}(a) \,\mathrm{d}a, \\& j^{*}(0)= \int_{0}^{\infty}\gamma_{2}(a)e^{*}(a) \,\mathrm{d}a+ \int _{0}^{\infty}\xi(a)i^{*}(a)\, \mathrm{d}a. \end{aligned}
(4.1)

Solving the third, fourth and fifth equations of (4.1) yields

$${e^{*}}(a) = {e^{*}}(0){B_{1}}(a), \qquad {i^{*}}(a) = {i^{*}}(0){B_{2}}(a), \qquad {j^{*}}(a) = {j^{*}}(0){B_{3}}(a).$$

From the first and second equations of (4.1), it is easy to get

\begin{aligned}& \Lambda-\mu S^{*}-(\mu+\rho)V^{*}-S^{*} \bigl( \Lambda-\mu S^{*}-(\mu +\rho)V^{*} \bigr)f_{1} \bigl(S^{*},V^{*} \bigr) \\& \quad {}-V^{*} \bigl(\Lambda-\mu S^{*}-(\mu+ \rho)V^{*} \bigr)f_{2} \bigl(S^{*},V^{*} \bigr)=0, \end{aligned}

where

\begin{aligned}& f_{1} \bigl(S^{*},V^{*} \bigr)= \int_{0}^{\infty} \biggl(\frac{K_{1}K_{3}(a)}{1+\alpha K_{1} (\Lambda-\mu S^{*}-(\mu+\rho)V^{*} )B_{2}(a)}+(K_{2}+K_{1}K_{7})K_{4}(a) \biggr)\,\mathrm{d}a, \\& f_{2} \bigl(S^{*},V^{*} \bigr)= \int_{0}^{\infty} \biggl(\frac{K_{1}K_{5}(a)}{1+\alpha K_{1} (\Lambda-\mu S^{*}-(\mu+\rho)V^{*} )B_{2}(a)}+(K_{2}+K_{1}K_{7})K_{6}(a) \biggr)\,\mathrm{d}a, \end{aligned}

and

\begin{aligned}& K_{1}= \int_{0}^{\infty}\gamma_{1}(a)B_{1}(a) \,\mathrm{d}a,\qquad K_{2}= \int _{0}^{\infty}\gamma_{2}(a)B_{1}(a) \,\mathrm{d}a, \\& K_{3}(a)=\beta_{1}(a)B_{2}(a),\qquad K_{4}(a)=\beta_{3}(a)B_{3}(a),\qquad K_{5}(a)=\beta_{2}(a)B_{2}(a), \\& K_{6}(a)=\beta_{4}(a)B_{3}(a),\qquad K_{7}= \int_{0}^{\infty}\xi(a)B_{2}(a)\, \mathrm{d}a. \end{aligned}

Let

\begin{aligned} F \bigl(S^{*},V^{*} \bigr) =&\Lambda-\mu S^{*}-( \mu+\rho)V^{*}-S^{*} \bigl(\Lambda-\mu S^{*}-( \mu+\rho)V^{*} \bigr)f_{1} \bigl(S^{*},V^{*} \bigr) \\ &{}-V^{*} \bigl(\Lambda-\mu S^{*}-(\mu+ \rho)V^{*} \bigr)f_{2} \bigl(S^{*},V^{*} \bigr). \end{aligned}

Let $$V^{*}=V_{0}$$. It is easy to see that $$F(0,V_{0})=\Lambda-(\mu+\rho )V_{0}-V_{0} (\Lambda-(\mu+\rho)V_{0} )f_{2}(0,V_{0})$$ and $$F(S_{0},V_{0})=0$$. When $$0\leq S^{*}\leq S_{0}$$, $$\Lambda-\mu S^{*}-(\mu+\rho)V_{0}\neq0$$, then we get

$$S^{*}f_{1} \bigl(S^{*},V_{0} \bigr)+V_{0}f_{2} \bigl(S^{*},V_{0} \bigr)-1=0.$$

Let $$g(S^{*})=S^{*}f_{1}(S^{*},V_{0})+V_{0}f_{2}(S^{*},V_{0})-1$$, where $$g(0) = {V_{0}}{f_{2}}(0,{V_{0}}) - 1 < {V_{0}}{f_{2}}({S^{*}},{V_{0}}) - 1 < 0$$ and $$g(S_{0})=S_{0}f_{1}(S_{0},V_{0})+V_{0}f_{2}(S_{0},V_{0})-1$$. Note that

$$g' \bigl(S^{*} \bigr)=f_{1} \bigl(S^{*},V_{0} \bigr)+S^{*} \frac{\mathrm {d}f_{1}(S^{*},V_{0})}{\mathrm{d}S^{*}}+V_{0}\frac{\mathrm {d}f_{2}(S^{*},V_{0})}{\mathrm{d}S^{*}}>0,$$

for

\begin{aligned}& \frac{\mathrm{d}f_{1}(S^{*},V_{0})}{\mathrm{d}S^{*}}= \int_{0}^{\infty }\frac{\mu\alpha K_{1}^{2}K_{3}(a)B_{2}(a)}{(1+\alpha K_{1} (\Lambda -\mu S^{*}-(\mu+\rho)V_{0} )B_{2}(a))^{2}}\,\mathrm{d}a>0, \\& \frac{\mathrm{d}f_{2}(S^{*},V_{0})}{\mathrm{d}S^{*}}= \int_{0}^{\infty }\frac{\mu\alpha K_{1}^{2}K_{5}(a)B_{2}(a)}{(1+\alpha K_{1} (\Lambda -\mu S^{*}-(\mu+\rho)V_{0} )B_{2}(a))^{2}}\,\mathrm{d}a>0. \end{aligned}

It is easy to show that if $$g(S_{0})>0$$, $$g(S^{*})=0$$ has a unique positive root. Define the basic reproduction number as

$$R_{0}=g(S_{0})+1=S_{0}f_{1}(S_{0},V_{0})+V_{0}f_{2}(S_{0},V_{0}),$$

which means the number of newly infected individuals produced by one infected individual during its period of disease. Therefore, if $$R_{0}>1$$, there exists a unique positive steady state $$E^{*}$$ of system (1.5), where $$E^{*}=(S^{*},V^{*},e^{*}(a),i^{*}(a),j^{*}(a))$$.

From the above discussions, we have the following theorem.

### Theorem 4.1

System (1.5) always has a steady state $$E_{0}(S_{0},V_{0},0,0,0)$$, where $$S_{0}=\Lambda(\mu+\rho+\eta)/ [(\mu+p)(\mu+\rho+\eta)-p\eta ]$$, $$V_{0}=p\Lambda/ [(\mu+p)(\mu+\rho+\eta)-p\eta ]$$; system (1.5) has a unique positive steady state $$E^{*}(S^{*},V^{*},e^{*}(\cdot ),i^{*}(\cdot),j^{*}(\cdot))$$ if and only if $$R_{0}>1$$.

## 5 Local stability

This section is mainly used to prove the local stability of steady states, and to verify that the basic reproduction number is related to the stability of the steady states.

### Theorem 5.1

The steady state $$E_{0}$$ is locally asymptotically stable if $$R_{0}<1$$.

### Proof

First, we introduce the change of variables as follows:

\begin{aligned}& s_{1}(t)=S(t)-S_{0},\qquad v_{1}(t)=V(t)-V_{0}, \qquad e_{1}(t,a)=e(t,a), \\& i_{1}(t,a)=i(t,a),\qquad j_{1}(t,a)=j(t,a). \end{aligned}

Linearizing system (1.5) at the steady state $$E_{0}$$ yields the following system:

\begin{aligned}& \dot{s}_{1}(t)=-(\mu+p)s_{1}(t)+\eta v_{1}(t)-S_{0} \int_{0}^{\infty } \bigl(\beta_{1}(a)i_{1}(t,a)+ \beta_{3}(a)j_{1}(t,a) \bigr)\,\mathrm {d}a, \\& \dot{v}_{1}(t)=ps_{1}(t)-(\mu+\rho+\eta)v_{1}(t)-V_{0} \int_{0}^{\infty } \bigl(\beta_{2}(a)i_{1}(t,a)+ \beta_{4}(a)j_{1}(t,a) \bigr)\,\mathrm {d}a, \\& \frac{\partial e_{1}(t,a)}{\partial a}+\frac{\partial e_{1}(t,a)}{\partial t}=-\theta_{1}(a)e_{1}(t,a), \\& \frac{\partial i_{1}(t,a)}{\partial a}+\frac{\partial i_{1}(t,a)}{\partial t}=-\theta_{2}(a)i_{1}(t,a), \\& \frac{\partial j_{1}(t,a)}{\partial a}+\frac{\partial j_{1}(t,a)}{\partial t}=-\theta_{3}(a)j_{1}(t,a), \\& e_{1}(t,0)=S_{0} \int_{0}^{\infty} \bigl(\beta_{1}(a)i_{1}(t,a)+ \beta _{3}(a)j_{1}(t,a) \bigr)\,\mathrm{d}a \\& \hphantom{e_{1}(t,0)={}}{}+V_{0} \int_{0}^{\infty} \bigl(\beta_{2}(a)i_{1}(t,a)+ \beta _{4}(a)j_{1}(t,a) \bigr)\,\mathrm{d}a, \\& i_{1}(t,0)= \int_{0}^{\infty}\gamma_{1}(a)e_{1}(t,a) \,\mathrm{d}a, \\& j_{1}(t,0)= \int_{0}^{\infty}\gamma_{2}(a)e_{1}(t,a) \,\mathrm{d}a+ \int _{0}^{\infty}\xi(a)i_{1}(t,a)\, \mathrm{d}a. \end{aligned}

Set

\begin{aligned}& s_{1}(t)=s_{1}^{0}e^{\lambda t},\qquad v_{1}(t)=v_{1}^{0}e^{\lambda t},\qquad e_{1}(t,a)=e_{1}^{0}(a)e^{\lambda t}, \\& i_{1}(t,a)=i_{1}^{0}(a)e^{\lambda t},\qquad j_{1}(t,a)=j_{1}^{0}(a)e^{\lambda t}, \end{aligned}

where $$s_{1}^{0}$$, $$v_{1}^{0}$$, $$e_{1}^{0}(a)$$, $$i_{1}^{0}(a)$$, $$j_{1}^{0}(a)$$ will be determined later. We can get

\begin{aligned}& \lambda s_{1}^{0}=-(\mu+p)s_{1}^{0}+ \eta v_{1}^{0}-S_{0} \int _{0}^{\infty} \bigl(\beta_{1}(a)i_{1}^{0}(a)+ \beta _{3}(a)j_{1}^{0}(a) \bigr)\,\mathrm{d}a, \\& \lambda v_{1}^{0}=ps_{1}^{0}-(\mu+ \rho+\eta)v_{1}^{0}-V_{0} \int _{0}^{\infty} \bigl(\beta_{2}(a)i_{1}^{0}(a)+ \beta _{4}(a)j_{1}^{0}(a) \bigr)\,\mathrm{d}a, \\& \frac{\mathrm{d}e_{1}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta _{1}(a) \bigr)e_{1}^{0}(a), \\& \frac{\mathrm{d}i_{1}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta _{2}(a) \bigr)i_{1}^{0}(a), \\& \frac{\mathrm{d}j_{1}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta _{3}(a) \bigr)j_{1}^{0}(a), \\& e_{1}^{0}(0)=S_{0} \int_{0}^{\infty} \bigl(\beta_{1}(a)i_{1}^{0}(a)+ \beta _{3}(a)j_{1}^{0}(a) \bigr)\,\mathrm{d}a \\& \hphantom{e_{1}^{0}(0)={}}{}+V_{0} \int_{0}^{\infty} \bigl(\beta_{2}(a)i_{1}^{0}(a)+ \beta _{4}(a)j_{1}^{0}(a) \bigr)\,\mathrm{d}a \\& i_{1}^{0}(0)= \int_{0}^{\infty}\gamma_{1}(a)e_{1}^{0}(a) \,\mathrm{d}a, \\& j_{1}^{0}(0)= \int_{0}^{\infty}\gamma_{2}(a)e_{1}^{0}(a) \,\mathrm {d}a+ \int_{0}^{\infty}\xi(a)i_{1}^{0}(a) \,\mathrm{d}a. \end{aligned}
(5.1)

Integrating the third, fourth and fifth equations of (5.1) from 0 to a yields

\begin{aligned}& e_{1}^{0}(a)=e_{1}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{1}( \tau) \bigr)\,\mathrm{d}\tau \biggr), \\& i_{1}^{0}(a)=i_{1}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{2}( \tau) \bigr)\,\mathrm{d}\tau \biggr), \\& j_{1}^{0}(a)=j_{1}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{3}( \tau) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}
(5.2)

If $$e_{1}^{0}(0)=0$$, then $$i_{1}^{0}(0)=0$$, $$j_{1}^{0}(0)=0$$. Plugging it into (5.1), we have

$$(\lambda+\mu+p)s_{1}^{0}-\eta v_{1}^{0}=0, \qquad (\lambda+\mu+\rho+\eta )v_{1}^{0}-ps_{1}^{0}=0.$$

For $$s_{1}^{0}\neq0$$ and $$v_{1}^{0}\neq0$$, it is easy to get

$$\lambda^{2}+(2\mu+p+\rho+\eta)\lambda+(\mu+p) (\mu+\rho+\eta)-p \eta=0.$$

Then

$$\lambda=\frac{-b_{1}\pm\sqrt{b_{1}^{2}-4b_{2}}}{2},$$

where $$b_{1}=2\mu+p+\rho+\eta$$, $$b_{2}=(\mu+p)(\mu+\rho+\eta)-p\eta$$.

If $$e_{1}^{0}(0)\neq0$$,

$$(\lambda+\mu)s_{1}^{0}+(\lambda+\mu+\rho)v_{1}^{0}+e_{1}^{0}(0)=0,$$

where

\begin{aligned}& s_{1}^{0}=\frac{S_{0}(\lambda+\mu+\rho+\eta)f_{3}(\lambda)+\eta V_{0}f_{4}(\lambda)}{p\eta-(\lambda+\mu+p)(\lambda+\mu+\rho+\eta)} e_{1}^{0}(0), \\& v_{1}^{0}=\frac{p S_{0}f_{3}(\lambda)+ V_{0}(\lambda+\mu +p)f_{4}(\lambda)}{p\eta-(\lambda+\mu+p)(\lambda+\mu+\rho+\eta)}e_{1}^{0}(0), \end{aligned}

where

\begin{aligned}& f_{3}(\lambda)= \int_{0}^{\infty} \bigl(\beta_{1}(a)B_{2}( \lambda ,a)U_{1}(\lambda)+\beta_{3}(a)B_{3} ( \lambda,a) \bigl(U_{2}(\lambda)+U_{1}(\lambda)U_{7}( \lambda) \bigr) \bigr)\,\mathrm {d}a, \\& f_{4}(\lambda)= \int_{0}^{\infty} \bigl(\beta_{2}(a)B_{2}( \lambda ,a)U_{1}(\lambda)+\beta_{4}(a)B_{3} ( \lambda,a) \bigl(U_{2}(\lambda)+U_{1}(\lambda)U_{7}( \lambda) \bigr) \bigr)\,\mathrm {d}a, \end{aligned}

where

\begin{aligned}& U_{1}(\lambda)= \int_{0}^{\infty}\gamma_{1}(a)B_{1}( \lambda,a)\,\mathrm{d}a,\qquad U_{2}(\lambda)= \int_{0}^{\infty}\gamma_{2}(a)B_{1}( \lambda,a)\,\mathrm {d}a, \\& U_{7}(\lambda)= \int_{0}^{\infty}\xi(a)B_{2}(\lambda,a)\, \mathrm{d}a,\qquad B_{1}(\lambda,a)=\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta_{1}( \tau) \bigr)\, \mathrm{d}\tau \biggr), \\& B_{2}(\lambda,a)=\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta_{2}( \tau) \bigr)\, \mathrm{d}\tau \biggr),\qquad B_{3}(\lambda,a)=\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta_{3}( \tau) \bigr)\, \mathrm{d}\tau \biggr). \end{aligned}

That is,

$$S_{0}f_{3}(\lambda) +V_{0}f_{4}( \lambda)=1.$$
(5.3)

Assume that $$\operatorname{Re}{\lambda}\geq0$$, then $$|f_{3}(\lambda)|\leq f_{1}(S_{0},V_{0})$$ and $$|f_{4}(\lambda)|\leq f_{2}(S_{0},V_{0})$$ hold. Hence, the modulus of the left-hand side of Eq. (5.3) satisfies

\begin{aligned}& \bigl\vert {S_{0}f_{3}(\lambda)+V_{0}f_{4}( \lambda)} \bigr\vert \\& \quad \leq \bigl\vert {S_{0}f_{3}(\lambda)} \bigr\vert + \bigl\vert {V_{0}f_{4}(\lambda)} \bigr\vert < S_{0}f_{1}(S_{0},V_{0})+V_{0}f_{2}(S_{0},V_{0})=R_{0}. \end{aligned}

It follows from (5.3) that there is a contradiction. Thus, all the roots of Eq. (5.3) have a negative real part if and only if $$R_{0}<1$$. Therefore, the steady state $$E_{0}$$ is locally asymptotically stable if $$R_{0}<1$$. This completes the proof. □

### Theorem 5.2

The steady state $$E^{*}$$ is locally asymptotically stable if $$R_{0}>1$$.

### Proof

Linearizing system (1.5) at the steady state $$E^{*}$$ under introducing the perturbation variables

\begin{aligned}& s_{2}(t)=S(t)-S^{*},\qquad v_{2}(t)=V(t)-V^{*}, \qquad e_{2}(t,a)=e(t,a)-e^{*}(a), \\& i_{2}(t,a)=i(t,a)-i^{*}(a),\qquad j_{2}(t,a)=j(t,a)-j^{*}(a), \end{aligned}

we obtain the following system:

\begin{aligned}& \dot{s}_{2}(t)=-(\mu+p)s_{2}(t)+\eta v_{2}(t)-s_{2}(t) \int_{0}^{\infty }\frac{\beta_{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\, \mathrm{d}a-S^{*} \int _{0}^{\infty}\frac{\beta_{1}(a)i_{2}(t,a)}{(1+\alpha i^{*}(a))^{2}}\, \mathrm{d}a \\& \hphantom{\dot{s}_{2}(t)={}}{}-s_{2}(t) \int_{0}^{\infty}\beta_{3}(a)j^{*}(a) \,\mathrm{d}a -S^{*} \int_{0}^{\infty}\beta_{3}(a)j_{2}(t,a) \,\mathrm{d}a, \\& \dot{v}_{2}(t)=ps_{2}(t)-(\mu+\rho+\eta)v_{2}(t)-v_{2}(t) \int _{0}^{\infty}\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\,\mathrm {d}a-V^{*} \int_{0}^{\infty}\frac{\beta_{2}(a) i_{2}(t,a)}{(1+\alpha i^{*}(a))^{2}}\,\mathrm{d}a \\& \hphantom{\dot{v}_{2}(t)={}}{}-v_{2}(t) \int_{0}^{\infty}\beta_{4}(a)j^{*}(a) \,\mathrm{d}a-V^{*} \int _{0}^{\infty}\beta_{4}(a)j_{2}(t,a) \,\mathrm{d}a, \\& \frac{\partial e_{2}(t,a)}{\partial a}+\frac{\partial e_{2}(t,a)}{\partial t}=-\theta_{1}(a)e_{2}(t,a), \\& \frac{\partial i_{2}(t,a)}{\partial a}+\frac{\partial i_{2}(t,a)}{\partial t}=-\theta_{2}(a)i_{2}(t,a), \\& \frac{\partial j_{2}(t,a)}{\partial a}+\frac{\partial j_{1}(t,a)}{\partial t}=-\theta_{3}(a)j_{2}(t,a), \\& e_{2}(t,0)=s_{2}(t) \int_{0}^{\infty}\frac{\beta_{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\, \mathrm{d}a+S^{*} \int_{0}^{\infty}\frac{\beta _{1}(a)i_{2}(t,a)}{(1+\alpha i^{*}(a))^{2}}\, \mathrm{d}a+s_{2}(t) \int _{0}^{\infty}\beta_{3}(a)j^{*}(a) \,\mathrm{d}a \\& \hphantom{e_{2}(t,0)={}}{}+S^{*} \int_{0}^{\infty}\beta_{3}(a)j_{2}(t,a) \,\mathrm{d}a+v_{2}(t) \int _{0}^{\infty}\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\,\mathrm {d}a+V^{*} \int_{0}^{\infty}\frac{\beta_{2}(a) i_{2}(t,a)}{(1+\alpha i^{*}(a))^{2}}\,\mathrm{d}a \\& \hphantom{e_{2}(t,0)={}}{}+v_{2}(t) \int_{0}^{\infty}\beta_{4}(a)j^{*}(a) \,\mathrm{d}a+V^{*} \int _{0}^{\infty}\beta_{4}(a)j_{2}(t,a) \,\mathrm{d}a, \\& i_{2}(t,0)= \int_{0}^{\infty}\gamma_{1}(a)e_{2}(t,a) \,\mathrm{d}a, \\& j_{2}(t,0)= \int_{0}^{\infty}\gamma_{2}(a)e_{2}(t,a) \,\mathrm{d}a+ \int _{0}^{\infty}\xi(a)i_{2}(t,a)\, \mathrm{d}a. \end{aligned}

Set

\begin{aligned}& s_{2}(t)=s_{2}^{0}e^{\lambda t},\qquad v_{2}(t)=v_{2}^{0}e^{\lambda t},\qquad e_{2}(t,a)=e_{2}^{0}(a)e^{\lambda t}, \\& i_{2}(t,a)=i_{2}^{0}(a)e^{\lambda t},\qquad j_{2}(t,a)=j_{2}^{0}(a)e^{\lambda t}, \end{aligned}

where $$s_{2}^{0}$$, $$v_{2}^{0}$$, $$e_{2}^{0}(a)$$, $$i_{2}^{0}(a)$$, $$j_{2}^{0}(a)$$ will be determined later. We get

\begin{aligned}& \lambda s_{2}^{0}=-(\mu+p)s_{2}^{0}+ \eta v_{2}^{0}-S^{*} \int _{0}^{\infty}\frac{\beta_{1}(a)i_{2}^{0}(a)}{(1+\alpha i^{*}(a))^{2}}\, \mathrm{d}a-s_{2}^{0} \int_{0}^{\infty}\frac{\beta _{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \\& \hphantom{\lambda s_{2}^{0}={}}{}-S^{*} \int_{0}^{\infty}\beta_{3}(a)j_{2}^{0}(a) \,\mathrm{d}a -s_{2}^{0} \int_{0}^{\infty}\beta_{3}(a)j^{*}(a) \,\mathrm{d}a, \\& \lambda v_{2}^{0}=ps_{2}^{0}-(\mu+ \rho+\eta)v_{2}^{0}-V^{*} \int _{0}^{\infty}\frac{\beta_{2}(a)i_{2}^{0}(a)}{(1+\alpha i^{*}(a))^{2}}\, \mathrm{d}a -v_{2}^{0} \int_{0}^{\infty}\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \\& \hphantom{\lambda v_{2}^{0}={}}{}-V^{*} \int_{0}^{\infty}\beta_{4}(a)j_{2}^{0}(a) \,\mathrm{d}a -v_{2}^{0} \int_{0}^{\infty}\beta_{4}(a)j^{*}(a) \,\mathrm{d}a, \\& \frac{\mathrm{d}e_{2}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta _{1}(a) \bigr)e_{2}^{0}(a), \\& \frac{\mathrm{d}i_{2}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta _{2}(a) \bigr)i_{2}^{0}(a), \\& \frac{\mathrm{d}j_{2}^{0}(a)}{\mathrm{d}a}=- \bigl(\lambda+\theta_{3}(a) \bigr)j_{2}^{0}(a), \\& e_{2}^{0}(0)=s_{2}^{0} \int_{0}^{\infty}\frac{\beta _{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\, \mathrm{d}a+S^{*} \int_{0}^{\infty }\frac{\beta_{1}(a)i_{2}^{0}(a)}{(1+\alpha i^{*}(a))^{2}}\,\mathrm {d}a \\& \hphantom{e_{2}^{0}(0)={}}{}+s_{2}^{0} \int_{0}^{\infty}\beta_{3}(a)j^{*}(a) \,\mathrm{d}a+S^{*} \int_{0}^{\infty}\beta_{3}(a)j_{2}^{0}(a) \,\mathrm {d}a \\& \hphantom{e_{2}^{0}(0)={}}{}+v_{2}^{0} \int_{0}^{\infty}\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\, \mathrm{d}a+V^{*} \int_{0}^{\infty}\frac{\beta_{2}(a) i_{2}^{0}(a)}{(1+\alpha i^{*}(a))^{2}}\,\mathrm{d}a \\& \hphantom{e_{2}^{0}(0)={}}{}+v_{2}^{0} \int_{0}^{\infty}\beta_{4}(a)j^{*}(a) \,\mathrm{d}a+V^{*} \int _{0}^{\infty}\beta_{4}(a)j_{2}^{0}(a) \,\mathrm{d}a, \\& i_{2}^{0}(0)= \int_{0}^{\infty}\gamma_{1}(a)e_{2}^{0}(a) \,\mathrm{d}a, \\& j_{2}^{0}(0)= \int_{0}^{\infty}\gamma_{2}(a)e_{2}^{0}(a) \,\mathrm {d}a+ \int_{0}^{\infty}\xi(a)i_{2}^{0}(a) \,\mathrm{d}a. \end{aligned}
(5.4)

Integrating the third, fourth and fifth equations of (5.4) from 0 to a yields

\begin{aligned}& e_{2}^{0}(a)=e_{2}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{1}( \tau) \bigr)\,\mathrm{d}\tau \biggr), \\& i_{2}^{0}(a)=i_{2}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{2}( \tau) \bigr)\,\mathrm{d}\tau \biggr), \\& j_{2}^{0}(a)=j_{2}^{0}(0)\exp \biggl(- \int_{0}^{a} \bigl(\lambda+\theta _{3}( \tau) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}
(5.5)

The characteristic equation is

$$(\lambda+\mu)s_{2}^{0}+(\lambda+\mu+\rho)v_{2}^{0}+e_{2}^{0}(0)=0,$$

where

\begin{aligned}& s_{2}^{0}=\frac{S^{*}(\lambda+\mu+\rho+\eta+f_{6}(e^{*}))f_{3}(e^{*},\lambda )+\eta V^{*}f_{4}(e^{*},\lambda)}{p\eta-(\lambda+\mu+p+f_{5}(e^{*}))(\lambda +\mu+\rho+\eta+f_{6}(e^{*}))} e_{2}^{0}, \\& v_{2}^{0}=\frac{p S^{*}f_{3}(e^{*},\lambda)+ V^{*}(\lambda+\mu +p+f_{5}(e^{*}))f_{4}(e^{*},\lambda)}{p\eta-(\lambda+\mu +p+f_{5}(e^{*}))(\lambda+\mu+\rho+\eta+f_{6}(e^{*}))}e_{2}^{0}, \end{aligned}

where

\begin{aligned}& f_{3} \bigl(e^{*},\lambda \bigr)= \int_{0}^{\infty} \biggl(\frac{\beta _{1}(a)B_{2}(\lambda,a)U_{1}(\lambda)}{(1+\alpha K_{1}e^{*}(0)B_{2}(a))^{2}}+ \beta_{3}(a)B_{3}(\lambda,a) \bigl(U_{2}( \lambda)+U_{1}(\lambda)U_{7}(\lambda) \bigr) \biggr)\, \mathrm {d}a, \\& f_{4} \bigl(e^{*},\lambda \bigr)= \int_{0}^{\infty} \biggl(\frac{\beta _{2}(a)B_{2}(\lambda,a)U_{1}(\lambda)}{(1+\alpha K_{1}e^{*}(0)B_{2}(a))^{2}}+ \beta_{4}(a)B_{3}(\lambda,a) \bigl(U_{2}( \lambda)+U_{1}(\lambda)U_{7}(\lambda) \bigr) \biggr)\, \mathrm {d}a, \\& f_{5} \bigl(e^{*} \bigr)= \int_{0}^{\infty} \biggl(\frac{K_{1}\beta _{1}(a)B_{2}(a)}{1+\alpha K_{1}e^{*}(0)B_{2}(a)}+\beta _{3}(a)B_{3}(a) (K_{2}+K_{1}K_{7}) \biggr)\,\mathrm{d}a, \\& f_{6} \bigl(e^{*} \bigr)= \int_{0}^{\infty} \biggl(\frac{K_{1}\beta _{2}(a)B_{2}(a)}{1+\alpha K_{1}e^{*}(0)B_{2}(a)}+\beta _{4}(a)B_{3}(a) (K_{2}+K_{1}K_{7}) \biggr)\,\mathrm{d}a. \end{aligned}

That is,

\begin{aligned}& \frac{(\lambda+\mu)(\lambda+\mu+\rho+\eta+f_{6}(e^{*}))+p(\lambda+\mu +\rho)}{(\lambda+\mu+f_{5}(e^{*}))(\lambda+\mu+\rho+\eta+f_{6}(e^{*})) +p(\lambda+\mu +\rho+f_{6}(e^{*}))} S^{*}f_{3} \bigl(e^{*},\lambda \bigr) \\& \qquad {}+\frac{(\lambda+\mu+p+f_{5}(e^{*}))(\lambda+\mu+\rho)+\eta(\lambda+\mu )}{(\lambda+\mu+p+f_{5}(e^{*}))(\lambda+\mu+\rho+f_{6}(e^{*})) +\eta(\lambda+\mu +f_{5}(e^{*})) }V^{*}f_{4} \bigl(e^{*},\lambda \bigr) \\& \quad =1. \end{aligned}
(5.6)

Assume that $$\operatorname{Re}{\lambda}\geq0$$, then $$|f_{3}(e^{*},\lambda)|\leq f_{5}(e^{*})$$ and $$|f_{4}(e^{*},\lambda)|\leq f_{6}(e^{*})$$ hold. Hence, the modulus of the left-hand side of Eq. (5.6) satisfies

\begin{aligned}& \biggl\vert \frac{(\lambda+\mu)(\lambda+\mu+\rho+\eta+f_{6}(e^{*}))+p(\lambda +\mu+\rho)}{(\lambda+\mu+f_{5}(e^{*}))(\lambda+\mu+\rho+\eta +f_{6}(e^{*}))+p(\lambda+\mu +\rho+f_{6}(e^{*}))} S^{*}f_{3} \bigl(e^{*}, \lambda \bigr) \\& \qquad {} +\frac{(\lambda+\mu+p+f_{5}(e^{*}))(\lambda+\mu+\rho)+\eta(\lambda +\mu)}{(\lambda+\mu+p+f_{5}(e^{*}))(\lambda+\mu+\rho +f_{6}(e^{*}))+\eta(\lambda+\mu +f_{5}(e^{*})) }V^{*}f_{4} \bigl(e^{*},\lambda \bigr) \biggr\vert \\& \quad < \bigl\vert {S^{*}f_{3} \bigl(e^{*},\lambda \bigr) +V^{*}f_{4} \bigl(e^{*},\lambda \bigr)} \bigr\vert \leq \bigl\vert {S^{*}f_{3} \bigl(e^{*},\lambda \bigr)} \bigr\vert + \bigl\vert {V^{*}f_{4} \bigl(e^{*},\lambda \bigr)} \bigr\vert \\& \quad \leq S^{*}f_{5} \bigl(e^{*} \bigr)+V^{*}f_{6} \bigl(e^{*} \bigr)=1. \end{aligned}

It follows from (5.6) that there is a contradiction. Therefore, $$\operatorname{Re}\lambda<0$$. This means that all the roots of (5.6) have negative real parts. Consequently, if $$R_{0}>1$$, the steady state $$E^{*}$$ is locally asymptotically stable. This completes the proof. □

## 6 Uniform persistence

In this section, we investigate the uniform persistence of system (1.5) by using the persistence theory for infinite dimensional dynamics system. Define

\begin{aligned}& \bar{a} = \inf \biggl\{ a: \int_{a}^{\infty}{{\theta_{1}}(u)\,\mathrm{d}u = 0} \biggr\} , \qquad \bar{b} = \inf \biggl\{ b: \int_{b}^{\infty}{{\theta_{2}}(u)\, \mathrm{d}u = 0} \biggr\} , \\& \bar{c} = \inf \biggl\{ c: \int_{c}^{\infty}{{\theta_{3}}(u)\, \mathrm{d}u = 0} \biggr\} . \end{aligned}

Since $${\theta_{1}}(a),{\theta_{2}}(a),{\theta_{3}}(a) \in L_{+} ^{1}(0,\infty)$$, we have $$\bar{a},\bar{b}, \bar{c}>0$$. Furthermore, let

\begin{aligned}& \tilde{X} = L_{+} ^{1}(0,\infty) \times L_{+} ^{1}(0,\infty) \times L_{+} ^{1}(0,\infty), \\& \tilde{Y} = \biggl\{ \bigl( {e(t, \cdot),i(t, \cdot),j(t, \cdot )} \bigr)^{T} \in\tilde{X}: \int_{0}^{\bar{a}} {e(t, x)\,\mathrm {d}x > 0} , \int_{0}^{\bar{b}} {i(t, x)\,\mathrm{d}x > 0} , \int _{0}^{\bar{c}} {j(t, x)\,\mathrm{d}x > 0} \biggr\} , \end{aligned}

and

$${ Y} = {{R}^{+} } \times{{R}^{+} } \times{{\tilde{Y}}},\qquad \partial{ Y} = {X} \backslash{ Y},\qquad \partial{{\tilde{Y}}} = {{\tilde{X}}}\backslash {{\tilde{Y}}}.$$

It is not difficult to verify the following proposition.

### Proposition 6.1

The subsets Y and ∂Y are both positively invariant under the semi-flow $${\{ \Phi(t)\} _{t \ge0}}$$, namely, $$\Phi(t,{Y}) \subset{Y}$$ and $$\Phi(t,\partial{Y}) \subset\partial{Y}$$ for $$t\ge 0$$.

Furthermore, the following result is useful for the proof of uniform persistence.

### Theorem 6.1

The disease-free steady state $$E_{0}$$ of system (1.5) is globally asymptotically stable for the semi-flow $${\{ \Phi(t)\} _{t \ge0}}$$ restricted to ∂Y.

### Proof

Letting $$({S_{0}},{V_{0}},{e_{0}}( \cdot),{i_{0}}( \cdot),{j_{0}}( \cdot)) \in \partial{Y}$$, namely, $$({e_{0}}( \cdot),{i_{0}}( \cdot),{j_{0}}( \cdot)) \in\partial\tilde{Y}$$, we consider the following system:

\begin{aligned}& \frac{{\partial e(t,a)}}{{\partial a}} + \frac{{\partial e(t,a)}}{{\partial t}} = - {\theta_{1}}(a)e(t,a), \\& \frac{{\partial i(t,a)}}{{\partial a}} + \frac{{\partial i(t,a)}}{{\partial t}} = - {\theta_{2}}(a)i(t,a), \\& \frac{{\partial j(t,a)}}{{\partial a}} + \frac{{\partial j(t,a)}}{{\partial t}} = - {\theta_{3}}(a)j(t,a), \\& e(t,0) = S(t) \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{1}}(a)i(t,a)}}{{1 + \alpha i(t,a)}} + { \beta_{3}}(a)j(t,a)} \biggr)} \,\mathrm{d}a \\& \hphantom{ e(t,0) ={}}{}+ V(t) \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{2}}(a)i(t,a)}}{{1 + \alpha i(t,a)}} + { \beta_{4}}(a)j(t,a)} \biggr)} \,\mathrm{d}a, \\& i(t,0) = \int_{0}^{\infty}{{\gamma_{1}}} (a)e(t,a){\, \mathrm{d}}a, \\& j(t,0) = \int_{0}^{\infty}{{\gamma_{2}}} (a)e(t,a){\, \mathrm{d}}a + \int _{0}^{\infty}\xi (a)i(t,a){\,\mathrm{d}}a, \\& e(0,a) = {\varphi_{e}}(a),\qquad i(0,a) = {\varphi_{i}}(a),\qquad j(0,a) = {\varphi_{j}}(a). \end{aligned}

Since $$S(t) \le S_{0}$$ and $$V(t) \le V_{0}$$ as t tends to infinity, by comparison, we have $$e(t,a) \le\tilde{e}(t,a),i(t,a) \le\tilde{i}(t,a),j(t,a) \le\tilde{j}(t,a)$$, where $$\tilde{e}(t,a)$$, $$\tilde{i}(t,a)$$ and $$\tilde{j}(t,a)$$ satisfy the following auxiliary system:

\begin{aligned} &\frac{{\partial\tilde{e}(t,a)}}{{\partial a}} + \frac{{\partial\tilde{e}(t,a)}}{{\partial t}} = - {\theta_{1}}(a)\tilde{e}(t,a), \\ &\frac{{\partial\tilde{i}(t,a)}}{{\partial a}} + \frac{{\partial\tilde{i}(t,a)}}{{\partial t}} = - {\theta_{2}}(a)\tilde{i}(t,a), \\ &\frac{{\partial\tilde{j}(t,a)}}{{\partial a}} + \frac{{\partial\tilde{j}(t,a)}}{{\partial t}} = - {\theta_{3}}(a)\tilde{j}(t,a), \\ &\tilde{e}(t,0) = S_{0} \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{1}}(a)\tilde{i}(t,a)}}{{1 + \alpha\tilde{i}(t,a)}} + { \beta_{3}}(a)\tilde{j}(t,a)} \biggr)} \,\mathrm{d}a \\ &\hphantom{\tilde{e}(t,0) ={}}{} + V_{0} \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{2}}(a)\tilde{i}(t,a)}}{{1 + \alpha\tilde{i}(t,a)}} + { \beta_{4}}(a)\tilde{j}(t,a)} \biggr)} \, \mathrm{d}a, \\ &\tilde{i}(t,0) = \int_{0}^{\infty}{{\gamma_{1}}} (a)\tilde{e}(t,a){\,\mathrm {d}}a, \\ &\tilde{j}(t,0) = \int_{0}^{\infty}{{\gamma_{2}}} (a)\tilde{e}(t,a){\,\mathrm {d}}a + \int_{0}^{\infty}\xi (a)\tilde{i}(t,a){\,\mathrm{d}}a, \\ &\tilde{e}(0,a) = {\varphi_{e}}(a), \tilde{i}(0,a) = { \varphi_{i}}(a), \tilde{j}(0,a) = {\varphi_{j}}(a). \end{aligned}
(6.1)

Similar to (2.2)–(2.4), solving the first three equations of (6.1) yields

\begin{aligned}& \tilde{e}(t,a) = \textstyle\begin{cases} {{\tilde{L}}_{1}}(t - a){B_{1}}(a), &0 \le a \le t, \\ {\varphi_{e}}(a - t)\frac{{{B_{1}}(a)}}{{{B_{1}}(a - t)}},& 0 \le t \le a, \end{cases}\displaystyle \end{aligned}
(6.2)
\begin{aligned}& \tilde{i}(t,a) = \textstyle\begin{cases} {{\tilde{L}}_{2}}(t - a){B_{2}}(a),& 0 \le a \le t, \\ {\varphi_{i}}(a - t)\frac{{{B_{2}}(a)}}{{{B_{2}}(a - t)}}, &0 \le t \le a, \end{cases}\displaystyle \end{aligned}
(6.3)
\begin{aligned}& \tilde{j}(t,a) = \textstyle\begin{cases} {{\tilde{L}}_{3}}(t - a){B_{3}}(a) &\mbox{for }0 \le a \le t, \\ {\varphi_{j}}(a - t)\frac{{{B_{3}}(a)}}{{{B_{3}}(a - t)}} &\mbox{for }0 \le t \le a, \end{cases}\displaystyle \end{aligned}
(6.4)

where

\begin{aligned}& {{\tilde{L}}_{1}}(t) = S_{0} \int_{0}^{\infty}{ \biggl( {\frac{{{\beta _{1}}(a)\tilde{i}(t,a)}}{{1 + \alpha\tilde{i}(t,a)}} + { \beta_{3}}(a)\tilde{j}(t,a)} \biggr)} \,\mathrm{d}a \\& \hphantom{{{\tilde{L}}_{1}}(t) ={}}{}+ V_{0} \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{2}}(a)\tilde{i}(t,a)}}{{1 + \alpha\tilde{i}(t,a)}} + { \beta_{4}}(a)\tilde{j}(t,a)} \biggr)} \, \mathrm{d}a, \\& {{\tilde{L}}_{2}}(t) = \int_{0}^{\infty}{{\gamma_{1}}} (a)\tilde{e}(t,a){\, \mathrm{d}}a, \\& {{\tilde{L}}_{3}}(t) = \int_{0}^{\infty}{{\gamma_{2}}} (a)\tilde{e}(t,a){\, \mathrm{d}}a + \int_{0}^{\infty}\xi (a)\tilde{i}(t,a){\,\mathrm{d}}a. \end{aligned}

It follows from (6.2)–(6.4) that

\begin{aligned} &{{\tilde{L}}_{1}}(t) = S_{0} \int_{0}^{t} { \biggl( {\frac{{{\beta_{1}}(a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)}}{{1 + \alpha{{\tilde{L}}_{2}}(t - a){B_{2}}(a)}} + { \beta_{3}}(a){{\tilde{L}}_{3}}(t - a){B_{3}}(a)} \biggr)} \,\mathrm{d}a \\ &\hphantom{{{\tilde{L}}_{1}}(t) ={}}{} + V_{0} \int_{0}^{t} { \biggl( {\frac{{{\beta_{2}}(a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)}}{{1 + \alpha{{\tilde{L}}_{2}}(t - a){B_{2}}(a)}} + {\beta _{4}}(a){{\tilde{L}}_{3}}(t - a){B_{3}}(a)} \biggr)} \,\mathrm{d}a + {G_{1}}(t), \\ &{{\tilde{L}}_{2}}(t) = \int_{0}^{t} {{\gamma_{1}}} (a){{\tilde{L}}_{1}}(t - a){B_{1}}(a)\,\mathrm{d}a + {G_{2}}(t), \\ &{{\tilde{L}}_{3}}(t) = \int_{0}^{t} {{\gamma_{2}}} (a){{\tilde{L}}_{1}}(t - a){B_{1}}(a)\,\mathrm{d}a + \int_{0}^{t} \xi (a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)\,\mathrm{d}a + {G_{3}}(t), \end{aligned}
(6.5)

where

\begin{aligned} &{G_{1}}(t) = S_{0} \int_{t}^{\infty}{ \biggl( {\frac{{{\beta_{1}}(a){\varphi _{i}}(a - t){B_{2}}(a)}}{{{B_{2}}(a - t) + \alpha{\varphi_{i}}(a - t){B_{2}}(a)}} + { \beta_{3}}(a){\varphi_{j}}(a - t)\frac {{{B_{3}}(a)}}{{{B_{3}}(a - t)}}} \biggr)} \,\mathrm{d}a \\ &\hphantom{{G_{1}}(t) ={}}{} + V_{0} \int_{t}^{\infty}{ \biggl( {\frac{{{\beta_{2}}(a){\varphi_{i}}(a - t){B_{2}}(a)}}{{{B_{2}}(a - t) + \alpha{\varphi_{i}}(a - t){B_{2}}(a)}} + { \beta_{4}}(a){\varphi_{j}}(a - t)\frac{{{B_{3}}(a)}}{{{B_{3}}(a - t)}}} \biggr)} \,\mathrm{d}a, \\ &{G_{2}}(t) = \int_{t}^{\infty}{{\gamma_{1}}} (a){ \varphi_{e}}(a - t)\frac {{{B_{1}}(a)}}{{{B_{1}}(a - t)}}\,\mathrm{d}a, \\ &{G_{3}}(t) = \int_{t}^{\infty}{{\gamma_{2}}} (a){ \varphi_{e}}(a - t)\frac {{{B_{1}}(a)}}{{{B_{1}}(a - t)}}\,\mathrm{d}a + \int_{t}^{\infty}\xi (a){\varphi_{i}}(a - t) \frac{{{B_{2}}(a)}}{{{B_{2}}(a - t)}}\,\mathrm{d}a. \end{aligned}

Since $$( {{\varphi_{e}}( \cdot),{\varphi_{i}}( \cdot),{\varphi _{j}}( \cdot)} )\in\partial\tilde{Y}$$, we have $${G_{i}}(t) \equiv 0$$ ($$i = 1,2,3$$) for all $$t \ge0$$. From (6.5), we obtain

\begin{aligned} &{{\tilde{L}}_{1}}(t) = S_{0} \int_{0}^{t} { \biggl( {\frac{{{\beta_{1}}(a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)}}{{1 + \alpha{{\tilde{L}}_{2}}(t - a){B_{2}}(a)}} + { \beta_{3}}(a){{\tilde{L}}_{3}}(t - a){B_{3}}(a)} \biggr)} \,\mathrm{d}a \\ &\hphantom{{{\tilde{L}}_{1}}(t) ={}}{} + V_{0} \int_{0}^{t} { \biggl( {\frac{{{\beta_{2}}(a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)}}{{1 + \alpha{{\tilde{L}}_{2}}(t - a){B_{2}}(a)}} + {\beta _{4}}(a){{\tilde{L}}_{3}}(t - a){B_{3}}(a)} \biggr)} \,\mathrm{d}a, \\ &{{\tilde{L}}_{2}}(t) = \int_{0}^{t} {{\gamma_{1}}} (a){{\tilde{L}}_{1}}(t - a){B_{1}}(a)\,\mathrm{d}a, \\ &{{\tilde{L}}_{3}}(t) = \int_{0}^{t} {{\gamma_{2}}} (a){{\tilde{L}}_{1}}(t - a){B_{1}}(a)\,\mathrm{d}a + \int_{0}^{t} \xi (a){{\tilde{L}}_{2}}(t - a){B_{2}}(a)\,\mathrm{d}a. \end{aligned}
(6.6)

It is easy to show that system (6.6) has a unique solution $${{\tilde{L}}_{i}}(t) \equiv0$$ ($$i = 1,2,3$$). From (6.2)–(6.4), we have $$\tilde{e}(t,a) = 0$$, $$\tilde{i}(t,a) = 0$$, $$\tilde{j}(t,a) = 0$$. For $$a \ge t$$, it follows that

\begin{aligned} &{ \bigl\Vert {\tilde{e}(t,a)} \bigr\Vert _{{L^{1}}}} = { \biggl\Vert {{\varphi_{e}}(a - t)\frac{{{B_{1}}(a)}}{{{B_{1}}(a - t)}}} \biggr\Vert _{{L^{1}}}} \le{e^{ - {\mu_{0}}t}} { \Vert {{\varphi_{e}}} \Vert _{{L^{1}}}}, \\ &{ \bigl\Vert {\tilde{i}(t,a)} \bigr\Vert _{{L^{1}}}} = { \biggl\Vert {{\varphi_{i}}(a - t)\frac{{{B_{2}}(a)}}{{{B_{2}}(a - t)}}} \biggr\Vert _{{L^{1}}}} \le{e^{ - {\mu_{0}}t}} { \Vert {{\varphi_{i}}} \Vert _{{L^{1}}}}, \\ &{ \bigl\Vert {\tilde{j}(t,a)} \bigr\Vert _{{L^{1}}}} = { \biggl\Vert {{\varphi_{j}}(a - t)\frac{{{B_{3}}(a)}}{{{B_{3}}(a - t)}}} \biggr\Vert _{{L^{1}}}} \le{e^{ - {\mu_{0}}t}} { \Vert {{\varphi_{j}}} \Vert _{{L^{1}}}}, \end{aligned}

which implies that $$\tilde{e}(t,a) = 0$$, $$\tilde{i}(t,a) = 0$$, $$\tilde{j}(t,a) = 0$$ as $$t \to\infty$$. Noting that $$e(t,a) \le\tilde{e}(t,a)$$, $$i(t,a) \le\tilde{i}(t,a)$$ and $$j(t,a) \le\tilde{j}(t,a)$$, we have $$e(t,a) \to0$$, $$i(t,a) \to0$$ and $$j(t,a) \to0$$ as $$t \to\infty$$. It follows from the first two equations of system (1.5) that $$S(t) \to {S_{0}}$$ and $$V(t) \to{V_{0}}$$ as $$t \to\infty$$. Thus, $$E_{0}$$ is globally asymptotically stable in ∂Y. □

### Theorem 6.2

If $${R}_{0} > 1$$, then the semi-flow $${\{ \Phi(t)\} _{t \ge0}}$$ is uniformly persistent with respect to $$({ Y},\partial{Y})$$, i.e., there exists an $$\varepsilon> 0$$ which is independent of initial values such that $${\lim_{t \to\infty}}{ \Vert {\Phi(t,x)} \Vert _{X}} \ge \varepsilon$$ for $$x \in{Y}$$. Furthermore, there is a compact subset $${{A}_{0}} \subset{Y}$$ which is a global attractor for $$\{\Phi(t,x)\} _{t\ge0}$$ in Y.

### Proof

It follows from Theorem 6.1 that $$E_{0}$$ is globally asymptotically stable in ∂Y. Applying Theorem 4.2 in [25], we need only to show that $${W^{s}}({E_{0}}) \cap{Y} = \emptyset$$, where

$${W^{s}}({E_{0}}) = \bigl\{ x \in{Y}: \lim _{t \to\infty} \Phi(t,x) = {E_{0}} \bigr\} .$$

Otherwise, there exists a solution $$y \subset{Y}$$ such that $$\Phi (t,y) \to{E_{0}}$$ as $$t \to\infty$$. In this case, there exists a sequence $$\{y_{n}\} \subset{Y}$$ such that $${ \Vert {\Phi(t,y_{n}) - {E_{0}}} \Vert _{X}} < 1/n$$ for $$t \ge0$$. Denote $$\Phi(t,{y_{n}}) = ( {{S_{n}}(t),{V_{n}}(t),e(t, \cdot),i(t, \cdot),j(t, \cdot)} )$$ and $${y_{n}} = ( {{S_{n}}(0),{V_{n}}(0),e(t, \cdot),i(t, \cdot),j(t, \cdot)} )$$. Since $${R}_{0} > 1$$, we can choose n sufficiently large satisfying $${S_{0}}> 1/n$$ and $${V_{0}}>1/n$$,

\begin{aligned}& \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{3}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{4}}(a)} \bigr]} {\,\mathrm{d}}a \\& \quad {}+ \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{5}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{6}}(a)} \bigr]} {\,\mathrm{d}}a > 1. \end{aligned}
(6.7)

For such a $$n > 0$$, there exists a $$T > 0$$ such that, for $$t > T$$, $${S_{0}} - 1/n < {S_{n}}(t) < {S_{0}} + 1/n$$ and $${V_{0}} - 1/n < {V_{n}}(t) < {V_{0}} + 1/n$$. Consider the following auxiliary system:

\begin{aligned}& \frac{{\partial\hat{e}(t,a)}}{{\partial a}} + \frac{{\partial\hat{e}(t,a)}}{{\partial t}} = - {\theta_{1}}(a)\hat{e}(t,a), \\& \frac{{\partial\hat{i}(t,a)}}{{\partial a}} + \frac{{\partial\hat{i}(t,a)}}{{\partial t}} = - {\theta_{2}}(a)\hat{i}(t,a), \\& \frac{{\partial\hat{j}(t,a)}}{{\partial a}} + \frac{{\partial\hat{j}(t,a)}}{{\partial t}} = - {\theta_{3}}(a)\hat{j}(t,a), \\& \hat{e}(t,0) = \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{1}}(a)\hat{i}(t,a)}}{{1 + \alpha\hat{i}(t,a)}} + { \beta _{3}}(a)\hat{j}(t,a)} \biggr)} \,\mathrm{d}a \\& \hphantom{\hat{e}(t,0) ={}}{} + \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \biggl( {\frac {{{\beta_{2}}(a)\hat{i}(t,a)}}{{1 + \alpha\hat{i}(t,a)}} + { \beta _{4}}(a)\hat{j}(t,a)} \biggr)} \,\mathrm{d}a, \\& \hat{i}(t,0) = \int_{0}^{\infty}{{\gamma_{1}}} (a)\hat{e}(t,a){\,\mathrm {d}}a, \\& \hat{j}(t,0) = \int_{0}^{\infty}{{\gamma_{2}}} (a)\hat{e}(t,a){\,\mathrm {d}}a + \int_{0}^{\infty}\xi (a)\hat{i}(t,a){\,\mathrm{d}}a. \end{aligned}
(6.8)

Looking for solutions of system (6.8) of the form $$\hat{e}(t,a) = \hat{e}(a){e^{\lambda t}}$$, $$\hat{i}(t,a) = \hat{i}(a){e^{\lambda t}}$$ and $$\hat{j}(t,a) = \hat{j}(a){e^{\lambda t}}$$, where the functions $$\hat{e}(a)$$, $$\hat{i}(a)$$ and $$\hat{j}(a)$$ will be determined later, we obtain the following linear eigenvalue problem:

\begin{aligned}& \frac{{\partial\hat{e}(a)}}{{\partial a}} = - \bigl( {\lambda + {\theta_{1}}(a)} \bigr)\hat{e}(a), \\& \frac{{\partial\hat{i}(a)}}{{\partial a}} = - \bigl( {\lambda + {\theta_{2}}(a)} \bigr)\hat{i}(a), \\& \frac{{\partial\hat{j}(a)}}{{\partial a}} = - \bigl( {\lambda + {\theta_{3}}(a)} \bigr)\hat{j}(a), \\& \hat{e}(0) = \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \biggl( {\frac{{{\beta_{1}}(a)\hat{i}(a)}}{{1 + \alpha\hat{i}(a){e^{\lambda t}}}} + { \beta_{3}}(a)\hat{j}(a)} \biggr)} \,\mathrm{d}a \\& \hphantom{\hat{e}(0) ={}}{}+ \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \biggl( {\frac {{{\beta_{2}}(a)\hat{i}(a)}}{{1 + \alpha\hat{i}(a){e^{\lambda t}}}} + { \beta_{4}}(a)\hat{j}(a)} \biggr)} \,\mathrm{d}a, \\& \hat{i}(0) = \int_{0}^{\infty}{{\gamma_{1}}} (a)\hat{e}(a){ \,\mathrm{d}}a, \\& \hat{j}(0) = \int_{0}^{\infty}{{\gamma_{2}}} (a)\hat{e}(a){ \,\mathrm{d}}a + \int_{0}^{\infty}\xi (a)\hat{i}(a){\,\mathrm{d}}a. \end{aligned}
(6.9)

Solving the first three equations of system (6.9) yields

\begin{aligned} &\hat{e}(a) = \hat{e}(0)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{1}}(s)} \bigr)\,\mathrm{d}s} } \biggr],\qquad \hat{i}(a) = \hat{i}(0)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{2}}(s)} \bigr)\,\mathrm{d}s} } \biggr], \\ &\hat{j}(a) = \hat{j}(0)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{3}}(s)} \bigr)\,\mathrm{d}s} } \biggr]. \end{aligned}
(6.10)

Substituting (6.10) into the last three equations of (6.9), we obtain the characteristic equation of system (6.8) at the steady state $$E_{0}$$ as follows:

$$f(\lambda)=1,$$
(6.11)

where

\begin{aligned} f(\lambda) =& \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \\ &{}\times\int_{0}^{\infty}{\frac {{{\beta_{1}}(a)\int_{0}^{\infty}{{\gamma_{1}}} (a)\exp [ { - \int_{0}^{a} { ( {\lambda + {\theta_{1}}(s)} )\,\mathrm{d}s} } ]{\, \mathrm{d}}a\exp [ { - \int_{0}^{a} { ( {\lambda + {\theta _{2}}(s)} )\,\mathrm{d}s} } ]}}{{1 + \alpha\hat{i}(0)\exp [ {\lambda t - \int_{0}^{a} { ( {\lambda + {\theta_{2}}(s)} )\,\mathrm{d}s} } ]}}\,\mathrm{d}a} \\ &{} + \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}\biggl\{ {\beta _{3}}(a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{3}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{}\times\int_{0}^{\infty}{{\gamma_{2}}} (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{1}}(s)} \bigr)\,\mathrm {d}s} } \biggr]\,\mathrm{d}a \biggr\} \, \mathrm{d}a \\ &{} + \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{\beta _{3}}(a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{3}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{}\times\int_{0}^{\infty}\xi (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{2}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{} \times \int_{0}^{\infty}{{\gamma_{1}}} (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{1}}(s)} \bigr)\,\mathrm{d}s} } \biggr]\,\mathrm {d}a\, \mathrm{d}a\,\mathrm{d}a \\ &{} + \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \\ &{}\times \int_{0}^{\infty}{\frac{{{\beta _{2}}(a)\exp [ { - \int_{0}^{a} { ( {\lambda + {\theta_{2}}(s)} )\,\mathrm{d}s} } ]\int_{0}^{\infty}{{\gamma_{1}}} (a)\exp [ { - \int_{0}^{a} { ( {\lambda + {\theta_{1}}(s)} )\,\mathrm {d}s} } ]\,\mathrm{d}a}}{{1 + \alpha\hat{i}(0)\exp [ {\lambda t - \int_{0}^{a} { ( {\lambda + {\theta_{2}}(s)} )\, \mathrm{d}s} } ]}}} \,\mathrm{d}a \\ &{} + \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}\biggl\{ {\beta _{4}}(a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{3}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{}\times\int_{0}^{\infty}{{\gamma_{2}}} (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{1}}(s)} \bigr)\,\mathrm {d}s} } \biggr]\,\mathrm{d}a \biggr\} \, \mathrm{d}a \\ &{} + \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{\beta _{4}}(a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{3}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{}\times \int_{0}^{\infty}\xi (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{2}}(s)} \bigr)\,\mathrm{d}s} } \biggr] \\ &{}\times \int_{0}^{\infty}{{\gamma_{1}}} (a)\exp \biggl[ { - \int_{0}^{a} { \bigl( {\lambda + { \theta_{1}}(s)} \bigr)\,\mathrm{d}s} } \biggr]\,\mathrm {d}a\, \mathrm{d}a\,\mathrm{d}a. \end{aligned}

Clearly, we have $${\lim_{\lambda\to\infty}}f(\lambda) = 0$$ and

\begin{aligned} f(0) < & \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{3}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{4}}(a)} \bigr]} {\,\mathrm{d}}a \\ &{}+ \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{5}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{6}}(a)} \bigr]} {\,\mathrm{d}}a. \end{aligned}

From (6.7), there exist a $$n > 0$$ and a $$T > 0$$ such that

\begin{aligned} & \biggl( {{S_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{3}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{4}}(a)} \bigr]} {\,\mathrm{d}}a \\ &\quad {}+ \biggl( {{V_{0}} - \frac{1}{n}} \biggr) \int_{0}^{\infty}{ \bigl[ {{K_{1}} {K_{5}}(a) + ({K_{2}} + {K_{1}} {K_{7}}){K_{6}}(a)} \bigr]} {\,\mathrm{d}}a > 1. \end{aligned}

Hence, if $${R}_{0} > 1$$, Eq. (6.11) has at least one positive root. This implies that the solution $$(\hat{e}(t,\cdot), \hat{i}(t,\cdot),\hat{j}(t,\cdot) )$$ of system (6.8) is unbounded. By comparison, the solution $$\Phi(t,{y_{n}})$$ of system (1.5) is unbounded, which contradicts Proposition 2.1. Therefore, the semi-flow $${\Phi(t)}_{t \ge0}$$ generated by system (1.5) is uniformly persistent. Furthermore, there is a compact subset $${A}_{0} \subset{Y}$$ which is a global attractor for $${\Phi(t)}_{t \ge0}$$ in Y. This completes the proof. □

## 7 Global stability

This section is devoted to the global stability of equilibria. Before going into details, we make some preparations.

First, we introduce an important function which is obtained from the linear combination of Volterra-type functions of the form

$$g(x)=x-1-\ln x.$$

Obviously, $$g(x)\geq0$$ for $$x>0$$ and $$g'(x)=1-1/x$$. Then $$g(x)$$ has a global minimum at $$x=1$$ and $$g(1)=0$$.

### Theorem 7.1

If $$R_{0}<1$$, the disease-free steady state $$E_{0}$$ is globally asymptotically stable.

### Proof

Define a Lyapunov functional as

$$V_{1}=V_{11}+V_{12}+V_{13},$$

where

\begin{aligned}& V_{11}=S_{0}g \bigl(S(t)/S_{0} \bigr)+V_{0}g \bigl(V(t)/V_{0} \bigr),\qquad V_{12}= \int_{0}^{\infty }\omega_{1}(a)e(t,a)\, \mathrm{d}a, \\& V_{13}= \int_{0}^{\infty}\omega_{2}(a)i(t,a)\, \mathrm{d}a, \qquad V_{14}= \int _{0}^{\infty}\omega_{3}(a)j(t,a)\, \mathrm{d}a, \end{aligned}

where

\begin{aligned}& \omega_{1}(a)= \int_{a}^{\infty} \bigl(\omega_{3}(0) \gamma_{2}(x) +\omega_{2}(0)\gamma_{1}(x) \bigr) \exp \biggl(- \int_{a}^{x}\theta_{1}(\tau)d\tau \biggr)dx, \\& \omega_{2}(a)= \int_{a}^{\infty} \bigl(S_{0} \beta_{1}(x)+V_{0}\beta_{2}(x) + \omega_{3}(0)\xi(x) \bigr)\exp \biggl(- \int_{a}^{x}\theta_{2}(\tau)d\tau \biggr)dx, \\& \omega_{3}(a)= \int_{a}^{\infty} \bigl(S_{0} \beta_{3}(x) +V_{0}\beta_{4}(x) \bigr)\exp \biggl(- \int_{a}^{x}\theta_{3}(\tau)d\tau \biggr)dx, \end{aligned}

then

\begin{aligned}& \omega_{2}(0)\gamma_{1}(a)+\omega_{3}(0) \gamma_{2}(a)+\omega '_{1}(a)- \theta_{1}(a)\omega_{1}(a)=0, \\& S_{0}\beta_{1}(a)+V_{0}\beta_{2}+ \omega_{3}(0)\xi(a)+\omega '_{2}(a)- \theta_{2}(a)\omega_{2}(a)=0, \\& S_{0}\beta_{3}(a)+V_{0}\beta_{4}+ \omega'_{3}(a)-\theta_{3}(a)\omega _{3}(a)=0. \end{aligned}

The derivative of $$V_{11}$$ along with the solution of system (1.5) can be calculated as

\begin{aligned} \frac{\mathrm{d}V_{11}}{\mathrm{d}t} =& \biggl(1-\frac{S_{0}}{S(t)} \biggr) \biggl(\Lambda-(\mu+ p) S(t)+\eta V(t)-S(t) \int_{0}^{\infty}\frac{\beta _{1}(a)i(t,a)}{1+\alpha i(t,a)}\,\mathrm{d}a \\ &{}+S(t) \int_{0}^{\infty}\beta_{3}(a)j(t,a)\, \mathrm{d}a \biggr) \\ &{}+ \biggl(1-\frac{V_{0}}{V(t)} \biggr) \biggl(pS(t)-(\mu+\rho+\eta)V(t) -V(t) \int_{0}^{\infty}\frac{\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}\, \mathrm{d}a \\ &{}-V(t) \int_{0}^{\infty}\beta_{4}(a)j(t,a)\, \mathrm{d}a \biggr) \\ =&\mu S_{0} \biggl(2-\frac{S(t)}{S_{0}}-\frac{S_{0}}{S(t)} \biggr)+(\mu +\rho)V_{0} \biggl(3-\frac{V(t)}{V_{0}}- \frac{S_{0}}{S(t)} -\frac{S(t)V_{0}}{S_{0}V(t)} \biggr) \\ &{}+\eta V_{0} \biggl(2-\frac{S(t)V_{0}}{S_{0}V(t)}-\frac {S_{0}V(t)}{S(t)V_{0}} \biggr) \\ &{}- \bigl(S(t)-S_{0} \bigr) \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}- \bigl(V(t)-V_{0} \bigr) \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a)j(t,a) \biggr)\,\mathrm{d}a. \end{aligned}
(7.1)

The derivative of $$V_{12}$$ along with the solution of system (1.5) can be calculated as

\begin{aligned} \frac{dV_{12}}{dt} =&\frac{d}{dt} \int_{0}^{t}\omega_{1}(a)e(t-a,0)\exp \biggl(- \int_{0}^{a}\theta_{1}(\tau)d\tau \biggr)\,\mathrm{d}a \\ &{}+\frac{d}{dt} \int_{t}^{\infty}\omega_{1}(a) \varphi_{e}(a-t)\exp \biggl(- \int_{a-t}^{a}\theta_{1}(\tau)d\tau \biggr)\,\mathrm{d}a. \end{aligned}

Let $$r=t-a$$, then

\begin{aligned} \frac{\mathrm{d}V_{12}}{\mathrm{d}t} =&\frac{\mathrm{d}}{\mathrm{d}t} \int _{0}^{t}\omega_{1}(t-r)e(r,0)\exp \biggl(- \int_{0}^{t-r}\theta_{1}(\tau )d\tau \biggr)\,\mathrm{d}r \\ &{}+\frac{\mathrm{d}}{\mathrm{d}t} \int_{t}^{\infty}\omega_{1}(t+r)\varphi _{e}(r)\exp \biggl(- \int_{r}^{t+r}\theta_{1}(\tau)d\tau \biggr)\,\mathrm{d}r \\ =&\omega_{1}(0)e(t,0)+ \int_{0}^{\infty} \bigl(\omega'(a)-\theta _{1}(a)\omega_{1}(a) \bigr)e(t,a)\,\mathrm{d}a. \end{aligned}
(7.2)

Similarly, we can get

\begin{aligned} &\frac{\mathrm{d}V_{13}}{\mathrm{d}t}=\omega_{2}(0) \int_{0}^{\infty }\gamma_{1}(a)e(t,a)\, \mathrm{d}a+ \int_{0}^{\infty} \bigl(\omega'_{2}(a) -\theta_{2}(a)\omega_{2}(a) \bigr)i(t,a)\,\mathrm{d}a, \\ &\frac{\mathrm{d}V_{14}}{\mathrm{d}t}=\omega_{3}(0) \int_{0}^{\infty } \bigl(\gamma_{2}(a)e(t,a)+ \xi(a)i(t,a) \bigr)\,\mathrm{d}a \\ &\hphantom{\frac{\mathrm{d}V_{14}}{\mathrm{d}t}={}}{}+ \int_{0}^{\infty} \bigl(\omega'_{3}(a) -\theta_{3}(a)\omega_{3}(a) \bigr)j(t,a)\,\mathrm{d}a. \end{aligned}
(7.3)

Combining the (7.1)–(7.3), it is easy to get

\begin{aligned} \frac{\mathrm{d}V_{1}}{\mathrm{d}t} =&\mu S_{0} \biggl(2-\frac {S_{0}}{S(t)}- \frac{S(t)}{S_{0}} \biggr)+(\mu+\rho)V_{0} \biggl(3- \frac {V(t)}{V_{0}}-\frac{S_{0}}{S(t)} -\frac{S(t)V_{0}}{S_{0}V(t)} \biggr) \\ &{}+\eta V_{0} \biggl(2-\frac{S(t)V_{0}}{S_{0}V(t)}-\frac {S_{0}V(t)}{S(t)V_{0}} \biggr) -e(t,0)+\omega_{1}(0)e(t,0) \\ &{}+ \biggl(S_{0}\frac{\beta_{1}(a)}{1+\alpha i(t,a)}+V_{0} \frac{\beta _{2}(a)}{1+\alpha i(t,a)}+\omega_{2}(0)\xi(a)+\omega'_{1}(a) -\theta_{1}(a)\omega_{1}(a) \biggr)i(t,a)\,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S_{0} \beta_{3}(a)+V_{0}\beta_{4}(a)+ \omega'_{3}(a) -\theta_{3}(a) \omega_{3}(a) \bigr)j(t,a)\,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(\omega_{2}(0) \gamma_{1}(a)+\omega_{3}(0)\gamma _{2}(a)+ \omega'(a) -\theta_{1}(a)\omega_{1}(a) \bigr)e(t,a)\,\mathrm{d}a \\ \leq&\mu S_{0} \biggl(2-\frac{S_{0}}{S(t)}-\frac{S(t)}{S_{0}} \biggr)+(\mu +\rho)V_{0} \biggl(3-\frac{V(t)}{V_{0}}- \frac{S_{0}}{S(t)} -\frac{S(t)V_{0}}{S_{0}V(t)} \biggr) \\ &{}+\eta V_{0} \biggl(2-\frac{S(t)V_{0}}{S_{0}V(t)}-\frac {S_{0}V(t)}{S(t)V_{0}} \biggr)+(R_{0}-1)e(t,0). \end{aligned}

Therefore, $$R_{0}\leq1$$ ensures that $$\mathrm{d}V_{1}/\mathrm{d}t\leq0$$ holds. Furthermore, the strict equality holds if and only if $$S=S_{0}$$, $$V=V_{0}$$, $$e(t,a)=0$$, $$i(t,a)=0$$ and $$j(t,a)=0$$, simultaneously. Thus, $$M_{0}={E_{0}}\subset\Upsilon$$ is the largest invariant subset of $${\mathrm{d}V_{1}/\mathrm{d}t=0}$$, and by the Lyapunov–LaSalle invariance principle, the steady state $$E_{0}$$ is globally asymptotically stable when $$R_{0}\leq1$$. □

### Theorem 7.2

If $$R_{0}>1$$, the steady state $$E^{*}$$ is globally asymptotically stable.

### Proof

Constructing a Lyapunov functional as follows

$$V_{2}=V_{21}+V_{22}+V_{23}+V_{24},$$

where

\begin{aligned}& V_{21}=g \bigl(S(t)/S^{*} \bigr)+g \bigl(V(t)/V^{*} \bigr), \\& V_{22}= \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \int_{0}^{\infty }q_{1}(a)g \bigl(e(t,a)/e^{*}(a) \bigr)\,\mathrm{d}a \\& \hphantom{V_{22}={}}{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \biggl( \int_{0}^{\infty }q_{2}(a)g \bigl(e(t,a)/e^{*}(a) \bigr)\,\mathrm{d}a \\& \hphantom{V_{22}={}}{} + \int_{0}^{\infty}q_{3}(a)g \bigl(i(t,a)/i^{*}(a) \bigr)\,\mathrm{d}a \biggr), \\& V_{23}= \int_{0}^{\infty}q_{i}(a)g \bigl(i(t,a) \bigl(1+\alpha i^{*}(a) \bigr)/i^{*}(a) \bigl(1+\alpha i(t,a) \bigr) \bigr)\,\mathrm{d}a, \\& V_{24}= \int_{0}^{\infty}q_{j}(a)g \bigl(j(t,a)/j^{*}(a) \bigr)\,\mathrm{d}a, \end{aligned}

where

\begin{aligned}& q_{1}(a)= \int_{a}^{\infty}\gamma_{1}( \sigma)e^{*}(\sigma)\,\mathrm {d}\sigma,\qquad q_{2}(a)= \int_{a}^{\infty}\gamma_{2}( \sigma)e^{*}(\sigma)\,\mathrm {d}\sigma, \\& q_{3}(a)= \int_{a}^{\infty}\xi(\sigma)i^{*}(\sigma)\, \mathrm{d}\sigma,\qquad q_{i}(a)= \int_{a}^{\infty}\frac{\beta_{1}(\sigma)S^{*}+\beta_{2}(\sigma )V^{*}}{1+\alpha i^{*}(\sigma)}i^{*}( \sigma)\, \mathrm{d}\sigma, \\& q_{j}(a)= \int_{a}^{\infty} \bigl(\beta_{3}( \sigma)S^{*}+\beta_{4}(\sigma )V^{*} \bigr)j^{*}(\sigma)\, \mathrm{d}\sigma. \end{aligned}

Calculating the derivative of $$V_{21}$$ along with the solution of system (1.5), we have

\begin{aligned} \frac{\mathrm{d}V_{21}}{\mathrm{d}t} =& \biggl({1-\frac{S^{*}}{S(t)}} \biggr) \biggl(\Lambda- ( \mu+ p) S(t)+\eta V(t) \\ &{}- S(t) \int_{0}^{\infty}\biggl(\frac {\beta_{1}(a) i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a) j(t,a) \biggr) \, \mathrm{d}a \biggr) \\ &{}+ \biggl({1-\frac{V^{*}}{V(t)}} \biggr) \biggl(p S(t)- (\mu+ \rho+\eta) V(t)-V(t) \int_{0}^{\infty}\frac{\beta_{2}(a) i(t,a)}{1+\alpha i(t,a)} \, \mathrm{d}a \\ &{}-V(t) \int_{0}^{\infty}\beta_{4}(a) j(t,a) \, \mathrm{d}a \biggr) \\ =&\mu S^{*} \biggl(2-\frac{S(t)}{S^{*}}-\frac{S^{*}}{S(t)} \biggr)+ \biggl(\mu+\rho+ \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a \biggr) V^{*} \\ &{}\times \biggl(3-\frac{V(t)}{V^{*}}-\frac{S^{*}}{S(t)} -\frac{S(t)V^{*}}{S^{*}V(t)} \biggr) +\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+S^{*} \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)j^{*}(a) \biggr)\,\mathrm{d}a \\ &{}-S(t) \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}-\frac{S^{*}}{S(t)}S^{*} \int_{0}^{\infty} \biggl(\frac{\beta _{1}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)j^{*}(a) \biggr)\,\mathrm {d}a \\ &{}+S^{*} \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)j(t,a) \biggr)\,\mathrm{d}a-V^{*} \int_{0}^{\infty } \biggl(\frac{\beta_{2}i^{*}(a)}{1+\alpha i^{*}(a)} \\ &{}+\beta_{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a+V^{*} \int_{0}^{\infty} \biggl(\frac {\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a)j(t,a) \biggr)\,\mathrm {d}a \\ &{}-V(t) \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)i(t,a)}{1+\alpha i(t,a)}+ \beta_{4}(a)j(t,a) \biggr)\,\mathrm{d}a \\ &{}+V(t) \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a. \end{aligned}

It follows that

\begin{aligned} \frac{\mathrm{d}V_{21}}{\mathrm{d}t} =&\mu S^{*} \biggl(2-\frac {S(t)}{S^{*}}- \frac{S^{*}}{S(t)} \biggr)+ \biggl(\mu+\rho+ \int_{0}^{\infty } \biggl(\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)} \\ &{}+\beta_{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a \biggr) V^{*} \biggl(3-\frac{V(t)}{V^{*}}-\frac{S^{*}}{S(t)} - \frac{S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr)+ \int_{0}^{\infty} \biggl(\frac{\beta _{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)S^{*}j^{*}(a) \biggr)\,\mathrm{d}a \\ &{}-\frac{S(t)}{S^{*}} \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)S^{*}i(t,a)}{1+\alpha i(t,a)}+ \beta_{3}(a)S^{*}j(t,a) \biggr)\,\mathrm{d}a \\ &{}-\frac{S^{*}}{S(t)} \int_{0}^{\infty} \biggl(\frac{\beta _{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{3}(a)S^{*}j^{*}(a) \biggr)\,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \biggl(\frac{\beta_{1}(a)S^{*}i(t,a)}{1+\alpha i(t,a)} + \beta_{3}(a)S^{*}j(t,a) \biggr)\,\mathrm{d}a- \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \\ &{}+\beta_{4}(a)V^{*}j^{*}(a) \biggr)\, \mathrm{d}a + \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)V^{*}i(t,a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)V^{*}j(t,a) \biggr)\,\mathrm{d}a \\ &{}-\frac{V(t)}{V^{*}} \int_{0}^{\infty} \biggl(\frac{\beta _{2}(a)V^{*}i(t,a)}{1+\alpha i(t,a)} + \beta_{4}(a)V^{*}j(t,a) \biggr)\,\mathrm{d}a \\ &{}+\frac{V(t)}{V^{*}} \int_{0}^{\infty} \biggl(\frac{\beta _{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)V^{*}j^{*}(a) \biggr)\,\mathrm{d}a. \end{aligned}
(7.4)

The derivative of $$V_{22}$$, $$V_{23}$$ and $$V_{24}$$ can be calculated as follows:

\begin{aligned}& \frac{\mathrm{d}V_{22}}{\mathrm{d}t}= \int_{0}^{\infty}\frac{S^{*}\beta _{1}(a)B_{2}(a)+V^{*}\beta_{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm {d}a \int_{0}^{\infty}\gamma_{1}(a)e^{*}(a) \biggl\{ \frac {e(t,0)}{e^{*}(0)}-\frac{e(t,a)}{e^{*}(a)} \\& \hphantom{\frac{\mathrm{d}V_{22}}{\mathrm{d}t}={}}{}+\ln\frac{e(t,a)}{e^{*}(a)}-\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \, \mathrm{d}a \\& \hphantom{\frac{\mathrm{d}V_{22}}{\mathrm{d}t}={}}{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\gamma_{2}(a)e^{*}(a) \biggl\{ \frac {e(t,0)}{e^{*}(0)}-\frac{e(t,a)}{e^{*}(a)} \\& \hphantom{\frac{\mathrm{d}V_{22}}{\mathrm{d}t}={}}{}+\ln\frac{e(t,a)}{e^{*}(a)}-\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \, \mathrm{d}a \\& \hphantom{\frac{\mathrm{d}V_{22}}{\mathrm{d}t}={}}{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\xi(a)i^{*}(a) \biggl\{ \frac{i(t,0)}{i^{*}(0)}-\frac{i(t,a)}{i^{*}(a)} \\& \hphantom{\frac{\mathrm{d}V_{22}}{\mathrm{d}t}={}}{}+\ln\frac{i(t,a)}{i^{*}(a)}-\ln\frac{i(t,0)}{i^{*}(0)} \biggr\} \, \mathrm{d}a, \end{aligned}
(7.5)
\begin{aligned}& \frac{\mathrm{d}V_{23}}{\mathrm{d}t}= \int_{0}^{\infty}\frac{(\beta _{1}(a)S^{*}+\beta_{2}(a)V^{*})i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ \frac {i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))}-\frac {i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))} \\& \hphantom{\frac{\mathrm{d}V_{23}}{\mathrm{d}t}={}}{}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \, \mathrm{d}a, \end{aligned}
(7.6)
\begin{aligned}& \frac{\mathrm{d}V_{24}}{\mathrm{d}t}= \int_{0}^{\infty} \bigl(\beta _{3}(a)S^{*}+ \beta_{4}(a)V^{*} \bigr)j^{*}(a) \biggl\{ \frac {j(t,0)}{j^{*}(0)}-\frac{j(t,a)}{j^{*}(a)} +\ln\frac{j(t,a)}{j^{*}(a)}-\ln \frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm{d}a. \end{aligned}
(7.7)

Combining (7.4)–(7.7), we get

\begin{aligned} \frac{\mathrm{d}V_{2}}{\mathrm{d}t} =&\mu S^{*} \biggl(2-\frac {S(t)}{S^{*}}- \frac{S^{*}}{S(t)} \biggr)+ \biggl(\mu+\rho+ \int_{0}^{\infty } \biggl(\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+\beta _{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a \biggr) V^{*} \\ &{}\times \biggl(3-\frac{V(t)}{V^{*}}-\frac{S^{*}}{S(t)} -\frac{S(t)V^{*}}{S^{*}V(t)} \biggr) +\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+ \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ 1- \frac{S(t)i(t,a)(1+\alpha i^{*}(a))}{S^{*}i^{*}(a)(1+\alpha i(t,a))} -\frac{S^{*}}{S(t)}+\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \\ &{}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ -1- \frac{V(t)i(t,a)(1+\alpha i^{*}(a))}{V^{*}i^{*}(a)(1+\alpha i(t,a))}+\frac{V^{*}}{V(t)} +\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \\ &{}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{3}(a)S^{*}j^{*}(a) \biggl\{ 1-\frac {S(t)j(t,a)}{S^{*}j^{*}(a)}-\frac{S^{*}}{S(t)} +\frac{j(t,0)}{j^{*}(0)} +\ln \frac{j(t,a)}{j^{*}(a)}-\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm {d}a \\ &{}+ \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a) \biggl\{ -1-\frac {V(t)j(t,a)}{V^{*}j^{*}(a)}+\frac{V^{*}}{V(t)} +\frac{j(t,0)}{j^{*}(0)} \\ &{}+\ln \frac{j(t,a)}{j^{*}(a)}-\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm {d}a \\ &{}+ \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \int_{0}^{\infty}\gamma _{1}(a)e^{*}(a) \biggl\{ \frac{e(t,0)}{e^{*}(0)}-\frac {e(t,a)}{e^{*}(a)} \\ &{}+\ln\frac{e(t,a)}{e^{*}(a)}-\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\gamma_{2}(a)e^{*}(a) \biggl\{ \frac {e(t,0)}{e^{*}(0)}-\frac{e(t,a)}{e^{*}(a)} \\ &{}+\ln\frac{e(t,a)}{e^{*}(a)}-\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \, \mathrm{d}a \end{aligned}
\begin{aligned} &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\xi(a)i^{*}(a) \biggl\{ \frac{i(t,0)}{i^{*}(0)}-\frac{i(t,a)}{i^{*}(a)} \\ &{}+\ln\frac{i(t,a)}{i^{*}(a)}-\ln\frac{i(t,0)}{i^{*}(0)} \biggr\} \, \mathrm{d}a \\ =&\mu S^{*} \biggl(2-\frac{S(t)}{S^{*}}-\frac{S^{*}}{S(t)} \biggr)+ \biggl(\mu+\rho+ \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)j^{*}(a) \biggr)\,\mathrm{d}a \biggr) V^{*} \\ &{}\times \biggl(3-\frac{V(t)}{V^{*}}-\frac{S^{*}}{S(t)} -\frac{S(t)V^{*}}{S^{*}V(t)} \biggr) +\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+ \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ 1- \frac{S^{*}}{S(t)}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))} \\ &{}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ -1+ \frac{V^{*}}{V(t)}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))} \\ &{}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{3}(a)S^{*}j^{*}(a) \biggl\{ 1-\frac {S^{*}}{S(t)}+\ln\frac{j(t,a)}{j^{*}(a)} -\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a) \biggl\{ -1+\frac {V^{*}}{V(t)}+\ln\frac{j(t,a)}{j^{*}(a)} -\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\gamma _{1}(a)e^{*}(a) \biggl\{ \ln\frac{e(t,a)}{e^{*}(a)} -\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\gamma _{2}(a)e^{*}(a) \biggl\{ \ln\frac{e(t,a)}{e^{*}(a)} -\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\xi(a)i^{*}(a) \biggl\{ \ln \frac{i(t,a)}{i^{*}(a)} -\ln\frac{i(t,0)}{i^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \int_{0}^{\infty}\gamma _{1}(a)e^{*}(a) \biggl\{ \frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a) \int_{0}^{\infty}\gamma _{2}(a)e^{*}(a) \biggl\{ \frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\xi(a)i^{*}(a) \biggl\{ \frac{i(t,0)}{i^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)}\frac{S(t)i(t,a)(1+\alpha i^{*}(a))}{S^{*}i^{*}(a)(1+\alpha i(t,a))}\, \mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{\beta_{2}V^{*}(a)i^{*}(a)}{1+\alpha i^{*}(a)}\frac{V(t)i(t,a)(1+\alpha i^{*}(a))}{V^{*}i^{*}(a)(1+\alpha i(t,a))}\, \mathrm{d}a \\ &{}- \int_{0}^{\infty}\beta_{3}(a)S^{*}j^{*}(a) \frac {S(t)j(t,a)}{S^{*}j^{*}(a)}\,\mathrm{d}a - \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a) \frac {V(t)j(t,a)}{V^{*}j^{*}(a)}\,\mathrm{d}a \\ &{}- \bigl(K_{3}S^{*}+K_{5}V^{*} \bigr)i(t,0) - \bigl(K_{4}S^{*} +K_{6}V^{*} \bigr)j(t,0) \\ &{}+ \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}+\beta_{2}(a)V^{*}}{1+\alpha i^{*}(a)}i^{*}(a) \frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))}\,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(\beta_{3}(a)S^{*}+ \beta_{4}(a)V^{*} \bigr)j^{*}(a) \frac {j(t,0)}{j^{*}(0)}\,\mathrm{d}a. \end{aligned}

It is easy to see that the last 11 terms of the above equation equal 0. Then we have

\begin{aligned} \frac{\mathrm{d}V_{2}}{\mathrm{d}t} \leq&\mu S^{*} \biggl(2-\frac {S(t)}{S^{*}}- \frac{S^{*}}{S(t)} \biggr)+ (\mu+\rho ) V^{*} \biggl(3- \frac{V(t)}{V^{*}}-\frac{S^{*}}{S(t)} -\frac{S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+ \int_{0}^{\infty} \biggl(\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)}+ \beta_{4}(a)V^{*}j^{*}(a) \biggr) \biggl\{ 3- \frac {V(t)}{V^{*}}-\frac{S^{*}}{S(t)} -\frac{S(t)V^{*}}{S^{*}V(t)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ 1- \frac{S^{*}}{S(t)}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))} \\ &{}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)} \biggl\{ -1+ \frac{V^{*}}{V(t)}+\ln\frac{i(t,a)(1+\alpha i^{*}(a))}{i^{*}(a)(1+\alpha i(t,a))} \\ &{}-\ln\frac{i(t,0)(1+\alpha i^{*}(0))}{i^{*}(0)(1+\alpha i(t,0))} \biggr\} \, \mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{3}(a)S^{*}j^{*}(a) \biggl\{ 1-\frac {S^{*}}{S(t)}+\ln\frac{j(t,a)}{j^{*}(a)} -\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a) \biggl\{ -1+\frac {V^{*}}{V(t)}+\ln\frac{j(t,a)}{j^{*}(a)} -\ln\frac{j(t,0)}{j^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\gamma _{1}(a)e^{*}(a) \biggl\{ \ln\frac{e(t,a)}{e^{*}(a)} -\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\gamma _{2}(a)e^{*}(a) \biggl\{ \ln\frac{e(t,a)}{e^{*}(a)} -\ln\frac{e(t,0)}{e^{*}(0)} \biggr\} \,\mathrm{d}a \\ &{}+ \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \\ &{}\times\int_{0}^{\infty}\xi(a)i^{*}(a) \biggl\{ \ln \frac{i(t,a)}{i^{*}(a)} -\ln\frac{i(t,0)}{i^{*}(0)} \biggr\} \,\mathrm{d}a. \end{aligned}

Consequently, we have

\begin{aligned} \frac{\mathrm{d}V_{2}}{\mathrm{d}t} \leq&\mu S^{*} \biggl(2-\frac {S(t)}{S^{*}}- \frac{S^{*}}{S(t)} \biggr)+(\mu+\rho)V^{*} \biggl(3- \frac {V(t)}{V^{*}} -\frac{S^{*}}{S(t)}-\frac{S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}+\eta V^{*} \biggl(2-\frac{S^{*}V(t)}{S(t)V^{*}}-\frac {S(t)V^{*}}{S^{*}V(t)} \biggr) \\ &{}- \int_{0}^{\infty} \biggl[\frac{\beta_{1}(a)S^{*}+\beta _{2}(a)V^{*}}{1+\alpha i^{*}(a)}i^{*}(a)+ \bigl(\beta_{3}(a)S^{*} +\beta_{4}(a)V^{*} \bigr)j^{*}(a) \biggr]g \biggl(\frac{S^{*}}{S(t)} \biggr)\, \mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{S^{*}\beta_{1}(a)B_{2}(a)+V^{*}\beta _{2}(a)B_{2}(a)}{1+\alpha i^{*}(a)}\,\mathrm{d}a \int_{0}^{\infty}\gamma _{1}(a)e^{*}(a)g \biggl(\frac{e(t,a)i^{*}(0)}{e^{*}(a)i(t,0)} \biggr)\, \mathrm{d}a \\ &{}- \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\gamma _{2}(a)e^{*}(a)g \biggl(\frac{e(t,a)j^{*}(0)}{e^{*}(a)j(t,0)} \biggr)\, \mathrm{d}a \\ &{}- \int_{0}^{\infty} \bigl(S^{*} \beta_{3}(a)B_{3}(a)+V^{*}\beta _{4}(a)B_{3}(a) \bigr)\,\mathrm{d}a \int_{0}^{\infty}\xi(a)i^{*}(a)g \biggl( \frac{i(t,a)j^{*}(0)}{i^{*}(a)j(t,0)} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{\beta_{1}(a)S^{*}i^{*}(a)}{1+\alpha i^{*}(a)}g \biggl( \frac{S(t)i(t,a)e^{*}(0)(1+\alpha i^{*}(a))}{S^{*}i^{*}(a)e(t,0)(1+\alpha i(t,a))} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)}g \biggl( \frac{V(t)i(t,a)e^{*}(0)(1+\alpha i^{*}(a))}{V^{*}i^{*}(a)e(t,0)(1+\alpha i(t,a))} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\beta_{3}(a)S^{*}j^{*}(a)g \biggl(\frac {S(t)j(t,a)e^{*}(0)}{S^{*}j^{*}(a)e(t,0)} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a)g \biggl(\frac {V(t)j(t,a)e^{*}(0)}{V^{*}j^{*}(a)e(t,0)} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\frac{\beta_{2}(a)V^{*}i^{*}(a)}{1+\alpha i^{*}(a)}g \biggl( \frac{S(t)V^{*}}{S^{*}V(t)} \biggr)\,\mathrm{d}a \\ &{}- \int_{0}^{\infty}\beta_{4}(a)V^{*}j^{*}(a)g \biggl(\frac {S(t)V^{*}}{S^{*}V(t)} \biggr)\,\mathrm{d}a. \end{aligned}

Hence, $$\mathrm{d}V_{2}/\mathrm{d}t\leq0$$ holds. Furthermore, the strict equality holds if and only if $$S=S^{*}$$, $$V=V^{*}$$, $$e(t,a)=e^{*}(a)$$, $$i(t,a)=i^{*}(a)$$, $$j(t,a)=j^{*}(a)$$. Thus, $$M^{*}=\{E^{*}\}\subset \Omega$$ is the largest invariant subset of $$\mathrm{d}V_{2}/\mathrm {d}t=0$$, and by the Lyapunov–LaSalle invariance principle, when $$R_{0}>1$$, the steady state $$E^{*}$$ is globally asymptotically stable. This completes the proof. □

## 8 Discussion

An age-structured HBV model with saturating incidence has been proposed here to incorporate patients with acute hepatitis B and chronic hepatitis B. By mathematical analysis, the dynamic behavior of system (1.5) was shown to be determined completely by the basic reproduction number $$R_{0}$$: disease free steady state $$E_{0}$$ is locally and globally asymptotically stable if $$R_{0}<1$$; endemic steady state $$E^{*}$$ is locally and globally asymptotically stable if $$R_{0}>1$$. To place the model on more sound biological grounds, we considered the fact that the vaccines may lose efficacy and the infection may reach a saturating state. Next, we will focus on the numerical simulations of such a complex partial differential equations (PDEs) model.

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## Acknowledgements

This work was supported by the National Natural Science Foundation of China (No. 11371368), the Natural Science Foundation of Hebei Province of China (No. A2014506015), Shanxi Scientific Data Sharing Platform for Animal Diseases (201605D121014), and the Science and Technology Innovation Team of Shanxi Province (201605D131044-06).

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YL and RX have contributed equally to the writing of this paper except for Sect. 6. JL made major contribution to the revised paper and Sect. 6. All authors read and approved the final manuscript.

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Correspondence to Rui Xu.

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Li, Y., Xu, R. & Lin, J. The stability analysis of an epidemic model with saturating incidence and age-structure in the exposed and infectious classes. Adv Differ Equ 2018, 202 (2018). https://doi.org/10.1186/s13662-018-1635-6