
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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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
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,
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-Hern
To study the role of the heroin vaccine in the control of heroin abuse, adding a compartment
Motivated by the development of the heroin vaccine, in this paper, we present an age structured heroin transmission
{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
{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
In system (1)-(2),
Throughout the paper, we make the following assumptions: (A1) there is a positive number of drug users not in treatment, i.e.,
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)=e−∫θ0(μ+δ2+p(τ))dτ, ∫∞0Φ1(θ)dθ=ϕ1, |
Φ(θ)=γp(θ)e−∫θ0(μ+δ2+p(τ))dτ, ∫∞0Φ(θ)dθ=ϕ,k1(a)=e−∫a0(μ+α(τ))dτ, ∫∞0k1(a)da=K1,k(a)=α(a)e−∫a0(μ+α(τ))dτ, ∫∞0k(a)da=K. |
By simple calculations, we have that
Naturally, system (1)-(2) has a unique drug-free steady state
{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)=ψS0e−∫a0(μ+α(τ))dτ=ψS0k1(a). | (4) |
Substituting Equation (4) into the first equation of (3), we have that
S0=Λμ+ψ(1−K)=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=βμ+δ1+γ−ϕ(S0+σ∫∞0V0(a)da) =βμ+δ1+γ(μ+δ2)ϕ1⋅Λμ(1+ψK1)⋅(1+σψK1). | (7) |
According to [24], any positive equilibrium (
{0=Λ−βS∗U∗1−(μ+ψ)S∗+∫∞0α(a)V∗(a)da,0=βS∗U∗1+σβU∗1∫∞0V∗(a)da−(μ+δ1+γ)U∗1+∫∞0p(θ)U∗2(θ)dθ,dU∗2(θ)dθ=−(μ+δ2+p(θ))U∗2(θ),U∗2(0)=γU∗1,dV∗(a)da=−(μ+α(a)+σβU∗1)V∗(a),V∗(0)=ψS∗. | (8) |
Let
π(a)=e−∫a0(μ+α(τ)+σβU∗1)dτ, Kπ=∫∞0π(a)da. | (9) |
By simple calculations, we have that
∫∞0ψα(a)π(a)da=ψ−ψ(μ+σβU∗1)Kπ. | (10) |
From the third and the forth equation of (8), we have that
U∗2(θ)=γU∗1Φ1(θ). | (11) |
From the fifth and the sixth equation of (8), we have that
V∗(a)=ψS∗π(a). | (12) |
Substituting (11) and (12) into the second and the first equation of (8) respectively, we have that the following equations
{0=Λ−βS∗U∗1+(ψ−ψ(μ+σβU∗1)Kπ)S∗−(μ+ψ)S∗0=βS∗U∗1+σβU∗1KπψS∗−(μ+δ1+γ)U∗1+ϕU∗1. | (13) |
It follows from the first equation of (13), we have that
S∗=ΛβU∗1+ψ(μ+σβU∗1)Kπ+μ. | (14) |
Substituting (14) into the second equation of (13) and eliminating
1=βΛ(1+σψKπ)(βU∗1+ψ(μ+σβU∗1)Kπ+μ)(μ+δ1+γ−ϕ). | (15) |
Define a function
Summarizing the above discussion, we have the following result.
Theorem 2.1. System (1) always has a drug-free steady state
Now, by setting
N(t)=S(t)+U1(t)+∫∞0U2(θ,t)dθ+∫∞0V(a,t)da, |
we deduce from (1) that
N′(t)=Λ−μN(t)−δ1U1(t)+δ2∫∞0U2(θ,t)dθ≤Λ−μN(t), | (16) |
and therefore
Ω={(S,U1,U2,V)∈R+×R+×L1+(0,∞)×L1+(0,∞):S+U1+∫∞0U2(θ,⋅)dθ+∫∞0V(a,⋅)da≤Λμ} |
Then
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+=X0∩X+. |
Let the linear operator
A(SU1(0U2)(0V))=(−(μ+ψ)S−(μ+δ1+γ)U1(−U2(0)−U′2−(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
F(t)=(Λ−βS(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)fort≥0,withu(0)=x∈X0+. | (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
In this section, by use of characteristic equation, we will prove the local and global stability of the drug-free steady state
For the sake of convenience, we give the following Laplace transforms of the corresponding functions
ˆΦ(λ)=∫∞0Φ(θ)e−λθdθ, ˆΦ1(λ)=∫∞0Φ1(θ)e−λθdθ,ˆK(λ)=∫∞0k(a)e−λada, ˆK1(λ)=∫∞0k1(a)e−λada. | (18) |
By direct calculations, we have the following relationships
ˆK(λ)=1−(λ+μ)ˆK1(λ) and ˆΦ(λ)=γ−γ(λ+μ+δ2)ˆΦ1(λ). | (19) |
Theorem 3.1. The drug-free steady state
Proof. By linearization of system (1) at the drug-free steady state
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
{d˜S(t)dt=−(μ+ψ)˜S(t)−βS0˜U1(t)+∫∞0α(a)˜V(a,t)da,d˜U1(t)dt=βS0˜U1(t)+σβ∫∞0V0(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
˜S(t)=¯xeλt, ˜U1(t)=¯yeλt, ˜U2(θ,t)=¯z(θ)eλt, and ˜V(a,t)=¯w(a)eλt |
where
{λ¯x=−(μ+ψ)¯x−βS0¯y(t)+∫∞0α(a)¯w(a)da,λ¯y=βS0¯y+σβ∫∞0V0(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)e−λθΦ1(θ)=γ¯ye−λθΦ1(θ). | (22) |
Solving the fifth equation of (21), we have
¯w(a)=¯w(0)e−λak1(a)−σβ¯y∫a0e−λ(a−s)k1(a)k1(s)V0(s)ds=ψ¯xe−λak1(a)−σβψS0k1(a)¯y∫a0e−λ(a−s)ds=ψ¯xe−λak1(a)−σβψS0k1(a)¯y1λ(1−e−λa). | (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(ˆK−ˆK(λ))),A22=λ+(μ+δ1+γ)−ˆΦ(λ)−σβψS0K1−βS0. |
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)−σβψS0K1−βS0=0. | (26) |
Obviously,
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
Assume that
x≥0⇒1=|H(λ)|≤|H(x)|≤H(0)=R0, i.e., R0≥1. |
Thus, we can have that
In the following, we will use the Fluctuation Lemma to establish the global stability of the drug-free steady state
g∞=lim inft→∞g(t) and g∞=lim supt→∞g(t). |
Then the Fluctuation Lemma is given as follows.
Lemma 3.2. (Fluctuation Lemma [13]) Let
Lemma 3.3. [15] Suppose
lim supt→∞∫t0h(θ)f(t−θ)dθ≤f∞‖h‖1, |
where
Using integration,
U2(θ,t)={γU1(t−θ)Φ1(θ), if t≥θ,U2(θ−t,0)Φ1(θ)Φ1(θ−t), if θ≥t. | (29) |
V(a,t)={ψS(t−a)e−∫a0(μ+α(τ)+σβU1(t−τ))dτ, if t≥a,V(a−t,0)e−∫t0(μ+α(a−τ)+σβU1(τ))dτ, if a≥t. | (30) |
Theorem 3.4. If
Proof. Theorem 3.1 shows that the drug-free steady state
{dSdt=Λ−βS(t)U1(t)+∫t0ψα(a)e−∫a0(μ+α(τ)+σβU1(t−τ))dτS(t−a)da −(μ+ψ)S(t)+FV(t)dU1dt=βS(t)U1(t)+σβU1(t)∫t0ψe−∫a0(μ+α(τ)+σβU1(t−τ))dτS(t−a)da −(μ+δ1+γ)U1(t)+∫t0Φ(θ)U1(t−θ)dθ+FU(t)+FUV(t), | (31) |
where
FV(t)=∫∞tψα(a)V(a−t,0)e−∫t0(μ+α(a−τ)+σβU1(τ))dτda,FU(t)=∫∞tp(θ)U2(θ−t,0)Φ1(θ)Φ1(θ−t)dθ,FUV(t)=σβU1(t)∫∞tV(a−t,0)e−∫t0(μ+α(a−τ)+σβU1(τ))dτda |
with
Choose the sequences
0=Λ−βS∞U1(t)−(μ+ψ)S∞+S∞ψK, |
and
S∞≤Λμ+ψ−ψK. |
Choose the sequences
0≤βS∞U∞1+σβψK1S∞U∞1−(μ+δ1+γ)U∞1+ϕU∞1=(βS∞(1+σψK1)−(μ+δ1+γ)+ϕ)U∞1≤(βΛ(1+σψK1)μ+ψ−ψK−(μ+δ1+γ)+ϕ)U∞1=(μ+δ1+γ−ϕ)(R0−1)U∞1. |
Thus we obtain that
Choose the sequences
0=Λ−βS∞U∞1−(μ+ψ)S∞+ψKS∞=Λ−(μ+ψ)S∞+ψKS∞. |
It follows that
Λμ+ψ−ψK=S∞≤S∞≤Λμ+ψ−ψK, |
which implies that
limt→∞V(a,t)=ψΛμ+ψ−ψKk1(a)=V0(a). |
Therefore,
This section aims to establish the stability of the drug spread steady state of system (1) in terms of the basic reproduction number
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.,
lim inft→∞S(t)≥ε, lim inft→∞U1(t)≥ε, lim inft→∞‖U2(⋅,t)‖L1+≥ε |
and
For the sake of convenience, we let
ˆKπ(λ):=∫∞0π(a)e−λada, and ˆKα(λ)=∫∞0α(a)π(a)e−λada. | (32) |
It then follows that
ˆKα(λ)=1−(λ+μ+σβU∗1)ˆKπ(λ). | (33) |
In the following, we try to study the stability of the drug spread steady state
S(t)=˜S(t)+S∗, U1(t)=˜U1(t)+U∗1, U2(θ,t)=˜U2(θ,t)+U∗2(θ) |
and
{d˜S(t)dt=−(μ+ψ)˜S(t)−βU∗1˜S(t)−βS∗˜U1(t)+∫∞0α(a)˜V(a,t)da,d˜U1(t)dt=βU∗1˜S(t)+βS∗˜U1(t)+σβU∗1∫∞0˜V(a,t)da+σβ∫∞0V∗(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)+σβU∗1)˜V(a,t)−σβV∗(a)˜U1(t),˜V(0,t)=ψ˜S(t). | (34) |
To analyze the asymptotic behaviors around
˜S(t)=¯xeλt, ˜U1(t)=¯yeλt, ˜U2(θ,t)=¯z(θ)eλt, and ˜V(a,t)=¯w(a)eλt |
where
{λ¯x=−(μ+ψ)¯x−βU∗1¯x−βS∗¯y(t)+∫∞0α(a)¯w(a)da,λ¯y=βU∗1¯x+βS∗¯y+σβU∗1∫∞0¯w(a)da+∫∞0p(θ)¯z(θ)dθ +σβ∫∞0V∗(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)+σβU∗1)¯w(a)−σβV∗(a)¯y,¯w(0)=ψ¯x. | (35) |
Solving the third equation of (35), we have
¯z(θ)=¯z(0)e−λθΦ1(θ)=γ¯ye−λθΦ1(θ). | (36) |
Solving the fifth equation of (35), we have
¯w(a)=¯w(0)e−λaπ(a)−σβ¯y∫a0e−λ(a−s)π(a)π(s)V∗(s)ds=ψ¯xe−λaπ(a)−σβ¯yf(a,λ), | (37) |
where
f(a,λ)=∫a0e−λ(a−s)π(a)π(s)V∗(s)ds=ψS∗π(a)∫a0e−λ(a−s)ds=ψS∗π(a)∫a0e−λsds=ψS∗π(a)1λ(1−e−λa). |
Substituting (37) into the first equation of (35), we have
¯x=−¯yG11(βS∗+σβ∫∞0α(a)f(a,λ)da)=−¯yG11[βS∗+σβψS∗1λ∫∞0α(a)π(a)(1−e−λa)da] | (38) |
where
G11=λ+(ψ+μ+βU∗1)−ψˆKα(λ). |
It follows from (33) we have that
G11=(λ+μ)(1+ψˆKπ(λ))+βU∗1(1+σψˆKπ(λ)). |
Substituting (36) and (37) into the second equation of (35), by use of
(λ+ϕ−Φ(λ)+σ2β2U∗1∫∞0f(a,λ)da)¯y=(βU∗1+σψβU∗1ˆKπ(λ))¯x. |
Substituting (38) into the above equation and dividing both sides by
λ+ϕ−Φ(λ)+σ2β2ψS∗U∗11λ∫∞0π(a)(1−e−λa)da=−(βU∗1+σψβU∗1ˆKπ(λ))1G11 ⋅[βS∗+σβψS∗1λ∫∞0α(a)π(a)(1−e−λa)da]. | (39) |
It follows from Equation (39) we have that
G11(λ+ϕ−ˆΦ(λ))+β2S∗U∗1(1+σψˆKπ(λ))2+1λσ2β2ψS∗U∗1G11(Kπ−ˆKπ(λ))=1λσβ2ψS∗U∗1(Kπ−ˆKπ(λ))(μ+σβU∗1)(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
Case (ⅰ) We assume that
{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
Proof. Linearizing system (41) at its positive steady state
|λ+μ+ψ+βU∗1βS∗0−βU∗1λ+ϕ−ˆΦ(λ)−σβU∗1−ψσβV∗λ+μ+σβU∗1|=0. | (42) |
By simple calculations, we have the following
B1(λ)B2(λ)(λ+ϕ−ˆΦ(λ))+B1(λ)σ2β2U∗1V∗+B2(λ)β2U∗1S∗+σψβ2U∗1S∗=0, | (43) |
where
B1(λ)=λ+μ+ψ+βU∗1,B2(λ)=λ+μ+σβU∗1. |
Obviously, if
λ+ϕ+Z=ˆΦ(λ), | (44) |
where
Z=σ2β2U∗1V∗B2(λ)+β2U∗1S∗B1(λ)+σψβ2U∗1S∗B1(λ)B2(λ). |
Now, assume that
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
Case (ⅱ) We assume that
{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
Proof. It then follows from (18) and (19) we have that Equation (40) can be modified as
(λ+μ+ψ+βU∗1−ˆSα(λ))(λ+ϕ−ˆΦ(λ))+β2S∗U∗1=0. | (46) |
Since
λ+μ+ψ+βU∗1+Z1=ˆSα(λ) | (47) |
where
Z1=β2S∗U∗1λ+ϕ−ˆΦ(λ). |
Assuming
ψ<|μ+ψ+βU∗1|≤|λ+μ+ψ+βU∗1+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
Based on the persistence results in Theorem 4.1 and the local stability results of
Set
ε1(θ)=∫∞θp(τ)e−∫τθ(μ+δ2+p(s))dsdτ and ε2(a)=∫∞aα(τ)V∗(τ)dτ. | (49) |
Note that
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)=S∗g(S(t)S∗),WU1(t)=U∗1g(U1(t)U∗1),WU2(t)=∫∞0ε1(θ)U∗2(θ)g(U2(θ,t)U∗2(θ))dθ,WV(t)=∫∞0ε2(a)g(V(a,t)V∗(a))da, | (52) |
Then
dWS(t)dt=−(μ+ψ)S∗(S∗S(t)+S(t)S∗−2)+βS∗U∗1(1−S∗S(t))(1−S(t)S∗U1(t)U∗1)+∫∞0α(a)V∗(a)(V(a,t)V∗(a)−V(a,t)V∗(a)S∗S(t)−1+S∗S(t))da,dWU1(t)dt=βS(t)U1(t)−βS∗U1(t)−βS(t)U∗1+βS∗U∗1+∫∞0p(θ)U∗2(θ)(U2(θ,t)U∗2(θ)−U1(t)U∗1−U∗1U1(t)U2(θ,t)U∗2(θ)+1)dθ,dWU2(t)dt=−ε1(θ)U∗2(θ)g(U2(θ,t)U∗2(θ))|θ=∞+ϕU∗1g(U1(t)U∗1)−∫∞0p(θ)U∗2(θ)g(U2(θ,t)U∗2(θ))dθ,dWV(t)dt=−ε2(a)g(V(a,t)V∗(a))|a=∞+∫∞0α(a)V∗(a)g(V(0,t)V∗(0))da−∫∞0α(a)V∗(a)g(V(a,t)V∗(a))da=+∫∞0α(a)V∗(a)(S(t)S∗−lnS∗S(t)−V(a,t)V∗(a)+lnV(a,t)V∗(a))da−ε2(a)g(V(a,t)V∗(a))|a=∞. |
By adding
dWS(t)dt+dWV(t)dt=βS∗U∗1(1−S∗S(t))(1−S(t)S∗U1(t)U∗1)−(μ+ψ)S∗(S∗S(t)+S(t)S∗−2)+∫∞0α(a)V∗(a)(S∗S(t)+S(t)S∗−2)da−∫∞0α(a)V∗(a)g(V(a,t)V∗(a)S∗S(t))da−ε2(a)g(V(a,t)V∗(a))|a=∞=βS∗U∗1(1−S∗S(t))(1−S(t)S∗U1(t)U∗1)−∫∞0α(a)V∗(a)g(V(a,t)V∗(a)S∗S(t))da−ε2(a)g(V(a,t)V∗(a))|a=∞+(S∗S(t)+S(t)S∗−2)(∫∞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(S∗S(t)+S(t)S∗−2)−ε1(θ)U∗2(θ)g(U2(θ,t)U∗2(θ))|θ=∞−∫∞0p(θ)U∗2(θ)g(U∗1U1(t)U2(θ,t)U∗2(θ))dθ−ε2(a)g(V(a,t)V∗(a))|a=∞≤0. |
Notice that equality holds only if
Case (ⅲ) We assume that
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
Recalling that the expression of
βΛμ(μ+δ1+γ(μ+δ2)ϕ1)1+σψK1(1+ψK1)≤βΛμ(μ+δ1+γ(μ+δ2)ϕ1). | (53) |
It implies that
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.,
α(a)={0.10(a−10)2e−0.35(a−10), 10<a≤40;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)e−0.2(θ+5), | (55) |
for
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=3.5×10−7, 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
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=7×10−7, μ=0.001, δ1=0.02, δ2=0.01, ψ=0.5, σ=0.85, γ=0.8. | (57) |
We obtain that
In the course of the proof of Lemma 4.2, we let
Z=σ2β2U∗1V∗B2(λ)+β2U∗1S∗B1(λ)+σψβ2U∗1S∗B1(λ)B2(λ)(A.1). |
To study the sign of
\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
To consider the roots of Equation (47), we let
Z_1 = \frac{\beta^2S^*U^*_1}{\lambda+\phi-\widehat{\Phi}(\lambda)}\quad\quad\quad\quad{(B.1)}. |
Let
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\geq x+\phi-\int_0^\infty\Phi(\theta)e^{-x\theta}d\theta\geq x\geq 0 |
we have that
If
\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
\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
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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