# Global Dynamics of a Discretized Heroin Epidemic Model with Time Delay

Authors: Xamxinur Abdurahman; Ling Zhang; Zhidong Teng
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We derive a discretized heroin epidemic model with delay by applying a nonstandard finite difference scheme. We obtain positivity of the solution and existence of the unique endemic equilibrium. We show that heroin-using free equilibrium is globally asymptotically stable when the basic reproduction numberR0<1, and the heroin-using is permanent when the basic reproduction number R0>1.


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## Body

## 1. Introduction

As we all know, the use of heroin and other drugs in Europe and more specifically in Ireland and the resulting prevalence are well documented [1–3]. It shows that the use of heroin is very popular and causes many preventable deaths. Heroin is so soluble in the fat cells that it crosses the blood-brain barrier within 15–20 seconds, rapidly achieving a high level syndrome in the brain and central nervous system which causes both the “rush” experience by users and the toxicity. Heroin-related deaths are associated with the use of alcohol or other drugs [4]. Treatment of heroin users is a huge burden on the health system of any country.We often study infectious diseases with mathematical and statistical techniques; see, for example, [5–11]; however, little has been done to apply this method to the heroin epidemics. In 1979, Mackintosh and Stewart [9] considered an exponential model which is simplified from infectious disease model of Kermack and McKendrick to illustrate how the heroin-using spreads in epidemic fashion. They arranged a numerical simulation to show how the dynamics of spread are influenced by parameters in the model. White and Comiskey [5] attempted to extend dynamic disease modeling to the drug-using career and formulated an ordinary differential equation. They divided the whole population into three classes, namely, susceptible, heroin users, and heroin users undergoing treatment. Their model allows a steady state (constant) solution which represents an equilibrium between the number of susceptible, heroin users, and heroin users in treatment. Furthermore, this ODE model was revisited by Mulone and Straughan [12]; the authors proved that this equilibrium solution is stable both linearly and nonlinearly under the realistic condition in which relapse rate of those in treatment returning to untreated drug use is greater than the prevalence rate of susceptible becoming drug users. Recently, the study of the global properties and permanence of continuous heroin epidemic models attracted the researchers and have some very good results; see [13–16]. Specially, Samanta [15] considered a model with time-dependent coefficients and with different removal rates for three different classes, introduced some new threshold values R* and R*, and obtained the permanence of heroin-using career.Motivated by Samanta [15] and Zhang and Teng [8], we alter a nonautonomous heroin epidemic model with time delay to an autonomous heroin epidemic model. For convenience, we replace U1 and U2 by U and V, respectively. Thus, we obtain the following continuous heroin epidemic model with a distributed time delay:
(1)S˙(t)=λ-β1(U)S(t)∫0h‍U(t-s)dη(s)mmki-μ1S(t)+ξ1U(t)+ξ2V(t),U˙(t)=β1(U)S(t)∫0h‍U(t-s)dη(s)imikm+β3U(t)V(t)-(μ2+P+ξ1)U(t),V˙(t)=PU(t)-β3U(t)V(t)-(μ3+ξ2)V(t),
where S(t), U(t), and V(t) represent the number of susceptible, heroin users not in treatment, and heroin users in treatment, respectively. We assume that the time taken to become heroin user is s. The function η(s):[0,h]→[0,∞) is nondecreasing and has bounded variation such that ∫0h‍η(s)ds=η(h)-η(0)=1.For understanding more realistic phenomenon of heroin, a little complicated epidemic model is helpful. By applying Micken’s nonstandard discretization method [17] to continuous heroin epidemics model with time delay (1), we derive the following discretized heroin epidemic model with a distributed time delay:
(2)Sn+1-Sn=λ-β1(Un)Sn+1∑k=0h‍Un-kηkmmmimm-μ1Sn+1+ξ1Un+1+ξ2Vn+1,Un+1-Un=β1(Un)Sn+1∑k=0h‍Un-kηkmmmmmii+β3Un+1Vn+1-(μ2+P+ξ1)Un+1,Vn+1-Vn=PUn+1-β3Un+1Vn+1-(μ3+ξ2)Vn+1,
where Sn is the susceptible class, Un is the class of heroin users not in treatment, and Vn is the class of heroin users in treatment at nth step. Since the sufficient condition can be obtained, independently of the choice of a time step-size, we let the time step-size be one for the sake of simplicity. The nonnegative constants μ1, μ2, and μ3 denote the death rate of the susceptible, heroin users not in treatment, and heroin users in treatment class, respectively. Throughout the paper, it is biologically natural to assume that μ1≤min⁡{μ2,μ3}. The constant λ>0 denotes the recruitment rate of susceptible population from the general population. Constant P>0 is the proportion of heroin users who enter the treatment class. The individuals in treatment who stop using heroin are susceptible at a constant rate ξ2≥0. Constant β3 represents the transmission rate from heroin users in treatment to untreated heroin users. β1(Un) is the probability per unit time and the transmission is used with the form β1(Un)Sn+1∑k=0h‍Un-kηk, which includes various delays. By a natural biological meaning, we assume that β1(U) is a positive function and that there exists a constant Uβ>0 such that β1(U) is nondecreasing on the interval [0,Uβ]. The integer h≥0 is the time delay. The sequence ηk:-∞<ηk<+∞(k=0,1,…,h) is nondecreasing and has bounded.The initial conditions of the system (2) are given by
(3)Sn=ψn(1),Un=ψn(2),Vn=ψn(3),forn=-h,-h+1,…,0,
where ψn(i)≥0(n=-h,-h+1,…,0,i=1,2,3). Again, by biological meaning, we further assume that ψ0(i)>0 for all i=1,2,3.The paper is organized as follows. In Section2, we prove the positivity and boundedness of the solution of system (2). In Section 3, we deal with the global asymptotic stability of the heroin-using free equilibrium. In Section 4, we consider the permanence of the discrete epidemic model applying Wang’s technique. In the discretized epidemic model, sufficient condition for global asymptotic stability and permanence are the same as for the original continuous epidemic model. We give some numerical examples and conclusion in Sections 5 and 6.


## 2. Basic Properties

For system (2), the heroin-using free equilibrium is given by
(4)E0=(S0,0,0),S0=λμ1.
Define a positive constant A≡∑k=0h‍ηk. The stability of E0 is studied by using the next generation method in [7]. The associated matrix F (of the new heroin-using terms) and the M-matrix V (of the remaining transfer terms) are given as follows, respectively:
(5)F=(β1(0)λAμ1000),V=(μ2+P+ξ10-Pμ3+ξ2).
Clearly, F is nonnegative, V is a nonsingular M-matrix, and V-F has Z sign pattern. The associated basic reproduction number, denoted by R0, is then given by R0=ρ(FV-1), where ρ is the spectral radius of FV-1. It follows that
(6)R0=β1(0)λAμ1(μ2+P+ξ1).Now, we will consider the positivity and boundedness of solution to system (2). For most continuous epidemic models, positivity of the solution is clear, but, for system (2), the positivity of the sequences Sn,Un, and Vn holds in some condition.Lemma 1.
Let(Sn,Un,Vn) be any solution of system (2) with initial condition (3); then (Sn,Un,Vn) is positive for any n∈N and V0<P(1+μ3+ξ2)/β3.Proof.
Let(Sn,Un,Vn) be any solution of system (2) with initial condition (3). It is evident that system (2) is equivalent to the following iteration system:
(7)Un+1=β1(Un)Sn+1∑k=0hUn-kηk+Un1-β3Vn+1+μ2+P+ξ1,Vn+1=Vn+PUn+11+β3Un+1+μ3+ξ2,Sn+1=λ+Sn+ξ1Un+1+ξ2Vn+11+β1(Un)∑k=0hUn-kηk+μ1.
In the following, we will use the induction to prove the positivity of solution. When n=0, we have
(8)U1=β1(U0)S1∑k=0hU-kηk+U01-β3V1+μ2+P+ξ2,(9)V1=V0+PU11+β3U1+μ3+ξ2,(10)S1=λ+S0+ξ1U1+ξ2V11+β1(U0)∑k=0hU-kηk+μ1.
From (8)–(10), we see that, as long as U1 is obtained, V1 and S1 will be obtained too.
IfU1>0, from (9), we directly obtain V1>0 and, from (10), we further obtain that S1>0. Furthermore, we also have N1=S1+U1+V1>0.
Letx=U1; then, from (8)–(10), we see that x satisfies the following equation:
(11)ψ(x)=x-((1+β1(U0)∑k=0hU-kηk+μ1)U0mmmimk+β1∑k=0hU-kηkW1(x))×((1+β1(U0)∑k=0hU-kηk+μ1)W2(x))-1,
where
(12)W1(x)=λ+S0+ξ1x+ξ2V1,W2(x)=1+μ2+P+ξ1-β3V1,V1=V0+Px1+β3x+μ3+ξ2.
Substituting V1 in W2(x) gets
(13)W2(x)=1+μ2+ξ1+P(1+μ3+ξ2)-β3V01+μ3+ξ2+β3x.
Since V0<P(1+μ3+ξ2)/β3, we have
(14)W2(0)=1+μ2+P+ξ1-β3V01+μ3+ξ2>0,
and because of W1(0)=λ+S0+ξ2V0/(1+μ3+ξ2)>0, thus
(15)ψ(0)=-((1+β1(U0)∑k=0hU-kηk+μ1)U0mmmm+β1(U0)∑k=0hU-kηkW1(0))×((1+β1(U0)∑k=0hU-kηk+μ1)W2(0))-1<0.
Substituting W1 and W2 in ψ(x) yields
(16)ψ(x)=ax2+bx-c(1+β1(U0)∑k=0hU-kηk+μ1)(mx+n).
Here, constants a, b, c, m, and n are as follows:
(17)a=(1+μ2)(1+β1(U0)∑k=0hU-kηk+μ1)+ξ1(1+μ1)>0,b=(1+μ3+ξ2)(1+μ1)(1+μ2+P+ξ1)mk+(1+μ1)(β1(U0)∑k=0hU-kηk+β3S0-β3N0)mk+β1(U0)∑k=0hU-kηkP(1+μ3)mk-β3β1(U0)∑k=0hU-kηk(λ+N0),c=β1(U0)∑k=0hU-kηk[(1+μ3+ξ2)(λ+N0)-(1+μ3)V0]mk-U0(1+μ1)ξ1,m=β3(1+μ2+ξ1)>0,n=(1+μ2+P+ξ1)(1+μ3+ξ2)-β3V0>0.
We take the limit on both sides of the above equation:
(18)lim⁡x→+∞ψ(x)=lim⁡x→+∞ax2+bx-c(1+β1(U0)∑k=0hU-kηk+μ1)(mx+n)=+∞;
this means that ψ(x)=0 has at least one positive solution x-∈(0,+∞). So, we have U1=x->0. Therefore, the positivity of S1>0, U1>0, and V1>0 is finally obtained. When n=2, we have
(19)U2=β1S2∑k=0hU1-kηk+U11-β3V2+μ2+P+ξ1,V2=V1+PU21+β3U2+μ3+ξ2,S2=λ+S1+ξ1U2+ξ2V21+β1∑k=0hU1-kηk+μ1,
a similar argument as in the above for U1, V1, and S1; we also can obtain that U2>0, V2>0, and S2>0. Lastly, by using the induction, we can finally obtain that Sn>0, Un>0, and Vn>0, for all n>0.
Now, we define the total population asNn=Sn+Un+Vn. Then, from system (2), we know that
(20)Nn+1-Nn=λ-μ1Sn+1-μ2Un+1-μ3Vn+1.
Notice the assumption that μ1≤min⁡(μ2,μ3); we obtain
(21)Nn≤λ+Nn-11+μ1≤λ1+μ1×{1+11+μ1+⋯+(11+μ1)n-1}+(11+μ1)nN0=λμ1{1-(11+μ1)n}+(11+μ1)nN0≤max⁡{λμ1,N0}.
If λ/μ1≥N0, it is easy to see that Nn≤λ/μ1=S0, for all large n. If λ/μ1<N0, from right hand side of system (2), we obtain
(22)N1≤λ+N01+μ1<N0.
Hence, we have N1<N0 and there exists i∈N such that Ni≤λ/μ1=S0. Then, we may use this Ni as a starting value instead of N0. This argument leads to the following result.Lemma 2.
For any solution(Sn,Un,Vn) of system (2), the total population Nn=Sn+Un+Vn satisfies
(23)limsup⁡n→+∞Nn≤S0=λμ1;
thus (Sn,Un,Vn) is ultimately bounded.LetΩ={(Sn,Un,Vn):Sn,Un,Vn≥0,Sn+Un+Vn≤λ/μ1}; then Ω is the positive invariant set to the solution of system (2).In the following, we will examine the existence of endemic equilibrium for a special case of system (2).Lemma 3.
Assume thatβ1(U)=β1>0 is a constant. If R0>1, system (2) admits a heroin-using equilibrium E*=(S*,U*,V*) when V<P/β3, where E* satisfies following equality:
(24)λ-Aβ1S*U*-μ1S*+ξ1U*+ξ2V*=0,Aβ1S*U*+β3U*V*-(μ2+P+ξ1)U*=0,PU*-β3U*V*-(μ3+ξ2)V*=0.Proof.
Consider the following equation:(25)λ-Aβ1SU-μ1S+ξ1U+ξ2V=0,Aβ1SU+β3UV-(μ2+P+ξ1)U=0,PU-β3UV-(μ3+ξ2)V=0.
From the first equation and the second equation of the system (25), we have
(26)λ-μ1S+ξ2V+β3UV-(μ2+P)U=0.
From the second equation and the third equation of the system (25), we obtain
(27)Aβ1SU-(μ2+ξ1)U-(μ3+ξ2)V=0;
thus,
(28)U=(μ3+ξ2)VAβ1S-μ2+ξ1.
Since U≠0, from the second equation of the system, we have
(29)S=μ2+P+ξ1-β3VAβ1.
Substituting S in (28), we obtain
(30)U=(μ3+ξ2)VP-β3V.
Substituting U and S in (26) yields a quadratic equation of V as follows:
(31)F(V)=aV2+bV+c=0,
where the coefficients are given by
(32)a=μ1β32-Aβ1β3μ3,c=μ1P(μ2+P+ξ1)-λAβ1P,b=λAβ1β3-μ1β3(μ2+ξ1)+Aβ1μ3(μ2+P)+Aβ1μ2ξ2.
Since R0=λAβ1/μ1(μ2+P+ξ1)>1, then it is easy to see that c<0 and b>0. According to Descartes’ rule of signs, if a≥0, then F(V)=0 has a positive solution; if a<0, then F(V)=0 has two positive solutions. From the expression of S and U, we note that V<P/β3. Since
(33)F(0)=c<0,F(Pβ3)=Aβ1Pμ2(μ3+ξ2)β3>0.
This means that F(V)=0 has a unique positive solution V*∈(0,P/β3). Therefore, there exists a unique positive solution (S*,U*,V*) of system (2).For the local stability of the equilibria, we refer to Theorem 2 in [7] and have the following results.Theorem 4.
Assume thatβ1(U)=β1, β1 is a positive constant. The heroin-using free equilibrium E0=(S0,0,0) of system (2) is locally asymptotically stable if R0<1 and unstable if R0>1.


## 3. Global Asymptotic Stability of the Heroin-Using Free Equilibrium

In this section, we still assume thatβ1(U)=β1>0, and obtain a sufficient condition for global asymptotic stability of the heroin-using free equilibrium E0 of system (2).Using a Lyapunov function similar to that in [11], we can easily prove the global asymptotic stability of the heroin-using free equilibrium E0.Theorem 5.
IfR0<1, the drug-using free equilibrium E0 of system (2) is globally asymptotically stable.Proof.
Let us take the following Lyapunov function:(34)Hn=Un+c1Vn+c2∑k=0h‍(∑l=n-kn‍Ul)ηk+c32(Sn+1-S0)2,
where ci>0(i=1,2,3) are the constants to be defined later and S0=λ/μ1. Using system (2), the difference of Hn satisfies
(35)ΔH=Hn+1-Hn=Un+1-Un+c1(Vn+1-Vn)m+c2∑k=0h‍(Un+1ηk-Un-kηk)m+c32{(Sn+1-S0)2-(Sn-S0)2}.
From Sn≤Nn≤S0, for all n≥0, we have
(36)ΔH≤∑k=0h‍Un-kηk{β1Sn+1-c2-c3β1Sn+1(Sn+1-S0)}+(1-c1)β3Un+1Vn+1+[c2A-μ2-P-ξ1+c1P+c3ξ1(Sn+1-S0)]Un+1-[c1(μ3+ξ2)-c3ξ2(Sn+1-S0)]Vn+1-c3μ1(Sn+1-S0)2≤∑k=0n‍Un-kηk{β1Sn+1-c2-c3β1Sn+1(Sn+1-S0)}+(1-c1)β3Un+1Vn+1+[c2A-μ2-P-ξ1+c1P]Un+1-c1(μ3+ξ2)Vn+1-c3μ1(Sn+1-S0)2.
Let us choose ci>0(i=1,2,3) such that these constants satisfy the following inequalities:
(37)β1Sn+1-c2-c3β1Sn+1(Sn+1-S0)<0,(38)1-c1<0,(39)c1P+c2A<μ2+P+ξ1.
From (37), we have c3β1Sn+12+(c3β1S0-β1)Sn+1+c2>0; since Sn+1>0, then the following inequality is true:
(40)β1(1+c3S0)2<4c2c3;
that is,
(41)β1(S0)2c32+(2β1S0-4c2)c3+β1<0.
Since R0<1, which implies that β1AS0<μ2+P+ξ1, we can choose c2=β1S0+ϵ; here, ϵ(0<ϵ<(μ2+ξ1-Aβ1S0)/A) is a sufficiently small positive number such that β1AS0+Aϵ<μ2+(1-c1)P+ξ1. Since β1S0-2c2<0 and (β1S0-2c2)2>(β1S0)2, we can choose c3>0 to satisfy (41). We may further choose c1>1 to satisfy (38). Therefore, ΔH is negative definite and is equal to zero if and only if Sn+1=S0, Un+1=0, and Vn+1=0. The proof is complete.


## 4. Permanence of System (2)

The system (2) is said to be permanent if there are positive constants m and M such that
(42)m≤liminf⁡n→∞Sn≤limsup⁡n→∞Sn≤M
hold for any sequence Sn of the system (2), and the same inequalities hold for Un and Vn. For each class Sn,Un, and Vn, m and M are independent of initial conditions.Following the method used by Wang in [6], we will prove the permanence of system (2) for the general case; that is, assume that β1(U) is related to U.Theorem 6.
IfR0>1, then system (2) is permanent for any initial condition (3).Proof.
Firstly, from system (2) and Lemmas 1 and 2, for any ϵ0>0, there exists sufficiently large n0>0 such that Un≤λ/μ1+ϵ0 as n≥n0+h. Then, we have
(43)Sn+1=λ+Sn+ξ1Un+1+ξ2Vn+11+β1(Un)∑k=0h‍Un-kηk+μ1>λ1+μ1+β1(Un)∑k=0h‍Un-kηk.
Let β1M(ϵ0)=max⁡U∈[0,λ/μ1+ϵ0]β1(U). Thus, we have
(44)Sn+1≥λ1+μ1+β1M∑k=0h‍Un-kηk≥λ1+μ1+β1M(λ/μ1+ϵ0)A.
Notice that ϵ0 can be arbitrarily small. Then, we have
(45)liminf⁡n→+∞Sn+1≥ms=λ1+μ1+β1M(Aλ/μ1),β1M=max⁡U∈[0,λ/μ1]β1(U).
Next, let us consider the positive sequences Sn and Un of (2). According to these sequences, we define
(46)Hn=Un+μ2+P+ξ1A∑k=0h‍(∑l=n-kn‍Ul)ηk.
Then, for n≥0, we obtain
(47)ΔH=Hn+1-Hn=Un+1-Un+μ2+P+ξ1A∑k=0h‍(Un+1ηk-Un-kηk)=(β1(Un)Sn+1-μ2+P+ξ1A)∑k=0h‍Un-kηk+β3Un+1Vn+1.
Since R0=β1(0)Aλ/μ1(μ2+P+ξ1)>1, there exist 0<α<Uβ and ρ>0 such that
(48)Aβ1(0)μ2+P+ξ1×λμ1+αβ1(α)A×{1-(11+μ1+αβ1(α)A)ρh}>1;
note that
(49)SΔ=λμ1+αβ1(α)A{1-(11+μ1+αβ1(α)A)ρh}.
We claim that it is impossible thatUn≤α holds for all n≥n1≥[ρh]. The function [x] gives the smallest integer not less than x. Suppose the contrary, for n≥n1+h. Consider
(50)Sn+1=λ+Sn+ξ1Un+1+ξ2Vn+11+μ1+β1(Un)∑k=0h‍Un-kηk>λ+Sn1+μ1+αβ1(α)A>λ1+μ1+αβ1(α)A{(11+μ1+αβ1(α)A)n-n1-h-11+11+μ1+αβ1(α)Ammmmmmmmmimm+(11+μ1+αβ1(α)A)n-n1-h-1}+(11+μ1+αβ1(α)A)n-n1-hSn1+h+1.
From Lemma 1, Sn satisfies
(51)Sn+1>λμ1+αβ1(α)A{1-(11+μ1+αβ1(α)A)n-n1-h},
and we have that, for n≥n1+h+[ρh], we have
(52)Sn+1>λμ1+αβ1(α)A{1-(11+μ1+αβ1(α)A)[ρh]}≥λμ1+αβ1(α)A{1-(11+μ1+αβ1(α)A)ρh}=SΔ.
Hence, for n≥n1+h+[ρh], we have
(53)ΔH≥(β1(Un)Sn+1-μ2+P+ξ1A)∑k=0h‍Un-kηk>(β1(0)SΔ-μ2+P+ξ1A)∑k=0h‍Un-kηk=μ2+P+ξ1A(Aβ1(0)SΔμ2+P+ξ1-1)∑k=0h‍Un-kηk.
Let ϵ=min⁡θ{Un1+[ρh]+h+θ;θ=-h,-h+1,…,0}. Now, we will show that Un≥ϵ for all n≥n1+[ρh]+h. In fact, there is an integer n->0 such that
(54)Un≥ϵ,n1+[ρh]+h≤n≤n1+[ρh]+h+n-,Un+1<ϵ,n=n1+[ρh]+h+n-.
However, for n=n1+[ρh]+h+n-, we have
(55)Un+1-ϵ=β1(Un)Sn+1∑k=0h‍Un-kηk+Un1+μ2+P+ξ1-β3Vn+1-ϵ≥β1(Un)Sn+1∑k=0h‍Un-kηk+Un1+μ2+P+ξ1-1+μ2+P+ξ11+μ2+P+ξ1ϵ≥β1(0)SΔA-(μ2+P+ξ1)1+μ1+P+ξ1ϵ=μ2+P+ξ11+μ2+p+ξ1{β1(0)SΔAμ2+P+ξ1-1}ϵ>0.
Which is a contradiction. Thus, Un≥ϵ for n≥n1+[ρh]+h. Therefore, for n≥n1+[ρh]+h,
(56)ΔH>(μ2+P+ξ1){Aβ1(0)SΔμ2+P+ξ1-1}ϵ>0,
which implies that Hn→+∞ as n→+∞. But, from Lemma 2 and (46), there exists a sufficiently large integer n1′>0 such that, for n>n1′,
(57)Hn≤λμ1+μ2+P+ξ1A∑k=0h‍(∑l=n-kn‍λμ1)ηk≤λμ1{1+(μ2+P+ξ1)(h+1)},
which is a contradiction. Hence, the claim is proved.
In the rest, we only need to consider the following two cases:(i)
Un≥α for all large n.(ii)
Un oscillates about α for all large n.
We show thatUn≥mu for all large n, where 0<mu≤α, is a constant which will be given later. Clearly, we only need to consider case (ii). Let positive integers n1 and n2 be sufficiently large that Un1≥α, Un2≥α, and Un<α, for n1<n<n2.
Ifn2-n1<h+[ρh], since
(58)Un=β1(Un-1)Sn∑k=0h‍Un-k-1ηk+Un-11+μ2+P+ξ1-β3Vn≥Un-11+μ2+P+ξ1,
we have
(59)Un≥(11+μ2+P+ξ1)n-n1Un1>(11+μ2+P+ξ1)n2-n1Un1>mu=(11+μ2+P+ξ1)h+[ρh]α.
Hence, Un>mu for n∈[n1,n2].
Ifn2-n1>h+[ρh], we can easily obtain that Un>mu for n∈[n1,n1+h+[ρh]]. Assume that there exists an integer n^>0 such that
(60)Un≥mu,n1+h+[ρh]≤n≤n1+h+[ρh]+n^,Un+1<mu,n=n1+h+[ρh]+n^.
However, for n=n1+h+[ρh]+n^,
(61)Un+1-mu≥μ2+P+ξ11+μ2+p+ξ1{β1(0)SΔAμ2+P+ξ1-1}mu>0.
This is a contradiction to the proposition that Un+1<mu. Therefore, Un≥mu for n∈[n1,n2]. Since these positive integers n1 and n2 are chosen in an arbitrary way, we conclude that Un≥mu for all large n in case (ii). Hence, liminf⁡n→∞Un≥mu.
Note that, from that third equation of system (2), we have
(62)liminf⁡n→∞Vn≥mv=Pμ3+ξ2+λβ3/μ1mu=Pμ3+ξ2+β3S0mu.
From Lemma 2 and the discussion above, we have
(63)ms≤liminf⁡n→∞Sn≤limsup⁡n→∞Sn≤S0,mu≤liminf⁡n→∞Un≤limsup⁡n→∞Un≤S0,mv≤liminf⁡n→∞Vn≤limsup⁡n→∞Vn≤S0.
The proof is completed.


## 5. Numerical Example

In order to confirm the validity of our results, we consider the following heroin epidemic model with a discrete time delay:(64)iSn+1-Sn=λ-β1Sn+1Un-h-μ1Sn+1+ξ1Un+1+ξ2Vn+1,Un+1-Un=β1Sn+1Un-h+β3Un+1Vn+1mmmmimi-(μ2+P+ξ1)Un+1,iVn+1-Vn=PUn+1-β3Un+1Vn+1-(μ3+ξ2)Vn+1.
Now, we present a numerical example. For the sake of simplicity, we choose the parameters as β1=0.9, β3=0.8, λ=2, μ1=0.1, μ2=0.2, μ3=0.1, P=0.4, ξ1=0.1, and ξ2=0.2; we get R0=25.7143<1. Figure 1 shows that the disease free equilibrium E0 of the system (64) is globally asymptotically stable when R0<1. Figure 2 shows that the system (64) is permanent when R0>1.Figure 1Figure 2


## 6. Conclusions

In this paper, we have modified the Samanta heroin epidemic model into an autonomous heroin epidemic model with distributed time delay. Further, we established a discretized heroin epidemic model with time delay, sufficient conditions have been obtained to ensure the global asymptotic stability of heroin-using free equilibrium whenR0≤1 and β1(U) is replaced by a positive constant. We also carried out some discussion about the heroin-using equilibrium, but our results are only restricted to the existence of this equilibrium for β1(U)=β1>0, a special case of system (2). The stability of heroin-using equilibrium is yet to be studied. As a main result of this paper, we obtained the permanence of the system (2). From the expression of R0=β1(0)λA/μ1(μ2+P+ξ1), we see that a decrease in β1 (transmission coefficient from susceptible population) will cause a decrease of the same proportion in R0. If the rate of migration or recruitment is restricted into susceptible community, the spread of the disease can also be kept under control by reducing λ and thereby decreasing R0. The spread of the heroin users can also be controlled by educators, epidemiologists, and treatment providers to increase the values of ξ (removal rate of heroin users not in treatment who stop using heroin but are susceptible) and P (proportion of heroin users who enter treatment) and thereby to decrease R0. This analysis tells us that prevention is better than cure; efforts to increase prevention are more effective in controlling the spread of habitual drug use than efforts to increase the numbers of individuals accessing treatment.



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*Source: 742385-2014-10-20.xml*