The global dynamics in a wild-type and drug-resistant HIV infection model with saturated incidence

Abstract

In this paper we investigate the global dynamics in an HIV virus infection model with saturated incidence. The model includes two viral strains, one is wild-type (i.e. drug sensitive) and another is drug-resistant. The wild-type strain can mutate and become drug-resistant during the process of reverse transcription. The nonnegativity and boundedness of solutions are established. The basic reproduction numbers of two strains and the existence of equilibria are also obtained. The threshold criteria on the local and global stability of equilibria and the uniform persistence of the model are established by using the linearization method, constructing suitable Lyapunov functions and the theory of persistence in dynamical systems. Moreover, the mathematical analysis and numerical examples show that model may have a positive equilibrium which is globally asymptotically stable.

1 Introduction

It is well known that mathematical models that describe the dynamical behaviors of virus infection play an important role in understanding the mechanism of the diffusion of virus. There has been much interest in mathematical modeling of viral dynamics within-host. So, the research of virus dynamics with specific immune response, which can control the virus propagation, has drawn significant attention [1–6]. A few years ago, Perelson et al. in [7] constructed a model that has been widely adopted to model the plasma viral load in HIV infected patients as follows:

$$ \textstyle\begin{cases} \frac{\mathrm {d}T(t)}{\mathrm {d}t}= \lambda -dT-kVT, \\ \frac{\mathrm {d}T_{s}(t)}{\mathrm {d}t}= kVT-\delta T_{s}, \\ \frac{\mathrm {d}V(t)}{\mathrm {d}t}= N\delta T_{s}-cV. \end{cases} $$

Treating HIV-infected patients with a combination of several antiretroviral drugs usually contributes to a substantial decline in viral load and an increase in $\mathit{CD}_{4}^{+}$ T cells. Nevertheless, there is a reasonable chance that drug-resistant variants of HIV preexist even before the initiation of therapy due to a single mutation, or a number of mutation combinations can result in drug resistance by Ribeiro and Bonhoeffer in 2000 (see [8, 9]). In order to study the mechanism of the emergence of drug resistance during the treatment of HIV-infected patients, a dynamical model including wild-type and drug-resistant strains was proposed by Rong et al. in [9] as follows:

$$ \textstyle\begin{cases} \frac{\mathrm {d}T(t)}{\mathrm {d}t}= \lambda -dT-k_{s}V_{s}T-k_{r}V_{r}T,\\ \frac{\mathrm {d}T_{s}(t)}{\mathrm {d}t}=(1-u)k_{s}V_{s}T-\delta T_{s},\\ \frac{\mathrm {d}V_{s}(t)}{\mathrm {d}t}=N_{s}\delta T_{s}-cV_{s},\\ \frac{\mathrm {d}T_{r}(t)}{\mathrm {d}t}=uk_{s}V_{s}T+k_{r}V_{r}T-\delta T_{r}, \\ \frac{\mathrm {d}V_{r}(t)}{\mathrm {d}t}=N_{r}\delta T_{r}-cV_{r}. \end{cases} $$

(1)

Usually the rate of infection in most HIV-1 models is assumed to be bilinear in the virus and the uninfected cells. However, the actual incidence rate is probably not linear over the entire range of virus and the uninfected cells. Thus, it is reasonable to assume that the infection rate of HIV-1 is given by the Beddington–DeAngelis functional response [10], which was introduced by Beddington [11] and DeAngelis et al. [12]. For a specific nonlinear incidence rate, we consider the following HIV-1 infection model with saturated incidence:

$$ \textstyle\begin{cases} \frac{\mathrm {d}T(t)}{\mathrm {d}t}= \lambda -dT-\frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}-\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}, \\ \frac{\mathrm {d}T_{s}(t)}{\mathrm {d}t}= (1-u)\frac{k_{s}V_{s}T}{1+\omega _{1}V _{s}}-\delta T_{s}, \\ \frac{\mathrm {d}V_{s}(t)}{\mathrm {d}t}= N_{s}\delta T_{s}-cV_{s}, \\ \frac{\mathrm {d}T_{r}(t)}{\mathrm {d}t}= u \frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}+\frac{k_{r}V_{r}T}{1+\alpha _{1}V _{r}}-\delta T_{r}, \\ \frac{\mathrm {d}V_{r}(t)}{\mathrm {d}t}= N_{r}\delta T_{r}-cV_{r}. \end{cases} $$

(2)

The biological significance of variables and parameters in model (2) is given in Table 1.

Table 1 Biological significance of variables and parameters

In model (2), the parameter u ($0< u<1$) is the conversion fraction at which cells infected by the wild-type mutate and become drug-resistant during the process of reverse transcription of viral RNA into proviral DNA (SR conversion, for short). It should be noted that the backward mutation from drug-resistant to wild-type strain is neglected since the wild-type virus dominates the population before the initiation of therapy (see [13, 14]). And the terms $\frac{k_{s}V_{s}T}{1+\omega _{1}V _{s}}$ and $\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}$ express the saturated incidence for virus $V_{s}$ and $V_{r}$, where $\omega _{1}$ and $\alpha _{1}$ are the nonnegative constants. When $\omega _{1}=0$ or $\alpha _{1}=0$, the corresponding incidence degrades into bilinear incidence for $V_{s}$ or $V_{r}$.

In [9], we see that model (1) with bilinear incidence is investigated. The authors only obtained the existence and local stability of the infection-free equilibrium, the equilibrium with only wild-type virus, drug-resistant virus, and the coexistence equilibrium (see Proposition 1 and Proposition 2 in [9]). We all know that in many realistic infectious diseases the nonlinear incidence rates play very important roles, and the global dynamics of the model, including the global asymptotic stability of equilibria, the uniform persistence, etc., also needs to be investigated in detail. In [5, 15], we see that the global dynamics for virus infection models with nonlinear incidence rates is discussed. Therefore, in this paper we carry out the research for a wild-type and drug-resistant HIV infection model with saturated incidence. We establish a series of threshold criteria for the local and global asymptotic stability of infection-free, drug-resistant strain infection equilibria, and the uniform persistence of HIV infection.

The organization of this paper is as follows. In Sect. 2, the nonnegativity and boundedness of solutions are established, and then the basic reproduction numbers of two strains and the existence of equilibria are obtained. In Sect. 3, the main theorems on the local and global stability of equilibria of model (2) are stated and proved. In Sect. 4, the uniform persistence of model (2) is also investigated. In Sect. 5, some numerical examples are given to illustrate our main results. In the last section, a brief conclusion is presented.

2 Preliminaries

For any integer $n>0$, denote $R^{n}_{+}=\{(x_{1},x_{2},\ldots,x_{n}) \in R^{n}: x_{i}\geq 0, i=1,2,\ldots,n\}$. The initial condition for model (2) is given by

$$ \bigl(T(0),T_{s}(0),V_{s}(0),T_{r}(0),V_{r}(0) \bigr)=(T_{0},T_{s0},V_{s0},T_{r0},V _{r0})\in R^{5}_{+}. $$

(3)

Firstly, on the positivity and boundedness of solutions for model (2), we have the following result.

Theorem 1

The solution $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t))$of model (2) with initial condition (3) is defined for all $t\in [0,\infty )$and is nonnegative and ultimately bounded.

Proof

On the nonnegativity of solutions, by the continuity of solutions with respect to initial values, we only need to prove that, for any positive initial value $(T_{0},T_{s0},V_{s0},T_{r0},V_{r0})$, the solution $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t))$ with initial condition (3) is also positive for any $t>0$ in the definition interval. From the first equation of model (2), we have

$$ \frac{\mathrm {d}T(t)}{\mathrm {d}t}>-\biggl(d+\frac{k_{s}V_{s}}{1+\omega _{1}V_{s}}+\frac{k _{r}V_{r}}{1+\alpha _{1}V_{r}}\biggr)T(t). $$

Hence, as $T_{0}>0$, we directly have $T(t)>0$ for any $t>0$ in the definition interval.

Define $m(t)=\min \{T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t)\}$. Obviously, $m(0)=\min \{T(0),T_{s}(0),V_{s}(0), T_{r}(0),V_{r}(0)\}>0$. By the continuity of solutions there exists $\delta >0$ such that $m(t)>0$, when $t\in [0,\delta )$. We only need to prove $m(t)>0$ for all $t\geq 0$ in the definition interval. Suppose that there exists $t^{*}>0$ such that $m(t^{*})=0$ and $m(t)>0$ for all $t\in [0,t^{*})$. Then there exist the following cases: (1) $m(t ^{*})=T_{s}(t^{*})$, (2) $m(t^{*})=V_{s}(t^{*})$, (3) $m(t^{*})=T _{r}(t^{*})$, and (4) $m(t^{*})=V_{r}(t^{*})$.

For case (1), according to $m(t)>0$ for all $t\in [0,t^{*})$, from the second equation of model (2), we know $\frac{\mathrm {d}T_{s}(t)}{\mathrm {d}t}>- \delta T_{s}$. Thus, $T_{s}(t)>T_{s}(0)e^{-\delta t}$ for any $t\in [0,t^{*})$. Taking $t \rightarrow t^{*}$, then $0=T_{s}(t^{*}) \geq T_{s}(0)e^{-\delta {t^{*}}}>0$, which leads to a contradiction. Similarly, we can get the contradiction for cases (2), (3) and (4). Therefore, $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t))$ is positive for all $t\geq 0$ in the definition interval.

Define a Lyapunov function

$$ W(t)=T(t)+T_{s}(t)+\frac{1}{2N_{s}}V_{s}(t)+T_{r}(t)+ \frac{1}{2N_{r}}V _{r}(t). $$

We have

$$\begin{aligned} \frac{\mathrm {d}W(t)}{\mathrm {d}t}={}&\lambda -dT-\frac{1}{2}\delta T_{s}- \frac{c}{2N _{s}}V_{s}-\frac{1}{2}\delta T_{r}- \frac{c}{2N_{r}}V_{r}\leq \lambda -nW(t), \end{aligned}$$

where $n=\min \{d,\frac{\delta }{2},c\}$. Since solution $U(t)$ of the comparison equation

$$ \frac{\mathrm {d}U(t)}{\mathrm {d}t}=\lambda -nU(t) $$

with initial condition $U(0)=U_{0}\geq 0$ is defined for all $t\in [0,\infty )$ and satisfies $\lim_{t\rightarrow \infty }U(t)=\frac{ \lambda }{n}$, by the comparison principle, we directly have that $W(t)$ is bounded, and hence solution $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V _{r}(t))$ is also bounded. Thus, $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V _{r}(t))$ can be defined for all $t\in [0,\infty )$. Furthermore, since $W(t)\leq U(t)$ as $W(0)\leq U(0)$, we obtain that $\limsup_{t\rightarrow \infty } W(t)\leq \lim_{t\rightarrow \infty }U(t)=\frac{ \lambda }{n}$. This implies that the solution $(T(t),T_{s}(t),V_{s}(t),T _{r}(t), V_{r}(t))$ is also ultimately bounded. This completes the proof. □

Following the concept of the basic reproductive number for an epidemic disease presented in [16], we define the wild-type strain infection reproduction number $R_{s}$ and the drug-resistant strain infection reproduction number $R_{r}$ as follows:

$$ R_{s}=\frac{k_{s}N_{s}\lambda }{dc},\quad\quad R_{r}= \frac{k_{r}N_{r}\lambda }{dc}. $$

The fraction $\frac{1}{c}$ gives the average life-span of a virus for strain i ($i=r,s$). $\frac{\lambda }{d}$ is the steady-state target cell density at the beginning of the strain i infection process (i.e. near the infection-free steady state). $k_{i}N_{i}$ gives the magnitude of virus particles produced by one strain i infectious (virus-producing) cell during its average survival time. Multiplying these quantities together gives the expected number of newly infected cells produced by a single newly for strain i infected cell, that is, $R_{i}$.

Now, we discuss the equilibrium of model (2). The equilibrium can be given from the following equations:

$$ \textstyle\begin{cases} \lambda -dT-\frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}-\frac{k_{r}V_{r}T}{1+ \alpha _{1}V_{r}}=0, \\ (1-u)\frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}-\delta T_{s}=0, \\ N_{s}\delta T_{s}-cV_{s}=0, \\ u\frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}+\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}-\delta T_{r}=0, \\ N_{r}\delta T_{r}-cV_{r}=0. \end{cases} $$

(4)

Obviously, model (2) always has a unique infection-free equilibrium $E_{0}=(\frac{\lambda }{d},0,0,0,0)$. When $T_{s}>0$ and $T_{r}=0$, from (4) we directly have $V_{r}=0$ and $V_{s}T=0$, and then $T_{s}=0$, which leads to a contradiction. When $T_{s}=0$ and $T_{r}>0$, from (4) we can obtain that if $R_{r}>1$, then model (2) has a unique boundary equilibrium $E_{r}=(T_{1},0,0,T_{r1},V_{r1})$ with

$$ T_{1}=\frac{\lambda (k_{r}+d\alpha _{1}R_{r})}{dR_{r}(k_{r}+d\alpha _{1})}, \quad\quad T_{r1}= \frac{dc(R_{r}-1)}{N_{r}\delta (k_{r}+d\alpha _{1})}, \quad\quad V_{r1}=\frac{d(R_{r}-1)}{k_{r}+d\alpha _{1}}, $$

and if $R_{r}\leq 1$, then $E_{r}$ does not exist. When $T_{s}>0$ and $T_{r}>0$, from (4) we can obtain that

$$ T_{r}=\frac{\lambda }{\delta }-T_{s}-\frac{\lambda (1+\omega _{1}\frac{ \delta }{c}N_{s}T_{s})}{\delta (1-u)R_{s}}:=T_{r}(T_{s}) $$

(5)

and

$$ T_{s}=\frac{\frac{\lambda }{\delta }((1-u)R_{s}-1)+(((1-u)R_{s}-1) \alpha _{1}\frac{\lambda }{c}N_{r}-R_{r})T_{r}}{R_{s}+\omega _{1}\frac{ \lambda }{c}N_{s} +((R_{s}+\omega _{1}\frac{\lambda }{c}N_{s})\alpha _{1}\frac{\delta }{c}N_{r}+\omega _{1}\frac{\delta }{c}N_{s}R_{r})T _{r}}:=T_{s}(T_{r}). $$

(6)

Clearly, functions $T_{r}(T_{s})$ and $T_{s}(T_{r})$ are decreasing in $T_{s}\geq 0$ and $T_{r}\geq 0$, respectively. We have

$$\begin{aligned}& T_{r}(0)=\frac{\lambda }{\delta R_{s}(1-u)}\bigl((1-u)R_{s}-1\bigr), \quad \quad T_{r}(+\infty )=-\infty , \\& T_{s}(0)=\frac{\frac{\lambda }{\delta }((1-u)R_{s}-1)}{R_{s}+\omega _{1}\frac{\lambda }{c}N_{s}}, \quad\quad T_{s}(+\infty )= \frac{((1-u)R_{s}-1) \alpha _{1}\frac{\lambda }{c}N_{r}-R_{r}}{(R_{s}+\omega _{1}\frac{ \lambda }{c}N_{s})\alpha _{1}\frac{\delta }{c}N_{r}+\omega _{1}\frac{ \delta }{c}N_{s}R_{r}}. \end{aligned}$$

Furthermore, from $T_{r}(T_{s})=0$ and $T_{s}(T_{r})=0$, we obtain

$$ T_{s}^{*}=\frac{\lambda ((1-u)R_{s}-1)}{\delta (1-u)R_{s}+\lambda \omega _{1}\frac{\delta }{c}N_{s}}, \quad\quad T_{r}^{*}=\frac{\frac{\lambda }{ \delta }((1-u)R_{s}-1)}{R_{r}-((1-u)R_{s}-1)\alpha _{1} \frac{\lambda }{c}N_{r}}. $$

It is easy to verify that $T_{s}(0)< T_{s}^{*}$ when $(1-u)R_{s}>1$. From $T_{r}(0)< T_{r}^{*}$ we let

$$ R_{r}=(1-u)R_{s}+\bigl((1-u)R_{s}-1\bigr)\alpha _{1}\frac{\lambda }{c}N_{r}:=R _{ru}(R_{s}). $$

This shows that curves $T_{r}(T_{s})$ and $T_{s}(T_{r})$ have a unique intersection point $(T_{sc},T_{rc})$ in the positive quadrant, which means $E_{c}=(T_{c},T_{sc},V_{sc},T_{rc},V_{rc})$ is the unique positive equilibrium of model (2). Thus, we finally have the following results.

Theorem 2

• (i) Model (2) always has a unique infection-free equilibrium $E_{0}$;

• (ii) If $R_{r}>\max \{1,(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{ \lambda }{c}N_{r}\}$, then model (2) only has equilibria $E_{0}$and $E_{r}$;

• (iii) If $(1-u)R_{s}>1\geq R_{r}$, then model (2) only has equilibria $E_{0}$and $E_{c}$;

• (iv) If $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}>R _{r}>1$, then model (2) has three equilibria $E_{0}$, $E_{r}$, and $E_{c}$.

The existence of equilibria for model (2) is also intuitively expressed in Fig. 1. From Theorem 2 and Fig. 1, we can find that the saturated coefficient $\omega _{1}$ of wild-type virus $V_{s}$ has no effect on Fig. 1. Along with the decreasing of saturated coefficient $\alpha _{1}$, the orange region will shrink, and it finally becomes the region $\{(R_{s},R_{r}): (1-u)R_{s}>R_{r}>1\}$ as $\alpha _{1}\to 0$. On the contrary, along with the increasing of saturated coefficient $\alpha _{1}$, the orange region will enlarge, and it finally becomes the region $\{(R_{s},R_{r}): (1-u)R_{s}>1, R_{r}>1\}$ as $\alpha _{1} \to +\infty$.

Figure 1

The existence of equilibria of model (2)

3 Stability of equilibrium

Let $E=(T,T_{s},V_{s},T_{r},V_{r})$ be any equilibrium of model (2). By calculating, we get that the Jacobian matrix at equilibria E is

$$ J(E)= \begin{pmatrix} -d-\frac{k_{s}V_{s}}{1+\omega _{1}V_{s}}-\frac{k_{r}V_{r}}{1+\alpha _{1}V_{r}} & 0 & -\frac{k_{s}T}{(1+\omega _{1}V_{s})^{2}} & 0 & -\frac{k _{r}T}{(1+\alpha _{1}V_{r})^{2}} \\ (1-u)\frac{k_{s}V_{s}}{1+\omega _{1}V_{s}} & -\delta & (1-u)\frac{k _{s}T}{(1+\omega _{1}V_{s})^{2}} & 0 & 0 \\ 0 & N_{s}\delta & -c & 0 & 0 \\ u\frac{k_{s}V_{s}}{1+\omega _{1}V_{s}}+\frac{k_{r}V_{r}}{1+\alpha _{1}V _{r}} & 0 & u\frac{k_{s}T}{(1+\omega _{1}V_{s})^{2}} & -\delta & \frac{k _{r}T}{(1+\alpha _{1}V_{r})^{2}} \\ 0 & 0 & 0 & N_{r}\delta & -c \end{pmatrix} . $$

(7)

Firstly, for the stability of equilibrium $E_{0}$, we have the following results.

Theorem 3

• (a) If $(1-u)R_{s}<1$and $R_{r}<1$, then infection-free equilibrium $E_{0}$is locally asymptotically stable.

• (b) If $R_{s}\leq 1$and $R_{r}\leq 1$, then $E_{0}$is globally asymptotically stable.

• (c) If $(1-u)R_{s}>1$or $R_{r}>1$, then $E_{0}$is unstable.

Proof

At equilibrium $E_{0}$, from (7) the characteristic equation of $J(E_{0})$ is

$$ f(X)=(X+d) \bigl(X^{2}+(\delta +c)X+\delta c \bigl(1-(1-u)R_{s}\bigr)\bigr) \bigl(X^{2}+(\delta +c)X+ \delta c(1-R_{r})\bigr)=0. $$

(8)

One root of (8) is $X_{1}=-d<0$. When $(1-u)R_{s}<1$ and $R_{r}<1$, by the Routh–Hurwitz criterion, all roots of the equations

$$ X^{2}+(\delta +c)X+\delta c\bigl(1-(1-u)R_{s} \bigr)=0 $$

(9)

and

$$ X^{2}+(\delta +c)X+\delta c(1-R_{r})=0 $$

(10)

have negative real parts, respectively. This implies that $E_{0}$ is locally asymptotically stable. When $(1-u)R_{s}>1$ or $R_{r}>1$, we easily see that equation (9) or (10) has at least a root with positive real part. This implies that $E_{0}$ is unstable.

For the global stability of $E_{0}$, we define Lyapunov function $L_{1}(t)$ as follows:

$$ L_{1}(t)=T_{0}\biggl(\frac{T}{T_{0}}-\ln \frac{T}{T_{0}}-1\biggr)+T_{s}+\frac{1}{N _{s}}V_{s}+T_{r}+ \frac{1}{N_{r}}V_{r}. $$

We have

$$\begin{aligned} \frac{\mathrm {d}L_{1}(t)}{\mathrm {d}t} ={}&\biggl(1-\frac{T_{0}}{T}\biggr) \biggl(\lambda -dT- \frac{k _{s}V_{s}T}{1+\omega _{1}V_{s}}-\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}\biggr)+\biggl((1-u)\frac{k _{s}V_{s}T}{1+\omega _{1}V_{s}}-\delta T_{s}\biggr) \\ & {} +\biggl(u\frac{k_{s}V_{s}T}{1+\omega _{1}V_{s}}+\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}-\delta T_{r} \biggr)+\frac{1}{N_{r}}(N_{r}\delta T_{r}-cV_{r})+ \frac{1}{N _{s}}(N_{s}\delta T_{s}-cV_{s}) \\ ={}&dT_{0}\biggl(2-\frac{T}{T_{0}}-\frac{T_{0}}{T}\biggr)+ \frac{c(R_{s}-1)}{(1+ \omega _{1}V_{s})N_{s}}V_{s}-\frac{c\omega _{1}V_{s}^{2}}{(1+\omega _{1}V _{s})N_{s}} \\ & {} +\frac{c(R_{r}-1)}{(1+\alpha _{1}V_{r})N_{r}}V_{r}-\frac{c\alpha _{1}V _{r}^{2}}{(1+\alpha _{1}V_{r})N_{r}}. \end{aligned}$$

When $R_{s}\leq 1$ and $R_{r}\leq 1$, then $\frac{\mathrm {d}L_{1}(t)}{\mathrm {d}t}\leq 0$ and the set $M=\{(T,T_{s},V_{s},T _{r},V_{r}): \frac{\mathrm {d}L_{1}(t)}{\mathrm {d}t}=0\}\subset \{(T,T_{s},V _{s},T_{r},V_{r}): T=T_{0},T_{s}\geq 0,V_{s}\geq 0,T_{r}\geq 0,V_{r} \geq 0\}$.

For any solution trajectory $\{(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t)): t\geq 0\}\subset M$, we have $T(t)\equiv T_{0}$. From the first equation of model (4), we obtain $\frac{k_{s}V_{s}(t)T_{0}}{1+\omega _{1}V_{s}(t)}+\frac{k_{r}V_{r}(t)T _{0}}{1+\alpha _{1}V_{r}(t)}\equiv 0$, which implies $V_{s}(t)=V_{r}(t) \equiv 0$. From the third and fifth equations of model (4), we also get $N_{s}\delta T_{s}(t)-cV_{s}(t)=0$ and $N_{r}\delta T_{r}(t)-cV_{r}(t)=0$, which further imply $T_{s}(t)=T _{r}(t)\equiv 0$. Hence, $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t)) \equiv E_{0}$. From LaSalle’s invariance principle [17], $E_{0}$ is globally asymptotically stable. This completes the proof. □

Remark 1

In Theorem 3, we only obtained the global asymptotic stability of $E_{0}$ under $R_{s}\leq 1$ and $R_{r}\leq 1$. Therefore, based on conclusion (a) of Theorem 3, an interesting open problem is whether we can establish the global asymptotic stability of $E_{0}$ when $(1-u)R_{s}\leq 1$ and $R_{r}\leq 1$.

Next, about the stability of equilibrium $E_{r}$, we have the following results.

Theorem 4

• (a) If $R_{r}>\max \{1,(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{ \lambda }{c}N_{r}\}$, then equilibrium $E_{r}$is locally asymptotically stable.

• (b) If $R_{r}>1$and $R_{r}<(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{ \lambda }{c}N_{r}$, then $E_{r}$is unstable.

• (c) If $R_{r}>\max \{1,R_{s}+\alpha _{1}\frac{\lambda }{c}N_{r}(R _{s}-1)\}$, then $E_{r}$is globally asymptotically stable.

Proof

At equilibrium $E_{r}$, from (7) the characteristic equation of $J(E_{r})$ is

$$ f(X)=\bigl(X^{2}+a_{1}X+a_{0} \bigr) \bigl(X^{3}+b_{2}X^{2}+b_{1}X+b_{0} \bigr)=0, $$

(11)

where

$$ \begin{aligned} & a_{1}=\delta +c, \quad\quad a_{0}=\delta c\frac{k_{r}(R_{r}-(1-u)R_{s}-((1-u)R _{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r})}{R_{r}(k_{r}+d\alpha _{1})}, \\ & b_{2}=d+c+\delta +\frac{dk_{r}(R_{r}-1)}{k_{r}+d\alpha _{1}}, \quad\quad b_{1}=d( \delta +c)+\frac{\delta c\alpha _{1}+k_{r}(\delta +c)}{k_{r}+d \alpha _{1}R_{r}}d(R_{r}-1), \\ & b_{0}=\frac{k_{r}+d\alpha _{1}}{k_{r}+d \alpha _{1}R_{r}}d\delta c(R_{r}-1). \end{aligned} $$

When $R_{r}>\max \{1,(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}\}$, we have $a_{i}>0$ and $b_{i}>0$ for $i=0,1,2$. Since

$$\begin{aligned} b_{1}b_{2}-b_{0}={}&\biggl(d(\delta +c)+ \frac{d\delta k_{r}(R_{r}-1)}{k_{r}+d \alpha _{1}R_{r}}\biggr) \biggl(d+c+\delta +\frac{dk_{r}(R_{r}-1)}{k_{r}+d\alpha _{1}R_{r}}\biggr) \\ & {} +\frac{dck_{r}(R_{r}-1)}{k_{r}+d\alpha _{1}R_{r}}\biggl(d+c+\frac{dk_{r}(R _{r}-1)}{k_{r}+d\alpha _{1}R_{r}}\biggr) \\ & {} +\frac{d\delta c\alpha _{1}(R_{r}-1)}{k_{r}+d\alpha _{1}R_{r}}\biggl(\delta +c+\frac{dk_{r}(R_{r}-1)}{k_{r}+d\alpha _{1}R_{r}}\biggr)>0. \end{aligned}$$

According to the Routh–Hurwitz criterion, all roots of equation (11) have negative real parts. Therefore, $E_{r}$ is locally asymptotically stable. When $R_{r}>1$ and $R_{r}<(1-u)R_{s}+((1-u)R _{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}$, the equation $X^{2}+a_{1}X+a _{0}=0$ has at least a positive real part root. This implies that $E_{r}$ is unstable.

To obtain the global stability of $E_{r}$, we define Lyapunov function $L_{2}(t)$ as follows:

$$ \begin{aligned} L_{2}(t)={}&T_{1}\biggl( \frac{T}{T_{1}}-\ln \frac{T}{T_{1}}-1\biggr)+T_{s}+ \frac{1}{N _{s}}V_{s}+T_{r1}\biggl(\frac{T_{r}}{T_{r1}}-\ln \frac{T_{r}}{T_{r1}}-1\biggr) \\ & {} +\frac{1}{N _{r}}V_{r1}\biggl(\frac{V_{r}}{V_{r1}}-\ln \frac{V_{r}}{V_{r1}}-1\biggr). \end{aligned} $$

We have

$$\begin{aligned} \frac{\mathrm {d}L_{2}(t)}{\mathrm {d}t} ={}&\biggl(1-\frac{T_{1}}{T}\biggr) \biggl(\lambda -dT- \frac{k _{s}V_{s}T}{1+\omega _{1}V_{s}}-\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}\biggr)+\biggl((1-u)\frac{k _{s}V_{s}T}{1+\omega _{1}V_{s}}-\delta T_{s}\biggr) \\ & {} +\frac{1}{N_{s}}(N_{s}\delta T_{s}-cV_{s})+ \biggl(1-\frac{T_{r1}}{T_{r}}\biggr) \biggl(u\frac{k _{s}V_{s}T}{1+\omega _{1}V_{s}}+\frac{k_{r}V_{r}T}{1+\alpha _{1}V_{r}}- \delta T_{r}\biggr) \\ & {} +\frac{1}{N_{r}}\biggl(1-\frac{V_{r1}}{V_{r}}\biggr) (N_{r} \delta T_{r}-cV_{r}) \\ ={}&dT_{1}\biggl(2-\frac{T}{T_{1}}-\frac{T_{1}}{T}\biggr)+ \frac{k_{r}V_{r1}T_{1}}{1+ \alpha _{1}V_{r1}}\biggl(4-\frac{T_{1}}{T}-\frac{T_{r}V_{r1}}{T_{r1}V_{r}} - \frac{T _{r1}V_{r}T}{T_{r}V_{r1}T_{1}}\frac{1+\alpha _{1}V_{r1}}{1+\alpha _{1}V _{r}} \\ & {} -\frac{1+\alpha _{1}V_{r}}{1+\alpha _{1}V_{r1}}\biggr) -\frac{k_{r}V_{r1}T _{1}}{1+\alpha _{1}V_{r1}}\frac{\alpha _{1}(V_{r}-V_{r1})^{2}}{(1+\alpha _{1}V_{r1})V_{r1}(1+\alpha _{1}V_{r})} \\ & {} -\frac{ck_{r}(R_{r}-R_{s}-\alpha _{1}\frac{\lambda }{c}N_{r}(R_{s}-1))}{R _{r}(k_{r}+d\alpha _{1})(1+\omega _{1}V_{s})N_{s}}V_{s}-u\frac{k_{s}V _{s}T}{1+\omega _{1}V_{s}}\frac{T_{r1}}{T_{r}}- \frac{c\omega _{1}}{(1+ \omega _{1}V_{s})N_{s}}V_{s}^{2}. \end{aligned}$$

Obviously, when $R_{r}>\max \{1,R_{s}+\alpha _{1}\frac{\lambda }{c}N _{r}(R_{s}-1)\}$, we have $\frac{\mathrm {d}L_{2}(t)}{\mathrm {d}t}\leq 0$ and the set $M=\{(T,T_{s},V_{s},T_{r},V_{r}): \frac{\mathrm {d}L_{2}(t)}{\mathrm {d}t}=0 \}\subseteq \{(T,T_{s},V_{s},T_{r},V_{r}): T=T_{1},T_{s}\geq 0,V_{s} \geq 0,T_{r}=T_{r1},V_{r}=V_{r1}\}$. From $T(t)\equiv T_{1}$, $T_{r}(t)\equiv T_{r1}$ and $V_{r}(t)\equiv V_{r1}$, we have $\lambda -dT_{1}-\frac{k_{s}V_{s}(t)T_{1}}{1+\omega _{1}V_{s}(t)}-\frac{k _{r}V_{r1}T_{1}}{1+\alpha _{1}V_{r1}}\equiv 0$, which implies $V_{s}(t)\equiv 0$. From the third equation of model (4), we get $N_{s}\delta T_{s}(t)-cV_{s}(t)\equiv 0$, which implies $T_{s}(t) \equiv 0$. Hence, $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t))\equiv E _{r}$. Thus, LaSalle’s invariance principle implies that $E_{r}$ is globally asymptotically stable. This completes the proof. □

Remark 2

In Theorem 4, we only obtained the global asymptotic stability of $E_{r}$ when $R_{r}>\max \{1,R_{s}+\alpha _{1}\frac{\lambda }{c}N_{r}(R _{s}-1)\}$. Therefore, combining conclusion (a) of Theorem 4 an interesting open problem is whether we can establish the global asymptotic stability of $E_{r}$ when $R_{r}>\max \{1,(1-u)R_{s}+((1-u)R _{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}\}$.

Remark 3

It is regretful that we here do not establish the corresponding criteria on the local and global stability for positive equilibrium $E_{c}$ of model (2). The reasons are that the analysis of the characteristic equation of $J(E_{c})$ is very complex, and the construction of a suitable Lyapunov function is also very difficult. However, in the next section we can establish the uniform persistence of model (2) when positive equilibrium $E_{c}$ exists.

4 Uniform persistence

Theorem 5

If $(1-u)R_{s}>1\geq R_{r}$or $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{ \lambda }{c}N_{r}>R_{r}>1$, then model (2) is uniformly persistent. That is, there exists a positive constantδsuch that, for any positive solution $(T(t),T_{s}(t),V_{s}(t),T_{r}(t),V_{r}(t))$of model (2),

$$\begin{aligned}& \liminf_{t\rightarrow \infty }T(t)\geq \delta , \quad\quad \liminf _{t\rightarrow \infty }T_{s}(t)\geq \delta , \quad\quad \liminf _{t\rightarrow \infty }V_{s}(t)\geq \delta , \\& \liminf_{t\rightarrow \infty }T_{r}(t)\geq \delta , \quad\quad \liminf_{t\rightarrow \infty }V_{r}(t)\geq \delta . \end{aligned}$$

Proof

For any $x_{0}=(T_{0},T_{s0},V_{s0},T_{r0},V_{r0})\in R^{5}_{+}$, let $u(t,x_{0})=(T(t,x_{0}), T_{s}(t,x_{0}), V_{s}(t,x_{0}), T_{r}(t,x_{0}), V_{r}(t,x_{0}))$ be the solution of model (2) with the initial condition $u(0,x_{0})=x_{0}$. From the proof of Theorem 1, we have $\limsup_{t\rightarrow \infty }u(t,x_{0})\leq \frac{\lambda }{n}$, where $n=\min \{d,\frac{\delta }{2},c\}$. Hence, for any constant $\epsilon >0$, there is $T_{0}>0$, when $t\geq T_{0}$ we get $u(t,x_{0})<\frac{\lambda }{n}+\epsilon$. Then, from the first equation of model (2), we have

$$ \frac{\mathrm {d}T(t,x_{0})}{\mathrm {d}t}\geq \lambda -dT-k_{s}V_{s}T-k_{r}V _{r}T\geq \lambda -\biggl(d+(k_{s}+k_{r}) \biggl(\frac{\lambda }{n}+\epsilon \biggr)\biggr)T(t,x _{0}). $$

From the comparison theorem and the arbitrariness of ϵ, we have

$$ \liminf_{t\rightarrow \infty }T(t,x_{0})\geq \frac{\lambda }{d+(k_{s}+k _{r})\frac{\lambda }{n}}. $$

This shows that $T(t,x_{0})$ is uniformly persistent.

Define

$$ X=\bigl\{ x=(T,T_{s},V_{s},T_{r},V_{r}) \in R^{5}_{+}: T\geq 0, T_{s}>0, V _{s}>0, T_{r}>0, V_{r}>0\bigr\} . $$

The boundary of X is

$$\begin{aligned} \partial X=\bigl\{ (T,T_{s},V_{s},T_{r},V_{r}) \in R^{5}_{+}: T\geq 0, T_{s}=0 \text{ or } V_{s}=0 \text{ or } T_{r}=0 \text{ or }V_{r}=0 \bigr\} . \end{aligned}$$

Denote

$$ M_{\partial }=\bigl\{ x_{0}\in R^{5}_{+}: u(t,x_{0})\in \partial X, \forall t\geq 0\bigr\} . $$

Let $\omega (x_{0})$ be the ω-limit set of solution $u(t,x_{0})$. Then we consider the following two cases.

Case (1): $(1-u)R_{s}>1\geq R_{r}$. From Theorem 2, model (2) has only two equilibria $E_{0}$ and $E_{c}$. Let $M_{0}=\{E_{0}\}$. It is clear that $M_{0}\subset \bigcup_{x_{0}\in M_{\partial }}\omega (x _{0})$. For any $x_{0}\in M_{\partial }$, let $x_{0}=(T_{0},T_{s0},V _{s0},T_{r0},V_{r0})$. Due to $u(t,x_{0})\in \partial X$ for all $t\geq 0$, we have $T_{s}(t,x_{0})\equiv 0$ or $V_{s}(t,x_{0})\equiv 0$ or $T_{r}(t,x_{0})\equiv 0$ or $V_{r}(t,x_{0})\equiv 0$. If $T_{s}(t,x_{0})\equiv 0$, then from the second equation of model (2), we have $V_{s}(t,x_{0})\equiv 0$. Thus, model (2) degenerates into the following form:

$$ \textstyle\begin{cases} \frac{\mathrm {d}T(t,x_{0})}{\mathrm {d}t}= \lambda -dT(t,x_{0})-\frac{k_{r}V _{r}(t,x_{0})T(t,x_{0})}{1+\alpha _{1}V_{r}(t,x_{0})}, \\ \frac{\mathrm {d}T_{r}(t,x_{0})}{\mathrm {d}t}= \frac{k_{r}V_{r}(t,x_{0})T(t,x _{0})}{1+\alpha _{1}V_{r}(t,x_{0})}-\delta T_{r}(t,x_{0}), \\ \frac{\mathrm {d}V_{r}(t,x_{0})}{\mathrm {d}t}= N_{r}\delta T_{r}(t,x_{0})-cV _{r}(t,x_{0}). \end{cases} $$

(12)

If $T_{r0}+V_{r0}=0$, then from system (12) we can obtain $T_{r}(t,x _{0})\equiv V_{r}(t,x_{0})\equiv 0$. Thus, model (2) can further degenerate into

$$ \frac{\mathrm {d}T(t,x_{0})}{\mathrm {d}t}=\lambda -dT(t,x_{0}). $$

It follows that $\lim_{t\rightarrow \infty }T(t,x_{0})=\frac{\lambda }{d}=T_{0}$. This shows that $\omega (x_{0})=E_{0}\subset M_{0}$.

If $T_{r0}+V_{r0}>0$, without loss of generality, we assume $T_{r0}>0$ and $V_{r0}\geq 0$. From the second equation of system (12), we can obtain $T_{r}(t,x_{0})\geq T_{r0}e^{-\delta t}>0$ for all $t\geq 0$, and then, from the third equation of (12), we further obtain $V_{r}(t,x_{0})>V_{r0}e^{-ct}\geq 0$ for all $t>0$. Choose a Lyapunov function as follows:

$$ U_{0}(t)=T_{0}\biggl(\frac{T}{T_{0}}-\ln \frac{T}{T_{0}}-1\biggr)+T_{r}+\frac{1}{N _{r}}V_{r}. $$

We obtain

$$ \frac{\mathrm {d}U_{0}(t)}{\mathrm {d}t}=dT_{0}\biggl(2-\frac{T_{0}}{T}- \frac{T}{T _{0}}\biggr)+\frac{c}{N_{r}}(R_{r}-1)V_{r}- \frac{c\alpha _{1}V_{r}^{2}}{(1+ \alpha _{1}V_{r})N_{r}}\leq 0 $$

and $\{(T,T_{r},V_{r}): \frac{\mathrm {d}U_{0}(t)}{\mathrm {d}t}=0\}\subset \{(T,T _{r},V_{r}): T=T_{0}\}$. If $T(t,x_{0})\equiv T_{0}$, then from the first equation of system (12), we have $V_{r}(t,x_{0})\equiv 0$; further, from the third equation of system (12), we have $T_{r}(t,x _{0})\equiv 0$. Thus, LaSalle’s invariance principle [17] implies that $(T(t,x_{0}),T_{r}(t,x_{0}),V_{r}(t,x_{0}))\rightarrow (T_{0},0,0)$ when $t\rightarrow \infty$. This shows that $\omega (x_{0})=E_{0}\subset M_{0}$.

If $V_{s}(t,x_{0})\equiv 0$, from the third equation of model (2), we have $T_{s}(t,x_{0})\equiv 0$. Similar to the above argument, we also get $\omega (x_{0})=E_{0}\subset M_{0}$.

If $T_{r}(t,x_{0})\equiv 0$, from the fourth equation of model (2), we have $V_{s}(t,x_{0})\equiv 0$ and $V_{r}(t,x_{0})\equiv 0$. Then, from the third equation of model (2), we have $T_{s}(t,x_{0})\equiv 0$. Thus, model (2) degenerates into

$$ \frac{\mathrm {d}T(t,x_{0})}{\mathrm {d}t}=\lambda -dT(t,x_{0}). $$

It follows that $\lim_{t\rightarrow \infty }T(t,x_{0})=T_{0}$. This shows that $\omega (x_{0})=E_{0}\subset M_{0}$.

If $V_{r}(t,x_{0})\equiv 0$, from the fifth equation of model (2), we get $T_{r}(t,x_{0})\equiv 0$. Similar to the above argument, we also get $\omega (x_{0})=E_{0}\subset M_{0}$.

Finally, we have $M_{0}=\bigcup_{x_{0}\in M_{\partial }}\omega (x_{0})$. Furthermore, it is clear that $M_{0}$ is isolated invariant and non-cycle in ∂X.

Now, we prove that $W^{s}(E_{0})\cap X=\emptyset$, where $W^{s}(E _{0})$ is the stable set of $E_{0}$. Suppose that there is an $x_{0}\in X$ such that $\lim_{t\rightarrow \infty }u(t,x_{0})=E_{0}$, then we have $\lim_{t\rightarrow \infty }T(t,x_{0})=T_{0}$. Hence, for any constant $\epsilon >0$, there is $T^{*}>0$ such that $T(t,x_{0}) \geq T_{0}-\epsilon$ and $V_{s}(t,x_{0})<\epsilon$ for any $t\geq T^{*}$. Define the function

$$ U_{1}(t)=T_{s}(t,x_{0})+\frac{1}{N_{s}}V_{s}(t,x_{0}). $$

We have $\lim_{t\rightarrow \infty }U_{1}(t,x_{0})=0$. When $t\geq T^{*}$, we have

$$\begin{aligned} \frac{\mathrm {d}U_{1}(t)}{\mathrm {d}t}={}&(1-u)\frac{k_{s}V_{s}(t,x_{0})T(t,x _{0})}{1+\omega _{1}V_{s}(t,x_{0})}-\delta T_{s}(t,x_{0})+ \frac{1}{N _{s}}\bigl(N_{s}\delta T_{s}(t,x_{0})-cV_{s}(t,x_{0}) \bigr) \\ \geq{}&\biggl((1-u)\frac{k_{s}(T_{0}-\epsilon )}{1+\omega _{1}\epsilon }-\frac{c}{N _{s}}\biggr)V_{s}(t,x_{0}). \end{aligned}$$

Due to $(1-u)R_{s}>1$, we choose enough small $\epsilon >0$ such that $(1-u)\frac{k_{s}(T_{0}-\epsilon )}{1+\omega _{1}\epsilon }-\frac{c}{N _{s}}>0$. Thus, $U_{1}(t)$ is increasing for $t\geq T^{*}$. Hence, we know that $U_{1}(t)$ does not tend to zero as $t\to \infty$, which leads to a contradiction. This shows that $W^{s}(E_{0})\cap X=\emptyset$. According to the theory of persistence in dynamical systems (see [18]), there is a constant $\delta >0$ such that, for any $x_{0} \in X$, one has

$$\begin{aligned}& \liminf_{t\rightarrow \infty }T_{s}(t,x_{0})\geq \delta , \quad\quad \liminf_{t\rightarrow \infty }V_{s}(t,x_{0}) \geq \delta , \quad\quad \liminf_{t\rightarrow \infty }T_{r}(t,x_{0}) \geq \delta , \\& \liminf_{t\rightarrow \infty }V_{r}(t,x_{0}) \geq \delta . \end{aligned}$$

This shows that model (2) is uniformly persistent.

Case (2): $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}>R_{r}>1$. From Theorem 2, model (2) has three equilibria $E_{0}$, $E_{r}$, and $E_{c}$. Denote $M_{0}=\{E_{0},E_{r}\}$. It is clear that $M_{0}\subset \bigcup_{x_{0}\in M_{\partial }}\omega (x_{0})$. For any $x_{0}\in M_{\partial }$, let $x_{0}=(T_{0},T_{s0},V_{s0},T _{r0},V_{r0})$. Due to $u(t,x_{0})\in \partial X$ for all $t\geq 0$, we have $T_{s}(t,x_{0})\equiv 0$ or $V_{s}(t,x_{0})\equiv 0$ or $T_{r}(t,x_{0})\equiv 0$ or $V_{r}(t,x_{0})\equiv 0$. If $T_{s}(t,x _{0})\equiv 0$, then, similar to the above argument, model (2) degenerates into system (12).

If $T_{r0}+V_{r0}=0$, from a similar argument as in case (1), we can obtain $\omega (x_{0})=E_{0}\subset M_{0}$.

If $T_{r0}+V_{r0}>0$, then we also can obtain $T_{r}(t,x_{0})>0$ and $V_{r}(t,x_{0})>0$ for all $t>0$. Choose the Lyapunov function

$$ U_{2}(t)=T_{1}\biggl(\frac{T}{T_{1}}-\ln \frac{T}{T_{1}}-1\biggr)+T_{r1}\biggl(\frac{T _{r}}{T_{r1}}-\ln \frac{T_{r}}{T_{r1}}-1\biggr)+\frac{1}{N_{r}}V_{r1}\biggl( \frac{V _{r}}{V_{r1}}-\ln \frac{V_{r}}{V_{r1}}-1\biggr). $$

Then we have

$$\begin{aligned} \frac{\mathrm {d}U_{2}(t)}{\mathrm {d}t}={}&dT_{1}\biggl(2-\frac{T_{1}}{T}- \frac{T}{T _{1}}\biggr)+\frac{k_{r}V_{r1}T_{1}}{1+\alpha _{1}V_{r1}}\biggl(4-\frac{T_{1}}{T} - \frac{T _{r}V_{r1}}{T_{r1}V_{r}}-\frac{T_{r1}V_{r}T}{T_{r}V_{r1}T_{1}}\frac{1+ \alpha _{1}V_{r1}}{1+\alpha _{1}V_{r}} \\ & {} -\frac{1+\alpha _{1}V_{r}}{1+\alpha _{1}V_{r1}}\biggr) -\frac{k_{r}V_{r1}T _{1}}{1+\alpha _{1}V_{r1}}\frac{\alpha _{1}(V_{r}-V_{r1})^{2}}{(1+\alpha _{1}V_{r1})V_{r1}(1+\alpha _{1}V_{r})}\leq 0 \end{aligned}$$

and the set $\{(T,T_{r},V_{r}): \frac{\mathrm {d}U_{2}(t)}{\mathrm {d}t}=0\}=\{(T _{1},T_{r1},V_{r1})\}$. Hence, LaSalle’s invariance principle [17] implies that $(T(t,x_{0}),T_{r}(t,x_{0}),V_{r}(t,x_{0}))\rightarrow (T _{1},T_{r1},V_{r1})$ as $t\rightarrow \infty$. This shows that $\omega (x_{0})=E_{r}\subset M_{0}$.

If $V_{s}(t,x_{0})\equiv 0$ or $T_{r}(t,x_{0})\equiv 0$ or $V_{r}(t,x _{0})\equiv 0$, then, following a similar argument as in case (1), we can also obtain $\omega (x_{0})=E_{0}$ or $\omega (x_{0})=E_{r}$, and hence $\omega (x_{0})\subset M_{0}$.

Finally, we have $M_{0}=\bigcup_{x_{0}\in M_{\partial }}\omega (x_{0})$. Furthermore, it is clear that $E_{0}$ and $E_{r}$ are isolated invariant and $M_{0}$ is non-cycle in ∂X.

Now, we prove that $W^{s}(E_{0})\cap X=\emptyset$ and $W^{s}(E_{r}) \cap X=\emptyset$. Similar to the above argument in case (1) we can get $W^{s}(E_{0})\cap X=\emptyset$. Suppose that there is $x_{0}\in X$ such that $\lim_{t\rightarrow \infty }u(t,x_{0})=E_{r}$, then we have $\lim_{t\rightarrow \infty }T(t,x_{0})=T_{1}$. Hence, for any constant $\epsilon >0$, there is $T^{*}>0$ such that $T(t,x_{0})\geq T_{1}- \epsilon$ and $V_{s}(t,x_{0})<\epsilon$ for any $t\geq T^{*}$. Define the function

$$ U_{3}(t)=T_{s}(t,x_{0})+\frac{1}{N_{s}}V_{s}(t,x_{0}). $$

We have $\lim_{t\rightarrow \infty }U_{3}(t,x_{0})=0$. When $t\geq T^{*}$, we have

$$\begin{aligned} \frac{\mathrm {d}U_{3}(t)}{\mathrm {d}t}={}&(1-u)\frac{k_{s}V_{s}(t,x_{0})T(t,x _{0})}{1+\omega _{1}V_{s}(t,x_{0})}-\delta T_{s}(t,x_{0})+ \frac{1}{N _{s}}\bigl(N_{s}\delta T_{s}(t,x_{0})-cV_{s}(t,x_{0}) \bigr) \\ \geq{}&\biggl((1-u)\frac{k_{s}(T_{1}-\epsilon )}{1+\omega _{1}\epsilon }-\frac{c}{N _{s}}\biggr)V_{s}(t,x_{0}). \end{aligned}$$

Due to $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}>R _{r}>1$, we choose enough small $\epsilon >0$ such that $(1-u)\frac{k _{s}(T_{1}-\epsilon )}{1+\omega _{1}\epsilon }-\frac{c}{N_{s}}>0$. Then $U_{3}(t)$ is increasing for $t\geq T^{*}$. Thus, we know that $U_{3}(t)$ does not tend to zero, which leads to a contradiction. Hence, $W^{s}(E_{r})\cap X=\emptyset$. According to the theory of persistence in dynamical systems (see [18]), there is a constant $\delta >0$ such that, for any $x_{0}\in X$, one has

$$\begin{aligned}& \liminf_{t\rightarrow \infty }T_{s}(t,x_{0})\geq \delta , \quad\quad \liminf_{t\rightarrow \infty }V_{s}(t,x_{0}) \geq \delta , \quad\quad \liminf_{t\rightarrow \infty }T_{r}(t,x_{0}) \geq \delta , \\& \liminf_{t\rightarrow \infty }V_{r}(t,x_{0}) \geq \delta . \end{aligned}$$

This shows that model (2) is also uniformly persistent. This completes the proof. □

Remark 4

An interesting open problem is whether the positive equilibrium $E_{c}$ is also globally asymptotically stable when the conditions in Theorem 5 are satisfied.

5 Numerical examples

In this section, we provide the numerical examples to illustrate the global asymptotic stability of the equilibria for model (2), and Examples 1 and 2 can further verify Remarks 1 and 2, respectively.

Example 1

In model (2), we take the parameters $\lambda =10^{5}$, $d=0.1$, $k_{s}=1.0\times 10^{-8}$, $k_{r}=1.0 \times 10^{-8}$, $u=0.6$, $\delta =1$, $N_{s}=2000$, $N_{r}=900$, $c=11$, $\omega _{1}=10^{-5}$, and $\alpha _{1}=10^{-4}$. By calculating, we have $R_{s}\approx 1.8182>1$, $(1-u)R_{s}\approx 0.7273<1$, and $R_{r}\approx 0.8182<1$. Furthermore, we also have the infection-free equilibrium $E_{0}=(10^{6}, 0, 0, 0, 0)$. We give three different groups of initial values in Table 2.

Table 2 Initial values of model (2)

The numerical simulations given in Fig. 2 illustrate that equilibrium $E_{0}$ may be globally asymptotically stable. This shows that the open problem given in Remark 1 may be right.

Figure 2

(a) Dynamical behaviors of uninfected target cells; (b) Dynamical behaviors of infected cells infected by wild-type virus; (c) Dynamical behaviors of drug-resistant virus; (d) Dynamical behaviors of infected cells infected by wild-type virus; (e) Dynamical behaviors of drug-resistant virus

Example 2

In model (2), we take the parameters $\lambda =10^{5}$, $d=0.005$, $k_{s}=1.2\times 10^{-9}$, $k_{r}=1.0 \times 10^{-8}$, $u=0.6$, $\delta =1$, $N_{s}=2000$, $N_{r}=250$, $c=10$, $\omega _{1}=10^{-3}$, and $\alpha _{1}=10^{-7}$. By calculating, we have $R_{r}=5$, $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}=2.15$ and $R_{s}+(R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}=5.75$. Hence, $\max \{1,(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r} \}< R_{r}< R_{s}+(R_{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}$. Furthermore, we also have the boundary equilibrium $E_{r}=(4.76\times 10^{6}, 0, 0, 7.62\times 10^{4}, 1.90\times 10^{6})$. We give three different groups of initial values in Table 3.

Table 3 Initial values of model (2)

The numerical simulations given in Fig. 3 illustrate that equilibrium $E_{r}$ may be globally asymptotically stable. This shows that the open problem given in Remark 2 may be right.

Figure 3

(a) Dynamical behaviors of uninfected target cells; (b) Dynamical behaviors of infected cells infected by wild-type virus; (c) Dynamical behaviors of drug-resistant virus; (d) Dynamical behaviors of infected cells infected by wild-type virus; (e) Dynamical behaviors of drug-resistant virus

Example 3

In model (2), we take the parameters $\lambda =10^{5}$, $d=0.005$, $k_{s}=1.2\times 10^{-9}$, $k_{r}=4.0 \times 10^{-10}$, $u=3\times 10^{-5}$, $\delta =1$, $N_{s}=2000$, $N_{r}=1000$, $c=10$, $\omega _{1}=10^{-8}$, and $\alpha _{1}=10^{-2}$. By calculating, we have $(1-u)R_{s}\approx 4.80$, $R_{r}=0.8$, and $(1-u)R_{s}>1\geq R_{r}$, and model (2) has a coexistence equilibrium $E_{c}\approx (4.80\times 10^{6}, 1.6\times 10^{6}, 1.520\times 10^{7}, 2.416, 241.577)$. We give three different groups of initial values in Table 4.

Table 4 Initial values of model (2)

The numerical simulations given in Fig. 4 illustrate that equilibrium $E_{c}$ may be globally asymptotically stable. This shows that the open problem given in Remark 4 may be right.

Figure 4

(a) Dynamical behaviors of uninfected target cells; (b) Dynamical behaviors of infected cells infected by wild-type virus; (c) Dynamical behaviors of drug-resistant virus; (d) Dynamical behaviors of infected cells infected by wild-type virus; (e) Dynamical behaviors of drug-resistant virus

Example 4

In model (2), we take the parameters $\lambda =10^{5}$, $d=0.005$, $k_{s}=1.2\times 10^{-8}$, $k_{r}=1.0 \times 10^{-8}$, $u=3\times 10^{-5}$, $\delta =1$, $N_{s}=2000$, $N_{r}=1000$, $c=10$, $\omega _{1}=10^{-8}$, and $\alpha _{1}=10^{-8}$. By calculating, we have $(1-u)R_{s}+((1-u)R_{s}-1)\alpha _{1}\frac{ \lambda }{c}N_{r}\approx 52.798$, $R_{r}=20$, and $(1-u)R_{s}+((1-u)R _{s}-1)\alpha _{1}\frac{\lambda }{c}N_{r}>R_{r}>1$, and model (2) has a coexistence equilibrium $E_{c}\approx (4.979\times 10^{5}, 9.750 \times 10^{4}, 1.950\times 10^{7}, 5.826, 582.635)$. We give three different groups of initial values in Table 5.

Table 5 Initial values of model (2)

The numerical simulations given in Fig. 5 illustrate that equilibrium $E_{c}$ may be globally asymptotically stable. This shows that the open problem given in Remark 4 may be right.

Figure 5

(a) Dynamical behaviors of uninfected target cells; (b) Dynamical behaviors of infected cells infected by wild-type virus; (c) Dynamical behaviors of drug-resistant virus; (d) Dynamical behaviors of infected cells infected by wild-type virus; (e) Dynamical behaviors of drug-resistant virus

6 Conclusion

In this paper, we study the global dynamics for a two-strain HIV infection model with saturated incidence which includes wild-type (i.e. drug sensitive) and drug-resistant strains. The wild-type strain can mutate and become drug-resistant during the process of reverse transcription. The main results are presented in Theorems 1–5. Concretely, the nonnegativity and boundedness of solutions are obtained in Theorem 1; the existence of wild-type strain-free equilibrium and coexistence equilibrium is also obtained in Theorem 2; Theorems 3 and 4 show the sufficient and necessary threshold conditions for the local and global asymptotic stability of infection-free and wild-type strain-free equilibria; and the uniform persistence of HIV infection model is established in Theorem 5.

There are some problems waiting for further investigation. Firstly, Remarks 1 and 2 consider an interesting open problem is whether we can establish the global asymptotic stability of equilibria under the appropriate conditions. And it is meaningful to study more complex models (see [19]), for example, a two-strain infection model with delayed saturation incidence (see [20]) and general nonlinear incidence (see [15, 21]), etc. Furthermore, it is more reasonable to consider the dynamical behaviors of a virus infection model with spatial diffusion and age-dependence (see [22–25]). We will leave these problems for future investigation.