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Research article

Dynamics of an age-structured heroin transmission model with vaccination and treatment

  • Received: 04 April 2018 Accepted: 04 September 2018 Published: 14 December 2018
  • Based on the development of heroin vaccine, in this paper, we propose an age structured heroin transmission model with treatment and vaccination. The model allows the drug reuse rate of the individuals in treatment to depend on a treatment-age and the vaccine waning rate of the vaccinated to depend on a vaccination age. Meanwhile, the model allows that the heroin vaccine provides an imperfect protection (i.e., the vaccinated individuals can also become drug addicted). We derive the basic reproduction number which dependents on vaccination. The basic reproduction number completely determines the persistence and extinction of heroin spread, i.e., if the basic reproduction number is less than one the drug-free steady state is globally asymptotically stable (i.e., the heroin spread dies out), if the basic reproduction number is larger than one, there exists an unique positive steady state and it is locally and globally stable in some special cases. Finally, some numerical simulations are carried out to illustrate the stability of the positive steady state.

    Citation: Xi-Chao Duan, Xue-Zhi Li, Maia Martcheva. Dynamics of an age-structured heroin transmission model with vaccination and treatment[J]. Mathematical Biosciences and Engineering, 2019, 16(1): 397-420. doi: 10.3934/mbe.2019019

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  • Based on the development of heroin vaccine, in this paper, we propose an age structured heroin transmission model with treatment and vaccination. The model allows the drug reuse rate of the individuals in treatment to depend on a treatment-age and the vaccine waning rate of the vaccinated to depend on a vaccination age. Meanwhile, the model allows that the heroin vaccine provides an imperfect protection (i.e., the vaccinated individuals can also become drug addicted). We derive the basic reproduction number which dependents on vaccination. The basic reproduction number completely determines the persistence and extinction of heroin spread, i.e., if the basic reproduction number is less than one the drug-free steady state is globally asymptotically stable (i.e., the heroin spread dies out), if the basic reproduction number is larger than one, there exists an unique positive steady state and it is locally and globally stable in some special cases. Finally, some numerical simulations are carried out to illustrate the stability of the positive steady state.


    Heroin is a highly abused opioid and incurs a significant detriment to society worldwide. Heroin usually appears as a white or brown powder or as a black sticky substance, known as "black tar heroin" [1], and its most frequent routes of delivery were intravenous injection (25%) and inhalation [12]. It crosses the blood-brain barrier within 15–20 seconds, rapidly achieving a high level syndrome in the brain and the central nervous system which causes both the 'rush' experienced by users and the toxicity [25]. Heroin users are at high risk for addiction. It is estimated that about 23% of individuals who use heroin become dependent on it.

    More recently, a heroin conjugate vaccine attracted much attentions. It was developed through comprehensive evaluation of hapten structure, carrier protein, adjuvant and dosing, which can generate a significant and sustained antidrug IgG titers in each subject and it is effective in rhesus monkeys [3]. Also, it is found that immunization of mice with an optimized heroin-tetanus toxoid (TT) conjugate can reduce heroin potency by >15% and the vaccine effects proved to be durable and persisting for over eight months. Although it is unknown what will happen if the heroin vaccine is used in clinical setting, the heroin vaccine brings much hope for the defence against and control of heroin abuse.

    In fact, the spread of heroin habituation and addiction can be well modeled by epidemic type models as "transmission" occurs in the form of peer pressure where established users recruit susceptible individuals into trying and using the drug [4,9,16], that is, mathematical modelling is a means to provide a general insight for how classes of drug takers behave, and as such, could hopefully becomes a useful device to aid specialist teams in devising treatment strategies. Modeling heroin addiction and spread in epidemic fashion is not new [21]. Recently, Fang et al. [10] proposed a age-structured heroin transmission model and proved its global dynamics behaviors. Usually, the population is divided into three classes, namely the number of susceptibles, S(t), the number of drug users not in treatment, U1(t) and the number of drug users in treatment, U2(t), respectively. Naturally, we wonder how the heroin vaccine effects the heroin transmission process.

    The study of vaccination has been the subject of intense theoretical analysis [2,7,8,14,17,18,19,20,27,29]. Based on the study of classical epidemic models, Kribs-Zaleta and Velasco-Hernˊandez [17] added a compartment V into an SIS model and studied the vaccination of disease such as pertussis and tuberculosis; Kribs-Zaleta and Martcheva [18] studied the effects of a vaccination campaign upon spread of a non-fatal disease which features both acute and chronic infective stages, as well as variable infectivity and recovery rates in the chronic stage. Interestingly, Xiao and Tang [29] developed a simple SIV epidemic model including susceptible, infected and imperfectly vaccinated classes, with a nonlinear incidence rate and there would be backward bifurcations; Arino et al. [2] also showed that, if vaccines are imperfect, i.e., vaccinated individuals can be infected, there could be backward bifurcations.

    To study the role of the heroin vaccine in the control of heroin abuse, adding a compartment V(t) into a heroin transmission model is necessary. To our best knowledge, there is no related work in this field by now. Hopefully, mathematical modeling can provide a new insight into the interaction mechanism between the vaccinated, the susceptibles, the drug users and the individuals in treatment.

    Motivated by the development of the heroin vaccine, in this paper, we present an age structured heroin transmission SU1U2V model which incorporates a drug reuse rate α(a) dependant on treat-age (i.e., the time since the host has been in treatment) and a vaccine waning rate dependant on vaccination age (i.e., the time since the host has been vaccinated). We also assume the susceptible population is vaccinated at a constant rate ψ, the vaccinated individuals can be infected at reduced rate σβ with 0σ1. Obviously, σ=0 means the vaccine is completely effective in preventing infection, while σ=1 means that the vaccine is utterly ineffective. As a result, the system is as follows:

    {dSdt=Λ(μ+ψ)S(t)βS(t)U1(t)+0α(a)V(a,t)da,dU1dt=βS(t)U1(t)+σβU1(t)0V(a,t)da+0p(θ)U2(θ,t)dθ              (μ+δ1+γ)U1(t),U2(θ,t)θ+U2(θ,t)t=(μ+δ2+p(θ))U2(θ,t),V(a,t)a+V(a,t)t=(μ+α(a)+σβU1(t))V(a,t) (1)

    for t>0, with the initial and boundary conditions:

    {U2(0,t)=γU1(t),     V(0,t)=ψS(t),S(0)=S0, U1(0)=U10, U2(θ,0)=U20(θ), V(a,0)=V0(a), (2)

    where U20(θ),V0(a)L1+(0,).

    In system (1)-(2), θ is the treat-age, that is the time that has elapsed since a drug user is in treatment; U2(θ,t) is the density of drug users in treatment with age θ at time t; S(t) is the density of the susceptibles at time t; U1(t) is the density of drug users not in treatment, initial and relapsed drug users. The positive constant Λ is the recruitment of susceptible, μ the natural death rate of the general population, β is the force of drug use per contact with the susceptible per unit time, σβ is the force of drug use per contact with the vaccinated individuals per unit time, γ is the rate of drug users who enter treatment, δ1 a removal rate that includes drug-related deaths of users not in treatment and a spontaneous recovery rate, individuals not in treatment who stop using drugs but are no longer susceptible, δ2 a removal rate that includes the drug-related deaths of users in treatment and a rate of successful "cure" that corresponds to recovery to a drug free life and immunity to drug addiction for the duration of the modelling time period. The function p(θ) is the probability of a drug user in treatment with the treatment-age θ relapsing to the untreated users.

    Throughout the paper, we make the following assumptions: (A1) there is a positive number of drug users not in treatment, i.e., U10>0; (A2) the initial conditions U20(θ) and V0(a) are uniformly bounded respectively for θ,a(0,+); (A3) the maps θp(θ) and aα(a) are almost everywhere bounded and belong to L+((0,+),R){0L}.

    The paper is organized as follows. In the next section, we present some preliminary results of system (1)-(2). In Section 3, we prove the local and global stability of the drug-free steady state of system (1). In Section 4, we present the existence and the stability results of the drug spread steady state of system (1) when the basic reproduction number is larger than one. Finally, in Section 5, a brief discussion and some numerical examples are presented.

    In this section, we give some basic results prepared for the further study of system (1).

    For the sake of convenience, we let

    Φ1(θ)=eθ0(μ+δ2+p(τ))dτ,     0Φ1(θ)dθ=ϕ1,
    Φ(θ)=γp(θ)eθ0(μ+δ2+p(τ))dτ,  0Φ(θ)dθ=ϕ,k1(a)=ea0(μ+α(τ))dτ,      0k1(a)da=K1,k(a)=α(a)ea0(μ+α(τ))dτ,      0k(a)da=K.

    By simple calculations, we have that K=1μK1 and ϕ=γγ(μ+δ2)ϕ1.

    Naturally, system (1)-(2) has a unique drug-free steady state E0(S0,0,0,V0(a)) which satisfies that

    {0=Λ(μ+ψ)S0+0α(a)V0(a)da,dV0(a)da=(μ+α(a))V0(a),V0(0)=ψS0. (3)

    Solving the last two equations, we have that

    V0(a)=ψS0ea0(μ+α(τ))dτ=ψS0k1(a). (4)

    Substituting Equation (4) into the first equation of (3), we have that

    S0=Λμ+ψ(1K)=Λμ(1+ψK1). (5)

    Thus, we have that

    V0(a)=ψΛμ(1+ψK1)k1(a). (6)

    According to the definition of the basic reproduction number in existing literatures [5,6,28], we define the basic reproduction number R0 as:

    R0=βμ+δ1+γϕ(S0+σ0V0(a)da)    =βμ+δ1+γ(μ+δ2)ϕ1Λμ(1+ψK1)(1+σψK1). (7)

    According to [24], any positive equilibrium (S, U1, U2(θ), V(a)) of system (1), if it exists, must be a constant solution of the following equations

    {0=ΛβSU1(μ+ψ)S+0α(a)V(a)da,0=βSU1+σβU10V(a)da(μ+δ1+γ)U1+0p(θ)U2(θ)dθ,dU2(θ)dθ=(μ+δ2+p(θ))U2(θ),U2(0)=γU1,dV(a)da=(μ+α(a)+σβU1)V(a),V(0)=ψS. (8)

    Let

    π(a)=ea0(μ+α(τ)+σβU1)dτ,  Kπ=0π(a)da. (9)

    By simple calculations, we have that

    0ψα(a)π(a)da=ψψ(μ+σβU1)Kπ. (10)

    From the third and the forth equation of (8), we have that

    U2(θ)=γU1Φ1(θ). (11)

    From the fifth and the sixth equation of (8), we have that

    V(a)=ψSπ(a). (12)

    Substituting (11) and (12) into the second and the first equation of (8) respectively, we have that the following equations

    {0=ΛβSU1+(ψψ(μ+σβU1)Kπ)S(μ+ψ)S0=βSU1+σβU1KπψS(μ+δ1+γ)U1+ϕU1. (13)

    It follows from the first equation of (13), we have that

    S=ΛβU1+ψ(μ+σβU1)Kπ+μ. (14)

    Substituting (14) into the second equation of (13) and eliminating U1, we have that

    1=βΛ(1+σψKπ)(βU1+ψ(μ+σβU1)Kπ+μ)(μ+δ1+γϕ). (15)

    Define a function F(U1) to be the right hand side of Equation (15). Obviously, F(0)=R0. It follows from (9) we have that F(U1) is a decreasing function of U1 and F(U1)0 as U1. Thus, there exists an unique positive root of Equation (15) only if R0>1.

    Summarizing the above discussion, we have the following result.

    Theorem 2.1. System (1) always has a drug-free steady state E0, besides that, it has an unique drug spread steady state E only if R0>1.

    Now, by setting

    N(t)=S(t)+U1(t)+0U2(θ,t)dθ+0V(a,t)da,

    we deduce from (1) that N(t) satisfies the following ordinary differential equation:

    N(t)=ΛμN(t)δ1U1(t)+δ20U2(θ,t)dθΛμN(t), (16)

    and therefore lim suptN(t)Λμ. Denote

    Ω={(S,U1,U2,V)R+×R+×L1+(0,)×L1+(0,):S+U1+0U2(θ,)dθ+0V(a,)daΛμ}

    Then Ω is the maximum positively invariant set of system (1) that attracts all positive solutions of of (1). Therefore, we restrict our attention to solutions of (1) with initial conditions in Ω.

    In the following, we use the approach introduced by Thieme [26]. Consider

    X=R×R×R×L1((0,),R)×R×L1((0,),R),X0=R×R×{0}×L1((0,),R)×{0}×L1((0,),R),X+=R+×R+×R+×L1+((0,),R)×R+×L1+((0,),R)

    and

    X0+=X0X+.

    Let the linear operator A:Dom(A)XX defined by

    A(SU1(0U2)(0V))=((μ+ψ)S(μ+δ1+γ)U1(U2(0)U2(μ+δ2+p(θ))U2)(V(0)V(α(a)+μ)V))

    with

    Dom(A)=R×R×{0}×W1,1((0,+),R)×{0}×W1,1((0,+),R),

    where W1,1 is a Sobolev space. Then ¯Dom(A)=X0 is not dense in X. We consider a nonlinear map F:¯Dom(A)X which is defined by

    F(t)=(ΛβS(t)U1(t)+0α(a)v(a,t)daβS(t)U1(t)+σβU1(t)0V(a,t)da+0p(θ,t)U2(θ,t)dθ(ψS(t)0L1)(γU1(t)(σβU1(t)V(a,t))L1)),

    and let

    u(t)=(S(t), U1(t), (0U2(,t)), (0V(,t)))T.

    Then, we can reformulate system (2.3) as the following abstract Cauchy problem:

    du(t)dt=Au(t)+F(t)fort0,withu(0)=xX0+. (17)

    By applying the results in Hale [11], Magal [22], and Magal and Thieme [23], we obtain the following theorem.

    Theorem 2.2. System (1) generates a unique continuous semiflow {U(t)}t0 on X0+ that is bounded dissipative and asymptotically smooth. Furthermore, the semiflow {U(t)}t0 has a global compact attractor A in X0+, which attracts the bounded sets of X0+.

    In this section, by use of characteristic equation, we will prove the local and global stability of the drug-free steady state E0.

    For the sake of convenience, we give the following Laplace transforms of the corresponding functions

    ˆΦ(λ)=0Φ(θ)eλθdθ,  ˆΦ1(λ)=0Φ1(θ)eλθdθ,ˆK(λ)=0k(a)eλada,  ˆK1(λ)=0k1(a)eλada. (18)

    By direct calculations, we have the following relationships

    ˆK(λ)=1(λ+μ)ˆK1(λ)  and  ˆΦ(λ)=γγ(λ+μ+δ2)ˆΦ1(λ). (19)

    Theorem 3.1. The drug-free steady state E0 is locally asymptotically stable if R0<1, and unstable if R0>1.

    Proof. By linearization of system (1) at the drug-free steady state E0, we can obtain the corresponding linearized system. We let

    S(t)=˜S(t)+S0, U1(t)=˜U1(t), U2(θ,t)=˜U2(θ,t) and V(a,t)=˜V(a,t)+V0(a).

    By linearization of system (1) at the drug free steady state E0, we obtain the following system

    {d˜S(t)dt=(μ+ψ)˜S(t)βS0˜U1(t)+0α(a)˜V(a,t)da,d˜U1(t)dt=βS0˜U1(t)+σβ0V0(a)da˜U1(t)+0p(θ)˜U2(θ,t)dθ              (μ+δ1+γ)˜U1(t),˜U2(θ,t)θ+˜U2(θ,t)t=(μ+δ2+p(θ))˜U2(θ,t),˜U2(0,t)=γ˜U1(t),˜V(a,t)a+˜V(a,t)t=(μ+α(a))˜V(a,t)σβV0(a)˜U1(t),˜V(0,t)=ψ˜S(t). (20)

    To analyze the asymptotic behaviors around E0, we let

    ˜S(t)=¯xeλt, ˜U1(t)=¯yeλt, ˜U2(θ,t)=¯z(θ)eλt, and  ˜V(a,t)=¯w(a)eλt

    where ¯x, ¯y, ¯z(θ) and ¯w(a) can be determined. Thus, we consider the following eigenvalue problem

    {λ¯x=(μ+ψ)¯xβS0¯y(t)+0α(a)¯w(a)da,λ¯y=βS0¯y+σβ0V0(a)da¯y(μ+δ1+γ)¯y+0p(θ)¯z(θ)dθ,d¯z(θ)dθ=(λ+μ+δ2+p(θ))¯z(θ),¯z(0)=γ¯y,d¯w(a)da=(λ+μ+α(a))¯w(a)σβV0(a)¯y,¯w(0)=ψ¯x. (21)

    Solving the third equation of (21), we have

    ¯z(θ)=¯z(0)eλθΦ1(θ)=γ¯yeλθΦ1(θ). (22)

    Solving the fifth equation of (21), we have

    ¯w(a)=¯w(0)eλak1(a)σβ¯ya0eλ(as)k1(a)k1(s)V0(s)ds=ψ¯xeλak1(a)σβψS0k1(a)¯ya0eλ(as)ds=ψ¯xeλak1(a)σβψS0k1(a)¯y1λ(1eλa). (23)

    Substituting (23) and (22) into the first and the second equation of (21), we have the characteristic equation

    det(Δ(λ))=| A11    A12   0      A22 |=0, (24)

    where

    A11=(λ+μ)(1+ψˆK1(λ)),A12=βS0(1+1λσψ(ˆKˆK(λ))),A22=λ+(μ+δ1+γ)ˆΦ(λ)σβψS0K1βS0.

    Then the roots of Equation (24) are determined by the following equations

    (λ+μ)(1+ψˆK1(λ))=0 (25)

    and

    λ+(μ+δ1+γ)ˆΦ(λ)σβψS0K1βS0=0. (26)

    Obviously, λ=μ is the root of Equation (25). Then we need only to consider the root of Equation (26) which can be rewritten as

    λ+μ+δ1+γ(λ+μ+δ2)ˆΦ1(λ)=(σψK1+1)βS0. (27)

    We also have that

    1=(σψK1+1)βS0λ+μ+δ1+γ(λ+μ+δ2)ˆΦ1(λ). (28)

    Define a function H(λ) to be the right-hand side of Equation (28). It follows from the definition of the basic reproduction number R0 (see (7)), we have that H(0)=R0. By direct computing, it is easy to show that H(λ)<0, that is, H(λ) is a decreasing function of λ with limt+H(λ)=0.

    Assume that λ=x+iy is a root of Equation (28). Then it follows from (28) that

    x01=|H(λ)||H(x)|H(0)=R0, i.e.,  R01.

    Thus, we can have that (λ) is negative if R0<1, and therefore the steady state E0 is locally asymptotically stable if R0<1 and it is unstable if R0>1.

    In the following, we will use the Fluctuation Lemma to establish the global stability of the drug-free steady state E0. To this end, we first introduce the notation

    g=lim inftg(t)  and  g=lim suptg(t).

    Then the Fluctuation Lemma is given as follows.

    Lemma 3.2. (Fluctuation Lemma [13]) Let g: R+R be a bounded and continuously differentiable function. Then there exist sequences {sn} and {tn} such that sn, tn, g(sn)g, g(sn)0, g(tn)g and g(tn)0 as n.

    Lemma 3.3. [15] Suppose f: R+R be a bounded function. Then

    lim suptt0h(θ)f(tθ)dθfh1,

    where h1=0h(s)ds.

    Using integration, U2(θ,t) and V(a,t) satisfy the following Volterra formulation:

    U2(θ,t)={γU1(tθ)Φ1(θ),           if  tθ,U2(θt,0)Φ1(θ)Φ1(θt),     if  θt. (29)
    V(a,t)={ψS(ta)ea0(μ+α(τ)+σβU1(tτ))dτ,     if  ta,V(at,0)et0(μ+α(aτ)+σβU1(τ))dτ,     if  at. (30)

    Theorem 3.4. If R0<1, then the drug-free steady state E0 is the unique steady state of system (1), and it is globally stable.

    Proof. Theorem 3.1 shows that the drug-free steady state E0 of system (1)) is locally stable if R0<1. To use the Fluctuation Lemma, substituting the expressions of V(a,t) and U2(θ,t) into the first two equations of system (1), we have that

    {dSdt=ΛβS(t)U1(t)+t0ψα(a)ea0(μ+α(τ)+σβU1(tτ))dτS(ta)da         (μ+ψ)S(t)+FV(t)dU1dt=βS(t)U1(t)+σβU1(t)t0ψea0(μ+α(τ)+σβU1(tτ))dτS(ta)da         (μ+δ1+γ)U1(t)+t0Φ(θ)U1(tθ)dθ+FU(t)+FUV(t), (31)

    where

    FV(t)=tψα(a)V(at,0)et0(μ+α(aτ)+σβU1(τ))dτda,FU(t)=tp(θ)U2(θt,0)Φ1(θ)Φ1(θt)dθ,FUV(t)=σβU1(t)tV(at,0)et0(μ+α(aτ)+σβU1(τ))dτda

    with limtFV(t)=0, limtFU(t)=0 and limtFUV(t)=0.

    Choose the sequences t1n such that S(t1n)S and S(t1n)0. Then FV(t)0 as n. With the assistance of the Fluctuation Lemma, it follows from the first equation of (31) we have that

    0=ΛβSU1(t)(μ+ψ)S+SψK,

    and

    SΛμ+ψψK.

    Choose the sequences t2n such that U1(t2n)U1 and U1(t2n)0. Then FU(t)0 and FUV(t)0 as n. With the assistance of the Fluctuation Lemma, it follows from the second equation of (31) we have that

    0βSU1+σβψK1SU1(μ+δ1+γ)U1+ϕU1=(βS(1+σψK1)(μ+δ1+γ)+ϕ)U1(βΛ(1+σψK1)μ+ψψK(μ+δ1+γ)+ϕ)U1=(μ+δ1+γϕ)(R01)U1.

    Thus we obtain that U10 if R0<1. It then follows from (29) we have that limtU2(θ,t)=0.

    Choose the sequences t3n such that S(t3n)S and S(t3n)0. Note that limnU1(t3n)=0 and limnFV(t3n)=0. It then follows from the first equation of (31) we have that

    0=ΛβSU1(μ+ψ)S+ψKS=Λ(μ+ψ)S+ψKS.

    It follows that

    Λμ+ψψK=SSΛμ+ψψK,

    which implies that limtS(t)=Λμ+ψψK. It follows from (30) we have that

    limtV(a,t)=ψΛμ+ψψKk1(a)=V0(a).

    Therefore, (S,U1,U2,V)E0 in R+×R+×L1+×L1+ as t. This completes the proof of Theorem 3.4.

    This section aims to establish the stability of the drug spread steady state of system (1) in terms of the basic reproduction number R0.

    By a similar discussion as Theorem 3.5 in [10], we have the uniform persistence result as following.

    Theorem 4.1. Suppose the heroin spread is initially present, i.e., U10>0. If R0>1, then the semiflow generated by system (1) is uniformly persistent, i.e., there exists ε>0 which is independent of initial values such that

    lim inftS(t)ε, lim inftU1(t)ε, lim inftU2(,t)L1+ε

    and lim inftV(,t)L1+ε.

    For the sake of convenience, we let

    ˆKπ(λ):=0π(a)eλada,  and  ˆKα(λ)=0α(a)π(a)eλada. (32)

    It then follows that

    ˆKα(λ)=1(λ+μ+σβU1)ˆKπ(λ). (33)

    In the following, we try to study the stability of the drug spread steady state E (S, U1, U2(θ), V(a)) of system (1). We let

    S(t)=˜S(t)+S, U1(t)=˜U1(t)+U1, U2(θ,t)=˜U2(θ,t)+U2(θ)

    and V(a,t)=˜V(a,t)+V(a). By linearization of system (1) at the steady state E, we obtain the following system

    {d˜S(t)dt=(μ+ψ)˜S(t)βU1˜S(t)βS˜U1(t)+0α(a)˜V(a,t)da,d˜U1(t)dt=βU1˜S(t)+βS˜U1(t)+σβU10˜V(a,t)da+σβ0V(a)da˜U1(t)             (μ+δ1+γ)˜U1(t)+0p(θ)˜U2(θ,t)dθ,˜U2(θ,t)θ+˜U2(θ,t)t=(μ+δ2+p(θ))˜U2(θ,t),˜U2(0,t)=γ˜U1(t),˜V(a,t)a+˜V(a,t)t=(μ+α(a)+σβU1)˜V(a,t)σβV(a)˜U1(t),˜V(0,t)=ψ˜S(t). (34)

    To analyze the asymptotic behaviors around E, we let

    ˜S(t)=¯xeλt, ˜U1(t)=¯yeλt, ˜U2(θ,t)=¯z(θ)eλt, and  ˜V(a,t)=¯w(a)eλt

    where ¯x, ¯y, ¯z(θ) and ¯w(a) can be determined. Then, we consider the following eigenvalue problem

    {λ¯x=(μ+ψ)¯xβU1¯xβS¯y(t)+0α(a)¯w(a)da,λ¯y=βU1¯x+βS¯y+σβU10¯w(a)da+0p(θ)¯z(θ)dθ       +σβ0V(a)da¯y(μ+δ1+γ)¯y,d¯z(θ)dθ=(λ+μ+δ2+p(θ))¯z(θ),¯z(0)=γ¯y,d¯w(a)da=(λ+μ+α(a)+σβU1)¯w(a)σβV(a)¯y,¯w(0)=ψ¯x. (35)

    Solving the third equation of (35), we have

    ¯z(θ)=¯z(0)eλθΦ1(θ)=γ¯yeλθΦ1(θ). (36)

    Solving the fifth equation of (35), we have

    ¯w(a)=¯w(0)eλaπ(a)σβ¯ya0eλ(as)π(a)π(s)V(s)ds=ψ¯xeλaπ(a)σβ¯yf(a,λ), (37)

    where

    f(a,λ)=a0eλ(as)π(a)π(s)V(s)ds=ψSπ(a)a0eλ(as)ds=ψSπ(a)a0eλsds=ψSπ(a)1λ(1eλa).

    Substituting (37) into the first equation of (35), we have

    ¯x=¯yG11(βS+σβ0α(a)f(a,λ)da)=¯yG11[βS+σβψS1λ0α(a)π(a)(1eλa)da] (38)

    where

    G11=λ+(ψ+μ+βU1)ψˆKα(λ).

    It follows from (33) we have that

    G11=(λ+μ)(1+ψˆKπ(λ))+βU1(1+σψˆKπ(λ)).

    Substituting (36) and (37) into the second equation of (35), by use of σβψSKπ=(μ+δ1+γ)βSϕ (i.e., the second equation of (13)), we have that

    (λ+ϕΦ(λ)+σ2β2U10f(a,λ)da)¯y=(βU1+σψβU1ˆKπ(λ))¯x.

    Substituting (38) into the above equation and dividing both sides by ¯y leads to the characteristic equation

    λ+ϕΦ(λ)+σ2β2ψSU11λ0π(a)(1eλa)da=(βU1+σψβU1ˆKπ(λ))1G11   [βS+σβψS1λ0α(a)π(a)(1eλa)da]. (39)

    It follows from Equation (39) we have that

    G11(λ+ϕˆΦ(λ))+β2SU1(1+σψˆKπ(λ))2+1λσ2β2ψSU1G11(KπˆKπ(λ))=1λσβ2ψSU1(KπˆKπ(λ))(μ+σβU1)(1+σψˆKπ(λ)). (40)

    Due to the fact that the characteristic equation (39) is too complex, it is very difficult to determine the distribution of the eigenvalues. In the following, we will study the stability of the drug spread steady state E in three special cases of system (1) respectively.

    Case (ⅰ) We assume that α(a)=0, that is, the vaccine does not wane once the host is vaccinated. which is reasonable since the heroin does not change the toxic (pathogenic) substance. Mathematically, letting 0V(a,t)da=V(t) in this case, we can rewrite system (1) as

    {dSdt=Λ(μ+ψ)S(t)βS(t)U1(t),dU1dt=βS(t)U1(t)+σβU1(t)V(t)(μ+δ1+γ)U1(t)+0p(θ)U2(θ,t)dθ,U2(θ,t)θ+U2(θ,t)t=(μ+δ2+p(θ))U2(θ,t),dV(t)dt=ψS(t)(μ+σβU1(t))V(t),U2(0,t)=γU1(t),S(0)=S0, U1(0)=U10, U2(θ,0)=U20(θ)L1+(0,), V(0)=V0. (41)

    Then we have the following result.

    Lemma 4.2. If α(a)0, the drug spread steady state E of system (41) is locally asymptotically stable if it exists.

    Proof. Linearizing system (41) at its positive steady state (S,U1,U2(),V), we have the following characteristic equation

    |λ+μ+ψ+βU1βS0βU1λ+ϕˆΦ(λ)σβU1ψσβVλ+μ+σβU1|=0. (42)

    By simple calculations, we have the following

    B1(λ)B2(λ)(λ+ϕˆΦ(λ))+B1(λ)σ2β2U1V+B2(λ)β2U1S+σψβ2U1S=0, (43)

    where

    B1(λ)=λ+μ+ψ+βU1,B2(λ)=λ+μ+σβU1.

    Obviously, if σ0, λ=(μ+ψ+βU1) and λ=(μ+σβU1) are not the roots of Equation (43). Dividing both side of Equation (43) by B1(λ)B2(λ) leads to

    λ+ϕ+Z=ˆΦ(λ), (44)

    where

    Z=σ2β2U1VB2(λ)+β2U1SB1(λ)+σψβ2U1SB1(λ)B2(λ).

    Now, assume that λ is a root of Equation (44) with (λ)0. If we can prove the real part of Z is positive (see Appendix A for its detailed proof), then by using |ˆΦ(λ)|ϕ, we have

    ϕ|λ+ϕ|<|λ+ϕ+Z|=|ˆΦ(λ)|ϕ

    This contradiction implies that all roots of Equation (44) have negative real parts. Hence, the positive steady state (S,U1,U2(),V) of system (41) is locally stable if it exists.

    Case (ⅱ) We assume that σ=0, i.e., the heroin vaccine can provide a prefect protection for the vaccinated individuals to avoid the heroin drug addiction. It then follows that π(a)=k1(a). Mathematically, in this case, system (1) can be rewritten as

    {dSdt=Λ(μ+ψ)S(t)βS(t)U1(t)+0α(a)V(a,t)da,dU1dt=βS(t)U1(t)(μ+δ1+γ)U1(t)+0p(θ)U2(θ,t)dθ,U2(θ,t)θ+U2(θ,t)t=(μ+δ2+p(θ))U2(θ,t),V(a,t)a+V(a,t)t=(μ+α(a))V(a,t),U2(0,t)=γU1(t),     V(0,t)=ψS(t),S(0)=S0, U1(0)=U10, U2(θ,0)=U20(θ), V(a,0)=V0(a). (45)

    Lemma 4.3. If σ=0, the drug spread steady state E of system (45) is globally asymptotically stable if it exists.

    Proof. It then follows from (18) and (19) we have that Equation (40) can be modified as

    (λ+μ+ψ+βU1ˆSα(λ))(λ+ϕˆΦ(λ))+β2SU1=0. (46)

    Since λ=0 is not the roots of Equation (46). Both sides of Equation (46) are divided by (λ+ϕˆΦ(λ)), we obtain

    λ+μ+ψ+βU1+Z1=ˆSα(λ) (47)

    where

    Z1=β2SU1λ+ϕˆΦ(λ).

    Assuming λ is a root of Equation (47), we can prove that (λ) (i.e., the real part of λ) is negative. Supposed (λ) is nonnegative, then the real part of Z is nonnegative (see Appendix B). It follows from (10) and (47) we have that

    ψ<|μ+ψ+βU1||λ+μ+ψ+βU1+Z1|=|ˆSα(λ)|<ψ, (48)

    which is a contradiction. So, all roots of Equation (47) have negative real part, and therefore all roots of Equation (46) have negative real part and the drug spread steady state E of system (45) is locally asymptotically stable if it exists.

    Based on the persistence results in Theorem 4.1 and the local stability results of E of system (45), we will use a suitable Lyapunov functional to prove the global stability of E.

    Set g(x)=x1lnx, for xR+. The function g(x) has a global minimum at x=1 with g(1)=0. For our presentation here, we define

    ε1(θ)=θp(τ)eτθ(μ+δ2+p(s))dsdτ  and  ε2(a)=aα(τ)V(τ)dτ. (49)

    Note that ε1(θ), ε2(a)> for all θ>0 and a>0 respectively. We can easily check that ε1(0)=ϕ/γ, ε2(0)=0α(a)V(a)da and

    dε1(θ)dθ=ε1(θ)(μ+δ2+p(θ))p(θ)  and  dε2(a)da=α(a)V(a). (50)

    Now, we define the following Lyapunov functional

    W(t)=WS(t)+WU1(t)+WU2(t)+WV(t), (51)

    where

    WS(t)=Sg(S(t)S),WU1(t)=U1g(U1(t)U1),WU2(t)=0ε1(θ)U2(θ)g(U2(θ,t)U2(θ))dθ,WV(t)=0ε2(a)g(V(a,t)V(a))da, (52)

    Then W is bounded. Then we calculate the time derivative of W(t) along with the solutions of system (45). Following the results in the proof of Theorem 2.2 in [10] and the proof of Theorem 3.11 in [30], we have that

    dWS(t)dt=(μ+ψ)S(SS(t)+S(t)S2)+βSU1(1SS(t))(1S(t)SU1(t)U1)+0α(a)V(a)(V(a,t)V(a)V(a,t)V(a)SS(t)1+SS(t))da,dWU1(t)dt=βS(t)U1(t)βSU1(t)βS(t)U1+βSU1+0p(θ)U2(θ)(U2(θ,t)U2(θ)U1(t)U1U1U1(t)U2(θ,t)U2(θ)+1)dθ,dWU2(t)dt=ε1(θ)U2(θ)g(U2(θ,t)U2(θ))|θ=+ϕU1g(U1(t)U1)0p(θ)U2(θ)g(U2(θ,t)U2(θ))dθ,dWV(t)dt=ε2(a)g(V(a,t)V(a))|a=+0α(a)V(a)g(V(0,t)V(0))da0α(a)V(a)g(V(a,t)V(a))da=+0α(a)V(a)(S(t)SlnSS(t)V(a,t)V(a)+lnV(a,t)V(a))daε2(a)g(V(a,t)V(a))|a=.

    By adding dWS(t)dt and dWV(t)dt together, though some simple calculations, we have that

    dWS(t)dt+dWV(t)dt=βSU1(1SS(t))(1S(t)SU1(t)U1)(μ+ψ)S(SS(t)+S(t)S2)+0α(a)V(a)(SS(t)+S(t)S2)da0α(a)V(a)g(V(a,t)V(a)SS(t))daε2(a)g(V(a,t)V(a))|a==βSU1(1SS(t))(1S(t)SU1(t)U1)0α(a)V(a)g(V(a,t)V(a)SS(t))daε2(a)g(V(a,t)V(a))|a=+(SS(t)+S(t)S2)(0α(a)V(a)da(μ+ψ)S).

    Combining the above four compartments of the Lyapunov functionals, through some simple calculations, we obtain

    dW(t)dt=Λ(SS(t)+S(t)S2)ε1(θ)U2(θ)g(U2(θ,t)U2(θ))|θ=0p(θ)U2(θ)g(U1U1(t)U2(θ,t)U2(θ))dθε2(a)g(V(a,t)V(a))|a=0.

    Notice that equality holds only if S(t)=S, U1(t)=U1 and U2(θ,t)=U2(θ). Thus we conclude that the largest positive invariant is the singleton {E}. By Lyapunov-LaSalle invariance principle, we conclude that the drug spread steady state E is globally asymptotically stable when it exists.

    Case (ⅲ) We assume that σ=1, i.e., the heroin vaccine is noneffective and cannot provide any protection from heroin drug addicted, the vaccination makes no sense. In this case, mathematically, the compartments S and V can be combined, system (1) can be rewritten as the system in [10] (see Appendix C for more details) and the drug spread steady state E is locally and globally stable.

    In this paper, we have studied an age structured heroin transmission model with treatment and vaccination, in which the vaccination can only provide an imperfect protection and the vaccinated wanes the protection as vaccination age goes.

    The basic reproduction number R0 of our system (1) has been found by the definition. When R0<1, system (1) has only the drug free steady state E0 and it is globally asymptotically stable, which implies that the heroin drug will die out eventually. Meanwhile, from the expression of R0, we find that the vaccination, although it is imperfect, plays an important role in the control of heroin spread. When R0>1, system (1) has only the drug spread steady state E and it is uniformly persistent provided that the heroin spread is initially present. Due to the fact that the characteristic equation of system (1) at the drug spread steady state is very complex, it is difficult to discuss the distribution of its eigenvalues. From the biology angle, we have recast system (1) into three special cases and obtained the local and global stability of the drug spread steady state E if it exists (see Lemmas 4.2-4.3).

    Recalling that the expression of R0 in (7), it follows from 0σ1 we have that

    βΛμ(μ+δ1+γ(μ+δ2)ϕ1)1+σψK1(1+ψK1)βΛμ(μ+δ1+γ(μ+δ2)ϕ1). (53)

    It implies that R0(ψ)R0(0), i.e., the vaccination plays an important role in the basic reproduction number which can reduce the reproduction number, although the vaccine provides an imperfect protection. Thus, heroin vaccine will definitely benefit the people.

    In the following, we will present some numerical simulations to study the dynamic behaviors of system (1) under the condition that the basic reproduction number is lager than one, i.e., R0>1. Before that, for simplicity, we take one month as the unit time. Note that the function θp(θ) and aα(a) are both almost everywhere bounded and belong to L+((0,+),R){0L}. In this section we assume that the vaccine waning rate of the vaccinated individuals is

    α(a)={0.10(a10)2e0.35(a10),    10<a40;0.0025,                40<a<¯a;0,                      otherwise,  (54)

    and the drug reuse rate of the individuals in treatment is

    p(θ)=0.8(θ+2)e0.2(θ+5), (55)

    for θ[0,¯θ], where ¯a, ¯θ are the maximum value of vaccination age and treat-age respectively. For simplify, we adopt that ¯a=¯θ=50 (months).

    To study the stability of the drug spread steady state of system (1), we adopt the other parameters in system (1) as follows

    Λ=103, β=3.5×107, μ=0.001,δ1=0.02, δ2=0.01, ψ=0.1, σ=0.85, γ=4. (56)

    It follows from the expression of the basic reproduction number R0 that R0=1.0117>1. By use of the parameter values adopted in (54)-(56) and appropriate initial conditions, we will perform some numerical simulations with the help of Matlab. The numerical simulations show that the drug spread steady state E is asymptotically stable if it exists (see Figure 1).

    Figure 1.  If R0=1.0117>1, the drug spread steady state E of system (1) is asymptotically stable with initial conditions S0=15000, U10=100, U20(0)=10, U20(θ)=0 for θ(0,¯θ], V0(0)=3000, V0(a)=0 for a[0,¯a].

    Furthermore, we want to illustrate that the stability of the drug spread steady state is not dependent on the initial conditions by numerical simulations. Let

    Λ=103, β=7×107, μ=0.001, δ1=0.02, δ2=0.01,  ψ=0.5, σ=0.85, γ=0.8. (57)

    We obtain that R0=7.6102>1 and simulate the solutions of system (1) under four pairs of initial values (see Figure 2). The numerical simulation results show that the stability of the drug spread steady state is not dependent on the initial conditions. In this case, we may conjecture that the drug spread steady state is globally asymptotically stable whenever it exists.

    Figure 2.  If R0=7.6102>1, the solutions of system (1) approach the drug spread steady state E of system (1) with four different initial conditions.

    In the course of the proof of Lemma 4.2, we let (λ)0 and want to prove that (Z)>0, where

    Z=σ2β2U1VB2(λ)+β2U1SB1(λ)+σψβ2U1SB1(λ)B2(λ)(A.1).

    To study the sign of \Re(Z), we let \lambda = x + iy with x\geq0 which is assumed nonnegative in the proof of Lemma 4.2. Substituting \lambda = x + iy into (A.1), by some calculations, we can have that

    \begin{align*} \Re\left(\frac{\sigma^2\beta^2U_1^*V^*}{B_2(\lambda)}\right) = &\frac{(x+B_{22})\sigma^2\beta^2U_1^*V^*}{(x+B_{22})^2+y^2}, \\ \Re\left(\frac{\beta^2U_1^*S^*}{B_1(\lambda)}\right) = &\frac{(x+B_{11})\beta^2U_1^*S^*}{(x+B_{11})^2+y^2}, \\ \Re\left(\frac{\sigma\psi\beta^2U_1^*S^*}{B_1(\lambda)B_2(\lambda)}\right) = &\frac{[(x+B_{11})(x+B_{22})-y^2]\sigma\psi\beta^2U_1^*S^*}{[(x+B_{11})^2+y^2][(x+B_{22})^2+y^2]}, \end{align*}

    where

    B_{11}(\lambda) = \mu+\psi+\beta U_1^*, \quad B_{22} = \mu+\sigma\beta U_1^*.

    Summing the above three terms, we have that

    \begin{align*} \Re(Z) = &\Re\left(\frac{\sigma^2\beta^2U_1^*V^*}{B_2(\lambda)}\right)+\Re\left(\frac{\beta^2U_1^*S^*}{B_1(\lambda)}\right) +\Re\left(\frac{\sigma\psi\beta^2U_1^*S^*}{B_1(\lambda)B_2(\lambda)}\right)\\ = &\frac{1}{C}\Big\{(x+B_{22})\sigma^2\beta^2U_1^*V^*[(x+B_{11})^2+y^2]\\ &+(x+B_{11})\beta^2U_1^*S^*[(x+B_{22})^2+y^2]\\ &+[(x+B_{11})(x+B_{22})-y^2]\sigma\psi\beta^2U_1^*S^*\Big\}\\ = &\frac{1}{C}\Big\{D_1+D_2y^2\Big\}, \end{align*}

    where

    \begin{align*} C = &[(x+B_{11})^2+y^2][(x+B_{22})^2+y^2] \gt 0, \\ D_1 = &(x+B_{22})\sigma^2\beta^2U_1^*V^*(x+B_{11})^2+(x+B_{11})\beta^2U_1^*S^*(x+B_{22})^2\\ &+(x+B_{11})(x+B_{22})\sigma\psi\beta^2U_1^*S^* \gt 0, \\ D_2 = &(x+B_{22})\sigma^2\beta^2U_1^*V^*+(x+B_{11})\beta^2U_1^*S^*-\sigma\psi\beta^2U_1^*S^*\\ = &(x+B_{22})\sigma^2\beta^2U_1^*V^*+\left(x+\mu+\beta U_1^*+(1-\sigma)\psi\right)\beta^2U_1^*S^* \gt 0. \end{align*}

    It then follows that the real part of Z is positive, i.e., \Re(Z)>0.

    To consider the roots of Equation (47), we let \Re(\lambda)\geq0 and want to prove that \Re(Z_1)>0, where

    Z_1 = \frac{\beta^2S^*U^*_1}{\lambda+\phi-\widehat{\Phi}(\lambda)}\quad\quad\quad\quad{(B.1)}.

    Let \lambda = x + iy with x>0. By substituting \lambda = x + iy into (B.1), we have that

    Z_1 = \frac{\beta^2S^*U^*_1}{x+\phi-\int_0^\infty\Phi(\theta)e^{-(x+iy)\theta}d\theta+iy} = \frac{\beta^2S^*U^*_1}{E_1+iE_2},

    and

    \Re(Z_1) = \frac{E_1}{E_1^2+E_2^2}\beta^2S^*U^*_1.

    where E_1 = x+\phi-\int_0^\infty\Phi(\theta)e^{-x\theta}\cos (y\theta) d\theta and E_2 = y+\int_0^\infty\Phi(\theta)e^{-x\theta}\sin (y\theta) d\theta. It follows from

    E_1 = x+\phi-\int_0^\infty\Phi(\theta)e^{-x\theta}\cos (y\theta) d\theta\geq x+\phi-\int_0^\infty\Phi(\theta)e^{-x\theta}d\theta\geq x\geq 0

    we have that \Re(Z_1) is nonnegative.

    If \sigma = 1. Set V(t) = \int_0^\infty V(a, t)da. It follows from the last equation of system (1) that we have

    \begin{align*} &\frac{dV(t)}{dt} = \int_0^\infty\frac{ \partial V(a, t)}{\partial t}da\\ = &\int_0^\infty\left(-\frac{\partial V(a, t)}{\partial a}-(\mu+\alpha(a)+\beta U_1(t))V(a, t))\right)da\\ = &V(a, t)\Big|_{a = \infty}^{a = 0}-(\mu+\beta U_1(t))\int_0^\infty V(a, t)da-\int_0^\infty\alpha(a)V(a, t)da\\ = &\psi S(t)- (\mu+\beta U_1(t))V(t)-\int_0^\infty\alpha(a)V(a, t)da. \end{align*}

    Denote that \widehat{S}(t): = S(t)+V(t). By dropping the hat, we have that

    \frac{dS(t)}{dt} = \Lambda-\mu S(t)-\beta S(t) U_1(t).

    Then system (1) can be rewritten as the main system in [10] and the drug spread steady state E^* is locally stable.

    We would be very grateful to anonymous referees for their comments and suggestions that helped to improve this paper. This work is supported partially by China Postdoctoral Science Foundation 2017M621523; X. Li is supported partially by the National Natural Science Foundation of China (11771017); M. Martcheva is supported partially through grant DMS-1515661. Part of this work was done when XD was a visiting scholar at the Department of Mathematics, University of Florida. XD would like to thank the Department for kind hospitality he received there.

    The authors declare there is no conflict of interest.



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