Influence of Relapse in a Giving Up Smoking Model

Abstract

Smoking subject is an interesting area to study. The aim of this paper is to derive and analyze a model taking into account light smokers compartment, recovery compartment, and two relapses in the giving up smoking model. Stability of the model is obtained. Some numerical simulations are also provided to illustrate our analytical results and to show the effect of controlling the rate of relapse on the giving up smoking model.

1. Introduction

As early as 1889, people have established a model for the spread of infectious diseases. Then the spread rule and trend of the model were studied by analying the stability of the solutions ([1–4] and the references cited therein). De la Sen and Alonso-Quesada [3] present several simple linear vaccination-based control strategies for a SEIR propagation disease model and study the stability of this model. De La Sen et al. [4] discuss a generalized time-varying SEIR propagation disease model subject to delays which potentially involves mixed regular and impulsive vaccination rules, and in this paper the authors were using the good methods to study the dynamic behavior of the model especially the positivity of the model. There are many methods to discuss the stability, one of the most powerful techniques for qualitative analysis of a dynamical system is Direct Lyapunov Method [5]. This method employs an appropriate auxiliary function, called a Lyapunov function. For example, [6–8] use this method to discuss the stability of the model. In addition, there are a number of articles which use Routh-Hurwitz theory to explore the stability, see for example [9, 10].

In recent years, many types of epidemic models are discussed, such as virus dynamics models [11, 12], tuberculosis models [13, 14], and HIV models [15, 16].

Due to the increasing in the number of smokers, tobacco use is also as a disease to be treated. In order to explore the spread rule of smoking, quit smoking model is developed. Castillo-Garsow et al. [17] proposed a simple mathematical model for giving up smoking in the first time. In this model, a total constant population was divided into three classes: potential smokers, that is, people who do not smoke yet but might become smokers in the future (𝑃), smokers (𝑆), and quit smokers (𝑄). Zaman [18] extended the work of Castillo-Garsow et al. [17] by adding the population of occasional smokers in the model, and presented qualitative behavior of the model. Zaman [19] presented the optimal campaigns in the smoking dynamics. They consider two possible control variables in the form of education and treatment campaigns oriented to decrease the attitude towards smoking and first showed the existence of an optimal control for the control problem.

However, in real life, the usual quit smokers are only temporary quit smokers. Some of them may relapse since they contact with smokers again, and the others may become permanent quit smokers. Statistics also show that 15% quit smokers may relapse when they contact with smokers. Enlightening by the previously mentioned cases, we present a model, which extend the models in [17–19] by taking into account the temporary quit smoker compartment (𝑅) and two kinds of relapses, that is, once a smoker temporary quits smoking he/she may become a light or occasion smoker or a persistent smoker again. First, we derive the basic reproductive number, and discuss the positivity of the solution for the giving up smoking model. Then, we analyze the stability of equilibria by Lyapunov Method and Routh-Hurwitz theory. Finally, by estimating of parameter, we present the numerical simulation. Moreover, the numerical simulation shows that we can greatly improve the effect of quit smoking by using some methods, such as treatment and education, controlling the rate of relapse.

The organization of this paper is as follow: the model is given under some assumption in Section 2. The basic reproductive number, existence, and the stability of equilibria are investigated in Section 3. Some numerical simulations are given in Section 4. The paper ends with a discussion in Section 5.

2. The Model Formulation

2.1. System Description

In this paper, we establish the giving up smoking model as Figure 1. Table 1 presents the parameters description of the model.

Table 1 

The parameters description of giving up smoking model.

Figure 1 

Transfer diagram for the dynamics of giving up smoking model.

From Figure 1, the total population is divided into five compartments, namely, the potential smokers compartment (𝑃), light or occasion smokers compartment (𝐿), persistent smokers compartment (𝑆), temporary quit smokers (𝑅) that is the people who did some efforts to stop smoking, and quit smokers forever group (𝑄). As we know that if you smoke more, the harm of nicotine on the body will be greater. So the death rate is also higher. Hence, we can further assume that 𝑑₂<𝑑₃. The total population size is 𝑁(𝑡), where 𝑁(𝑡)=𝑃(𝑡)+𝐿(𝑡)+𝑆(𝑡)+𝑅(𝑡)+𝑄(𝑡). (1)

The transfer diagram leads to the following system of ordinary differential equations: 𝑑𝑃(𝑡)𝑑𝑡=𝑏−𝛽₁𝑃(𝑡)𝐿(𝑡)−𝑑₁𝑃(𝑡),(𝑡)+𝜇𝑑𝐿𝑑𝑡=𝛽₁𝑃(𝑡)𝐿(𝑡)−𝛽₂𝐿(𝑡)𝑆(𝑡)+𝜌₂𝐿(𝑡)𝑅(𝑡)−𝑑₂𝐿(𝑡),(𝑡)+𝜇𝑑𝑆𝑑𝑡=𝛽₂𝐿(𝑡)𝑆(𝑡)+𝜌₁𝑆(𝑡)𝑅(𝑡)−𝜔+𝑑₃𝑆(𝑡),(𝑡)+𝜇𝑑𝑅(𝑡)𝑑𝑡=𝜔𝑆−𝜌₁𝑆(𝑡)𝑅(𝑡)−𝜌₂𝐿(𝑡)𝑅(𝑡)−𝛾+𝑑₄𝑅(𝑡),(𝑡)+𝜇𝑑𝑄(𝑡)−𝑑𝑑𝑡=𝛾𝑅₅+𝜇𝑄(𝑡). (2)

2.2. Positivity and Boundedness of Solutions

For system (2), to ensure that the solutions of the system with positive initial conditions remain positive for all 𝑡>0, it is necessary to prove that all the state variables are nonnegative. Similar to the proof of [3, 4, 13], we have the following lemma.

Lemma 1. If 𝑃(0)>0,𝐿(0)>0,𝑆(0)>0,𝑅(0)>0,𝑄(0)>0, the solutions 𝑃(𝑡), 𝐿(𝑡), 𝑆(𝑡), 𝑅(𝑡), 𝑄(𝑡) of system (2) are positive for all 𝑡>0.

Proof. If the conclusion does not hold, then at least one of 𝑃(𝑡), 𝐿(𝑡), 𝑆(𝑡), 𝑅(𝑡), 𝑄(𝑡) is not positive. Thus, we have one of the following five cases.(1)There exists a first time 𝑡₁ such that 𝑃𝑡₁=0,𝑃^′𝑡₁(𝑡)(𝑡)𝑅(𝑡)(𝑡)<0,𝐿≥0,𝑆≥0,≥0,𝑄≥0,0≤𝑡≤𝑡₁. (3)(2)There exists a first time 𝑡₂ such that 𝐿𝑡₂=0,𝐿^′𝑡₂(𝑡)(𝑡)𝑅(𝑡)(𝑡)<0,𝑃≥0,𝑆≥0,≥0,𝑄≥0,0≤𝑡≤𝑡₂. (4)(3)There exists a first time 𝑡₃ such that 𝑆𝑡₃=0,𝑆^′𝑡₃(𝑡)(𝑡)𝑅(𝑡)(𝑡)<0,𝑃≥0,𝐿≥0,≥0,𝑄≥0,0≤𝑡≤𝑡₃. (5)(4)There exists a first time 𝑡₄ such that 𝑅𝑡₄=0,𝑅^′𝑡₄(𝑡)(𝑡)𝑆(𝑡)(𝑡)<0,𝑃≥0,𝐿≥0,≥0,𝑄≥0,0≤𝑡≤𝑡₄. (6)(5)There exists a first time 𝑡₅ such that 𝑄𝑡₅=0,𝑄^′𝑡₅(𝑡)(𝑡)𝑆(𝑡)(𝑡)<0,𝑃≥0,𝐿≥0,≥0,𝑅≥0,0≤𝑡≤𝑡₅. (7)
In case (1), we have 𝑃^′𝑡₁=𝑏>0, (8) which is a contradiction to 𝑃^′(𝑡₁)<0.
In case (2), we have 𝐿^′𝑡₂=0, (9) which is a contradiction to 𝐿^′(𝑡₂)<0.
In case (3), we have 𝑆^′𝑡₃=0, (10) which is a contradiction to 𝑆^′(𝑡₃)<0.
In case (4), we have 𝑅^′𝑡₄=𝜔𝑆𝑡₄>0, (11) which is a contradiction to 𝑅^′(𝑡₄)<0.
In case (5), we have 𝑄^′𝑡₅=𝛾𝑅𝑡₅>0, (12) which is a contradiction to 𝑄^′(𝑡₅)<0.
Thus, the solutions 𝑃(𝑡), 𝐿(𝑡), 𝑆(𝑡), 𝑅(𝑡), 𝑄(𝑡) of system (2) remain positive for all 𝑡>0.

Lemma 2. All feasible solution of the system (2) are bounded and enter the region Ω=(𝑃,𝐿,𝑆,𝑅,𝑄)∈𝑅⁵₊𝑏∶𝑃+𝐿+𝑆+𝑅+𝑄≤𝜇. (13)

Proof. Let (𝑃,𝐿,𝑆,𝑅,𝑄)∈𝑅⁵₊ be any solution with nonnegative initial condition: adding the first four equations of (2), we have 𝑑()𝑑𝑑𝑡𝑃+𝐿+𝑆+𝑅+𝑄=𝑏−₁+𝜇𝑃−𝑑₂+𝜇𝐿−𝑑₃𝑆−𝑑+𝜇₄+𝜇𝑅−𝑑₅𝑄()−𝑑+𝜇=𝑏−𝜇𝑃+𝐿+𝑆+𝑅+𝑄₁𝑃+𝑑₂𝐿+𝑑₃𝑆+𝑑₄𝑅+𝑑₅𝑄≤𝑏−𝜇𝑁. (14) It follows that (𝑡)≤𝑏0≤𝑁𝜇+𝑁(0)𝑒^−𝜇𝑡, (15) where 𝑁(0) represents initial values of the total population. Thus 0≤𝑁(𝑡)≤𝑏/𝜇, as 𝑡→∞. Therefore all feasible solutions of system (2) enter the region Ω=(𝑃,𝐿,𝑆,𝑅,𝑄)∈𝑅⁵₊𝑏∶𝑃+𝐿+𝑆+𝑅+𝑄≤𝜇. (16) Hence, Ω is positively invariant, and it is sufficient to consider solutions of system (2) in Ω. Existence, uniqueness, and continuation results of system (2) hold in this region. It can be shown that 𝑁(𝑡) is bounded and all the solutions starting in Ω approach enter or stay in Ω.

3. Analysis of the Model

In this section, we will analyze the existence of equilibria of system (2).

3.1. The Existence of Equilibria and the Basic Reproduction Number

The model (2) has a smoking-free equilibrium given by (see Theorem 3 (1)) 𝐸₀=𝑏𝑑₁.+𝜇,0,0,0,0 (17)

In the following, the basic reproduction number of system (2) will be obtained by the next generation matrix method formulated in [21].

Let 𝑥=(𝐿,𝑆,𝑅,𝑄,𝑃)^𝑇; then system (2) can be written as 𝑑𝑥(𝑥)(𝑥),𝑑𝑡=ℱ−𝒱 (18) where ℱ(𝑥)=𝛽₁0000,𝒱(𝑥)=𝛽𝑃𝐿₂𝐿𝑆+𝑑₂+𝜇𝐿−𝜌₂𝐿𝑅−𝛽₂𝐿𝑆−𝜌₁𝑆𝑅+𝜔+𝑑₃+𝜇𝑆−𝜔𝑆+𝜌₁𝑆𝑅+𝜌₂𝐿𝑅+𝛾+𝑑₄𝑅𝑑+𝜇−𝛾𝑅+₅+𝜇𝑄−𝑏+𝛽₁𝑃𝐿+𝑑₁𝑃.+𝜇 (19) The Jacobian matrices of ℱ(𝑥) and 𝒱(𝑥) at the smoking-free equilibrium 𝐸₀ are, respectively, 𝐷ℱ𝐸₀=𝐹_4×40,𝐸00𝐷𝒱₀=𝑉_4×40𝛽₁𝑏𝑑₁𝑑+𝜇000₁,+𝜇 (20) where 𝐹_4×4=𝛽₁𝑏𝑑₁,𝑉+𝜇000000000000000_4×4=𝑑₂+𝜇0000𝜔+𝑑₃+𝜇000−𝜔𝛾+𝑑₄+𝜇000−𝛾𝑑₅.+𝜇 (21) In order to simplify the calculation, letting 𝑏₁=𝑑₂+𝜇,𝑏₂=𝜔+𝑑₃+𝜇,𝑏₃=𝛾+𝑑₄+𝜇,𝑏₄=𝑑₅+𝜇, we obtain 𝑉⁻¹=1𝑏₁01000𝑏₂0𝜔00𝜔+𝑏₂𝛾+𝑏₃1𝛾+𝑏₃00𝜔𝛾𝜔+𝑏₂𝛾+𝑏₃𝑏₄𝛾𝛾+𝑏₃𝑏₄1𝑏₄. (22) The basic reproduction number, denoted by 𝑅₀, is thus given by 𝑅₀=𝜌𝐹𝑉⁻¹=𝛽₁𝑏𝑑₁+𝜇𝑏₁=𝛽₁𝑏𝑑₁+𝜇𝑑₂.+𝜇 (23) Throughout this paper, we denote 𝛼₃=𝜔+𝑑₃+𝜇,𝛼₄=𝛾+𝑑₄+𝜇. (24)

Theorem 3. For the giving up smoking model (2), there exist the following three types of equilibrium.(1) For all parameter values, system (2) exists the smoking-free equilibrium 𝐸₀(𝑏/(𝑑₁+𝜇),0,0,0,0).(2) If 𝑅₀>1, there exists the occasion smoking equilibrium 𝐸_𝐿((𝑑₂+𝜇)/𝛽₁,𝑏/(𝑑₂+𝜇)−(𝑑₁+𝜇)/𝛽₁,0,0,0), and there exists no occasion smoking equilibrium if 𝑅₀≤1.(3) If 𝑅₂=1/𝑅₀+(𝑑₂+𝜇)𝛼₃/𝑏𝛽₂<1, 𝑑₃𝜌₂−𝑑₂<0 and 𝜌₂>max{𝑑₂𝜌₁𝜔/𝑑₃𝛼₃,𝜌₁𝜔/𝛼₃}, then system (2) has positive smoking-present equilibrium 𝐸^∗(𝑃^∗,𝐿^∗,𝑆^∗,𝑅^∗,𝑄^∗) where 𝑅^∗∈(0,𝜔/𝜌₁).
Moveover, 𝐸^∗(𝑃^∗,𝐿^∗,𝑆^∗,𝑅^∗,𝑄^∗) satisfies the following equality: 𝐿^∗=𝛼₃−𝜌₁𝑅^∗𝛽₂,𝑆^∗=𝜌₁𝜌₂𝑅^∗2−𝛼₄𝛽₂𝑅^∗−𝛼₃𝜌₂𝑅^∗𝛽₂𝜌₁𝑅^∗,𝑃−𝜔_∗=1𝛽₁−𝛼₄𝛽₂𝑅^∗−𝑑₃+𝜇𝜌₂𝑅^∗+𝜌₁𝑅^∗−𝜔𝑑₂+𝜇𝜌₁𝑅^∗,𝑄−𝜔_∗=𝛾𝑑₅𝑅+𝜇^∗, (25) where 𝑅^∗ is a positive solution of 𝑓(𝑅)=0, where 𝑓(𝑅)=𝑑₁𝜌+𝜇¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₁𝜌₁−𝑑𝑅−𝜔₁𝜌+𝜇²𝛽₁𝑑𝑅+¹+𝜇𝑑₂+𝜇𝛽₁+𝑑₂𝛼+𝜇³−𝜌_1𝑅𝛽₂+𝑑₃𝜌+𝜇¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₂𝜌₁+𝑑𝑅−𝜔⁴+𝜇𝑅+𝛾𝑅−𝑏. (26)

Proof. It follows from system (2) that 𝑏−𝛽₁𝑃(𝑡)𝐿(𝑡)−𝑑₁𝑃(𝑡)𝛽+𝜇=0,₁𝑃(𝑡)𝐿(𝑡)−𝛽₂𝐿(𝑡)𝑆(𝑡)+𝜌₂𝐿(𝑡)𝑅(𝑡)−𝑑₂𝐿(𝑡)𝛽+𝜇=0,₂𝐿(𝑡)𝑆(𝑡)+𝜌₁𝑆(𝑡)𝑅(𝑡)−𝜔+𝑑₃𝑆(𝑡)(𝑡)+𝜇=0,𝜔𝑆−𝜌₁𝑆(𝑡)𝑅(𝑡)−𝜌₂𝐿(𝑡)𝑅(𝑡)−𝛾+𝑑₄𝑅(𝑡)(𝑡)−𝑑+𝜇=0,𝛾𝑅₅+𝜇𝑄(𝑡)=0. (27)(1) Letting 𝐿=𝑆=𝑅=𝑄=0 in (27), we can obtain the smoking-free equilibrium 𝐸₀(𝑏/(𝑑₁+𝜇),0,0,0,0).(2) If 𝑅₀>1, letting 𝑆=𝑅=𝑄=0 in (27), we can obtain the occasion smoking equilibrium 𝐸_𝐿((𝑑₂+𝜇)/𝛽₁,𝑏/(𝑑₂+𝜇)−(𝑑₁+𝜇)/𝛽₁,0,0,0).(3)From third equation of the system (27), we obtain 𝛼𝐿=³−𝜌₁𝑅𝛽₂. (28)By adding the third equation and the fourth equation, we get 𝛽₂𝐿(𝑡)𝑆(𝑡)−𝜔+𝑑₃𝑆(𝑡)(𝑡)−+𝜇+𝜔𝑆𝛾+𝑑₄+𝜇𝑅(𝑡)−𝜌₂𝐿(𝑡)𝑅(𝑡)=0. (29) From this we have 𝜌𝑆=¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₂𝜌₁.𝑅−𝜔 (30) From the second equation of the system (27) we get 𝛽₁𝑃(𝑡)=𝛽₂𝑆(𝑡)−𝜌₂𝑅(𝑡)+𝑑₂+𝜇. (31) Substituting 𝑆 into (31), we get 𝜌𝑃=¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₁𝜌₁−𝜌𝑅−𝜔𝛽²₁𝑑𝑅+²+𝜇𝛽₁=1𝛽₁−𝛼₄𝛽₂𝑅−𝑑₃+𝜇𝜌₂𝑅+𝜌₁𝑅−𝜔𝑑₂+𝜇𝜌₁.𝑅−𝜔 (32) It follows from the fifth equation that 𝛾𝑄=𝑑₅+𝜇𝑅. (33) Let 𝑓(𝑅)=𝑑₁+𝜇𝑃+𝑑₂+𝜇𝐿+𝑑₃𝑆+𝑑+𝜇₄+𝜇𝑅+𝑑₅=𝑑+𝜇𝑄−𝑏₁𝜌+𝜇¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₁𝜌₁−𝑑𝑅−𝜔₁𝜌+𝜇²𝛽₁𝑑𝑅+¹+𝜇𝑑₂+𝜇𝛽₁+𝑑₂𝛼+𝜇³−𝜌_1𝑅𝛽₂+𝑑₃𝜌+𝜇¹𝜌₂𝑅²−𝛼₄𝛽₂𝑅−𝛼₃𝜌₂𝑅𝛽₂𝜌₁+𝑑𝑅−𝜔₄+𝜇𝑅+𝛾𝑅−𝑏. (34) From (28) we can see that if 𝑅<𝛼₃/𝜌₁, then 𝐿>0. From (30) we can see that if 𝑅<𝜔/𝜌₁, then 𝜌₁𝑅−𝜔<0,𝜌₁𝜌₂𝑅−𝛼₄𝛽₂−𝛼₃𝜌₂<0; these show that 𝑆>0. From (32), we can see that if 𝑅<𝜔/𝜌₁, then 𝑃>0. Therefore, if 𝑅<𝜔/𝜌₁, 𝑃>0, 𝐿>0, 𝑆>0, 𝑄>0. In the following, we prove that the existence of positive solutions of 𝑓(𝑅)=0. From (34), we know lim_{𝑅→𝜔/𝜌1−}𝑓(𝑅)=+∞. (35) If (𝑑₁+𝜇)(𝑑₂+𝜇)/𝛽₁+(𝑑₂+𝜇)𝛼₃/𝛽₂<𝑏, then 𝑓(0)=𝑑₁+𝜇𝑑₂+𝜇𝛽₁+𝑑₂+𝜇𝛼₃𝛽₂𝑓−𝑏<0,_′(𝑅)=𝑑₁+𝜇𝛽₁𝑔(𝑅)𝜌₁𝑅−𝜔²−𝑑₁𝜌+𝜇²𝛽₁−𝑑₂𝜌+𝜇¹𝛽₂+𝑑₃+𝜇𝛽₂𝑔(𝑅)𝜌₁𝑅−𝜔²+𝑑₄+𝜇+𝛾, (36) where 𝑔(𝑅)=𝜌²₁𝜌₂𝑅²−2𝜌₁𝜌₂𝜔𝑅+𝛼₄𝛽₂𝜔+𝛼₃𝜌₂𝜔. (37) Clearly, 𝑅=𝜔/𝜌₁ is the minimum point of 𝑔(𝑅). For 𝑅∈(0,𝜔/𝜌₁), we have 𝑔(𝑅)𝜔≥𝑔𝜌₁=−𝜌₂𝜔+𝛼₄𝛽₂+𝛼₃𝜌₂𝜔>0. (38) Hence, 𝑓^′(𝑅)=𝑑₁+𝜇𝛽₁𝛼₄𝛽₂𝜔+𝛼₃𝜌₂𝜔−𝜌₂𝜔²𝜌₁𝑅−𝜔²+ℎ(𝑅)𝛽₂𝜌₁𝑅−𝜔²+𝑑₄+𝜇+𝛾, (39) where ℎ(𝑅)=𝑑₃𝜌²₁𝜌₂𝑅²−2𝑑₃𝜌₁𝜌₂𝜔𝑅+𝑑₃𝜔𝛼₄𝛽₂+𝑑₃𝜔𝛼₃𝜌₂+𝜇𝜌²₁𝜌₂𝑅²−2𝜇𝜌₁𝜌₂𝜔𝑅+𝜇𝜔𝛼₄𝛽₂+𝜇𝜔𝛼₃𝜌₂−𝑑₂𝜌³₁𝑅²+2𝑑₂𝜌²₁𝜔𝑅−𝑑₂𝜌₁𝜔²−𝜇𝜌³₁𝑅²+2𝜇𝜌²₁𝜔𝑅−𝜇𝜌₁𝜔². (40) Then ℎ^′(𝑅)=2𝑑₃𝜌₁𝜌₂𝜌₁𝑅−𝜔+2𝜇𝜌₁𝜌₂𝜌₁𝑅−𝜔−2𝑑₂𝜌₁𝜌₁𝑅−𝜔−2𝜇𝜌₁𝑅−𝜔=2𝜌₁𝜌₁𝑅−𝜔𝑑₃𝜌₂−𝑑₂+2𝜇𝜌₁𝜌₁𝑅−𝜔𝜌₂−1, (41) as we know that 𝜌₂<1 and 𝜌₁𝑅−𝜔<0. If 𝑑₃𝜌₂−𝑑₂<0, then for any 𝑅∈(0,𝜔/𝜌₁), we have ℎ^′(𝑅)>0, so ℎ(𝑅) is a strictly monotone increasing function on (0,𝜔/𝜌₁). As we know that 𝜌₂>max{𝑑₂𝜌₁𝜔/𝑑₃𝛼₃,𝜌₁𝜔/𝛼₃},𝑑₃>𝑑₂ and 𝛼₃>𝜔, so ℎ(0)=𝑑₃𝜔𝛼₄𝛽₂+𝑑₃𝜔𝛼₃𝜌₂+𝜇𝜔𝛼₄𝛽₂+𝜇𝜔𝛼₃𝜌₂−𝑑₂𝜌₁𝜔²−𝜇𝜌₁𝜔²=𝑑₃𝜔𝛼₄𝛽₂+𝜇𝜔𝛼₄𝛽₂+𝜔𝑑₃𝛼₃𝜌₂−𝑑₂𝜌₁𝜔×𝜇𝜔𝛼₃𝜌₂−𝜌₁𝜔>0. (42) Hence, ℎ(𝑅)>0 for any 𝑅∈(0,𝜔/𝜌₁), so 𝑓^′(𝑅)>0; that is, 𝑓(𝑅) is a strictly monotone increasing function on (0,𝜔/𝜌₁). Therefore, 𝑓(𝑅)=0 has an unique positive solutions on (0,𝜔/𝜌₁). The proof is completed.

3.2. Qualitative Analysis

In this part we will discuss the qualitative behavior of the giving up smoking model (2).

3.2.1. Stability of the Smoking-Free Equilibrium

Theorem 4. If 𝑅₀≤1, the smoking-free equilibrium 𝐸₀ is globally asymptotically stable.

Proof. We introduce the following Lyapunov function: 𝑉(𝑃(𝑡),𝐿(𝑡),𝑆(𝑡),𝑅(𝑡),𝑄(𝑡))=𝑃₀𝑃𝑃₀𝑃−ln𝑃₀+𝐿+𝑆+𝑅+𝑄. (43) The derivative of 𝑉 is given by 𝑉^′=𝑃^′−𝑃₀𝑃𝑃^′+𝐿^′+𝑆^′+𝑅^′+𝑄^′=𝑏−𝛽₁𝑃𝐿−𝑑₁𝑑+𝜇𝑃−𝑏/₁+𝜇𝑃×𝑏−𝛽₁𝑃𝐿−𝑑₁+𝜇𝑃+𝛽₁𝑃𝐿−𝛽₂𝐿𝑆+𝜌₂𝐿𝑅−𝑑₂+𝜇𝐿+𝛽₂𝐿𝑆+𝜌₁−𝑆𝑅𝜔+𝑑₃+𝜇𝑆+𝜔𝑆−𝜌₁𝑆𝑅−𝜌₂𝐿𝑅−𝛾+𝑑₄+𝜇𝑅+𝛾𝑅−𝑑₅𝑄𝑏+𝜇=2𝑏−²𝑑₁𝑃−𝑑+𝜇¹𝛽+𝜇𝑃+¹𝑏𝑑₁𝐿−𝑑+𝜇₂+𝜇𝐿−𝑑₃+𝜇𝑆−𝑑₄+𝜇𝑅−𝑑₅𝑄=𝑏+𝜇2−𝑑₁𝑃−𝑑+𝜇¹+𝜇𝑃𝑏𝑏+𝛽₁𝑏𝑑₁+𝜇−𝑑₂−𝜇𝐿−𝑑₃𝑆−𝑑+𝜇₄+𝜇𝑅−𝑑₅𝑄=𝑏+𝜇2−𝑑₁𝑃−𝑑+𝜇¹+𝜇𝑃𝑏𝑏+𝛽₁𝑏−𝑑₁+𝜇𝑑₂+𝜇𝑑₁𝑑+𝜇𝐿−₃+𝜇𝑆−𝑑₄𝑅−𝑑+𝜇₅+𝜇𝑄. (44) If 𝑅₀≤1, then 𝛽₁𝑏≤(𝑑₁+𝜇)(𝑑₂+𝜇), so we get (𝛽₁𝑏−(𝑑₁+𝜇)(𝑑₂+𝜇)/(𝑑₁+𝜇))𝐿≤0.
As we know, 2−𝑏/(𝑑₁+𝜇)𝑃−(𝑑₁+𝜇)𝑃/𝑏≤0, so we obtain 𝑉^′≤0 with equality only if 2−𝑏/(𝑑₁+𝜇)𝑃−(𝑑₁+𝜇)𝑃/𝑏=0, 𝑅₀=1 and 𝑆=𝑅=𝑄=0. By LaSalle invariance principle [20, 22], 𝐸₀ is globally asymptotically stable. Thus, for system (2), the smoking-free equilibrium 𝐸₀ is globally asymptotically stable if 𝑅₀≤1.

3.2.2. Stability of the Occasion Smoking Equilibrium

In this part, we will consider an occasional smoking equilibrium 𝐸_𝐿((𝑑₂+𝜇)/𝛽₁,𝑏/(𝑑₂+𝜇)−(𝑑₁+𝜇)/𝛽₁,0,0,0); that is, only potential smokers and occasionally smokers are not zero, and the other compartments are zero.

Theorem 5. If 𝑅₀>1, the occasion smoking equilibrium 𝐸_𝐿 is locally asymptotically stable.

Proof. The Jacobian matrix of the giving up smoking model (2) around 𝐸_𝐿 is given by 𝐽_𝐸𝐿=−𝛽₁𝑏𝑑₂+𝜇−𝑑₂𝛽−𝜇₁𝑏𝑑₂−𝑑+𝜇¹+𝜇0. (45) The characteristic polynomial is 𝜓(𝑥)=𝑥²+𝑐₁𝑥+𝑐₂, where 𝑐₁=𝛽₁𝑏/(𝑑₂+𝜇) and 𝑐₂=𝛽₁𝑏−(𝑑₁+𝜇)(𝑑₂+𝜇). Therefore by Routh-Hurwitz criteria we deduce that the roots of the polynomial 𝜓(𝑥) have negative real part when 𝛽₁𝑏>(𝑑₁+𝜇)(𝑑₂+𝜇), which shows that the system is locally asymptotically stable if 𝑅₀>1.

Theorem 6. If 𝑅₀>1 and 𝑅₁=𝑏𝛽₂/(𝑑₂+𝜇)(𝑑₃+𝜇)−(𝑑₁+𝜇)𝛽₂/𝛽₁(𝑑₃+𝜇)≤1, the occasion smoking equilibrium 𝐸_𝐿 is globally asymptotically stable.

Proof. We introduce the following Lyapunov function: 𝑉(𝑃(𝑡),𝐿(𝑡),𝑆(𝑡),𝑅(𝑡),𝑄(𝑡))=𝑃_𝐿𝑃𝑃_𝐿𝑃−ln𝑃_𝐿𝐿+𝐿𝐿_𝐿𝐿−ln𝐿_𝐿+𝑆+𝑅+𝑄. (46) The derivative of 𝑉 is given by 𝑉^′=𝑃^′−𝑃_𝐿𝑃𝑃^′+𝐿^′−𝐿_𝐿𝐿𝐿^′+𝑆^′+𝑅^′+𝑄^′𝑑=2𝑏−²+𝜇𝑏𝛽₁𝑃−𝛽₁𝑃𝑏𝑑₂+𝑏+𝜇𝑑₂−𝑑+𝜇¹+𝜇𝛽₁𝛽₂−𝑑₃𝑆−𝑏+𝜇𝑑₂−𝑑+𝜇¹+𝜇𝛽₁𝜌₂𝑅−𝑑₄+𝜇𝑅−𝑑₅𝑄=𝑑+𝜇2−²+𝜇𝛽₁𝑃−𝛽₁𝑃𝑑₂𝑏+𝑏+𝜇𝑑₂−𝑑+𝜇¹+𝜇𝛽₁𝛽₂−𝑑₃𝑆−𝑏+𝜇𝑑₂−𝑑+𝜇¹+𝜇𝛽₁𝜌₂𝑅−𝑑₄+𝜇𝑅−𝑑₅+𝜇𝑄. (47) If 𝑅₀>1, we can obtain 𝑏/(𝑑₂+𝜇)−(𝑑₁+𝜇)/𝛽₁>0. As we know 2−(𝑑₂+𝜇)/𝛽₁𝑃−𝛽₁𝑃/(𝑑₂+𝜇)≤0, if (𝑏/(𝑑₂+𝜇)−(𝑑₁+𝜇)/𝛽₁)𝛽₂≤𝑑₃+𝜇, so we obtain 𝑉^′≤0 with equality only if 𝑅₁=1 and 𝑅=𝑄=0. By LaSalle invariance principle [20, 22], 𝐸_𝐿 is globally asymptotically stable. This completes the proof.

3.2.3. Stability of the Smoking-Present Equilibrium

Theorem 7. Under the condition (3) of the Theorem 3, if 𝑆/𝑆^∗,𝑅/𝑅^∗,𝑄/𝑄^∗ satisfy one of four relations as follows: 𝑅𝑅^∗𝑄<1,𝑄^∗𝑆<1,𝑆^∗≥𝑅𝑅^∗≥𝑄𝑄^∗,𝑅𝑅^∗𝑄>1,𝑄^∗𝑆>1,𝑆^∗≤𝑅𝑅^∗≤𝑄𝑄^∗,𝑅𝑅^∗𝑄>1,𝑄^∗𝑆<1,𝑆^∗≤𝑅𝑅^∗,𝑅𝑅^∗≥𝑄𝑄^∗,𝑅𝑅^∗𝑄<1,𝑄^∗𝑆>1,𝑆^∗≥𝑅𝑅^∗,𝑅𝑅^∗≤𝑄𝑄^∗. (48) Then smoking-present equilibrium 𝐸^∗(𝑃^∗,𝐿^∗,𝑆^∗,𝑅^∗,𝑄^∗) is globally asymptotically stable.

Proof. At the equilibrium point, the expressions on the right-hand side of system (2) give us following relations: 𝑑₁=𝑏+𝜇𝑃^∗−𝛽₁𝐿^∗,𝑑₂+𝜇=𝛽₁𝑃^∗−𝛽₂𝑆^∗+𝜌₂𝑅^∗,𝜔+𝑑₃+𝜇=𝛽₂𝐿^∗+𝜌₁𝑅^∗,𝛾+𝑑₄𝑆+𝜇=𝜔^∗𝑅^∗−𝜌₁𝑆^∗−𝜌₂𝐿^∗,𝑑₅𝑅+𝜇=𝛾^∗𝑄^∗. (49) We now consider a candidate Lyapunov function 𝑉 such that 𝑉=𝑃−𝑃^∗−𝑃^∗𝑃ln𝑃^∗+𝐿−𝐿^∗−𝐿^∗𝐿ln𝐿^∗+𝑆−𝑆^∗−𝑆^∗𝑆ln𝑆^∗+𝑅−𝑅^∗−𝑅^∗𝑅ln𝑅^∗+𝑄−𝑄^∗−𝑄^∗𝑄ln𝑄^∗. (50) Then 𝑉>0 and the derivative of 𝑉 are given by 𝑉^′=𝑃1−^∗𝑃𝑃^′+𝐿1−^∗𝐿𝐿^′+𝑆1−^∗𝑆𝑆^′+𝑅1−^∗𝑅𝑅^′+𝑄1−^∗𝑄𝑄^′=𝑃1−^∗𝑃𝑏−𝛽₁𝐿𝑃−𝑑₁𝑃+𝐿+𝜇1−^∗𝐿𝛽₁𝐿𝑃−𝛽₂𝐿𝑆+𝜌₂𝑅𝐿−𝑑₂𝐿+𝑆+𝜇1−^∗𝑆𝛽₂𝐿𝑆+𝜌₁𝑅𝑆−𝜔+𝑑₃𝑆+𝑅+𝜇1−^∗𝑅𝜔𝑆−𝜌₁𝑅𝑆−𝜌₂𝑅𝐿−𝛾+𝑑₄𝑅+𝑄+𝜇1−^∗𝑄𝛾𝑅−𝑑₅+𝜇𝑄. (51) Using the relations in (49) we have𝑉^′=𝑃1−^∗𝑃𝑏−𝛽₁𝑃𝐿𝑃−^∗𝑃𝑏+𝛽₁𝑃𝐿^∗+𝐿1−^∗𝐿𝛽₁𝐿𝑃−𝛽₂𝐿𝑆+𝜌₂𝑅𝐿−𝛽₁𝑃^∗𝐿+𝛽₂𝐿𝑆^∗−𝜌₂𝑅^∗𝐿+𝑆1−^∗𝑆𝛽₂𝐿𝑆+𝜌₁𝑅𝑆−𝛽₂𝐿^∗𝑆−𝜌₁𝑅^∗𝑆+𝑅1−^∗𝑅𝜔𝑆−𝜌₁𝑅𝑆−𝜌₂𝑅𝑅𝐿−𝜔𝑅^∗𝑆^∗+𝜌₁𝑅𝑆^∗+𝜌₂𝑅𝐿^∗+𝑄1−^∗𝑄𝑄𝛾𝑅−𝛾𝑄^∗𝑅^∗. (52) We consider the following variable substitutions by letting𝑃𝑃^∗𝐿=𝑥,𝐿^∗𝑆=𝑦,𝑆^∗𝑅=𝑧,𝑅^∗𝑄=𝑢,𝑄^∗=𝑣. (53) The derivative of 𝑉 reduces to 𝑉^′=11−𝑥𝑏−𝑥𝑦𝛽₁𝑃^∗𝐿^∗−𝑥𝑏+𝑥𝛽₁𝑃^∗𝐿^∗+11−𝑦𝑥𝑦𝛽₁𝑃^∗𝐿^∗+𝑦𝑢𝜌₂𝑅^∗𝐿^∗−𝑦𝑧𝛽₂𝐿^∗𝑆^∗−𝑦𝛽₁𝑃^∗𝐿^∗+𝑦𝛽₂𝐿^∗𝑆^∗−𝑦𝜌₂𝑅^∗𝐿^∗+11−𝑧𝑦𝑧𝛽₂𝐿^∗𝑆^∗+𝑧𝑢𝜌₁𝑅^∗𝑆^∗−𝑧𝛽₂𝐿^∗𝑆^∗−𝑧𝜌₁𝑅^∗𝑆^∗+11−𝑢𝑧𝜔𝑆^∗−𝑧𝑢𝜌₁𝑅^∗𝑆^∗−𝑦𝑢𝜌₂𝑅^∗𝐿^∗−𝑢𝜔𝑆^∗+𝑢𝜌₁𝑅^∗𝑆^∗+𝑢𝜌₂𝑅^∗𝐿^∗+11−𝑣𝑢𝛾𝑅^∗−𝑣𝛾𝑅^∗=12−𝑥−𝑥1𝑏+1−𝑥+1𝑥−𝑥𝑦1−𝑦𝑥𝑦−𝑦𝛽₁𝑃^∗𝐿^∗+11−𝑦+1𝑦𝑢−𝑦1−𝑢𝑢−𝑦𝑢𝜌₂𝑅^∗𝐿^∗+11−𝑦+1𝑦−𝑦𝑧1−𝑧𝑦𝑧−𝑧𝛽₂𝐿^∗𝑆^∗+11−𝑧()+1𝑧𝑢−𝑧1−𝑢(𝑢−𝑧𝑢)𝜌₁𝑅^∗𝑆^∗+11−𝑢(𝑧−𝑢)𝜔𝑆^∗+11−𝑣(𝑢−𝑣)𝛾𝑅^∗=12−𝑥−𝑥1𝑏+1−𝑢(𝑧−𝑢)𝜔𝑆^∗+11−𝑣(𝑢−𝑣)𝛾𝑅^∗, (54) as we know that (2−𝑥−1/𝑥)≤0, when 𝑧,𝑢,𝑣 satisfy one of four relations as follows: 𝑢<1,𝑣<1,𝑧≥𝑢≥𝑣,𝑢>1,𝑣>1,𝑧≤𝑢≤𝑣,𝑢>1,𝑣<1,𝑧≤𝑢,𝑢≥𝑣,𝑢<1,𝑣>1,𝑧≥𝑢,𝑢≤𝑣. (55) This implies that 𝑉^′≤0 with equality only if 𝑃=𝑃^∗ and 𝑆^∗/𝑆=𝑅^∗/𝑅=𝑄^∗/𝑄, that is, 𝑥=1, 𝑧=𝑢=𝑣. By LaSalle invariance principle [20, 22], 𝐸^∗ is globally asymptotically stable. This completes the proof.

Remark 8. It is possible for condition (48) to fail, in which case the global stability of the interior equilibrium of system (2) has not been established. Figure 2, however, seems to support the idea that the interior equilibrium of system (2) is still globally asymptotically stable even in this case.

Figure 2 

The smoking-present equilibrium 𝐸∗ is globally asymptotically stable.

4. Numerical Simulation

In this section, some numerical results of system (2) are presented for supporting the analytic results obtained previously. Our data are taken from [18], we also consider the data from Statistical Yearbook of the World Health [23] and Report of the Global Tobacco Epidemic [24]. Now, we give the data in Table 2.

Table 2 

The parameter values of giving up smoking model.

According to the survey, the world population over the age of 15 is about 5.5 billion; this population is recorded as potential smokers, the smoking rate was 34%. Hence, we will consider 5.5 billion, 2.2 billion, 1.87 billion, 0.058 billion, and 0.01 billion as the initial values of the five compartments.

Using the data in Table 2 we can get images in Figure 2.

The data in Table 2 satisfy the condition (3) of Theorem 3; we can see that the smoking-present equilibrium 𝐸^∗ is globally asymptotically stable.

For appropriate adjustment parameters, we choose 𝛽₁=0.0038, then the smoking-free equilibrium 𝐸₀ is globally asymptotically stable (Figure 3).

Figure 3 

When 𝑅0<1, the smoking-free equilibrium 𝐸0 is globally asymptotically stable.

If we choose 𝛽₁=0.4,𝛽₂=0.5,𝜇=0.1,𝑑₁=0.1,𝑑₂=0.2,𝑑₃=0.23, numerical simulation gives 𝑅₀>1 and 𝑅₁<1; the occasion smoking equilibrium 𝐸_𝐿 is globally asymptotically stable (Figure 4).

Figure 4 

When 𝑅0>1 and 𝑅1<1, the occasion smoking equilibrium 𝐸𝐿 is globally asymptotically stable.

At last, we choose 𝛽₁=0.000078125,𝑏=0.02,𝜇+𝑑₁=0.01,𝜇+𝑑₂=0.01, numerical simulation gives 𝑅₀=1; then the smoking-free equilibrium 𝐸₀ is globally asymptotically stable (Figure 5).

Figure 5 

When 𝑅0=1, the smoking-free equilibrium 𝐸0 is globally asymptotically stable.

5. Discussion

We have formulated a giving up smoking model with relapse and investigate their dynamical behaviors. By means of the next generation matrix, we obtain their basic reproduction number, 𝑅₀, which plays a crucial role. By constructing Lyapunov function, we prove the global stability of their equilibria: when the basic reproduction number is less than or equal to one, all solutions converge to the smoking-free equilibrium; that is, the smoking dies out eventually; when the basic reproduction number exceeds one, the occasion smoking equilibrium is stable; that is, the smoking will persist in the population, and the number of infected individuals tends to a positive constant.

In this paper, we consider two relapses. One is relapsed into light smokers and the other is relapsed into persistent smokers. If we employ some ways, such as medical care or education, to reduce the relapse rate, then, the number of the quit smokers will increase. We choose 𝜌₁=𝜌₂=0.003, we can get images in Figure 6.

Figure 6 

When 𝜌1=𝜌2=0.003, the smoking-present equilibrium 𝐸∗ is globally asymptotically stable.

Comparison of Figures 2 and 6, we can see the difference between them. In Figure 6, the number of recovery and quit smokers is increasing obviously. 𝑃(𝑡), 𝐿(𝑡), 𝑆(𝑡), 𝑅(𝑡) and 𝑄(𝑡) are approaching the stable state earlier than the case of Figure 2.

Through numerical simulation, we clearly recognize that if we control the rate of relapse, then the efficiency of giving up smoking will be greatly improved.

For system (2), 𝑏 reflects recruitment number, 𝛽₁ reflects the contact rate between potential smokers and occasion smokers, and 𝑑₁ denotes the deaths rate of potential smokers. 𝑑₂ denotes the deaths rate of light or occasion smokers. These four parameters will directly affect the values of the basic reproductive number. Furthermore, when 𝑏 and 𝛽₁ increase, the number smokers will increase, that is, 𝑅₀ increases. When we reduce the mortality caused by medical treatment, the number of permanent quit smokers will increase. Hence, 𝑅₀ will decrease. Figure 7 shows the relation between the basic reproduction number 𝑅₀ and 𝑑₁, Figure 8 shows the relation between the basic reproduction number 𝑅₀ and 𝑑₂, Figure 9 shows the relation between the basic reproduction number 𝑅₀ and 𝑏, Figure 10 shows the relation between the basic reproduction number 𝑅₀ and 𝛽₁. From Figures 11, 12, and 13, we can also see that if 𝑑₁ and 𝑑₂ increase, then 𝑅₀ will decrease. And if 𝑏 and 𝛽₁ increase, then 𝑅₀ will increase. Biologically, this means that to reduce the relapse rate and the deaths rate of nicotine by medical treatment, education and legal constraints are very important.

Figure 7 

The relationship between 𝑅0 and 𝑑1.

Figure 8 

The relationship between 𝑅0 and 𝑑2.

Figure 9 

The relationship between 𝑅0 and 𝑏.

Figure 10 

The relationship between 𝑅0 and 𝛽1.

Figure 11 

The relationship between 𝑅0, 𝑑1, and 𝑑2.

Figure 12 

The relationship between 𝑅0, 𝑏, and 𝑑1.

Figure 13 

The relationship between 𝑅0, 𝑏, and 𝑑2.

Compared with [18], in this paper we add two bilinear relapse rates. Hence, our model is more closer to real life. In [18], the author only discussed the local asymptotic stability of occasional smoking equilibrium. In this paper, we give the proof of the global asymptotic stability of occasional smoking equilibrium, adding two bilinear relapse rates based on [18], our model becomes more complex. This brought difficulties to the discussion of the existence and stability of the endemic equilibrium. From (3) of Theorem 3, we know that if 𝑅₂=1/𝑅₀+(𝑑₂+𝜇)𝛼₃/𝑏𝛽₂<1, 𝑑₃𝜌₂−𝑑₂<0 and 𝜌₂>max{𝑑₂𝜌₁𝜔/𝑑₃𝛼₃,𝜌₁𝜔/𝛼₃}, then system (2) has positive smoking-present equilibrium 𝐸^∗(𝑃^∗,𝐿^∗,𝑆^∗,𝑅^∗,𝑄^∗) where 𝑅^∗∈(0,𝜔/𝜌₁). Hence, the numerical values of 𝜌₁ and 𝜌₂ are playing an important role in the existence of 𝐸^∗. Next, we simulate the relationship between 𝜌₁ and 𝜌₂ under the conditions 𝑅₂=1/𝑅₀+(𝑑₂+𝜇)𝛼₃/𝑏𝛽₂<1, 𝑑₃𝜌₂−𝑑₂<0. If 𝑑₂𝜌₁𝜔/𝑑₃𝛼₃>𝜌₁𝜔/𝛼₃, we plot the relationship between 𝜌₁ and 𝜌₂ in Figure 14. If 𝑑₂𝜌₁𝜔/𝑑₃𝛼₃<𝜌₁𝜔/𝛼₃, we plot the relationship between 𝜌₁ and 𝜌₂ in Figure 15. From Figures 14 and 15, we can know that the point which locates above the line is the positive smoking-present equilibrium.

Figure 14 

If 𝑑2𝜌1𝜔/𝑑3𝛼3>𝜌1𝜔/𝛼3, the relationship between 𝜌1 and 𝜌2.

Figure 15 

If 𝑑2𝜌1𝜔/𝑑3𝛼3<𝜌1𝜔/𝛼3, the relationship between 𝜌1 and 𝜌2.

Acknowledgments

This work was partially supported by the NNSF of China (10961018), the NSF of Gansu Province of China (1107RJZA088), the NSF for Distinguished Young Scholars of Gansu Province of China (1111RJDA003), the Special Fund for the Basic Requirements in the Research of University of Gansu Province of China, and the Development Program for HongLiu Distinguished Young Scholars in Lanzhou University of Technology.