# Dynamic of a TB-HIV Coinfection Epidemic Model with Latent Age

Authors: Xiaoyan Wang; Junyuan Yang; Fengqin Zhang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

A coepidemic arises when the spread of one infectious disease stimulates the spread of
another infectious disease. Recently, this has happened with human immunodeficiency virus (HIV)
and tuberculosis (TB). The density of individuals infected with latent tuberculosis is structured by age
since latency. The host population is divided into five subclasses of susceptibles, latent TB, active TB
(without HIV), HIV infectives (without TB), and coinfection class (infected by both TB and HIV). The
model exhibits three boundary equilibria, namely, disease free equilibrium, TB dominated equilibrium, and
HIV dominated equilibrium. We discuss the local or global stabilities of boundary equilibria. We
prove the persistence of our model. Our simple model of two synergistic infectious disease epidemics
illustrates the importance of including the effects of each disease on the transmission and progression
of the other disease. We simulate the dynamic behaviors of our model and give medicine explanations.


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## Body

## 1. Introduction

Coepidemics—the related spread of two or more infectious diseases—have afflicted mankind for centuries. Worldwide, there were an estimated 1.37 million coinfected HIV and TB patients globally in 2007. Around 80 percent of patients live in sub-Saharan Africa. 456000 people died of HIV-associated TB in 2007. HIV/AIDS and tuberculosis (TB) are commonly called the “deadly duo” and referred to as HIV/TB, despite biological differences. HIV is a retrovirus that is transmitted primarily by homosexual and heterosexual contact, needle sharing, and from mother to child. The disease eventually progresses to AIDS as the immune system weakens. HIV can be treated with highly active antiheroical therapy (HAART), but there is presently no cure [1]. Virtually all HIV-infected individuals can transmit the virus to others, and an infected individual's chance of spreading the virus generally increases as the disease progresses and damages the immune system [2]. Tuberculosis is caused by mycobacterium tuberculosis bacteria and is spread through the air. Some TB infections are “latent,” meaning that a person has the TB-causing bacteria but it is dormant. A person with latent TB is not sick and is not infectious. However, latent TB can progress to “active” TB. “Active” TB infection means that the TB bacteria are multiplying and spreading in the body. A person with active TB in their lungs or throat can transmit the bacteria to others. Symptoms of active TB include a cough that lasts several weeks, weight loss, loss of appetite, fever, night sweats, and coughing up blood.HIV weakens the immune system and so people are more susceptible to catching TB if they are exposed. At least one-third of people living with HIV worldwide are infected with TB and are 20–30 times more likely to develop TB than those without HIV. TB bacteria accelerate the progression of HIV to AIDS. Persons coinfected with TB and HIV may spread the disease not only to the other HIV-infected persons, but also to members of the general population who do not participate in any of the high risk behaviors associated with HIV. People living with HIV and displaying early diagnosis need treatment in time. If TB is present, they should receive TB preventive treatment (IPT). The treatments are not expensive. Therefore, it is essential that adequate attention must be paid to study the transmission dynamics of HIV-TB coinfection in the population. Current treatment for HIV is known as highly active antiretroviral therapy (HAART) [3].Some authors have developed simulation models to investigate HIV-TB co-epidemic dynamics [4–7]. West and Thompson [8] developed models which reflect the transmission dynamics of both TB and HIV and discussed the magnitude and duration of the effect that the HIV epidemic may have on TB. Naresh and Tripathi [9] presented a model for HIV-TB coinfection with constant recruitment of susceptibles and found that TB will be eradicated from the population if more than 90 percent TB infectives are recovered due to effective treatment. Naresh et al. [10] studied the effect of tuberculosis on the spread of HIV infection in a logistically growing human population. They found that as the number of TB infectives decreases due to recovery, the number of HIV infectives also decreases and endemic equilibrium tends to TB free equilibrium. Long et al. [11] discussed two synergistic infectious disease epidemics illustrating the importance of including the effects of each disease on the transmission and progression of the other disease.However, these models may exclude coinfection, may greatly simplify infection dynamics, and may include few disease states (e.g., the important distinction between latent and active TB was absent). They assume that HIV treatment is unavailable. Bifurcation will appear in some models [12, 13]. We perform the additional latent age, analyze the coexistence equilibria, and extend the model to include disease recovery in this paper. At the same time, we obtain the persistence of the system and global stabilities of the equilibria under some conditions.This paper is organized as follows. In Section2, we introduce the TB-HIV coinfection model. In Section 3, we introduce the reproduction numbers of TB and HIV R1,R2 and discuss the existence of the equilibria. The values of the disease-free equilibrium, the two boundary equilibria, and the coexistence equilibrium are given explicitly. Section 4 focuses on local and global stabilities of the equilibria. In Section 5, we discuss the persistence of the system in suitable period. In Section 6, we simulate and illustrate our results. We give some biological explanations in this section. In Section 7, we conclude our results and discuss the defect of our model.


## 2. The Model Formulation

Two diseases mentioned in the introduction are spreading in a population of total sizeN(t). We classify the total population into four classes: the susceptible S(t); the individuals infected by the tuberculosis, e(a,t) and I1(t) denotes the latent TB class which has no infectious ability and active TB class who can infect the susceptible class, respectively; the individuals infected by HIV I2(t); the individuals coinfected by tuberculosis and HIV J(t). The individuals infected with active TB infect the susceptible and then develop into the latent individuals at β1. The individuals infected by HIV infect the susceptible and become the individuals infected by HIV at β2. An individual already infected with TB can be coinfected with HIV at δ and thus become jointly infected individuals J(t).Figure1 presents a schematic flow diagram of the mathematical model as follows:
(1)dSdt=Λ-μS-β1SI1-β2SI2+∫0∞‍α0(a)e(a,t)da+α1I1+α2I2,∂e(a,t)∂a+∂e(a,t)∂t=-(μ+γ(a)+α0(a))e(a,t),e(0,t)=β1SI1,dI1dt=∫0∞‍γ(a)e(a,t)da-μI1-δI1I2-α1I1,dI2dt=β2SI2-(μ+α2)I2,dJdt=δI1I2-(μ+ν)J,
where μ is natural death rate and Λ is birth rate. γ(a) is rate of endogenous reactivation of latent TB. We assume that the individuals separately infected by TB and HIV are not lethal but the coinfection can lead to the extra death at ν. Specifically, we assume that jointly infected individuals do not recover. Individuals infected with latent TB, active TB, or HIV alone may be potentially treated at rates α0(a),α1, and α2, respectively.Figure 1
The flow diagram of the model (1). Two yellow rectangulars denote the individuals infected by latent tuberculosis and active tuberculosis. The red rectangular denotes the individuals infected by HIV. The orange rectangular denotes the individuals coinfected by TB and HIV.Assumption 1.
Suppose that(a)
Λ,μ,δ,α1,α2,β1,β2∈(0,+∞);(b)
ν∈[0,+∞);(c)
γ(a),α0(a)∈CB,U([0,+∞),R)∩C+([0,+∞),R) and for each a≥0, where CBU(R) denotes the space of bounded and uniformly continuous map from [0,+∞) into R.We assume (1) with the initial conditions:
(2)S(0)=S0≥0,e(a,0)=e0(a)∈L+1(0,+∞),I1(0)=I10≥0,I2(0)=I20≥0,J(0)=J0≥0.Set(3)X=R×Y×R2,X+=R+×Y+×R2+,X0=R×Y0×R2,X0+=X0∩X+.
with
(4)Y=R×L1(0,+∞),Y+=[0,+∞)×L+1(0,+∞),Y0={0}×L1(0,+∞).LetD(𝒜)=R×Z×R2, with Z={0R}×W1,1(0,∞), for each x=(S,0R,e,I1,I2)∈X0. Define 𝒜:D(𝒜)⊂X0⊂X→X and ℱ:X0→X by
(5)𝒜x=(-μS-e(0)-de(·)da-(μ+α0(·)+γ(·))e(·)-μI1-α1I1-μI2-α2I2-(μ+ν)J)ifx∈D(𝒜),(6)ℱ(x)=(μ-β1SI1-β2SI2+∫0∞‍α0(a)e(a)da+α1I1+α2I2β1SI10L1∫0∞‍γ(a)e(a)da-δI1I2β2SI2δI1I2).Rewrite problem (1) as an abstract Cauchy problem
(7)dux(t)dt=𝒜ux(t)+ℱ(ux(t)),t≥0,ux(0)=x∈X0+.It is well known that𝒜 is a Hille-Yosida operator. More precisely, we have (-μ,+∞)⊂ρ(𝒜) and for each λ>-μ,
(8)∥(λ-𝒜)-1∥≤1λ+μ.By applying the results in Magal et al. [14–16], we obtain the following proposition.Lemma 1.
There exists a uniquely determined semiflow{U(t)}t≥0 on X0+, such that for each x=(S0,0,e0,I10,I20,J0)T∈X0+, there exists a unique continuous map U∈C([0,+∞),X0+) which is an integrated solution of the Cauchy problem (1), that is to say that
(9)∫0t‍U(s)xds∈D(𝒜),∀t≥0,U(t)x=x+𝒜∫0t‍U(s)xds+∫0t‍ℱ(U(s)x)ds,∀t≥0.
The total population sizeN(t) is the sum of all individuals in all classes
(10)N=S(t)+∫0+∞‍e(a,t)da+I1(t)+I2(t)+J(t).
The total population size satisfies the equationN′(t)=Λ-μN-νJ. We introduce the notation
(11)π(a)=exp(-∫0a‍(α0(τ)+γ(τ))dτ),
To understand the biological meaning of the quantityπ(a) we note that π(a)e-μa is the probability to remain infected with TB time units after infection. In addition, we define the quantity
(12)B=∫0+∞‍α0(a)π(a)e-μada,
which gives the probability of treatment since the individuals can leave TB infectious period via treatment. 							(13)C=∫0+∞‍γ(a)π(a)e-μada,
which gives the probability of progression since the individuals can leave TB infectious period via progression. Since individuals can only leave the latent TB infected class through treatment, progression, or death, the sum of the probabilities of recovery, progression, and death equals one, that is,
(14)∫0∞‍α0(a)π(a)e-μada+∫0∞‍γ(a)π(a)e-μadamm+μ∫0∞‍π(a)e-μada=1.
It immediately follows thatB+C<1.


## 3. Equilibria of the Model with Coinfection

We introduce the reproduction numbers of the two diseases. The reproduction number of TB is(15)R1=Λβ1Cμ(μ+α1),
and the reproduction number of HIV is
(16)R2=Λβ2μ(μ+α2).We note that the coinfection rateδ does not affect the reproduction numbers since coinfection does not lead to additional infections. Setting the derivatives with respect to time to zero we obtain a system of algebraic equations and one ODE for the equilibria of (1). For convenience we consider s,e,i1,i2,j as the equilibria of the model. Therefore equilibria satisfy the following equations:
(17)Λ-β1si1-β2si2-μsΛ+∫0+∞‍α0(a)e(a,t)da+α1i1+α2i2=0,deda=-α0(a)e-μe-γ(a)e,e(0)=β1si1,∫0+∞‍γ(a)e(a)da-μi1-δi1i2-α1i1=0,β2si2-(μ+α2)i2=0,δi1i2-(μ+ν)j=0.The ODE in the system can be solved to result in(18)e(a)=e(0)π(a)e-μa.Substituting fori in the integrals, one obtains
(19)∫0+∞‍α0(a)e(a)da=e(0)∫0+∞‍α0(a)π(a)e-μada=e(0)B,∫0+∞‍γ(a)e(a)da=e(0)∫0+∞‍γ(a)π(a)e-μada=e(0)C.With this notation the system for the equilibria becomes(20)Λ-β1si1-β2si2-μs+e(0)B+α1i1+α2i2=0,deda=-α0(a)e-μe-γ(a)e,e(0)=β1si1,e(0)C-μi1-δi1i2-α1i1=0,β2si-(μ+α2)i2=0,δi1i2-(μ+ν)j=0.This system has three boundary equilibria.(1)
The disease-free equilibriumE0=(Λ/μ,0,0,0,0). The disease-free equilibrium always exists.(2)
The TB dominated equilibrium exists if and only ifR1>1. The steady distribution of infectives in the TB equilibrium is given by
(21)s1=ΛμR1,e=e(0)π(a)e-μa,i1=Λ(1-1/R1)(1-B)μ/C+(1-B-C)α1/C,e(0)=β1s1i1.
Thus, the equilibrium is
(22)E1=(s1,e,i1,0,0).(3)
The HIV dominated equilibrium exists if and only ifR2>1 and is given by
(23)E2=(s2,0,0,i2,0),
where s2=1/R2,i2=0,e2=0,i2=(Λ/μ)(1-1/R2),j2=0.(4)
The coexistence equilibrium exists if and only ifR1>R2, and [1-(1/R2+μ(μ+α1)/Λδ)][(Λβ1/μα1R2)(1-B)-1]>0 and it is given by
(24)E2=(s,e,i1,i2,j),
where s=Λ/μR2, e=e(0)π(a)e-μa, e(0)=β1si1, i1=(Λ/α1)(1-(1/R2+μ(μ+α1)/Λδ))/((Λβ1/μα1R2)(1-B)-1), i2=((μ+δ)/δ)(R1/R2-1), j=δi1i2/(μ+ν).Notice that the values of the two dominance equilibria do not depend on the coinfection. These exact same equilibria are present even ifδ=0.


## 4. Stability of Equilibria

In this section we investigate local and global stabilities of equilibria. In particular, we derive conditions for the stability of the disease-free equilibrium, of the TB dominance equilibrium and of the HIV dominance equilibrium. The stability of equilibria determines whether both diseasess will be eliminated, one of the diseases will be dominated, and both diseases will persist or not.To investigate the stability of the equilibria, we linearize the model (1). In particular, let x(t),y(a,t),z(t),  u(t), and w(t) be the perturbations, respectively, of s,e(a),i1,i2,j. That is, S=x+s,e=y+e,I1=z+i1,I2=u+i2,J=w+j. Thus the perturbations satisfy a linear system. Further we consider the eigenvalue problem for the linearized system. We will denote the eigenvector again with x,y(a),z,u, and w. These satisfy the following linear eigenvalue problem (here s,e,i1,i2, and j are the corresponding equilibria):
(25)λx=-β1sz-β1i1x-β2su-β2i2x-μx+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz+β1i1x,λz=∫0∞‍γ(a)y(a)da-μz-δi2z-δi1u-α1z,λu=β2su+β2i2x-(μ+α2)u,λw=δi2z+δi1u-(μ+ν)w.In the following we discuss local stability of the equilibria through the characteristic equation (25). Since the last equation has no relation with the other equations in (1) and (25), we just discuss the first four equations of them in the following.


### 4.1. Stability of the Disease-Free Equilibrium

For the disease-free equilibrium we havee(0)=i1=i2=j=0, and s=Λ/μ. Thus the system above simplifies to the following system:
(26)λx=-β1sz-β2su-μs+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz,λz=∫0∞‍γ(a)y(a)da-μz-δi2z-δi1u-α1z,λu=β2su-(μ+α2)u.From this system we will establish the following result regarding the local stability of the disease-free equilibriumE0.Theorem 2.
IfR1<1 and R2<1, then the disease-free equilibrium E0 is locally asymptotically stable. If R1>1 or R2>1 then the disease-free equilibrium E0 is unstable.Proof.
To see this, from the second to last equation we haveλu=[β2s-(μ+α2)]u, where either λ=β2s-(μ+α2), or u=0. This eigenvalue λ=(μ+α2)(R2-1)<0, if and only if R2<1. Thus, if R2>1, the disease-free equilibrium E0 is unstable because this eigenvalue is positive. Further, from the second equation we have that the remaining eigenvalues satisfying the equation,
(27)y(a)=y(0)e-(λ+μ)aπ(a)=β1sze-(λ+μ)aπ(a).
Further, from the fourth equation we have that the remaining eigenvalues satisfying the equation, also referred to as the characteristic equation, λz=β1s∫0∞‍γ(a)π(a)e-(λ+μ)adaz-(μ+α1)z. This eigenvalue λ=(μ+α1)(β1s/(μ+α1))∫0∞‍γ(a)π(a)e-(λ+μ)ada-1), or z=0.
We denote the left hand side of the equation above byℱ1(λ)=λ, and ℱ2(λ)=(μ+α1)(R1(λ)-1), where R1(λ)=(β1s/(μ+α1))∫0∞‍γ(a)π(a)e-(λ+μ)ada, and
(28)R1′(λ)<0,limt→+∞R1(λ)=0,limt→-∞R1(λ)=+∞.
Ifℱ1(λ)=ℱ2(λ) has a root with Reλ≥0, then Reℱ1(λ)≥0. But |ℱ2(λ)|≤(μ+α1)(R1-1)<0, when R1<1. This is a contradiction.
Thus, if bothR1<1 and R2<1 all eigenvalues have negative real part and the disease-free equilibrium E0 is locally asymptotically stable. If only R1>1 then if we consider ℱ1(λ)=ℱ2(λ) for λ is real, we see that ℱ2(λ) is a decreasing function of λ approaching zero as λ approaches infinity. Since ℱ2(0)=(μ+α1)(R1-1)>0 and ℱ1(0)=0 that implies that there is a positive eigenvalue λ*>0 and the disease-free equilibrium E0 is unstable. This concludes the proof.In what follows, we show that the diseases vanish ifR1<1,R2<1.Theorem 3.
IfRi<1,i=1,2, then E0 is a global attractor, that is, limt→∞e(a,t)=0, I1→0, I2→0, J→0 as t→∞.Proof.
SinceN′(t)=Λ-μN-νJ≤Λ-μN, then N∞≤Λ/μ. Hence S∞≤Λ/μ. Let ℬ(t)=e(0,t). Integrating this inequality along the characteristic lines we have
(29)e(a,t)={ℬ(t-a)π(a)e-μa,t<a,e0(a-t)π(a)π(a-t)e-μt,t≥a.
Sinceℬ(t)=β1SI1≤β1ΛI1/μ and
(30)I1′≤β1Λμ∫0t‍γ(a)I1(t-a)π(a)e-μada+F1(t)-μI1-α1I1,
where F1(t)=∫t∞‍γ(a)e0(a-t)(π(a)/π(a-t))e-μtda and limt→∞F1(t)=0, from the equation for I1 then we have the following inequality:
(31)I1≤I1(0)e-(μ+α1)t+β1Λμ∫0t‍e-(μ+α1)τ∫0t-τ‍γ(a)I1(t-a)π(a)e-μadadτ+∫0t‍e-(μ+α1)τF1(t-τ)dτ,
whereγ(a) is bounded and the integral of F1(t) goes to zero as t→∞. Consequently, taking a limsup of both sides as t→∞ we obtain
(32)I1∞≤R1I1∞.
SinceR1<1, this inequality can only be satisfied if I1∞→0, as t→∞.
Sinceℬ(t)≤β1(Λ/μ)I1, it is easy to obtain ℬ(t)→0, as t→∞.
From the equationI2 then we have the following inequality:
(33)I2′≤β2ΛμI2-(μ+α2)I2.
Solving the inequality we get(34)I2≤I2(0)e-(μ+α2)t+β2Λμ∫0∞‍I2(t-τ)e-(μ+α2)τdτ.
Consequently, taking a limsup of both sides we obtain(35)I2∞≤R2I2∞.
IfR2<1, then I2(t)→0, as t→∞.
From the fluctuations lemma, we can choose sequencetn such that S(tn)→S∞ and S′(tn)→0 when tn→+∞. It follows from the first equation of (1) that we have S∞≥Λ/μ. Then Λ/μ≤S∞≤S∞≤Λ/μ. Hence S→Λ/μ when t→∞. This completes the theorem.


### 4.2. Stability of the TB Dominated Equilibrium

In this subsection we discuss stabilities of the equilibriumE1 and derive conditions for domination of TB. We show that the equilibrium E1 can lose stability, and dominance of the TB is possible in the form of sustained oscillation. In this case i2=0, j=0, s=Λ/(μR1), i1=Λ(1-1/R1)/((1-B)μ/C+(1-B-C)α1/C), e(a)=e(0)π(a)e-μa.The eigenvalue problem takes the form(36)λx=-β1sz-β1i1x-β2su-μs+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz+β1i1x,λz=∫0∞‍γ(a)y(a)da-μz-δi1u-α1z,λu=β2su-(μ+α2)u.From the last equation we have(37)λ=(μ+α2)(R2R1-1),oru=0.HenceR1>R2 the partial characteristic root of (36) is negative. Substituting
(38)y(a)=y(0)π(a)e-(λ+μ)a,y(0)=β1sz+β1i1x
in the first and fourth equations and cancelling x,z, we arrive at the following characteristic equation:
(39)(λ+μ+α1-β1sC(λ))(λ+μ+β1i1(1-B(λ)))-β1i1C(λ)(α1+β1s(B(λ)-1))=0.Substituting the formula(40)1-B(λ)=C(λ)+(λ+μ)E(λ),
where
(41)E(λ)=∫0∞‍π(a)e-(λ+μ)a
into (39), we obtain the equivalent characteristic equation with (39) as follows:
(42)(λ+μ+α1)(1+β1i1E(λ))+β1i1C(λ)μ+α1=C(λ)C.It is easy to see if (42) has the roots with Reλ>0, the left side mode of (42) is larger than 1, while the right side mode of (42) is less than 1, which leads to a contradiction. Hence the characteristic roots of (42) have negative real parts, then the TB dominated equilibrium E1 is locally asymptotically stable.Therefore we are ready to establish the first result.Theorem 4.
LetR1>1 and R1>R2. Then, the TB dominated equilibrium E1 is locally asymptotically stable.For allx=(S,0R,e,I1,I2,J)∈X0, we define PS:X0→R,PI1,PI2:X0→R and PI12:X0→R by(43)PS(x)=S,PI1(x)=I1,PI2(x)=I2.Set(44)MI12=X0+,MI120={x∈MI12∣PI1x≠0,PI2x≠0},∂MI120=MI12MI120,MI1=∂MI120,MI10={x∈MI1∣PI1x≠0,PI2x=0},∂MI10=MI1MI10,MI2=∂MI10,MI20={x∈MI2∣PI2x≠0,PI1x≠0},∂MI20=MI2MI20,MS=∂MI20(see Figure2).Figure 2
Zoning the area.Lemma 5.
If
R
1
>
1, liminfI1(t)≥ε in MI20.Proof.
Letx=(S0,0R,e,0R,0R)∈MI10. We assume there is a t0>0 such that I1<ε for all t≥t0. From the first equation of (1) in MI10, it is easy to get
(45)dSdt≥Λ-εβ1S-μS,S(0)=S0.
We solve them as follows:(46)S(t)≥Λ(1-e-(μ+β1ε)t)μ≐S*(t).
AndS*(t)→Λ/μ, as t→+∞. Since R1>1, there exists δ>0 and T1>0 which satisfy β1(Λ/μ)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)π(a)e-μadads>1. From the second and third equations of (1), and integrating them from the characteristic line t-a=c, we obtain
(47)e(a,t)={β1S(t-a)I1(t-a)π(a)e-μa,t≥a,e0(a-t)π(a)π(a-t)e-μt,t<a.
From the fourth equation of (1), we get
(48)I1′(t)=β1∫0t‍γ(a)S(t-a)I1(t-a)π(a)e-μada+β1∫t∞‍γ(a)e0(a-t)π(a)π(a-t)e-μtda-(μ+α1)I1.
Solving it, we have(49)I1(t)≥β1Λμ∫0t‍e-(μ+α1)sI1(t)≥×∫0t-s‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
There exist aT1>0, such that
(50)I1(t+T1)≥β1Λμ∫0t+T1‍e-(μ+α1)s≥×∫0t+T1-s‍γ(a)I1(t-s-a)π(a)e-μadads,≥β1Λμ∫0t‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
Thus, fort≥δ, we have
(51)I1(t+T1)≥β1Λμ∫0δ‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
By Assumption1, there exists t1≥0, such that I1(t+T1)≥0, for all t+T1≥t1. Hence, there exists ξ>0, such that I1(t+T1)≥ξ, for all t+T1∈[2t1,2t1+δ]. Set
(52)t2=sup{t+T1≥2t1+δ:I1(l)≥ξ,∀l∈[2t1+δ,t]}.
Assuming thatt2<∞, it follows that
(53)I1(t2)≥β1Λμ∫0δ‍e-(μ+α1)sI1(t2)≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,I1(t2)≥β1Λμ∫0δ‍e-(μ+α1)sI1(t2)≥×∫0T1‍γ(a)π(a)e-μadadsξ.
Thus,I1(t2)>ξ. By the continuity for t→I1(t), it follows that there exists an ε1>0, such that I1(t)≥ξ, for all t∈[t2,t2+ε1] which contradicts the definition of t2. Therefore, I1(t)≥ξ, for all t>2t1. Denote I1∞=liminft→+∞I1(t)≥ξ>0. Using (51), it follows that I1∞≥I1∞β1(Λ/μ)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)π(a)e-μadads, which is impossible. Thus, I1(t)≥ε.Theorem 6.
IfR1>1, system (1) is permanent in MI10. Moreover there exists AI10 a compact subset of MI10 which is a global attractor for {U(t)}t≥0 in MI10.Proof.
Suppose thatx=(S(t),0,e,I1,0,0)∈MI10 be any solution of (1). From the first equation of (1), we have
(54)S′(t)>μ-(β1+μ)S(t).
Consider the comparison equation.(55)u′(t)=μ-(β1+μ)u(t),u(0)=S(0).
Similarly, (55) exists as a positive unique steady state. So we have S(t)≥u*-ε1≐m1, for t large enough, where u*=μ/(μ+β1). From the second and third equations of (1) and Volterra's formulation, we have e(a,t)≥β1m1π(a)e-μa≐m2, for t large enough.


### 4.3. Stability of the HIV Dominated Equilibrium

In this subsection we establish that the equilibriumE2 is locally stable whenever it exists. In this case e(0)=0,i1=0,j=0,s=Λ/μR2, i2=(Λ/μ)(1-1/R2). The linear eigenvalue problem becomes
(56)λx=-β1sz-β2su-β2i2x-μx+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz,λz=∫0+∞‍γ(a)y(a)da-μz-δi2z-α1z,λu=β2su+β2i2x-(μ+α2)u.From the equation fory(a) we have
(57)y(a)=y(0)π(a)e-(λ+μ)a.Substitutingy(a) in the equation for the initial condition y(0) and assuming that y(0)≠0 we obtain the following characteristic equation:
(58)λ+δi2=β1s∫0∞‍γ(a)π(a)e-(λ+μ)ada-(μ+α1)=(μ+α1)(R1(λ)R2-1).Denoting the left hand side of the equation above byℱ1(λ)=λ+δi2, and ℱ2(λ)=(μ+α1)((R1(λ)/R2)-1), and also noting that
(59)R1′(λ)<0,limt→+∞R1(λ)=0,limt→-∞R1(λ)=+∞,it is easy to see thatℱ1(λ) is an increasing function with λ. Note that ℱ1(0)=δI2≥0. But ℱ2(0)≤(μ+α1)((R1/R2)-1)<0, when R1<R2. Therefore the characteristic roots of this equation have only negative real part. Furthermore, for y(0)=y(a)=0, and z=0,
(60)λx=-β2su-β2i2x-μx+α2u,λu=β2su+β2i2x-(μ+α2)u.We expressx from the first equation
(61)x=(-β2s+α2)uλ+μ+β2i2,
and substitute it into the second equation. In addition, assuming that u is nonzero, we cancel it and obtain the following characteristic equation:
(62)λ+μ+α2=β2s+β2i2(-β2s+α2)λ+μ+β2i2,
that is,
(63)(λ+μ)(λ+μ+α2+β2i2-β2s)=0.Noticing thatβ2s=μ+α2, we obtain the following eigenvalues: λ=-μ and λ=-β2i2, which are both negative. Consequently, the equilibrium E2 is locally asymptotically stable.Therefore, we have the following theorem for equilibriumE2.Theorem 7.
LetR2>1. Assume that tuberculosis cannot invade the equilibrium of HIV, that is, R1<R2. Then the equilibrium E2 is locally asymptotically stable.Lemma 8.
IfR2>1, then liminfI2(t)≥ε in MI20.Proof.
Letx=(S0,0R,0L1,0R,I2,0R)∈MI20. We assume there is a t0>0 such that I2<ε for all t≥t0. From the first equation of (1) in MI20, it is easy to get
(64)dSdt≥Λ-εβ2S-μS,S(0)=S0.
We solve them,(65)S(t)≥Λ(1-e-(μ+β2ε)t)μ≐S*(t).
ThenS(t)≥S*(t), and S(t)→Λ/μ, as t→+∞. From the fourth of equation (1), ∃t0>0, we have
(66)I2′(t)≥β2ΛμI2(t)-(μ+α2)I2(t),∀t≥t0.
Solving it, we have(67)I2(t)≥β2Λμ∫0t‍e-(μ+α2)(t-s)I2(t-s)ds.
We do lim inf on both side of (67), and denote I2∞=liminft→+∞I2(t)>0, thus
(68)I2∞≥R2I2∞,
which is impossible. Thus, I2(t)≥ε.Theorem 9.
IfR2>1, system (1) is permanent in MI20. Moreover there exists AI20 a compact subset of MI20 which is a global attractor for {U(t)}t≥0 in MI20.Proof.
Suppose thatx=(S(t),0,0,I2(t),0)∈MI20 be any solution of (1). From the first equation of (1), we have
(69)S′(t)>Λ-(β2+μ)S(t).
Consider the comparison equation(70)u′(t)=Λ-(β2+μ)u(t),u(0)=S(0).
Similarly, (70) exists as a positive unique solution. So we have S(t)≥u*-ε1≐m2, for t large enough, where u*=Λ/(β2+μ).


### 4.4. Stability of Coexistence EquilibriumE*

In this subsection we establish that the equilibriumE* is locally stable whenever it exists. In this case s=Λ/μR2,e=e(0)π(a)e-μa,e(0)=β1si1,i1=(Λ/α1)((1-(1/R2+μ(μ+α1)/Λδ))/((Λβ1/μα1R2)(1-B)-1)),i2=((μ+δ)/δ)(R1/R2-1),j=δi1i2/(μ+ν). The characteristic equations at the coexistence equilibrium are as follows:
(71)(λ+μ+α1+δi2-β1sC(λ))λ×(λ+β1i1+β2i2+μ-β1i1B(λ)+β2i2μλ)=(β1sB(λ)-β1s)(β1i1-δβ2i1i2λ).Theorem 10.
IfR1>R2,  [1-(1/R2+(μ(μ+α1)/Λδ))][((λβ1/μα1R2)(1-B))-1]>0, and the characteristic equation (71) has only negative real parts roots, then the coexistence equilibrium E* is locally asymptotically stable.


## 4.1. Stability of the Disease-Free Equilibrium

For the disease-free equilibrium we havee(0)=i1=i2=j=0, and s=Λ/μ. Thus the system above simplifies to the following system:
(26)λx=-β1sz-β2su-μs+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz,λz=∫0∞‍γ(a)y(a)da-μz-δi2z-δi1u-α1z,λu=β2su-(μ+α2)u.From this system we will establish the following result regarding the local stability of the disease-free equilibriumE0.Theorem 2.
IfR1<1 and R2<1, then the disease-free equilibrium E0 is locally asymptotically stable. If R1>1 or R2>1 then the disease-free equilibrium E0 is unstable.Proof.
To see this, from the second to last equation we haveλu=[β2s-(μ+α2)]u, where either λ=β2s-(μ+α2), or u=0. This eigenvalue λ=(μ+α2)(R2-1)<0, if and only if R2<1. Thus, if R2>1, the disease-free equilibrium E0 is unstable because this eigenvalue is positive. Further, from the second equation we have that the remaining eigenvalues satisfying the equation,
(27)y(a)=y(0)e-(λ+μ)aπ(a)=β1sze-(λ+μ)aπ(a).
Further, from the fourth equation we have that the remaining eigenvalues satisfying the equation, also referred to as the characteristic equation, λz=β1s∫0∞‍γ(a)π(a)e-(λ+μ)adaz-(μ+α1)z. This eigenvalue λ=(μ+α1)(β1s/(μ+α1))∫0∞‍γ(a)π(a)e-(λ+μ)ada-1), or z=0.
We denote the left hand side of the equation above byℱ1(λ)=λ, and ℱ2(λ)=(μ+α1)(R1(λ)-1), where R1(λ)=(β1s/(μ+α1))∫0∞‍γ(a)π(a)e-(λ+μ)ada, and
(28)R1′(λ)<0,limt→+∞R1(λ)=0,limt→-∞R1(λ)=+∞.
Ifℱ1(λ)=ℱ2(λ) has a root with Reλ≥0, then Reℱ1(λ)≥0. But |ℱ2(λ)|≤(μ+α1)(R1-1)<0, when R1<1. This is a contradiction.
Thus, if bothR1<1 and R2<1 all eigenvalues have negative real part and the disease-free equilibrium E0 is locally asymptotically stable. If only R1>1 then if we consider ℱ1(λ)=ℱ2(λ) for λ is real, we see that ℱ2(λ) is a decreasing function of λ approaching zero as λ approaches infinity. Since ℱ2(0)=(μ+α1)(R1-1)>0 and ℱ1(0)=0 that implies that there is a positive eigenvalue λ*>0 and the disease-free equilibrium E0 is unstable. This concludes the proof.In what follows, we show that the diseases vanish ifR1<1,R2<1.Theorem 3.
IfRi<1,i=1,2, then E0 is a global attractor, that is, limt→∞e(a,t)=0, I1→0, I2→0, J→0 as t→∞.Proof.
SinceN′(t)=Λ-μN-νJ≤Λ-μN, then N∞≤Λ/μ. Hence S∞≤Λ/μ. Let ℬ(t)=e(0,t). Integrating this inequality along the characteristic lines we have
(29)e(a,t)={ℬ(t-a)π(a)e-μa,t<a,e0(a-t)π(a)π(a-t)e-μt,t≥a.
Sinceℬ(t)=β1SI1≤β1ΛI1/μ and
(30)I1′≤β1Λμ∫0t‍γ(a)I1(t-a)π(a)e-μada+F1(t)-μI1-α1I1,
where F1(t)=∫t∞‍γ(a)e0(a-t)(π(a)/π(a-t))e-μtda and limt→∞F1(t)=0, from the equation for I1 then we have the following inequality:
(31)I1≤I1(0)e-(μ+α1)t+β1Λμ∫0t‍e-(μ+α1)τ∫0t-τ‍γ(a)I1(t-a)π(a)e-μadadτ+∫0t‍e-(μ+α1)τF1(t-τ)dτ,
whereγ(a) is bounded and the integral of F1(t) goes to zero as t→∞. Consequently, taking a limsup of both sides as t→∞ we obtain
(32)I1∞≤R1I1∞.
SinceR1<1, this inequality can only be satisfied if I1∞→0, as t→∞.
Sinceℬ(t)≤β1(Λ/μ)I1, it is easy to obtain ℬ(t)→0, as t→∞.
From the equationI2 then we have the following inequality:
(33)I2′≤β2ΛμI2-(μ+α2)I2.
Solving the inequality we get(34)I2≤I2(0)e-(μ+α2)t+β2Λμ∫0∞‍I2(t-τ)e-(μ+α2)τdτ.
Consequently, taking a limsup of both sides we obtain(35)I2∞≤R2I2∞.
IfR2<1, then I2(t)→0, as t→∞.
From the fluctuations lemma, we can choose sequencetn such that S(tn)→S∞ and S′(tn)→0 when tn→+∞. It follows from the first equation of (1) that we have S∞≥Λ/μ. Then Λ/μ≤S∞≤S∞≤Λ/μ. Hence S→Λ/μ when t→∞. This completes the theorem.


## 4.2. Stability of the TB Dominated Equilibrium

In this subsection we discuss stabilities of the equilibriumE1 and derive conditions for domination of TB. We show that the equilibrium E1 can lose stability, and dominance of the TB is possible in the form of sustained oscillation. In this case i2=0, j=0, s=Λ/(μR1), i1=Λ(1-1/R1)/((1-B)μ/C+(1-B-C)α1/C), e(a)=e(0)π(a)e-μa.The eigenvalue problem takes the form(36)λx=-β1sz-β1i1x-β2su-μs+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz+β1i1x,λz=∫0∞‍γ(a)y(a)da-μz-δi1u-α1z,λu=β2su-(μ+α2)u.From the last equation we have(37)λ=(μ+α2)(R2R1-1),oru=0.HenceR1>R2 the partial characteristic root of (36) is negative. Substituting
(38)y(a)=y(0)π(a)e-(λ+μ)a,y(0)=β1sz+β1i1x
in the first and fourth equations and cancelling x,z, we arrive at the following characteristic equation:
(39)(λ+μ+α1-β1sC(λ))(λ+μ+β1i1(1-B(λ)))-β1i1C(λ)(α1+β1s(B(λ)-1))=0.Substituting the formula(40)1-B(λ)=C(λ)+(λ+μ)E(λ),
where
(41)E(λ)=∫0∞‍π(a)e-(λ+μ)a
into (39), we obtain the equivalent characteristic equation with (39) as follows:
(42)(λ+μ+α1)(1+β1i1E(λ))+β1i1C(λ)μ+α1=C(λ)C.It is easy to see if (42) has the roots with Reλ>0, the left side mode of (42) is larger than 1, while the right side mode of (42) is less than 1, which leads to a contradiction. Hence the characteristic roots of (42) have negative real parts, then the TB dominated equilibrium E1 is locally asymptotically stable.Therefore we are ready to establish the first result.Theorem 4.
LetR1>1 and R1>R2. Then, the TB dominated equilibrium E1 is locally asymptotically stable.For allx=(S,0R,e,I1,I2,J)∈X0, we define PS:X0→R,PI1,PI2:X0→R and PI12:X0→R by(43)PS(x)=S,PI1(x)=I1,PI2(x)=I2.Set(44)MI12=X0+,MI120={x∈MI12∣PI1x≠0,PI2x≠0},∂MI120=MI12MI120,MI1=∂MI120,MI10={x∈MI1∣PI1x≠0,PI2x=0},∂MI10=MI1MI10,MI2=∂MI10,MI20={x∈MI2∣PI2x≠0,PI1x≠0},∂MI20=MI2MI20,MS=∂MI20(see Figure2).Figure 2
Zoning the area.Lemma 5.
If
R
1
>
1, liminfI1(t)≥ε in MI20.Proof.
Letx=(S0,0R,e,0R,0R)∈MI10. We assume there is a t0>0 such that I1<ε for all t≥t0. From the first equation of (1) in MI10, it is easy to get
(45)dSdt≥Λ-εβ1S-μS,S(0)=S0.
We solve them as follows:(46)S(t)≥Λ(1-e-(μ+β1ε)t)μ≐S*(t).
AndS*(t)→Λ/μ, as t→+∞. Since R1>1, there exists δ>0 and T1>0 which satisfy β1(Λ/μ)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)π(a)e-μadads>1. From the second and third equations of (1), and integrating them from the characteristic line t-a=c, we obtain
(47)e(a,t)={β1S(t-a)I1(t-a)π(a)e-μa,t≥a,e0(a-t)π(a)π(a-t)e-μt,t<a.
From the fourth equation of (1), we get
(48)I1′(t)=β1∫0t‍γ(a)S(t-a)I1(t-a)π(a)e-μada+β1∫t∞‍γ(a)e0(a-t)π(a)π(a-t)e-μtda-(μ+α1)I1.
Solving it, we have(49)I1(t)≥β1Λμ∫0t‍e-(μ+α1)sI1(t)≥×∫0t-s‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
There exist aT1>0, such that
(50)I1(t+T1)≥β1Λμ∫0t+T1‍e-(μ+α1)s≥×∫0t+T1-s‍γ(a)I1(t-s-a)π(a)e-μadads,≥β1Λμ∫0t‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
Thus, fort≥δ, we have
(51)I1(t+T1)≥β1Λμ∫0δ‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,∀t≥t0.
By Assumption1, there exists t1≥0, such that I1(t+T1)≥0, for all t+T1≥t1. Hence, there exists ξ>0, such that I1(t+T1)≥ξ, for all t+T1∈[2t1,2t1+δ]. Set
(52)t2=sup{t+T1≥2t1+δ:I1(l)≥ξ,∀l∈[2t1+δ,t]}.
Assuming thatt2<∞, it follows that
(53)I1(t2)≥β1Λμ∫0δ‍e-(μ+α1)sI1(t2)≥×∫0T1‍γ(a)I1(t-s-a)π(a)e-μadads,I1(t2)≥β1Λμ∫0δ‍e-(μ+α1)sI1(t2)≥×∫0T1‍γ(a)π(a)e-μadadsξ.
Thus,I1(t2)>ξ. By the continuity for t→I1(t), it follows that there exists an ε1>0, such that I1(t)≥ξ, for all t∈[t2,t2+ε1] which contradicts the definition of t2. Therefore, I1(t)≥ξ, for all t>2t1. Denote I1∞=liminft→+∞I1(t)≥ξ>0. Using (51), it follows that I1∞≥I1∞β1(Λ/μ)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)π(a)e-μadads, which is impossible. Thus, I1(t)≥ε.Theorem 6.
IfR1>1, system (1) is permanent in MI10. Moreover there exists AI10 a compact subset of MI10 which is a global attractor for {U(t)}t≥0 in MI10.Proof.
Suppose thatx=(S(t),0,e,I1,0,0)∈MI10 be any solution of (1). From the first equation of (1), we have
(54)S′(t)>μ-(β1+μ)S(t).
Consider the comparison equation.(55)u′(t)=μ-(β1+μ)u(t),u(0)=S(0).
Similarly, (55) exists as a positive unique steady state. So we have S(t)≥u*-ε1≐m1, for t large enough, where u*=μ/(μ+β1). From the second and third equations of (1) and Volterra's formulation, we have e(a,t)≥β1m1π(a)e-μa≐m2, for t large enough.


## 4.3. Stability of the HIV Dominated Equilibrium

In this subsection we establish that the equilibriumE2 is locally stable whenever it exists. In this case e(0)=0,i1=0,j=0,s=Λ/μR2, i2=(Λ/μ)(1-1/R2). The linear eigenvalue problem becomes
(56)λx=-β1sz-β2su-β2i2x-μx+∫0∞‍α0(a)y(a)da+α1z+α2u,y′(a)=-λy-α0(a)y-μy-γ(a)y,y(0)=β1sz,λz=∫0+∞‍γ(a)y(a)da-μz-δi2z-α1z,λu=β2su+β2i2x-(μ+α2)u.From the equation fory(a) we have
(57)y(a)=y(0)π(a)e-(λ+μ)a.Substitutingy(a) in the equation for the initial condition y(0) and assuming that y(0)≠0 we obtain the following characteristic equation:
(58)λ+δi2=β1s∫0∞‍γ(a)π(a)e-(λ+μ)ada-(μ+α1)=(μ+α1)(R1(λ)R2-1).Denoting the left hand side of the equation above byℱ1(λ)=λ+δi2, and ℱ2(λ)=(μ+α1)((R1(λ)/R2)-1), and also noting that
(59)R1′(λ)<0,limt→+∞R1(λ)=0,limt→-∞R1(λ)=+∞,it is easy to see thatℱ1(λ) is an increasing function with λ. Note that ℱ1(0)=δI2≥0. But ℱ2(0)≤(μ+α1)((R1/R2)-1)<0, when R1<R2. Therefore the characteristic roots of this equation have only negative real part. Furthermore, for y(0)=y(a)=0, and z=0,
(60)λx=-β2su-β2i2x-μx+α2u,λu=β2su+β2i2x-(μ+α2)u.We expressx from the first equation
(61)x=(-β2s+α2)uλ+μ+β2i2,
and substitute it into the second equation. In addition, assuming that u is nonzero, we cancel it and obtain the following characteristic equation:
(62)λ+μ+α2=β2s+β2i2(-β2s+α2)λ+μ+β2i2,
that is,
(63)(λ+μ)(λ+μ+α2+β2i2-β2s)=0.Noticing thatβ2s=μ+α2, we obtain the following eigenvalues: λ=-μ and λ=-β2i2, which are both negative. Consequently, the equilibrium E2 is locally asymptotically stable.Therefore, we have the following theorem for equilibriumE2.Theorem 7.
LetR2>1. Assume that tuberculosis cannot invade the equilibrium of HIV, that is, R1<R2. Then the equilibrium E2 is locally asymptotically stable.Lemma 8.
IfR2>1, then liminfI2(t)≥ε in MI20.Proof.
Letx=(S0,0R,0L1,0R,I2,0R)∈MI20. We assume there is a t0>0 such that I2<ε for all t≥t0. From the first equation of (1) in MI20, it is easy to get
(64)dSdt≥Λ-εβ2S-μS,S(0)=S0.
We solve them,(65)S(t)≥Λ(1-e-(μ+β2ε)t)μ≐S*(t).
ThenS(t)≥S*(t), and S(t)→Λ/μ, as t→+∞. From the fourth of equation (1), ∃t0>0, we have
(66)I2′(t)≥β2ΛμI2(t)-(μ+α2)I2(t),∀t≥t0.
Solving it, we have(67)I2(t)≥β2Λμ∫0t‍e-(μ+α2)(t-s)I2(t-s)ds.
We do lim inf on both side of (67), and denote I2∞=liminft→+∞I2(t)>0, thus
(68)I2∞≥R2I2∞,
which is impossible. Thus, I2(t)≥ε.Theorem 9.
IfR2>1, system (1) is permanent in MI20. Moreover there exists AI20 a compact subset of MI20 which is a global attractor for {U(t)}t≥0 in MI20.Proof.
Suppose thatx=(S(t),0,0,I2(t),0)∈MI20 be any solution of (1). From the first equation of (1), we have
(69)S′(t)>Λ-(β2+μ)S(t).
Consider the comparison equation(70)u′(t)=Λ-(β2+μ)u(t),u(0)=S(0).
Similarly, (70) exists as a positive unique solution. So we have S(t)≥u*-ε1≐m2, for t large enough, where u*=Λ/(β2+μ).


## 4.4. Stability of Coexistence EquilibriumE*

In this subsection we establish that the equilibriumE* is locally stable whenever it exists. In this case s=Λ/μR2,e=e(0)π(a)e-μa,e(0)=β1si1,i1=(Λ/α1)((1-(1/R2+μ(μ+α1)/Λδ))/((Λβ1/μα1R2)(1-B)-1)),i2=((μ+δ)/δ)(R1/R2-1),j=δi1i2/(μ+ν). The characteristic equations at the coexistence equilibrium are as follows:
(71)(λ+μ+α1+δi2-β1sC(λ))λ×(λ+β1i1+β2i2+μ-β1i1B(λ)+β2i2μλ)=(β1sB(λ)-β1s)(β1i1-δβ2i1i2λ).Theorem 10.
IfR1>R2,  [1-(1/R2+(μ(μ+α1)/Λδ))][((λβ1/μα1R2)(1-B))-1]>0, and the characteristic equation (71) has only negative real parts roots, then the coexistence equilibrium E* is locally asymptotically stable.


## 5. Persistence of the System

In this section, we consider persistence of the system inMI120 when ν=0. It is easy to check if the system (1) is dissipative and the dynamical system is asymptotically smooth. MI120,  MI10,  MI20, and  Ms are positively invariant. Define
(72)∂Γ=MI10∪MI20∪Ms,∂Γ=MI120.Assuming the boundary equilibria of (1) are globally asymptotically stable, we have
(73)A~∂MI120:=∪(S0,e0(·),I1,I2,J)∈∂Γω((S0,e0(·),I1,I2,J))={E0,E1,E2}.By the above conclusions, it follows thatA~∂MI120 is isolated and has an acyclic covering M={E0,E1,E2}. Since the orbit of any bounded set is bounded, and from Theorem 4.2 in [17], we only need to show that E0,E1,E2 are ejective in MI120 if globally stable conditions of E0,E1,E2 are not satisfied. Therefore, we have the following lemmas.Lemma 11.
Let Assumption1 be satisfied and let E0 be globally asymptotically stable. If R1>1, then E0 is ejective in MI120 for {UI12(t)}t≥0.Proof.
Letδ>0, T1>0 and ε∈(0,1) satisfy
(74)β1(Λμ-ε)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)π1(a)e-μadads>1.
Letx=(S0,0R,I10,I20,J0)∈MI120 with ∥x-xS∥<ε. Assume that
(75)∥U(t)x-xs∥≤ε,t≥0.
DefiningS(t)=PSU(t)x, and I1(t)=PI1U(t)x, for all t≥0. from (75) it follows that
(76)S(t)≥Λμ-ε,∀t≥0.
From the second and third equations of (1), and integrating them from the characteristic line t-a=c, we obtain
(77)e(a,t)≥{β1(Λμ-ε)I1(t-a)π1(a)e-μa,t≥a,e0(a-t)π1(a)π1(a-t)e-μt,t<a,
where from the fourth equation of (1), we get
(78)I1′(t)≥β1Λμ∫0t‍I1(t-a)π1(a)e-μada+β1∫t∞‍γ(a)e0(a-t)π1(a)π1(a-t)e-μtda-(μ+α1)I1.
Solving it, we have(79)I1(t)≥β1(Λμ-ε)∫0t‍e-(μ+α1)s×∫0t-s‍γ(a)I1(t-s-a)π1(a)e-μadads,∀t≥0.
There exist aT1>0, such that
(80)I1(t+T1)≥β1(Λμ-ε)∫0t+T1‍e-(μ+α1)s≥×∫0t+T1-s‍γ(a)I1(t-s-a)π1(a)e-μadads,≥β1(Λμ-ε)∫0t‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π1(a)e-μadads,∀t≥0.
Thus, fort≥δ, we have
(81)I1(t+T1)≥β1(Λμ-ε)∫0δ‍e-(μ+α1)s≥×∫0T1‍γ(a)I1(t-s-a)π1(a)e-μadads,∀t≥0.
By Assumption1, there exists t1≥0, such that I1(t+T1)≥0, for all t+T1≥t1. Hence, there exists ξ>0, such that I1(t+T1)≥ξ, for all t+T1∈[2t1,2t1+δ]. Set
(82)t2=sup{t+T1≥2t1+δ:I(l)≥ξ,∀l∈[2t1+δ,t]}.
Assume thatt2<∞. Then
(83)I1(t2)≥β1(Λμ-ε)∫0δ‍e-(μ+α1)s×∫0T1‍γ(a)I1(t-s-a)π1(a)e-μadads,≥β1(Λμ-ε)∫0δ‍e-(μ+α1)s×∫0T1‍γ(a)π1(a)e-μadadsξ.
Thus,I1(t2)>ξ. By the continuity for t→I1(t), it follows that there exists an ε2>0, such that I1(t)≥ξ, for all t∈[t2,t2+ε2] which contradicts the definition of t2. Therefore, I1(t)≥ξ, for all t>2t1. Denote I1∞=liminft→+∞I1(t)≥ξ>0. Using (81), it follows that I1∞≥I1∞β1(Λ/μ-ε)∫0δ‍e-(μ+α1)s∫0T1‍γ(a)I1(t-s-a)π1(a)e-μadads, which is impossible.Lemma 12.
Let Assumption1 be satisfied and let E1 and E2 be globally asymptotically stable. Then one has the following.(i)
IfR1>1, thenxI1=E1is ejective inMI120.(ii)
IfR2>1, thenxI2=E2is ejective inMI120.Theorem 13.
Let Assumption1 be satisfied and let E0, E1, and E2 be globally asymptotically stable. Assume R1>1, R2>1. Then there exists ε>0 such that for all x∈MI120,
(84)liminf∥PI12U(t)x∥≥ε.Proof.
It is a consequence of Theorem 4.2 in [18] applied with Ω(∂MI120)={E0}∪{E1}∪{E2}. Using Lemma 12, the result follows.


## 6. Simulation

In this section, we use (1) to examine how the prevalence of HIV impacts on TB dynamics. We also present some numerical results on the stability of E0 (the disease-free equilibrium), E1 (the TB dominated equilibrium), and E2 (the HIV dominated equilibrium). We perform a numerical analysis to exhibit the TB impact on HIV under different treatments with (1). We now give three examples to illustrate the main results mentioned in the above section.Example 14.
In (1), we set Λ=2,μ=0.6,δ=73.13,β1=0.49,β2=0.15,ν=0.9,
(85)γ(a)={0.02a≥1.4900≤a≤1.49,α0(a)={0.1a≥1.4900≤a≤1.49,α1=0.01, α2=0.03. We have R1=0.07438<1 and R2=0.79365<1, which satisfy the conditions of Theorems 2 and 3. E0 should be globally asymptotically stable (see Figure 3(a)). In this case, both TB and HIV will be eliminated.(a) show thatE0 is globally asymptotically stable. (b) shows that E1 is globally asymptotically stable.
(a)
(b)Example 15.
In (1), Λ=2,μ=1,δ=4,β1=50.49,β2=0.15,ν=0.9,
(86)γ(a)={0.1a≥1.4900≤a≤1.49,α0(a)={0.1a≥1.4900≤a≤1.49,α1=0.01,α2=0.03. We have R1=34.48770>1>R2=0.79365, which satisfy the conditions of Theorem 4. E1 should be globally asymptotically stable (see Figure 3(b)). In this case, TB is dominated in the coinfected dynamics.Example 16.
In (1), Λ=2,μ=0.6,δ=0.0073,β1=0.49,β2=4.15,ν=0.9,
(87)γ(a)={0.1a≥1.4900≤a≤1.49,α0(a)={0.1a≥1.4900≤a≤1.49,α1=0.01, α2=0.03. We have R2=11.37566>1>R1=0.33470, which satisfies the conditions of Theorem 7. E2 should be globally asymptotically stable (see left of Figure 4). In this case, HIV is dominated in the coinfected dynamics.Figure 4
shows thatE2 is globally asymptotically stable.


### 6.1. Effect of Treatment Parameterαi,i=0,1,2

We examined hypothetical treatment to gain insight into the underlying co-epidemic dynamics. We considered nine treatment scenarios: no treatmentα0=0 (latent TB treatment) (IPT), α0=0.5 (latent TB treatment) (IPT), α0=1 (latent TB treatment) (IPT), no treatment α2=0, α2=0.5 (HAART), and α2=1 (HAART), where αi,i=0,1 denotes the fraction of the eligible population receiving the corresponding treatment.First, we considered the effect of each type of treatment in isolation (e.g., IPT for individuals with treatment for latent or active TB, but not AIDS). Exclusively treating people with latent TB reduced the number of new active TB cases, which subsequently decreased the number of new latent TB infections (Figure5(a)). However, latent TB treatment had an adverse effect on the HIV epidemic: the number of new HIV cases increased because individuals coinfected with HIV and latent TB lived longer (due to latent TB treatment) and thus could infect more people with HIV(see Figure 5(c)). Similarly, active TB treatment reduced the number of people with infectious TB, which subsequently reduced the number of new latent and active TB cases (Figure 5(b)). Once again, new HIV cases increased due to longer life expectancy among those who were coinfected with HIV. Hence, the coinfected people have same effect of people infected by latent TB and active TB.We take different latent TB treatments while we take no treatment of HAART and active TB.
(a)
(b)
(c)
(d)Second, we considered the effect of each type of treatment in isolation (e.g., HAART for individuals with AIDS, but no treatment for latent or active TB). The provision of HAART significantly slowed the HIV epidemic (Figure6(c)). However, exclusively treating HIV-infected individuals with HAART adversely affects the TB epidemic (Figures 6(a) and 6(b)). Because HAART reduces AIDS-related mortality, treated individuals have a longer time to potentially infect others with TB, especially in the absence of any TB treatment. Thus, the numbers of new latent and active TB cases increased as more people were given HAART. However, the coinfected people have the same effect of people infected by HIV (see Figure 6(d)).We take different HAART while we take no treatment of IPT.
(a)
(b)
(c)
(d)


## 6.1. Effect of Treatment Parameterαi,i=0,1,2

We examined hypothetical treatment to gain insight into the underlying co-epidemic dynamics. We considered nine treatment scenarios: no treatmentα0=0 (latent TB treatment) (IPT), α0=0.5 (latent TB treatment) (IPT), α0=1 (latent TB treatment) (IPT), no treatment α2=0, α2=0.5 (HAART), and α2=1 (HAART), where αi,i=0,1 denotes the fraction of the eligible population receiving the corresponding treatment.First, we considered the effect of each type of treatment in isolation (e.g., IPT for individuals with treatment for latent or active TB, but not AIDS). Exclusively treating people with latent TB reduced the number of new active TB cases, which subsequently decreased the number of new latent TB infections (Figure5(a)). However, latent TB treatment had an adverse effect on the HIV epidemic: the number of new HIV cases increased because individuals coinfected with HIV and latent TB lived longer (due to latent TB treatment) and thus could infect more people with HIV(see Figure 5(c)). Similarly, active TB treatment reduced the number of people with infectious TB, which subsequently reduced the number of new latent and active TB cases (Figure 5(b)). Once again, new HIV cases increased due to longer life expectancy among those who were coinfected with HIV. Hence, the coinfected people have same effect of people infected by latent TB and active TB.We take different latent TB treatments while we take no treatment of HAART and active TB.
(a)
(b)
(c)
(d)Second, we considered the effect of each type of treatment in isolation (e.g., HAART for individuals with AIDS, but no treatment for latent or active TB). The provision of HAART significantly slowed the HIV epidemic (Figure6(c)). However, exclusively treating HIV-infected individuals with HAART adversely affects the TB epidemic (Figures 6(a) and 6(b)). Because HAART reduces AIDS-related mortality, treated individuals have a longer time to potentially infect others with TB, especially in the absence of any TB treatment. Thus, the numbers of new latent and active TB cases increased as more people were given HAART. However, the coinfected people have the same effect of people infected by HIV (see Figure 6(d)).We take different HAART while we take no treatment of IPT.
(a)
(b)
(c)
(d)


## 7. Discussion

We have developed a mathematical model for modelling the coepidemics, calculated the basic reproduction number, the disease-free equilibrium, the dominated equilibria (defined as a state where one disease is eradicated, while the other disease remains endemic), and got the conditions for local and global stability. We presented the sufficient conditions for the local stability of the TB dominated equilibrium in Theorem4. We also obtained the persistence. In the case of the HIV dominated equilibrium, we presented the sufficient condition for the local stability in Theorem 7. We also obtained the persistence of (1). We simulated and illustrated our analyzed results in Section 6.Our models have several limitations. We simplified the complicated infection dynamics of TB and HIV to develop a tractable framework, which helped us gain insights about the basic reproduction number and equilibria. We assumed uniform patterns within compartments, that is, the mixing between compartments is homogeneous. To appropriately guide policy recommendations, our coepidemic model would need to be significantly expanded. We also apply a coepidemic model to describe the HIV coinfected with other diseases, such as HIV and hepatitis C (HCV). Some 50%–90% of HIV-infected injection drug users are coinfected with HCV [19]. The successful HIV treatment may be adversely impacted by the presence of HCV, and HCV may cause liver damage to occur more quickly in HIV-infected individuals [19]. Modelling this kind of coinfection may play particularly important role on evaluating interventions time targeted to injection drug.Unfortunately, we are unable yet to study the models that introduce a discrete delay and an impulsive perturbation in our model. We will explore them in the future.



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*Source: 429567-2013-03-26.xml*