Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 525461, 12 pages
http://dx.doi.org/10.1155/2013/525461
Research Article
Influence of Relapse in a Giving Up Smoking Model
Hai-Feng Huo and Cheng-Cheng Zhu
Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, China
Correspondence should be addressed to Hai-Feng Huo; huohf70@sohu.com
Received 8 November 2012; Revised 7 December 2012; Accepted 20 December 2012
Academic Editor: Sanyi Tang
Copyright © 2013 H.-F. Huo and C.-C. Zhu. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
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 (𝑃𝑃), smokers (𝑆𝑆), and
quit smokers (𝑄𝑄). Zaman [18] extended the work of CastilloGarsow 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 (𝑅𝑅) 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.
2
Abstract and Applied Analysis
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.
From Figure 1, the total population is divided into five
compartments, namely, the potential smokers compartment
(𝑃𝑃), light or occasion smokers compartment (𝐿𝐿), persistent
smokers compartment (𝑆𝑆), temporary quit smokers (𝑅𝑅) that
is the people who did some efforts to stop smoking, and quit
smokers forever group (𝑄𝑄). 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
𝑑𝑑2 < 𝑑𝑑3 . The total population size is 𝑁𝑁𝑁𝑁𝑁𝑁, where
(1)
𝑁𝑁 (𝑡𝑡) = 𝑃𝑃 (𝑡𝑡) + 𝐿𝐿 (𝑡𝑡) + 𝑆𝑆 (𝑡𝑡) + 𝑅𝑅 (𝑡𝑡) + 𝑄𝑄 (𝑡𝑡).
The transfer diagram leads to the following system of
ordinary differential equations:
𝑃𝑃 𝑃𝑃𝑃1 ) = 0,
𝑃𝑃󸀠󸀠 (𝑡𝑡1 ) < 0,
𝑅𝑅 (𝑡𝑡) ≥ 0,
𝑄𝑄 (𝑡𝑡) ≥ 0,
𝐿𝐿 𝐿𝐿𝐿2 ) = 0,
𝐿𝐿󸀠󸀠 (𝑡𝑡2 ) < 0,
𝑅𝑅 (𝑡𝑡) ≥ 0,
𝑄𝑄 (𝑡𝑡) ≥ 0,
𝑆𝑆 𝑆𝑆𝑆3 ) = 0,
𝑆𝑆󸀠󸀠 (𝑡𝑡3 ) < 0,
𝑅𝑅 (𝑡𝑡) ≥ 0,
− (𝑑𝑑2 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡),
𝑑𝑑𝑑𝑑 (𝑡𝑡)
= 𝛽𝛽2 𝐿𝐿 (𝑡𝑡) 𝑆𝑆 (𝑡𝑡) + 𝜌𝜌1 𝑆𝑆 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡) − (𝜔𝜔 𝜔𝜔𝜔3 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡),
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 (𝑡𝑡)
= 𝜔𝜔𝜔𝜔 (𝑡𝑡) − 𝜌𝜌1 𝑆𝑆 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡) − 𝜌𝜌2 𝐿𝐿 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡)
𝑑𝑑𝑑𝑑
− (𝛾𝛾 𝛾𝛾𝛾4 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡),
(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 𝑡𝑡 𝑡 𝑡, 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 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃
0, the solutions 𝑃𝑃𝑃𝑃𝑃𝑃, 𝐿𝐿𝐿𝐿𝐿𝐿, 𝑆𝑆𝑆𝑆𝑆𝑆, 𝑅𝑅𝑅𝑅𝑅𝑅, 𝑄𝑄𝑄𝑄𝑄𝑄 of system (2) are
positive for all 𝑡𝑡 𝑡 𝑡.
𝑃𝑃 (𝑡𝑡) ≥ 0,
𝑃𝑃 (𝑡𝑡) ≥ 0,
𝑄𝑄 (𝑡𝑡) ≥ 0,
𝑅𝑅󸀠󸀠 (𝑡𝑡4 ) < 0,
𝑆𝑆 (𝑡𝑡) ≥ 0,
𝑄𝑄 (𝑡𝑡) ≥ 0,
𝑄𝑄 𝑄𝑄𝑄5 ) = 0,
𝑄𝑄󸀠󸀠 (𝑡𝑡5 ) < 0,
𝑅𝑅 (𝑡𝑡) ≥ 0,
𝑃𝑃 (𝑡𝑡) ≥ 0,
(5)
𝐿𝐿 (𝑡𝑡) ≥ 0,
0 ≤ 𝑡𝑡𝑡𝑡𝑡4 .
(5) There exists a first time 𝑡𝑡5 such that
(4)
𝐿𝐿 (𝑡𝑡) ≥ 0,
0 ≤ 𝑡𝑡𝑡𝑡𝑡3 .
𝑃𝑃 (𝑡𝑡) ≥ 0,
(3)
𝑆𝑆 (𝑡𝑡) ≥ 0,
0 ≤ 𝑡𝑡𝑡𝑡𝑡2 .
(4) There exists a first time 𝑡𝑡4 such that
𝑅𝑅 𝑅𝑅𝑅4 ) = 0,
𝑆𝑆 (𝑡𝑡) ≥ 0,
0 ≤ 𝑡𝑡𝑡𝑡𝑡1 .
(3) There exists a first time 𝑡𝑡3 such that
In case (1), we have
𝑑𝑑𝑑𝑑 (𝑡𝑡)
= 𝛽𝛽1 𝑃𝑃 (𝑡𝑡) 𝐿𝐿 (𝑡𝑡) −𝛽𝛽2 𝐿𝐿 (𝑡𝑡) 𝑆𝑆 (𝑡𝑡) + 𝜌𝜌2 𝐿𝐿 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡)
𝑑𝑑𝑑𝑑
𝐿𝐿 (𝑡𝑡) ≥ 0,
(2) There exists a first time 𝑡𝑡2 such that
𝑆𝑆 (𝑡𝑡) ≥ 0,
𝑑𝑑𝑑𝑑 (𝑡𝑡)
= 𝑏𝑏 𝑏 𝑏𝑏1 𝑃𝑃 (𝑡𝑡) 𝐿𝐿 (𝑡𝑡) − (𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡),
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 (𝑡𝑡)
= 𝛾𝛾𝛾𝛾 (𝑡𝑡) − (𝑑𝑑5 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡).
𝑑𝑑𝑑𝑑
Proof. If the conclusion does not hold, then at least one of
𝑃𝑃𝑃𝑃𝑃𝑃, 𝐿𝐿𝐿𝐿𝐿𝐿, 𝑆𝑆𝑆𝑆𝑆𝑆, 𝑅𝑅𝑅𝑅𝑅𝑅, 𝑄𝑄𝑄𝑄𝑄𝑄 is not positive. Thus, we have one of
the following five cases.
(1) There exists a first time 𝑡𝑡1 such that
(6)
𝐿𝐿 (𝑡𝑡) ≥ 0,
0 ≤ 𝑡𝑡𝑡𝑡𝑡5 .
(7)
𝑃𝑃󸀠󸀠 (𝑡𝑡1 ) = 𝑏𝑏 𝑏𝑏𝑏
(8)
𝐿𝐿󸀠󸀠 (𝑡𝑡2 ) = 0,
(9)
𝑆𝑆󸀠󸀠 (𝑡𝑡3 ) = 0,
(10)
𝑅𝑅󸀠󸀠 (𝑡𝑡4 ) = 𝜔𝜔𝜔𝜔𝜔𝜔𝜔4 )>0,
(11)
𝑄𝑄󸀠󸀠 (𝑡𝑡5 ) = 𝛾𝛾𝛾𝛾 (𝑡𝑡5 ) >0,
(12)
which is a contradiction to 𝑃𝑃󸀠󸀠 (𝑡𝑡1 ) < 0.
In case (2), we have
which is a contradiction to 𝐿𝐿󸀠󸀠 (𝑡𝑡2 ) < 0.
In case (3), we have
which is a contradiction to 𝑆𝑆󸀠󸀠 (𝑡𝑡3 ) < 0.
In case (4), we have
which is a contradiction to 𝑅𝑅󸀠󸀠 (𝑡𝑡4 ) < 0.
In case (5), we have
which is a contradiction to 𝑄𝑄󸀠󸀠 (𝑡𝑡5 ) < 0.
Thus, the solutions 𝑃𝑃𝑃𝑃𝑃𝑃, 𝐿𝐿𝐿𝐿𝐿𝐿, 𝑆𝑆𝑆𝑆𝑆𝑆, 𝑅𝑅𝑅𝑅𝑅𝑅, 𝑄𝑄𝑄𝑄𝑄𝑄 of system (2)
remain positive for all 𝑡𝑡 𝑡 𝑡.
Abstract and Applied Analysis
3
ρ2 L(t)R(t)
ρ1 S(t)R(t)
β1 P (t)L(t)
b
P
d1
β2 L(t)S(t)
L
μ
d2
μ
d3
γ
ω
S
μ
R
d4
Q
μ
d5
μ
Figure 1: Transfer diagram for the dynamics of giving up smoking model.
Table 1: The parameters description of giving up smoking model.
Parameter
Description
𝑏𝑏
The total recruitment number into this homogeneous social mixing community.
Transmission coefficient from the potential smokers compartment to the light or occasion smokers
compartment.
Transmission coefficient from the light or occasion smokers compartment to the persistent smokers
compartment.
The relapse rate of which temporary quit people contact with persistent smokers.
The relapse rate of which temporary quit people contact with light smokers.
The temporary quit smoking rate.
The permanent quit smoking rate.
Naturally death rate.
The smoking-related death rate.
𝛽𝛽1
𝛽𝛽2
𝜌𝜌1
𝜌𝜌2
𝜔𝜔
𝛾𝛾
𝜇𝜇
𝑑𝑑𝑖𝑖 , 𝑖𝑖 𝑖 𝑖𝑖 𝑖 𝑖 𝑖 𝑖 𝑖
Lemma 2. All feasible solution of the system (2) are bounded
and enter the region
Ω = {(𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃) ∈ 𝑅𝑅+5 : 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃
𝑏𝑏
}. (13)
𝜇𝜇
Proof. Let (𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 𝑃𝑃+5 be any solution with nonnegative initial condition: adding the first four equations of (2),
we have
𝑑𝑑
(𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃)
𝑑𝑑𝑑𝑑
= 𝑏𝑏 𝑏 𝑏𝑏𝑏1 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇3 + 𝜇𝜇𝜇 𝜇𝜇
−(𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇 𝜇𝜇
(14)
= 𝑏𝑏 𝑏 𝑏𝑏 (𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃)
−(𝑑𝑑1 𝑃𝑃 𝑃 𝑃𝑃2 𝐿𝐿𝐿𝐿𝐿3 𝑆𝑆𝑆𝑆𝑆4 𝑅𝑅𝑅𝑅𝑅5 𝑄𝑄𝑄
≤ 𝑏𝑏 𝑏 𝑏𝑏𝑏𝑏𝑏
𝑏𝑏
+𝑁𝑁 (0) 𝑒𝑒−𝜇𝜇𝜇𝜇,
𝜇𝜇
(15)
where 𝑁𝑁𝑁𝑁𝑁 represents initial values of the total population.
Thus 0 ≤𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁, as 𝑡𝑡 𝑡 𝑡. Therefore all feasible
solutions of system (2) enter the region
Ω = {(𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃𝑃 𝑃𝑃) ∈ 𝑅𝑅+5 : 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃 𝑃𝑃 𝑃
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))
𝐸𝐸0 = (
It follows that
0 ≤𝑁𝑁 (𝑡𝑡) ≤
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 𝑁𝑁𝑁𝑁𝑁𝑁 is bounded and all the solutions starting
in Ω approach enter or stay in Ω.
𝑏𝑏
}. (16)
𝜇𝜇
𝑏𝑏
, 0, 0, 0, 0).
𝑑𝑑1 + 𝜇𝜇
(17)
In the following, the basic reproduction number of
system (2) will be obtained by the next generation matrix
method formulated in [21].
Let 𝑥𝑥 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑇𝑇 ; then system (2) can be written as
𝑑𝑑𝑑𝑑
= F (𝑥𝑥) − V (𝑥𝑥),
𝑑𝑑𝑑𝑑
(18)
4
Abstract and Applied Analysis
Theorem 3. For the giving up smoking model (2), there exist
the following three types of equilibrium.
where
𝛽𝛽1 𝑃𝑃𝑃𝑃
0
F (𝑥𝑥) = ( 0 ),
0
0
𝛽𝛽2 𝐿𝐿𝐿𝐿 𝐿 𝐿𝐿𝐿2 + 𝜇𝜇𝜇 𝜇𝜇 𝜇 𝜇𝜇2 𝐿𝐿𝐿𝐿
−𝛽𝛽2 𝐿𝐿𝐿𝐿 𝐿𝐿𝐿1 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑆𝑆𝑆3 + 𝜇𝜇𝜇 𝜇𝜇
V (𝑥𝑥) = (−𝜔𝜔𝜔𝜔𝜔𝜔𝜔1 𝑆𝑆𝑆𝑆𝑆𝑆𝑆2 𝐿𝐿𝐿𝐿 𝐿 𝐿𝐿𝐿 𝐿 𝐿𝐿4 + 𝜇𝜇𝜇 𝜇𝜇).
−𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾5 + 𝜇𝜇𝜇 𝜇𝜇
−𝑏𝑏 𝑏𝑏𝑏1 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃1 + 𝜇𝜇𝜇 𝜇𝜇
(19)
The Jacobian matrices of F(𝑥𝑥𝑥 and V(𝑥𝑥𝑥 at the smoking-free
equilibrium 𝐸𝐸0 are, respectively,
𝐷𝐷F (𝐸𝐸0 ) = (
𝐹𝐹4 × 4 0
),
0 0
0
𝑉𝑉4 × 4
𝐷𝐷V (𝐸𝐸0 ) = ( 𝛽𝛽1 𝑏𝑏 0 0 0 𝑑𝑑 + 𝜇𝜇),
1
𝑑𝑑1 + 𝜇𝜇
where
𝐹𝐹4 × 4
𝑉𝑉4 × 4
𝛽𝛽1 𝑏𝑏
𝑑𝑑1 + 𝜇𝜇
=( 0
0
0
𝑃𝑃∗ =
0 0 0
(21)
𝑉𝑉−1
0
1
(0
0
0)
(
)
𝑏𝑏
2
(
)
=(
).
1
𝜔𝜔
(0
0)
(
)
𝛾𝛾𝛾𝛾𝛾3
(𝜔𝜔𝜔𝜔𝜔2 )(𝛾𝛾𝛾𝛾𝛾3 )
𝛾𝛾
𝜔𝜔𝜔𝜔
1
0
𝑏𝑏
(𝜔𝜔𝜔𝜔𝜔
)(𝛾𝛾𝛾𝛾𝛾
)
𝑏𝑏
(𝛾𝛾𝛾𝛾𝛾
)
𝑏𝑏
4)
2
3 4
3 4
(
(22)
The basic reproduction number, denoted by 𝑅𝑅0 , is thus given
by
𝛽𝛽1 𝑏𝑏
𝛽𝛽1 𝑏𝑏
=
.
𝑅𝑅0 = 𝜌𝜌 𝜌𝜌𝜌𝜌𝜌−1 ) =
(𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇1 (𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇𝜇2 + 𝜇𝜇𝜇
𝛼𝛼4 = (𝛾𝛾𝛾𝛾𝛾4 + 𝜇𝜇).
2
𝛼𝛼3 −𝜌𝜌1 𝑅𝑅∗
,
𝛽𝛽2
𝜌𝜌1 𝜌𝜌2 𝑅𝑅∗ − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅∗ − 𝛼𝛼3 𝜌𝜌2 𝑅𝑅∗
,
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅∗ −𝜔𝜔𝜔
∗
∗
∗
1 −𝛼𝛼4 𝛽𝛽2 𝑅𝑅 −(𝑑𝑑3 + 𝜇𝜇𝜇 𝜇𝜇2 𝑅𝑅 +(𝜌𝜌1 𝑅𝑅 −𝜔𝜔𝜔𝜔𝜔𝜔2 + 𝜇𝜇𝜇
,
𝛽𝛽1
(𝜌𝜌1 𝑅𝑅∗ −𝜔𝜔𝜔
𝑄𝑄∗ =
𝛾𝛾
𝑅𝑅∗,
𝑑𝑑5 + 𝜇𝜇
(25)
where 𝑅𝑅 is a positive solution of 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, where
𝑓𝑓 (𝑅𝑅) = (𝑑𝑑1 + 𝜇𝜇𝜇
𝜌𝜌1 𝜌𝜌2 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅
𝛽𝛽1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
+(𝑑𝑑2 + 𝜇𝜇𝜇
+(𝑑𝑑3 + 𝜇𝜇𝜇
+𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾
(𝑑𝑑 + 𝜇𝜇𝜇 𝜇𝜇𝜇2 + 𝜇𝜇𝜇
𝜌𝜌2
𝑅𝑅𝑅 1
𝛽𝛽1
𝛽𝛽1
𝛼𝛼3 −𝜌𝜌1𝑅𝑅
𝛽𝛽2
𝜌𝜌1 𝜌𝜌2 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅
+(𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
(26)
Proof . It follows from system (2) that
𝑏𝑏 𝑏𝑏𝑏1 𝑃𝑃 (𝑡𝑡) 𝐿𝐿 (𝑡𝑡) −(𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) = 0,
𝛽𝛽1 𝑃𝑃 (𝑡𝑡) 𝐿𝐿 (𝑡𝑡) − 𝛽𝛽2 𝐿𝐿 (𝑡𝑡) 𝑆𝑆 (𝑡𝑡) +𝜌𝜌2 𝐿𝐿 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡)
−(𝑑𝑑2 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) = 0,
(23)
𝛽𝛽2 𝐿𝐿 (𝑡𝑡) 𝑆𝑆 (𝑡𝑡) +𝜌𝜌1 𝑆𝑆 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡) −(𝜔𝜔𝜔𝜔𝜔3 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) = 0,
(24)
−(𝛾𝛾𝛾𝛾𝛾4 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) = 0,
Throughout this paper, we denote
𝛼𝛼3 = (𝜔𝜔𝜔𝜔𝜔3 + 𝜇𝜇) ,
𝐿𝐿∗ =
−(𝑑𝑑1 + 𝜇𝜇𝜇
In order to simplify the calculation, letting 𝑏𝑏1 = 𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇2 =
𝜔𝜔𝜔𝜔𝜔3 + 𝜇𝜇𝜇𝜇𝜇3 = 𝛾𝛾𝛾𝛾𝛾4 + 𝜇𝜇𝜇𝜇𝜇4 = 𝑑𝑑5 + 𝜇𝜇, we obtain
0
𝑆𝑆∗ =
∗
0 0 0),
0 0 0
0 0 0
0
Moveover, 𝐸𝐸∗ (𝑃𝑃∗ , 𝐿𝐿∗ , 𝑆𝑆∗ , 𝑅𝑅∗ , 𝑄𝑄∗ ) satisfies the following
equality:
(20)
𝜇𝜇𝜇
𝑑𝑑2 + 𝜇𝜇
𝜇𝜇
0
𝜔𝜔𝜔𝜔𝜔3 + 𝜇𝜇
).
=(
𝜇
0
−𝜔𝜔𝜔𝜔𝜔𝜔𝜔4 + 𝜇𝜇
0
0
−𝛾𝛾𝛾𝛾5 + 𝜇𝜇
1
𝑏𝑏1
(1) For all parameter values, system (2) exists the smokingfree equilibrium 𝐸𝐸0 (𝑏𝑏𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇.
(2) If 𝑅𝑅0 > 1, there exists the occasion smoking equilibrium
𝐸𝐸𝐿𝐿 ((𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇𝜇1 , 𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇 𝜇 𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇1 , 0, 0, 0), and
there exists no occasion smoking equilibrium if 𝑅𝑅0 ≤ 1.
(3) If 𝑅𝑅2 = 1/𝑅𝑅0 + (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇3 /𝑏𝑏𝑏𝑏2 < 1, 𝑑𝑑3 𝜌𝜌2 −𝑑𝑑2 <
0 and 𝜌𝜌2 > max{𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 , 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 }, then system (2) has positive smoking-present equilibrium
𝐸𝐸∗ (𝑃𝑃∗ , 𝐿𝐿∗ , 𝑆𝑆∗ , 𝑅𝑅∗ , 𝑄𝑄∗ ) where 𝑅𝑅∗ ∈ (0, 𝜔𝜔𝜔𝜔𝜔1 ).
𝜔𝜔𝜔𝜔 (𝑡𝑡) −𝜌𝜌1 𝑆𝑆 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡) −𝜌𝜌2 𝐿𝐿 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡)
𝛾𝛾𝛾𝛾 (𝑡𝑡) − (𝑑𝑑5 + 𝜇𝜇) 𝑄𝑄 (𝑡𝑡) = 0.
(27)
Abstract and Applied Analysis
5
(1) Letting 𝐿𝐿 𝐿 𝐿𝐿 𝐿 𝐿𝐿 𝐿 𝐿𝐿 𝐿 𝐿 in (27), we can obtain the
smoking-free equilibrium 𝐸𝐸0 (𝑏𝑏𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜇𝜇.
(2) If 𝑅𝑅0 > 1, letting 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 in (27), we can
obtain the occasion smoking equilibrium 𝐸𝐸𝐿𝐿 ((𝑑𝑑2 +
𝜇𝜇𝜇𝜇𝜇𝜇1 , 𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇 𝜇 𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇1 ,0,0,0).
(3) From third equation of the system (27), we obtain
𝐿𝐿 𝐿
𝛼𝛼3 − 𝜌𝜌1 𝑅𝑅
.
𝛽𝛽2
(28)
By adding the third equation and the fourth equation, we get
𝛽𝛽2 𝐿𝐿 (𝑡𝑡) 𝑆𝑆 (𝑡𝑡) − (𝜔𝜔 𝜔𝜔𝜔3 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) + 𝜔𝜔𝜔𝜔 (𝑡𝑡)
− (𝛾𝛾 𝛾𝛾𝛾4 + 𝜇𝜇𝜇 𝜇𝜇 (𝑡𝑡) − 𝜌𝜌2 𝐿𝐿 (𝑡𝑡) 𝑅𝑅 (𝑡𝑡) =0.
(29)
𝛼𝛼3 𝜌𝜌2 <0; these show that 𝑆𝑆𝑆𝑆. From (32), we can see that if
𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅1 , then 𝑃𝑃 𝑃𝑃. Therefore, if 𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅1 , 𝑃𝑃 𝑃𝑃, 𝐿𝐿 𝐿 𝐿,
𝑆𝑆𝑆𝑆, 𝑄𝑄𝑄𝑄. In the following, we prove that the existence of
positive solutions of 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓. From (34), we know
lim 𝑓𝑓 (𝑅𝑅) = +∞.
𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅1 −
If (𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇𝜇𝜇𝜇1 + (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇3 /𝛽𝛽2 < 𝑏𝑏, then
𝑓𝑓 (0) =
𝑓𝑓󸀠󸀠 (𝑅𝑅) =
From this we have
𝜌𝜌 𝜌𝜌 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅
𝑆𝑆𝑆 1 2
.
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
(30)
where
(31)
Substituting 𝑆𝑆 into (31), we get
𝑑𝑑 + 𝜇𝜇
𝜌𝜌 𝜌𝜌 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅 𝜌𝜌2
− 𝑅𝑅𝑅 2
𝑃𝑃 𝑃 1 2
𝛽𝛽1
𝛽𝛽1
𝛽𝛽1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
1 −𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅𝑅3 + 𝜇𝜇𝜇 𝜇𝜇2 𝑅𝑅𝑅𝑅𝑅𝑅1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 + 𝜇𝜇𝜇
=
.
𝛽𝛽1
(𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
(32)
Let
(𝑑𝑑 + 𝜇𝜇𝜇 𝜇𝜇𝜇2 + 𝜇𝜇𝜇
𝜌𝜌
− (𝑑𝑑1 + 𝜇𝜇𝜇 2 𝑅𝑅𝑅 1
𝛽𝛽1
𝛽𝛽1
+ (𝑑𝑑2 + 𝜇𝜇𝜇
Hence,
where
𝜔𝜔
) = (−𝜌𝜌2 𝜔𝜔 𝜔𝜔𝜔4 𝛽𝛽2 + 𝛼𝛼3 𝜌𝜌2 ) 𝜔𝜔 𝜔𝜔𝜔
𝜌𝜌1
(𝑑𝑑1 + 𝜇𝜇𝜇 𝛼𝛼4 𝛽𝛽2 𝜔𝜔 𝜔𝜔𝜔3 𝜌𝜌2 𝜔𝜔 𝜔𝜔𝜔2 𝜔𝜔2
2
𝛽𝛽1
(𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
+
ℎ (𝑅𝑅)
2
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
+ 𝜇𝜇𝜇𝜇12 𝜌𝜌2 𝑅𝑅2 − 2𝜇𝜇𝜇𝜇1 𝜌𝜌2 𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔4 𝛽𝛽2 + 𝜇𝜇𝜇𝜇𝜇𝜇3 𝜌𝜌2
−𝑑𝑑2 𝜌𝜌13 𝑅𝑅2 + 2𝑑𝑑2 𝜌𝜌12 𝜔𝜔𝜔𝜔𝜔𝜔𝜔2 𝜌𝜌1 𝜔𝜔2
(34)
𝜌𝜌 𝜌𝜌 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅
+ (𝑑𝑑3 + 𝜇𝜇𝜇 1 2
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
From (28) we can see that if 𝑅𝑅 𝑅 𝑅𝑅3 /𝜌𝜌1 , then 𝐿𝐿 𝐿 𝐿. From (30)
we can see that if 𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅1 , then 𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅1 𝜌𝜌2 𝑅𝑅𝑅𝑅𝑅4 𝛽𝛽2 −
(36)
(37)
Then
(38)
(39)
+ (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇
ℎ (𝑅𝑅) =𝑑𝑑3 𝜌𝜌12 𝜌𝜌2 𝑅𝑅2 − 2𝑑𝑑3 𝜌𝜌1 𝜌𝜌2 𝜔𝜔𝜔𝜔𝜔𝜔𝜔3 𝜔𝜔𝜔𝜔4 𝛽𝛽2 + 𝑑𝑑3 𝜔𝜔𝜔𝜔3 𝜌𝜌2
𝛼𝛼3 − 𝜌𝜌1𝑅𝑅
𝛽𝛽2
+ (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝜇 𝜇𝜇𝜇
(𝑑𝑑3 + 𝜇𝜇𝜇 𝑔𝑔 (𝑅𝑅)
+ (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇
𝛽𝛽2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅2
𝑔𝑔 (𝑅𝑅) ≥ 𝑔𝑔 𝑔
(33)
+ (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇 𝜇𝜇 𝜇 𝜇𝜇
𝜌𝜌1 𝜌𝜌2 𝑅𝑅2 − 𝛼𝛼4 𝛽𝛽2 𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 𝑅𝑅
𝛽𝛽1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
+
𝑓𝑓󸀠󸀠 (𝑅𝑅) =
𝑓𝑓 (𝑅𝑅) = (𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇3 + 𝜇𝜇𝜇 𝜇𝜇
= (𝑑𝑑1 + 𝜇𝜇𝜇
(𝑑𝑑1 + 𝜇𝜇𝜇 𝑔𝑔 (𝑅𝑅)
𝜌𝜌
𝜌𝜌
− (𝑑𝑑1 + 𝜇𝜇𝜇 2 − (𝑑𝑑2 + 𝜇𝜇𝜇 1
2
𝛽𝛽1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
𝛽𝛽1
𝛽𝛽2
Clearly, 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 is the minimum point of 𝑔𝑔𝑔𝑔𝑔𝑔. For 𝑅𝑅 𝑅
(0, 𝜔𝜔𝜔𝜔𝜔1 ), we have
It follows from the fifth equation that
𝛾𝛾
𝑄𝑄𝑄
𝑅𝑅𝑅
𝑑𝑑5 + 𝜇𝜇
(𝑑𝑑1 + 𝜇𝜇𝜇 𝜇𝜇𝜇2 + 𝜇𝜇𝜇 (𝑑𝑑2 + 𝜇𝜇𝜇 𝜇𝜇3
+
− 𝑏𝑏 𝑏𝑏𝑏
𝛽𝛽1
𝛽𝛽2
𝑔𝑔 (𝑅𝑅) = 𝜌𝜌12 𝜌𝜌2 𝑅𝑅2 − 2𝜌𝜌1 𝜌𝜌2 𝜔𝜔𝜔𝜔𝜔𝜔𝜔4 𝛽𝛽2 𝜔𝜔 𝜔𝜔𝜔3 𝜌𝜌2 𝜔𝜔𝜔
From the second equation of the system (27) we get
𝛽𝛽1 𝑃𝑃 (𝑡𝑡) =𝛽𝛽2 𝑆𝑆 (𝑡𝑡) − 𝜌𝜌2 𝑅𝑅 (𝑡𝑡) + (𝑑𝑑2 + 𝜇𝜇𝜇 𝜇
(35)
(40)
− 𝜇𝜇𝜇𝜇13 𝑅𝑅2 + 2𝜇𝜇𝜇𝜇12 𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔1 𝜔𝜔2 .
ℎ󸀠󸀠 (𝑅𝑅) = 2𝑑𝑑3 𝜌𝜌1 𝜌𝜌2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 𝜌𝜌2 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅
− 2𝑑𝑑2 𝜌𝜌1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 𝑅𝑅𝑅𝑅𝑅𝑅
= 2𝜌𝜌1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅3 𝜌𝜌2 −𝑑𝑑2 )
+ 2𝜇𝜇𝜇𝜇1 (𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 − 1),
(41)
as we know that 𝜌𝜌2 < 1 and 𝜌𝜌1 𝑅𝑅𝑅𝑅𝑅 𝑅 𝑅. If 𝑑𝑑3 𝜌𝜌2 −𝑑𝑑2 <0, then
for any 𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 ), we have ℎ󸀠󸀠 (𝑅𝑅𝑅𝑅𝑅, so ℎ(𝑅𝑅𝑅 is a strictly
6
Abstract and Applied Analysis
monotone increasing function on (0, 𝜔𝜔𝜔𝜔𝜔1 ). As we know that
𝜌𝜌2 > max{𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 , 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 }, 𝑑𝑑3 > 𝑑𝑑2 and 𝛼𝛼3 > 𝜔𝜔, so
ℎ (0) = 𝑑𝑑3 𝜔𝜔𝜔𝜔4 𝛽𝛽2 + 𝑑𝑑3 𝜔𝜔𝜔𝜔3 𝜌𝜌2 + 𝜇𝜇𝜇𝜇𝜇𝜇4 𝛽𝛽2
+ 𝜇𝜇𝜇𝜇𝜇𝜇3 𝜌𝜌2 − 𝑑𝑑2 𝜌𝜌1 𝜔𝜔2 − 𝜇𝜇𝜇𝜇1 𝜔𝜔2
= 𝑑𝑑3 𝜔𝜔𝜔𝜔4 𝛽𝛽2 + 𝜇𝜇𝜇𝜇𝜇𝜇4 𝛽𝛽2 + 𝜔𝜔 𝜔𝜔𝜔3 𝛼𝛼3 𝜌𝜌2 − 𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔
(42)
Hence, ℎ(𝑅𝑅𝑅𝑅𝑅 for any 𝑅𝑅 𝑅 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 ), so 𝑓𝑓󸀠󸀠 (𝑅𝑅𝑅𝑅𝑅; that is,
𝑓𝑓𝑓𝑓𝑓𝑓 is a strictly monotone increasing function on (0, 𝜔𝜔𝜔𝜔𝜔1 ).
Therefore, 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 has an unique positive solutions on
(0, 𝜔𝜔𝜔𝜔𝜔1 ). 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 𝑅𝑅0 ≤ 1, the smoking-free equilibrium 𝐸𝐸0 is
globally asymptotically stable.
Proof. We introduce the following Lyapunov function:
= 𝑃𝑃0 (
𝑃𝑃
𝑃𝑃
− ln ) + 𝐿𝐿 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑃𝑃0
𝑃𝑃0
The derivative of 𝑉𝑉 is given by
𝑃𝑃
𝑉𝑉󸀠󸀠 = 𝑃𝑃󸀠󸀠 − 0 𝑃𝑃󸀠󸀠 + 𝐿𝐿󸀠󸀠 + 𝑆𝑆󸀠󸀠 + 𝑅𝑅󸀠󸀠 + 𝑄𝑄󸀠󸀠
𝑃𝑃
= 𝑏𝑏 𝑏𝑏𝑏1 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃1 + 𝜇𝜇𝜇𝜇𝜇𝜇
𝑏𝑏𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇
𝑃𝑃
× (𝑏𝑏 𝑏𝑏𝑏1 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 𝑃𝑃𝑃𝑃𝑃𝑃𝑃2 𝐿𝐿𝐿𝐿
+ 𝜌𝜌2 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇2 𝐿𝐿𝐿𝐿𝐿𝐿𝐿1 𝑆𝑆𝑆𝑆
− (𝜔𝜔 𝜔𝜔𝜔3 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 𝑆𝑆𝑆𝑆
− 𝜌𝜌2 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐿𝐿𝐿4 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇𝜇𝜇
𝛽𝛽 𝑏𝑏
𝑏𝑏2
− (𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇𝜇 1 𝐿𝐿
= 2𝑏𝑏 𝑏
𝑑𝑑1 + 𝜇𝜇
(𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇
= [2 −
+(
(43)
(𝑑𝑑 + 𝜇𝜇𝜇𝜇𝜇
𝑏𝑏
− 1
] 𝑏𝑏
𝑏𝑏
(𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇
𝛽𝛽1 𝑏𝑏
− 𝑑𝑑2 − 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇3 + 𝜇𝜇𝜇𝜇𝜇
𝑑𝑑1 + 𝜇𝜇
− (𝑑𝑑4 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇𝜇𝜇
= [2 −
× 𝜇𝜇𝜇𝜇𝜇𝜇𝜇3 𝜌𝜌2 − 𝜌𝜌1 𝜔𝜔𝜔 𝜔𝜔𝜔
𝑉𝑉 (𝑃𝑃 (𝑡𝑡) , 𝐿𝐿 (𝑡𝑡) , 𝑆𝑆 (𝑡𝑡) , 𝑅𝑅 (𝑡𝑡) , 𝑄𝑄 (𝑡𝑡))
− (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇3 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇4 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇𝜇𝜇
+
(𝑑𝑑 + 𝜇𝜇𝜇𝜇𝜇
𝑏𝑏
− 1
] 𝑏𝑏
𝑏𝑏
(𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇
𝛽𝛽1 𝑏𝑏 𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇
𝐿𝐿 𝐿𝐿𝐿𝐿3 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇4 + 𝜇𝜇𝜇𝜇𝜇
𝑑𝑑1 + 𝜇𝜇
− (𝑑𝑑5 + 𝜇𝜇𝜇𝜇𝜇𝜇
(44)
If 𝑅𝑅0 ≤ 1, then 𝛽𝛽1 𝑏𝑏 𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇, so we get (𝛽𝛽1 𝑏𝑏 𝑏𝑏𝑏𝑏1 +
𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇.
As we know, 2 − 𝑏𝑏𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇, so we
obtain 𝑉𝑉󸀠󸀠 ≤ 0 with equality only if 2 − 𝑏𝑏𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 +
𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇, 𝑅𝑅0 = 1 and 𝑆𝑆 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆. By LaSalle invariance
principle [20, 22], 𝐸𝐸0 is globally asymptotically stable. Thus,
for system (2), the smoking-free equilibrium 𝐸𝐸0 is globally
asymptotically stable if 𝑅𝑅0 ≤ 1.
3.2.2. Stability of the Occasion Smoking Equilibrium. In this
part, we will consider an occasional smoking equilibrium
𝐸𝐸𝐿𝐿 ((𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇𝜇1 , 𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇1 , 0, 0, 0); that is, only
potential smokers and occasionally smokers are not zero, and
the other compartments are zero.
Theorem 5. If 𝑅𝑅0 > 1, the occasion smoking equilibrium 𝐸𝐸𝐿𝐿 is
locally asymptotically stable.
Proof. The Jacobian matrix of the giving up smoking model
(2) around 𝐸𝐸𝐿𝐿 is given by
𝛽𝛽1 𝑏𝑏
𝑑𝑑2 + 𝜇𝜇
−𝑑𝑑2 − 𝜇𝜇
).
𝐽𝐽𝐸𝐸𝐿𝐿 = ( 𝛽𝛽 𝑏𝑏
1
− (𝑑𝑑1 + 𝜇𝜇𝜇𝜇
𝑑𝑑2 + 𝜇𝜇
−
(45)
The characteristic polynomial is 𝜓𝜓𝜓𝜓𝜓𝜓𝜓 𝜓𝜓2 + 𝑐𝑐1 𝑥𝑥𝑥𝑥𝑥2 , where
𝑐𝑐1 = 𝛽𝛽1 𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇 and 𝑐𝑐2 = 𝛽𝛽1 𝑏𝑏 𝑏𝑏𝑏𝑏1 + 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇.
Therefore by Routh-Hurwitz criteria we deduce that the roots
of the polynomial 𝜓𝜓𝜓𝜓𝜓𝜓 have negative real part when 𝛽𝛽1 𝑏𝑏 𝑏
(𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇𝜇2 + 𝜇𝜇𝜇, which shows that the system is locally
asymptotically stable if 𝑅𝑅0 > 1.
Theorem 6. If 𝑅𝑅0 > 1 and 𝑅𝑅1 = 𝑏𝑏𝑏𝑏2 /(𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇𝜇3 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 +
𝜇𝜇𝜇𝜇𝜇2 /𝛽𝛽1 (𝑑𝑑3 + 𝜇𝜇𝜇𝜇𝜇, the occasion smoking equilibrium 𝐸𝐸𝐿𝐿 is
globally asymptotically stable.
Abstract and Applied Analysis
7
Proof. We introduce the following Lyapunov function:
𝑉𝑉 (𝑃𝑃 (𝑡𝑡) , 𝐿𝐿 (𝑡𝑡) , 𝑆𝑆 (𝑡𝑡) , 𝑅𝑅 (𝑡𝑡) , 𝑄𝑄 (𝑡𝑡))
= 𝑃𝑃𝐿𝐿 (
𝑃𝑃
𝑃𝑃
𝐿𝐿
𝐿𝐿
− ln ) + 𝐿𝐿 𝐿
− ln )
𝑃𝑃𝐿𝐿
𝑃𝑃𝐿𝐿
𝐿𝐿 𝐿𝐿
𝐿𝐿 𝐿𝐿
Proof. At the equilibrium point, the expressions on the righthand side of system (2) give us following relations:
(46)
+ 𝑆𝑆 𝑆𝑆𝑆𝑆𝑆𝑆𝑆
= 2𝑏𝑏 𝑏
+ [(
−(
(𝜔𝜔 𝜔𝜔𝜔3 + 𝜇𝜇𝜇 𝜇𝜇𝜇2 𝐿𝐿∗ + 𝜌𝜌1 𝑅𝑅∗ ,
𝑃𝑃𝐿𝐿 󸀠󸀠
𝐿𝐿
𝑃𝑃 + 𝐿𝐿󸀠󸀠 − 𝐿𝐿 𝐿𝐿󸀠󸀠 + 𝑆𝑆󸀠󸀠 + 𝑅𝑅󸀠󸀠 + 𝑄𝑄󸀠󸀠
𝑃𝑃
𝐿𝐿
(𝛾𝛾 𝛾𝛾𝛾4 + 𝜇𝜇𝜇 𝜇𝜇𝜇
(𝑑𝑑2 + 𝜇𝜇𝜇 𝜇𝜇
𝛽𝛽 𝑃𝑃𝑃𝑃
− 1
𝛽𝛽1 𝑃𝑃
𝑑𝑑2 + 𝜇𝜇
(𝑑𝑑5 + 𝜇𝜇𝜇 𝜇𝜇𝜇
𝑑𝑑 + 𝜇𝜇
𝑏𝑏
) 𝛽𝛽2 − (𝑑𝑑3 + 𝜇𝜇𝜇𝜇 𝜇𝜇
− 1
𝑑𝑑2 + 𝜇𝜇
𝛽𝛽1
𝑑𝑑 + 𝜇𝜇
𝑏𝑏
) 𝜌𝜌2 𝑅𝑅
− 1
𝑑𝑑2 + 𝜇𝜇
𝛽𝛽1
(47)
− (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇 𝜇𝜇
= (2 −
+ [(
−(
𝑑𝑑2 + 𝜇𝜇
𝛽𝛽 𝑃𝑃
− 1 ) 𝑏𝑏
𝛽𝛽1 𝑃𝑃
𝑑𝑑2 + 𝜇𝜇
𝑑𝑑 + 𝜇𝜇
𝑏𝑏
) 𝛽𝛽2 − (𝑑𝑑3 + 𝜇𝜇𝜇𝜇 𝜇𝜇
− 1
𝑑𝑑2 + 𝜇𝜇
𝛽𝛽1
3.2.3. Stability of the Smoking-Present Equilibrium
Theorem 7. Under the condition (3) of the Theorem 3, if
𝑆𝑆𝑆𝑆𝑆∗ , 𝑅𝑅𝑅𝑅𝑅∗ , 𝑄𝑄𝑄𝑄𝑄∗ satisfy one of four relations as follows:
𝑅𝑅
> 1,
𝑅𝑅∗
𝑅𝑅
> 1,
𝑅𝑅∗
𝑅𝑅
< 1,
𝑅𝑅∗
𝑄𝑄
< 1,
𝑄𝑄∗
𝑄𝑄
> 1,
𝑄𝑄∗
𝑄𝑄
< 1,
𝑄𝑄∗
𝑄𝑄
> 1,
𝑄𝑄∗
𝑆𝑆
𝑅𝑅
𝑄𝑄
≥
≥
,
𝑆𝑆∗ 𝑅𝑅∗ 𝑄𝑄∗
𝑆𝑆
𝑅𝑅
𝑄𝑄
≤ ∗ ≤ ∗,
∗
𝑆𝑆
𝑅𝑅
𝑄𝑄
𝑆𝑆
𝑅𝑅
≤
,
𝑆𝑆∗ 𝑅𝑅∗
𝑆𝑆
𝑅𝑅
≥
,
𝑆𝑆∗ 𝑅𝑅∗
∗
𝑅𝑅
𝑄𝑄
≥
,
𝑅𝑅∗ 𝑄𝑄∗
∗
𝑅𝑅
𝑄𝑄
≤
.
𝑅𝑅∗ 𝑄𝑄∗
∗
∗
∗
(48)
∗
Then smoking-present equilibrium 𝐸𝐸 (𝑃𝑃 , 𝐿𝐿 , 𝑆𝑆 , 𝑅𝑅 , 𝑄𝑄 ) is
globally asymptotically stable.
𝑅𝑅∗
.
𝑄𝑄∗
𝑆𝑆
𝑅𝑅
) + (𝑅𝑅 𝑅 𝑅𝑅∗ − 𝑅𝑅∗ ln ∗ )
∗
𝑆𝑆
𝑅𝑅
+ (𝑄𝑄 𝑄 𝑄𝑄∗ − 𝑄𝑄∗ ln
If 𝑅𝑅0 > 1, we can obtain 𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇𝜇𝜇𝜇𝜇1 + 𝜇𝜇𝜇𝜇𝜇𝜇1 > 0. As we
know 2 − (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇𝜇1 𝑃𝑃 𝑃𝑃𝑃1 𝑃𝑃𝑃𝑃𝑃𝑃2 + 𝜇𝜇𝜇 𝜇 𝜇, if (𝑏𝑏𝑏𝑏𝑏𝑏2 + 𝜇𝜇𝜇𝜇
(𝑑𝑑1 + 𝜇𝜇𝜇𝜇𝜇𝜇1 )𝛽𝛽2 ≤ 𝑑𝑑3 + 𝜇𝜇, so we obtain 𝑉𝑉󸀠󸀠 ≤ 0 with equality
only if 𝑅𝑅1 = 1 and 𝑅𝑅 𝑅𝑅𝑅𝑅𝑅. By LaSalle invariance principle
[20, 22], 𝐸𝐸𝐿𝐿 is globally asymptotically stable. This completes
the proof.
𝑆𝑆
− 𝜌𝜌1 𝑆𝑆∗ − 𝜌𝜌2 𝐿𝐿∗ ,
𝑅𝑅∗
𝑃𝑃
𝐿𝐿
) + (𝐿𝐿 𝐿 𝐿𝐿∗ − 𝐿𝐿∗ ln ∗ )
∗
𝑃𝑃
𝐿𝐿
+ (𝑆𝑆 𝑆 𝑆𝑆∗ − 𝑆𝑆∗ ln
− (𝑑𝑑4 + 𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇5 + 𝜇𝜇𝜇 𝜇𝜇𝜇
(49)
∗
We now consider a candidate Lyapunov function 𝑉𝑉 such that
𝑉𝑉 𝑉 𝑉𝑉𝑉𝑉𝑉𝑉∗ − 𝑃𝑃∗ ln
𝑑𝑑 + 𝜇𝜇
𝑏𝑏
) 𝜌𝜌2 𝑅𝑅
− 1
𝑑𝑑2 + 𝜇𝜇
𝛽𝛽1
𝑅𝑅
< 1,
𝑅𝑅∗
𝑏𝑏
− 𝛽𝛽1 𝐿𝐿∗ ,
𝑃𝑃∗
(𝑑𝑑2 + 𝜇𝜇𝜇 𝜇𝜇𝜇1 𝑃𝑃∗ − 𝛽𝛽2 𝑆𝑆∗ + 𝜌𝜌2 𝑅𝑅∗ ,
The derivative of 𝑉𝑉 is given by
𝑉𝑉󸀠󸀠 = 𝑃𝑃󸀠󸀠 −
(𝑑𝑑1 + 𝜇𝜇𝜇 𝜇
(50)
𝑄𝑄
).
𝑄𝑄∗
Then 𝑉𝑉 𝑉𝑉 and the derivative of 𝑉𝑉 are given by
𝑉𝑉󸀠󸀠 = (1 −
𝑃𝑃∗
𝐿𝐿∗
𝑆𝑆∗
) 𝑃𝑃󸀠󸀠 + (1 − ) 𝐿𝐿󸀠󸀠 + (1 − ) 𝑆𝑆󸀠󸀠
𝑃𝑃
𝐿𝐿
𝑆𝑆
+ (1 −
𝑅𝑅∗
𝑄𝑄∗
) 𝑅𝑅󸀠󸀠 + (1 −
) 𝑄𝑄󸀠󸀠
𝑅𝑅
𝑄𝑄
+ (1 −
𝐿𝐿∗
) [𝛽𝛽1 𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 + 𝜇𝜇𝜇 𝜇𝜇𝜇
𝐿𝐿
= (1 −
𝑃𝑃∗
) [𝑏𝑏 𝑏𝑏𝑏1 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿1 + 𝜇𝜇𝜇 𝜇𝜇𝜇
𝑃𝑃
+ (1 −
𝑆𝑆∗
) [𝛽𝛽2 𝐿𝐿𝐿𝐿𝐿𝐿𝐿1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅3 + 𝜇𝜇𝜇 𝜇𝜇𝜇
𝑆𝑆
+ (1 −
𝑄𝑄∗
) [𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾5 + 𝜇𝜇𝜇 𝜇𝜇𝜇 𝜇
𝑄𝑄
+ (1 −
𝑅𝑅∗
) [𝜔𝜔𝜔𝜔𝜔𝜔𝜔1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅4 + 𝜇𝜇𝜇 𝜇𝜇𝜇
𝑅𝑅
(51)
8
Abstract and Applied Analysis
Using the relations in (49) we have
𝑉𝑉󸀠󸀠 = (1 −
+ [(1 −
𝑃𝑃∗
𝑃𝑃∗
) [𝑏𝑏 𝑏 𝑏𝑏1 𝐿𝐿𝐿𝐿𝐿
𝑏𝑏 𝑏 𝑏𝑏1 𝑃𝑃𝑃𝑃∗ ]
𝑃𝑃
𝑃𝑃
+ (1 −
+ (1 −
+ [(1 −
𝑆𝑆∗
) [𝛽𝛽2 𝐿𝐿𝐿𝐿 𝐿 𝐿𝐿1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 𝐿𝐿∗ 𝑆𝑆𝑆𝑆𝑆1 𝑅𝑅∗ 𝑆𝑆𝑆
𝑆𝑆
+ (1 −
𝑅𝑅
𝑅𝑅∗
) [𝜔𝜔𝜔𝜔𝜔𝜔𝜔1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅 ∗ 𝑆𝑆∗
𝑅𝑅
𝑅𝑅
+ (1 −
𝑄𝑄∗
𝑄𝑄
) [𝛾𝛾𝛾𝛾𝛾 𝛾𝛾 ∗ 𝑅𝑅∗ ] .
𝑄𝑄
𝑄𝑄
(52)
𝑆𝑆
= 𝑧𝑧𝑧
𝑆𝑆∗
𝑄𝑄
= 𝑣𝑣𝑣
𝑄𝑄∗
𝑅𝑅
= 𝑢𝑢𝑢
𝑅𝑅∗
The derivative of 𝑉𝑉 reduces to
1
𝑉𝑉 = (1 − ) (𝑏𝑏 𝑏𝑏𝑏𝑏𝑏𝑏𝑏1 𝑃𝑃∗ 𝐿𝐿∗ − 𝑥𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥1 𝑃𝑃∗ 𝐿𝐿∗ )
𝑥𝑥
󸀠󸀠
+ (1 −
1
) (𝑥𝑥𝑥𝑥𝑥𝑥1 𝑃𝑃∗ 𝐿𝐿∗ + 𝑦𝑦𝑦𝑦𝑦𝑦2 𝑅𝑅∗ 𝐿𝐿∗ − 𝑦𝑦𝑦𝑦𝑦𝑦2 𝐿𝐿∗ 𝑆𝑆∗
𝑦𝑦
−𝑦𝑦𝑦𝑦1 𝑃𝑃∗ 𝐿𝐿∗ + 𝑦𝑦𝑦𝑦2 𝐿𝐿∗ 𝑆𝑆∗ − 𝑦𝑦𝑦𝑦2 𝑅𝑅∗ 𝐿𝐿∗ )
1
+ (1 − ) (𝑦𝑦𝑦𝑦𝑦𝑦2 𝐿𝐿∗ 𝑆𝑆∗ + 𝑧𝑧𝑧𝑧𝑧𝑧1 𝑅𝑅∗ 𝑆𝑆∗
𝑧𝑧
−𝑧𝑧𝑧𝑧2 𝐿𝐿∗ 𝑆𝑆∗ − 𝑧𝑧𝑧𝑧1 𝑅𝑅∗ 𝑆𝑆∗ )
1
+ (1 − ) (𝑧𝑧𝑧𝑧𝑧𝑧∗ − 𝑧𝑧𝑧𝑧𝑧𝑧1 𝑅𝑅∗ 𝑆𝑆∗ − 𝑦𝑦𝑦𝑦𝑦𝑦2 𝑅𝑅∗ 𝐿𝐿∗
𝑢𝑢
−𝑢𝑢𝑢𝑢𝑢𝑢∗ + 𝑢𝑢𝑢𝑢1 𝑅𝑅∗ 𝑆𝑆∗ + 𝑢𝑢𝑢𝑢2 𝑅𝑅∗ 𝐿𝐿∗ )
1
+ (1 − ) (𝑢𝑢𝑢𝑢𝑢𝑢∗ − 𝑣𝑣𝑣𝑣𝑣𝑣∗ )
𝑣𝑣
= (2 − 𝑥𝑥 𝑥
1
1
) 𝑏𝑏 𝑏 𝑏𝑏𝑏𝑏 ) (𝑥𝑥 𝑥 𝑥𝑥𝑥𝑥𝑥
𝑥𝑥
𝑥𝑥
+ (1 −
1
) (𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥1 𝑃𝑃∗ 𝐿𝐿∗
𝑦𝑦
1
+ (1 − ) (𝑢𝑢 𝑢𝑢𝑢) 𝛾𝛾𝛾𝛾∗
𝑣𝑣
1
1
) 𝑏𝑏 𝑏 𝑏𝑏𝑏 ) (𝑧𝑧 𝑧𝑧𝑧) 𝜔𝜔𝜔𝜔∗
𝑥𝑥
𝑢𝑢
1
+ (1 − ) (𝑢𝑢 𝑢𝑢𝑢) 𝛾𝛾𝛾𝛾∗ ,
𝑣𝑣
We consider the following variable substitutions by letting
𝐿𝐿
= 𝑦𝑦𝑦
𝐿𝐿∗
1
+ (1 − ) (𝑧𝑧 𝑧𝑧𝑧) 𝜔𝜔𝜔𝜔∗
𝑢𝑢
= (2 − 𝑥𝑥 𝑥
+𝜌𝜌1 𝑅𝑅𝑅𝑅∗ +𝜌𝜌2 𝑅𝑅𝑅𝑅∗ ]
𝑃𝑃
= 𝑥𝑥𝑥
𝑃𝑃∗
1
1
) (𝑦𝑦 𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ) (𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦2 𝐿𝐿∗ 𝑆𝑆∗
𝑦𝑦
𝑧𝑧
1
1
+ [(1 − ) (𝑧𝑧𝑧𝑧𝑧 𝑧𝑧) + (1 − ) (𝑢𝑢 𝑢𝑢𝑢𝑢𝑢)] 𝜌𝜌1 𝑅𝑅∗ 𝑆𝑆∗
𝑧𝑧
𝑢𝑢
𝐿𝐿∗
) [𝛽𝛽1 𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 𝐿𝐿𝐿𝐿 𝐿 𝐿𝐿2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅1 𝑃𝑃∗ 𝐿𝐿
𝐿𝐿
+𝛽𝛽2 𝐿𝐿𝐿𝐿∗ − 𝜌𝜌2 𝑅𝑅∗ 𝐿𝐿𝐿
1
1
) (𝑦𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ) (𝑢𝑢 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢2 𝑅𝑅∗ 𝐿𝐿∗
𝑦𝑦
𝑢𝑢
(53)
(54)
as we know that (2 − 𝑥𝑥 𝑥𝑥𝑥𝑥𝑥𝑥 𝑥 𝑥, when 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧 satisfy one of
four relations as follows:
𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢 𝑢 𝑢𝑢𝑢
𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢𝑢 𝑢𝑢 𝑢𝑢𝑢𝑢
𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢𝑢 𝑢𝑢𝑢
𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢𝑢𝑢 𝑢 𝑢𝑢𝑢
𝑢𝑢 𝑢 𝑢𝑢𝑢
(55)
𝑢𝑢 𝑢𝑢𝑢𝑢
This implies that 𝑉𝑉󸀠󸀠 ≤0 with equality only if 𝑃𝑃 𝑃 𝑃𝑃∗ and
𝑆𝑆∗ /𝑆𝑆𝑆𝑆𝑆∗ /𝑅𝑅 𝑅𝑅𝑅∗ /𝑄𝑄𝑄 that is, 𝑥𝑥 𝑥𝑥, 𝑧𝑧 𝑧𝑧𝑧𝑧𝑧𝑧. By LaSalle
invariance principle [20, 22], 𝐸𝐸∗ 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.
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.
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 𝐸𝐸∗ is
globally asymptotically stable.
100
90
80
70
60
50
40
30
20
10
0
9
Population size
Population size
Abstract and Applied Analysis
0
100
200
300
400
500
600
700
800
900
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
100
200
300
P (t)
L(t)
S(t)
400
500
600
700
800
900
Time
Time
P (t)
L(t)
S(t)
R(t)
Q(t)
Figure 2: The smoking-present equilibrium 𝐸𝐸∗ is globally asymptotically stable.
160
R(t)
Q(t)
Figure 4: When 𝑅𝑅0 > 1 and 𝑅𝑅1 < 1, the occasion smoking equilibrium 𝐸𝐸𝐿𝐿 is globally asymptotically stable.
150
120
Population size
Population size
140
100
80
60
40
100
50
20
0
0
100
200
300
400
500
600
700
800
900
0
0
500
Time
P (t)
L(t)
S(t)
R(t)
Q(t)
Figure 3: When 𝑅𝑅0 < 1, the smoking-free equilibrium 𝐸𝐸0 is globally
asymptotically stable.
For appropriate adjustment parameters, we choose 𝛽𝛽1 =
0.0038, then the smoking-free equilibrium 𝐸𝐸0 is globally
asymptotically stable (Figure 3).
If we choose 𝛽𝛽1 = 0.4, 𝛽𝛽2 = 0.5, 𝜇𝜇 𝜇𝜇𝜇𝜇𝜇 𝜇𝜇1 = 0.1, 𝑑𝑑2 =
0.2, 𝑑𝑑3 = 0.23, numerical simulation gives 𝑅𝑅0 > 1 and
𝑅𝑅1 < 1; the occasion smoking equilibrium 𝐸𝐸𝐿𝐿 is globally
asymptotically stable (Figure 4).
At last, we choose 𝛽𝛽1 = 0.000078125, 𝑏𝑏 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑏 𝑏𝑏1 =
0.01, 𝜇𝜇 𝜇 𝜇𝜇2 = 0.01, numerical simulation gives 𝑅𝑅0 = 1; then
the smoking-free equilibrium 𝐸𝐸0 is globally asymptotically
stable (Figure 5).
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, 𝑅𝑅0 , 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
1000
1500
Time
P (t)
L(t)
S(t)
R(t)
Q(t)
Figure 5: When 𝑅𝑅0 = 1, the smoking-free equilibrium 𝐸𝐸0 is globally
asymptotically stable.
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 𝜌𝜌1 = 𝜌𝜌2 = 0.003, we
can get images in Figure 6.
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. 𝑃𝑃𝑃𝑃𝑃𝑃, 𝐿𝐿𝐿𝐿𝐿𝐿, 𝑆𝑆𝑆𝑆𝑆𝑆, 𝑅𝑅𝑅𝑅𝑅𝑅 and
𝑄𝑄𝑄𝑄𝑄𝑄 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.
10
Abstract and Applied Analysis
Table 2: The parameter values of giving up smoking model.
180
160
140
120
100
80
60
40
20
0
Data estimated
0.2 year−1
0.038 year−1
0.0411 year−1
0.081 year−1
0.06 year−1
0.041 year−1
0.0169 year−1
0.0111 year−1
0.0019 year−1
0.0021 year−1
0.0037 year−1
0.0012 year−1
0.0001 year−1
R0
Population size
Parameter
𝑏𝑏
𝛽𝛽1
𝛽𝛽2
𝜌𝜌1
𝜌𝜌2
𝜔𝜔
𝛾𝛾
𝜇𝜇
𝑑𝑑1
𝑑𝑑2
𝑑𝑑3
𝑑𝑑4
𝑑𝑑5
0
100
200
300
400
500
600
700
800
900
Data sources
Estimate
Estimate
Estimate
Estimate
Estimate
Estimate
Estimate
Reference [20]
Reference [16]
Reference [16]
Reference [16]
Reference [16]
Estimate
500
450
400
350
300
250
200
150
100
50
0
0
0.1
0.2
0.3
0.4
Time
For system (2), 𝑏𝑏 reflects recruitment number, 𝛽𝛽1 reflects
the contact rate between potential smokers and occasion
smokers, and 𝑑𝑑1 denotes the deaths rate of potential smokers.
𝑑𝑑2 denotes the deaths rate of light or occasion smokers. These
four parameters will directly affect the values of the basic
reproductive number. Furthermore, when 𝑏𝑏 and 𝛽𝛽1 increase,
the number smokers will increase, that is, 𝑅𝑅0 increases. When
we reduce the mortality caused by medical treatment, the
number of permanent quit smokers will increase. Hence, 𝑅𝑅0
will decrease. Figure 7 shows the relation between the basic
reproduction number 𝑅𝑅0 and 𝑑𝑑1 , Figure 8 shows the relation
between the basic reproduction number 𝑅𝑅0 and 𝑑𝑑2 , Figure 9
shows the relation between the basic reproduction number
𝑅𝑅0 and 𝑏𝑏, Figure 10 shows the relation between the basic
reproduction number 𝑅𝑅0 and 𝛽𝛽1 . From Figures 11, 12, and
13, we can also see that if 𝑑𝑑1 and 𝑑𝑑2 increase, then 𝑅𝑅0 will
decrease. And if 𝑏𝑏 and 𝛽𝛽1 increase, then 𝑅𝑅0 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.
0.7
0.8
0.9
1
0.9
1
Figure 7: The relationship between 𝑅𝑅0 and 𝑑𝑑1 .
R(t)
Q(t)
Figure 6: When 𝜌𝜌1 = 𝜌𝜌2 = 0.003, the smoking-present equilibrium
𝐸𝐸∗ is globally asymptotically stable.
0.6
d1
600
500
400
R0
P (t)
L(t)
S(t)
0.5
300
200
100
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
d2
Figure 8: The relationship between 𝑅𝑅0 and 𝑑𝑑2 .
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 𝑅𝑅2 = 1/𝑅𝑅0 + (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇3 /𝑏𝑏𝑏𝑏2 < 1, 𝑑𝑑3 𝜌𝜌2 − 𝑑𝑑2 < 0
R0
200
180
160
140
120
100
80
60
40
20
0
11
R0
Abstract and Applied Analysis
0.8
0.6
d2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
12000
Figure 9: The relationship between 𝑅𝑅0 and 𝑏𝑏.
10000
8000
ρ2
6000
4000
2000
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.4
0.2
1
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
β1
1
b
0.5
ρ1
0.6
0.7
0.8
0.9
1
0.7
0.6
0.5
ρ2
R0
0.8
0.8
0.8
0.6
0.4
0.2
d2
0 0
0.2
0.4
0.6
0.8
1
d1
250
200
150
100
50
0
1
0.8
0.6
d1
0.4
0.2
0 0
0.2
0.4
0.4
0.3
0.2
0.1
0
Figure 11: The relationship between 𝑅𝑅0 , 𝑑𝑑1 , and 𝑑𝑑2 .
R0
0.6
Figure 14: If 𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 > 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 , the relationship between 𝜌𝜌1
and 𝜌𝜌2 .
Figure 10: The relationship between 𝑅𝑅0 and 𝛽𝛽1 .
20
16
12
8
4
0
1
0.2
0 0
0.4
Figure 13: The relationship between 𝑅𝑅0 , 𝑏𝑏, and 𝑑𝑑2 .
b
R0
300
250
200
150
100
50
0
1
0.6
0.8
0
0.1
0.2
0.3
0.4
0.5
ρ1
0.6
0.7
0.8
0.9
1
Figure 15: If 𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 < 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 , the relationship between 𝜌𝜌1
and 𝜌𝜌2 .
1
b
Figure 12: The relationship between 𝑅𝑅0 , 𝑏𝑏, and 𝑑𝑑1 .
and 𝜌𝜌2 > max{𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 , 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 }, then system (2) has
positive smoking-present equilibrium 𝐸𝐸∗ (𝑃𝑃∗ , 𝐿𝐿∗ , 𝑆𝑆∗ , 𝑅𝑅∗ , 𝑄𝑄∗ )
where 𝑅𝑅∗ ∈ (0, 𝜔𝜔𝜔𝜔𝜔1 ). Hence, the numerical values of 𝜌𝜌1 and
𝜌𝜌2 are playing an important role in the existence of 𝐸𝐸∗ . Next,
we simulate the relationship between 𝜌𝜌1 and 𝜌𝜌2 under the
conditions 𝑅𝑅2 = 1/𝑅𝑅0 + (𝑑𝑑2 + 𝜇𝜇𝜇𝜇𝜇3 /𝑏𝑏𝑏𝑏2 < 1, 𝑑𝑑3 𝜌𝜌2 − 𝑑𝑑2 < 0.
If 𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 > 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 , we plot the relationship between
𝜌𝜌1 and 𝜌𝜌2 in Figure 14. If 𝑑𝑑2 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 𝛼𝛼3 < 𝜌𝜌1 𝜔𝜔𝜔𝜔𝜔3 , we plot the
relationship between 𝜌𝜌1 and 𝜌𝜌2 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.
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
12
HongLiu Distinguished Young Scholars in Lanzhou University of Technology.
Abstract and Applied Analysis
[18]
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