Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 216913, 13 pages
http://dx.doi.org/10.1155/2013/216913
Research Article
Traveling Wave Solutions in a Reaction-Diffusion
Epidemic Model
Sheng Wang,1 Wenbin Liu,2 Zhengguang Guo,1 and Weiming Wang1
1
College of Mathematics and Information Science, Wenzhou University,
Wenzhou 325035, China
2
College of Physics and Electronic Information Engineering, Wenzhou University,
Wenzhou 325035, China
Correspondence should be addressed to Weiming Wang; weimingwang2003@163.com
Received 4 February 2013; Revised 17 March 2013; Accepted 17 March 2013
Academic Editor: Anke Meyer-Baese
Copyright © 2013 Sheng Wang et al. 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.
We investigate the traveling wave solutions in a reaction-diffusion epidemic model. The existence of the wave solutions is derived
through monotone iteration of a pair of classical upper and lower solutions. The traveling wave solutions are shown to be unique
and strictly monotonic. Furthermore, we determine the critical minimal wave speed.
1. Introduction
Recently, great attention has been paid to the study of the
traveling wave solutions in reaction-diffusion models [1–17].
In the sense of epidemiology, the traveling wave solutions
describe the transition from a disease-free equilibrium to an
endemic equilibrium; the existence and nonexistence of nontrivial traveling wave solutions indicate whether or not the
disease can spread [11]. The results contribute to predicting
the developing tendency of infectious diseases, to determining the key factors of the spread of infectious disease, and to
seeking the optimum strategies of preventing and controlling
the spread of the infectious diseases [18–21].
Some methods have been used to derive the existence of
traveling wave solutions in reaction-diffusion models, and the
monotone iteration method has been proved to be an effective
one. Such a method reduces the existence of traveling wave
solutions to that of an ordered pair of upper-lower solutions
[6, 7, 9, 10, 14, 15].
In [22], Berezovsky and coworkers introduced a simple
epidemic model through the incorporation of variable population, disease-induced mortality, and emigration into the
classical model of Kermack and McKendrick [23]. The total
population (𝑁) is divided into two groups of susceptible (𝑆)
and infectious (𝐼); that is to say, 𝑁 = 𝑆 + 𝐼. The model
describing the relations between the state variables is
𝑁
𝑆𝐼
𝑑𝑆
= 𝑟𝑁 (1 − ) − 𝛽 − (𝜇 + 𝑚) 𝑆,
𝑑𝑡
𝐾
𝑁
𝑑𝐼
𝑆𝐼
= 𝛽 − (𝜇 + 𝑑) 𝐼,
𝑑𝑡
𝑁
(1)
where the reproduction of susceptible follows a logistic equation with the intrinsic growth rate 𝑟 and the carrying capacity
𝐾, 𝛽 denotes the contact transmission rate (the infection rate
constant), 𝜇 is the natural mortality; 𝑑 denotes the diseaseinduced mortality, and 𝑚 is the per-capita emigration rate of
uninfected.
For model (1), the epidemic threshold, the so-called basic
reproduction number 𝑅0 , is then computed as 𝑅0 = 𝛽/(𝜇 +
𝑑). The disease will successfully invade when 𝑅0 > 1 but will
die out if 𝑅0 < 1. 𝑅0 = 1 is usually a threshold whether the
disease goes to extinction or goes to an endemic. Large values
of 𝑅0 may indicate the possibility of a major epidemic [19]. In
addition, the basic demographic reproductive number 𝑅𝑑 is
given by 𝑅𝑑 = 𝑟/(𝜇 + 𝑚). It can be shown that if 𝑅𝑑 > 1 the
population grows, while 𝑅𝑑 ≤ 1 implies that the population
does not survive [22].
2
Abstract and Applied Analysis
For simplicity, rescaling model (1) by letting 𝑆 → 𝑆/𝐾,
𝐼 → 𝐼/𝐾, and 𝑡 → 𝑡/(𝜇 + 𝑑) leads to the following model:
𝑆𝐼
𝑑𝑆
= ]𝑅𝑑 (𝑆 + 𝐼) [1 − (𝑆 + 𝐼)] − 𝑅0
− ]𝑆,
𝑑𝑡
𝑆+𝐼
𝑆𝐼
𝑑𝐼
= 𝑅0
− 𝐼,
𝑑𝑡
𝑆+𝐼
(2)
where ] = (𝜇+𝑚)/(𝜇+𝑑) is defined by the ratio of the average
life span of susceptibles to that of infectious.
For details, we refer the reader to [20, 22].
In this paper, we are interested in the existence of traveling
wave solutions in the following reaction-diffusion epidemic
model [20]:
𝑆𝐼
𝜕2 𝑆
𝜕𝑆
= ]𝑅𝑑 (𝑆 + 𝐼) [1 − (𝑆 + 𝐼)] − 𝑅0
− ]𝑆 + 𝑑 2 ,
𝜕𝑡
𝑆+𝐼
𝜕𝑥
𝜕𝐼
𝑆𝐼
𝜕2 𝐼
= 𝑅0
− 𝐼 + 𝑑 2,
𝜕𝑡
𝑆+𝐼
𝜕𝑥
𝑆 (𝑥, 0) = 𝑆0 (𝑥) ,
𝑅 −1 ̂ ̂
𝑅 −1 ̂ ̂
𝜕𝑆̂
− 𝑆 + 𝐼) [1 − ( 𝑑
− 𝑆 + 𝐼)]
= −]𝑅𝑑 ( 𝑑
𝜕𝑡
𝑅𝑑
𝑅𝑑
+ 𝑅0
(3)
𝑇
𝜉 = 𝑥 + 𝑐𝑡,
(𝑅0 − 1) (]𝑅0 𝑅𝑑 − ]𝑅0 − ] + 1)
,
]𝑅02 𝑅𝑑
(9)
(𝑅 − 1) (]𝑅0 𝑅𝑑 − 𝑅0 − ] + 1)
̂𝐼 = 0
.
]𝑅02 𝑅𝑑
∗
Obviously,
− 1) (𝑅0 − 1)
̂𝐼∗ − 𝑆̂∗ = (]
,
]𝑅0 𝑅𝑑
𝑅 −1
𝑆̂ = 𝑑
− 𝑆∗ ,
𝑅𝑑
∗
(10)
̂𝐼 = 𝐼 .
∗
For simplicity, we define the following functions and constants:
𝛼0 =
𝑅𝑑 − 1
,
𝑅𝑑
∗
∗
∗
𝜙 (𝐼) = 𝛼0̂𝐼 + (̂𝐼 − 𝑆̂ ) 𝐼;
∗ 2 𝑅0
𝛽0 = 𝛼0 (̂𝐼 )
∗
−1
(𝑅0 − ] + 1) ;
𝑅0
∗
∗
∗
𝛾0 = 2] (𝑅𝑑 − 1) (̂𝐼 − 𝑆̂ ) − []̂𝐼 + (𝑅0 − 1) 𝑆̂ ] ;
∗
(6)
(8)
̂2 =
̂1 = (0, 0) and 𝐸
and for model (7), the equilibria are 𝐸
∗ ∗
(𝑆̂ , ̂𝐼 ), where
∗ 2
∗
𝜓 (𝐼) = ]𝑅𝑑 (̂𝐼 − 𝑆̂ ) 𝐼2 + 𝛾0̂𝐼 𝐼 + 𝛽0 ;
In this section, we will establish the existence of traveling
wave solutions of model (3) by constructing a pair of ordered
upper-lower solutions. The definition of the upper solution
and the lower solution is standard. We assume that the inequality between two vectors throughout this paper is componentwise.
Setting
̂𝐼 = 𝐼,
𝐼∗ = (𝑅0 − 1) 𝑆∗ ,
∗
2. Existence and Asymptotic Decay Rates
𝑅 −1
− 𝑆,
𝑆̂ = 𝑑
𝑅𝑑
]𝑅0 𝑅𝑑 − 𝑅0 − ] + 1
,
]𝑅02 𝑅𝑑
(4)
where 𝐸1 , 𝐸2 are the equilibrium points of model (3).
This paper is arranged as follows. In Section 2, we construct a pair of ordered upper-lower solutions of model (3)
and establish the uniqueness and strict monotonicity of the
traveling wave solutions.
∗ 𝑇
For model (3), the equilibria are 𝐸1 = ((𝑅𝑑 −1)/𝑅𝑑 , 0) and
𝐸2 = (𝑆∗ , 𝐼∗ ), where
∗
𝑆̂ =
(𝑆 (+∞) , 𝐼 (+∞))𝑇 = 𝐸2 ,
(5)
∗
(7)
satisfying the following boundary value conditions:
(𝑆 (−∞) , 𝐼 (−∞))𝑇 = 𝐸1 ,
𝑇
̂ ̂𝐼) (+∞) = (𝑆̂ , ̂𝐼 ) .
(𝑆,
̂ ̂𝐼) (−∞) = (0, 0)𝑇 ,
(𝑆,
𝐼 (𝑥, 0) = 𝐼0 (𝑥) ,
𝐼 (𝑥, 𝑡) = 𝐼 (𝜉) ,
̂ ̂𝐼
((𝑅𝑑 − 1) /𝑅𝑑 − 𝑆)
𝑅 −1 ̂
𝜕2 𝑆̂
− 𝑆) + 𝑑 2 ,
+ ]( 𝑑
𝑅𝑑
𝜕𝑥
(𝑅𝑑 − 1) /𝑅𝑑 − 𝑆̂ + ̂𝐼
̂ ̂𝐼
((𝑅𝑑 − 1) /𝑅𝑑 − 𝑆)
𝜕̂𝐼
𝜕2̂𝐼
− ̂𝐼 + 𝑑 2 ,
= 𝑅0
𝜕𝑡
𝜕𝑥
(𝑅𝑑 − 1) /𝑅𝑑 − 𝑆̂ + ̂𝐼
𝑆∗ =
where ], 𝑅0 , 𝑅𝑑 are all positive constants, 𝑑 is the diffusion
coefficient, and (𝑥, 𝑡) ∈ 𝑅 × 𝑅+ .
We are looking for the traveling wave solutions of model
(3) with the following form:
𝑆 (𝑥, 𝑡) = 𝑆 (𝜉) ,
then model (3) can be written as
(11)
∗
𝜂0 = −
𝜑 (𝐼) = 1,
𝐵=
𝛾0̂𝐼
∗
∗
2]𝑅𝑑 (̂𝐼 − 𝑆̂ )
𝐼 > 0;
2
;
𝜑 (𝐼) = −1,
𝐼 ≤ 0;
̂𝐼∗
̂𝐼∗
[1 + 𝜑 ( − 𝜂0 )] .
2
2
And we will always assume the following hypotheses
throughout the rest of this paper:
Abstract and Applied Analysis
3
system, if the sign of 𝜕𝐹1 (𝑆, 𝐼)/𝜕𝐼 and 𝜕𝐹2 (𝑆, 𝐼)/𝜕𝑆 are both
nonpositive.
Since
[H1]
𝑅0 > 1,
1 < 𝑅𝑑 <
2𝑅02 + 2𝑅0 − 2
,
3𝑅02 − 2𝑅0
2
𝜕𝐹2 (𝑆, 𝐼)
𝐼
= −𝑅0 (
) ≤ 0,
𝜕𝑆
𝛼0 − 𝑆 + 𝐼
2
max {
27𝑅0 (𝑅𝑑 − 1)
𝑅 −1
, 0
}
3
𝑅0 𝑅𝑑 − 1
𝑅𝑑
<]<
(12)
−1
.
𝑅0 𝑅𝑑 − 𝑅0 − 1
(𝛼0 − 𝑆) 𝐼
𝜕 𝜕𝐹1 (𝑆, 𝐼)
≤ −2]𝑅𝑑 < 0.
(
) = −2]𝑅𝑑 − 2𝑅0
3
𝜕𝑆
𝜕𝐼
(𝛼0 − 𝑆 + 𝐼)
(16)
[H2]
] ≥ max {
𝛼 −𝑆 2
𝜕𝐹1 (𝑆, 𝐼)
= 𝑅0 ( 0
) + 2]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) − ]𝑅𝑑 ,
𝜕𝐼
𝛼0 − 𝑆 + 𝐼
𝑅03 − 2𝑅02 + 4𝑅0 − 2
𝑅0
, 2
},
2 − 𝑅𝑑 2𝑅0 + 2𝑅0 − 2 − (3𝑅02 − 2𝑅0 ) 𝑅𝑑
(13)
Then,
2
𝛼
𝜕𝐹1 (𝑆, 𝐼)
≤ 𝑅0 ( 0 ) + 2]𝑅𝑑 (𝛼0 + 𝐼) − ]𝑅𝑑 .
𝜕𝐼
𝛼0 + 𝐼
𝜓 (𝐵) ≤ 0.
Let
Then we can obtain the following.
̂2 are endemic points of
Lemma 1. If [H1] holds, then 𝐸2 and 𝐸
model (3) and model (7), respectively.
̂1 is unstable,
Lemma 2. For model (7), if [H1] holds, then 𝐸
̂
and 𝐸2 is stable.
̃ . For simplicity,
For the sake of convenience, let 𝑥 = √𝑑 𝑥
̂ ̂𝐼, and 𝑥
̃,
we still use the variables 𝑆, 𝐼, and 𝑥 instead of 𝑆,
respectively, then model (7) could be rewritten as
𝜕𝑆
= −]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
𝜕𝑡
(14)
(𝑆, 𝐼) (−∞) = (0, 0) ,
Lemma 3. Model (14) is a quasi-monotone decreasing system
̂2 ].
̂1 , 𝐸
in (𝑆, 𝐼) ∈ [𝐸
Proof. Let
𝐹2 (𝑆, 𝐼) = 𝑅0
𝐺󸀠 (𝑧) = 2]𝑅𝑑 −
2𝛼02 𝑅0
= 0,
𝑧3
(19)
󸀠
3
2
obviously, 𝑧∗ = √𝛼
0 𝑅0 /]𝑅𝑑 is the unique real root of 𝐺 (𝑧).
Since ] > 27𝑅0 (𝑅𝑑 − 1)2 /𝑅𝑑3 , consider 𝛼0 = (𝑅𝑑 − 1)/𝑅𝑑 ,
then we can get
(𝛼02 𝑅0 )
2/3
[(27𝛼02 𝑅0 )
1/3
1/3
− (]𝑅𝑑 )
]
(𝑧∗ )2
< 0.
(20)
And
lim 𝐺 (𝑧) = lim 𝐺 (𝑧) = +∞;
(21)
𝑧 → +∞
𝐺 (𝛼0 + 𝐼̂∗ ) = ]𝑅𝑑 − 2] + 2]𝑅𝑑 𝐼̂∗ + 𝑅0 (
(15)
(𝛼0 − 𝑆) 𝐼
− 𝐼.
𝛼0 − 𝑆 + 𝐼
From [17], we can know that the functions 𝐹1 (𝑆, 𝐼) and
𝐹2 (𝑆, 𝐼) are said to possess a quasi-monotone nonincreasing
𝛼0
𝛼0 + 𝐼̂∗
2
)
< ]𝑅𝑑 − 2] + 2]𝑅𝑑 𝐼̂∗ + 𝑅0
=
𝐹1 (𝑆, 𝐼) = −]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆) ,
𝛼0 − 𝑆 + 𝐼
(18)
hence, 𝐺(𝑧) has two positive roots.
Since ] ≥ 𝑅0 /(2 − 𝑅𝑑 ), thus 𝐺(𝛼0 ) = 𝑅0 + ]𝑅𝑑 − 2] ≤ 0.
According to conditions [𝐻1] and [𝐻2], we can get
𝑇
Following the definition of quasi-monotonicity [17], we
can obtain the following results.
+ 𝑅0
𝑧 ∈ [𝛼0 , 𝛼0 + 𝐼̂∗ ] ,
then
𝑧 → 0+
∗ ∗
(𝑆, 𝐼) (+∞) = (𝑆̂ , ̂𝐼 ) .
𝑇
𝛼02 𝑅0
+ 2]𝑅𝑑 𝑧 − ]𝑅𝑑 ,
𝑧2
𝐺 (𝑧 ) =
(𝛼 − 𝑆) 𝐼
𝜕𝐼
𝜕2 𝐼
= 𝑅0 0
− 𝐼 + 2,
𝜕𝑡
𝛼0 − 𝑆 + 𝐼
𝜕𝑥
𝑇
𝐺 (𝑧) =
∗
(𝛼 − 𝑆) 𝐼
𝜕2 𝑆
+ 𝑅0 0
+ ] (𝛼0 − 𝑆) + 2 ,
𝛼0 − 𝑆 + 𝐼
𝜕𝑥
𝑇
(17)
(3𝑅02 𝑅𝑑 − 2𝑅02 − 2𝑅0 𝑅𝑑 − 2𝑅0 + 2) ]
𝑅02
+
(22)
(𝑅03 − 2𝑅02 + 4𝑅0 − 2)
𝑅02
≤ 0.
Then, 𝐺([𝛼0 , 𝛼0 + 𝐼̂∗ ]) ≤ 0. Hence, 𝜕𝐹1 (𝑆, 𝐼)/𝜕𝐼 ≤ 0.
That is to say, model (14) is a quasi-monotone system in
̂1 , 𝐸
̂2 ].
(𝑆, 𝐼) ∈ [𝐸
4
Abstract and Applied Analysis
Since the traveling wave solution of model (14) has the
following form
𝑆 (𝜉) = 𝑆 (𝑥 + 𝑐𝑡) ,
𝐼 (𝜉) = 𝐼 (𝑥 + 𝑐𝑡) ,
𝜉 = 𝑥 + 𝑐𝑡,
𝑐 > 0;
(23)
substituting (23) into model (14), we can get the following
model:
𝑆󸀠󸀠 − 𝑐𝑆󸀠 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
+ 𝑅0
(24)
(𝛼 − 𝑆) 𝐼
𝐼 − 𝑐𝐼 + 𝑅0 0
− 𝐼 = 0,
𝛼0 − 𝑆 + 𝐼
𝑇
󸀠
𝑇
(𝑆, 𝐼) (−∞) = (0, 0) ,
Obviously, we can know the following.
Remark 4. Model (24) is also a quasi-monotone system in
̂1 , 𝐸
̂2 ].
(𝑆, 𝐼) ∈ [𝐸
Now we establish the existence of traveling wave solutions
of model (24) through monotone iteration of a pair of
smooth upper and lower solutions. Following [17], we give the
definitions of the upper and lower solutions of model (24) as
follows, respectively.
𝑇
Definition 5. A smooth function (𝑆(𝜉), 𝐼(𝜉)) (𝜉 ∈ R) is an
󸀠 󸀠
upper solution of model (24) if its derivatives (𝑆 , 𝐼 )𝑇 and
󸀠󸀠 󸀠󸀠
𝑇
(𝑆 , 𝐼 ) are continuous on R, and (𝑆, 𝐼) satisfies
𝑆󸀠󸀠 − 𝑐𝑆󸀠 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
+ 𝑅0
(25)
∗
𝑆
𝑆̂
( ) (+∞) ≥ ( ∗ ) .
̂𝐼
𝐼
(26)
Definition 6. A smooth function (𝑆(𝜉), 𝐼(𝜉)) (𝜉 ∈ R) is a
lower solution of model (24) if its derivatives (𝑆󸀠 , 𝐼󸀠 )𝑇 and
(𝑆󸀠󸀠 , 𝐼󸀠󸀠 ) are continuous on R, and (𝑆, 𝐼)𝑇 satisfies
𝑆󸀠󸀠 − 𝑐𝑆󸀠 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
(𝛼0 − 𝑆) 𝐼
− 𝐼 ≥ 0,
𝛼0 − 𝑆 + 𝐼
(29)
where 𝑓 ∈ 𝐶2 ([0, 𝑏]) and 𝑓 > 0 in the open interval (0, 𝑏)
with 𝑓(0) = 𝑓(𝑏) = 0, 𝑓󸀠 (0) = 𝑎1 > 0, and 𝑓󸀠 (𝑏) = −𝑏1 < 0
[15]. First, let us recall the following result.
(i) For the wave solution with noncritical speed 𝑐 > 2√𝑎1 ,
one has
√𝑐2 −4𝑎1 )/2)𝜉
𝑤 (𝜉) = 𝑎𝜔 𝑒((𝑐−
√𝑐2 −4𝑎1 )/2)𝜉
+ 𝑜 (𝑒((𝑐−
)
𝑎𝑠 𝜉 󳨀→ −∞,
(30)
√𝑐2 +4𝑏1 )/2)𝜉
𝑤 (𝜉) = 𝑏 − 𝑏𝜔 𝑒((𝑐−
√𝑐2 +4𝑏1 )/2)𝜉
+ 𝑜 (𝑒((𝑐−
)
𝑎𝑠 𝜉 󳨀→ +∞,
𝑤 (𝜉) = (𝑎𝑐 + 𝑑𝑐 𝜉) 𝑒√𝑎1 𝜉 + 𝑜 (𝜉𝑒√𝑎1 𝜉 )
+ 𝑜 (𝑒(√𝑎1 −√𝑎1 +𝑏1 )𝜉 )
𝑇
𝐼󸀠󸀠 − 𝑐𝐼󸀠 + 𝑅0
𝑤 (+∞) = 𝑏,
𝑎𝑠 𝜉 󳨀→ −∞,
𝑤 (𝜉) = 𝑏 − 𝑏𝑐 𝑒(√𝑎1 −√𝑎1 +𝑏1 )𝜉
with the following boundary value conditions
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆) ≥ 0,
𝛼0 − 𝑆 + 𝐼
𝑤󸀠󸀠 − 𝑐𝑤󸀠 + 𝑓 (𝑤) = 0,
where 𝑎𝜔 and 𝑏𝜔 are positive constants.
(𝛼 − 𝑆) 𝐼
𝐼󸀠󸀠 − 𝑐𝐼󸀠 + 𝑅0 0
− 𝐼 ≤ 0,
𝛼0 − 𝑆 + 𝐼
+ 𝑅0
The construction of the smooth upper-lower solution pair
is based on the solution of the following KPP equation:
(ii) For the wave with critical speed 𝑐 = 2√𝑎1 , one has
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆) ≤ 0,
𝛼0 − 𝑆 + 𝐼
𝑆
0
( ) (−∞) = ( ) ,
𝐼
0
(28)
Lemma 7 (see [1, 15]). Corresponding to every 𝑐 ≥ 2√𝑎1 ,
model (29) has a unique (up to a translation of the origin)
monotonically increasing traveling wave solution 𝑤(𝜉) for 𝜉 ∈
𝑅. The traveling wave solution 𝑤 has the following asymptotic
behaviors.
𝑇
∗ ∗
(𝑆, 𝐼) (+∞) = (𝑆̂ , ̂𝐼 ) .
𝑇
∗
𝑆
𝑆̂
( ) (+∞) ≤ ( ∗ ) .
̂𝐼
𝐼
𝑆
0
( ) (−∞) = ( ) ,
𝐼
0
𝑤 (−∞) = 0,
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆) = 0,
𝛼0 − 𝑆 + 𝐼
󸀠󸀠
with the following boundary value conditions
𝑎𝑠 𝜉 󳨀→ +∞,
where the constant 𝑑𝑐 is negative, 𝑏𝑐 is positive, and 𝑎𝑐 ∈ 𝑅.
For constructing the upper solution of the model (24), we
start with the following model:
∗
𝐼󸀠󸀠 − 𝑐𝐼󸀠 + 𝐼 (̂𝐼 − 𝐼)
𝐼 (−∞) = 0,
(27)
(31)
𝛼0 (𝑅0 − 1)
= 0,
∗
∗
𝛼0̂𝐼 + (̂𝐼 − 𝑆̂ ) 𝐼
∗
(32)
∗
𝐼 (+∞) = ̂𝐼 .
∗
∗
Define 𝑓(𝐼) = 𝐼(̂𝐼 − 𝐼)(𝛼0 (𝑅0 − 1)/𝜙(𝐼)), 𝐼 ∈ [0, ̂𝐼 ], one
can verify that all of the following conditions are satisfied:
∗
∗
(i) 𝑓(𝐼) = 𝐼(̂𝐼 − 𝐼)(𝛼0 (𝑅0 − 1)/𝜙(𝐼)) ∈ 𝐶2 ([0, ̂𝐼 ]);
Abstract and Applied Analysis
5
∗
∗
(ii) 𝑓(𝐼) > 0, for all 𝐼 ∈ (0, ̂𝐼 ) and 𝑓(0) = 𝑓(̂𝐼 ) = 0;
∗
∗
(iii) 𝑓󸀠 (0) = 𝑅0 − 1 > 0, 𝑓󸀠 (̂𝐼 ) = −𝛼0 (𝑅0 − 1)2 /𝑅0̂𝐼 < 0.
Thus we can get:
󸀠󸀠
From Lemma 7, we know that, for each 𝑐 ≥ 2√𝑅0 − 1,
equation (32) has a unique traveling wave solution 𝐼(𝜉) (up
to a translation of the origin), satisfying the given boundary
value conditions (26).
Define
∗
𝑆̂
𝑆 (𝜉)
𝐼 (𝜉)
),
) = ( ̂𝐼∗
(
𝐼 (𝜉)
𝐼 (𝜉)
𝜉 ∈ 𝑅,
󸀠
𝑆 − 𝑐𝑆 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
+ 𝑅0
=
(33)
=
Lemma 8. For each 𝑐 ≥ 2√𝑅0 − 1, (33) is a smooth upper
solution of model (24).
∗
𝑆̂
𝑆
( ) (+∞) ≥ ( ∗ ) .
̂𝐼
𝐼
󸀠󸀠
󸀠
(𝛼0 − 𝑆) 𝐼
𝛼0 − 𝑆 + 𝐼
∗
= −𝐼 (̂𝐼 − 𝐼)
(34)
∗
= −𝐼 (̂𝐼 − 𝐼)
∗
= −𝐼 (̂𝐼 − 𝐼)
+𝐼
𝜙 (𝐼)
𝛼0 (𝑅0 − 1)
𝜙 (𝐼)
+ 𝑅0
(𝛼0 − 𝑆) 𝐼
𝛼0 − 𝑆 + 𝐼
∗
+ 𝑅0
−𝐼
∗
𝛼0̂𝐼 − 𝑆̂ 𝐼
𝜙 (𝐼)
=
𝐼−𝐼
(𝛼0 − 𝑆) 𝐼
𝛼0 − 𝑆 + 𝐼
+ ] (𝛼0 − 𝑆)
𝛼0 (𝑅0 − 1)
𝜙 (𝐼)
∗
∗
(̂𝐼 − 𝑆̂ ) 𝐼𝜓 (𝐼)
2
∗
(̂𝐼 ) 𝜙 (𝐼)
≤ 0.
(36)
Hence, (𝑆, 𝐼) forms a smooth upper solution for model (24).
(35)
For constructing the lower solution of the model (24), we
start with the following model:
𝛼0 (𝑅0 − 1)
∗
𝐼󸀠󸀠 − 𝑐𝐼󸀠 + 𝐼 [̂𝐼 − (1 + 𝜀) 𝐼]
𝜙 (𝐼)
∗
+𝐼
∗
(𝛼0 − 𝑆) 𝐼
𝑆̂
+ 𝐼) − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼)
∗ (−𝑅0
̂𝐼
𝛼0 − 𝑆 + 𝐼
∗
∗
𝛼0̂𝐼 − 𝑆̂ 𝐼
+]
̂𝐼∗
𝜙 (𝐼)
∗
+ ] (𝛼0 − 𝑆)
∗
̂𝐼∗ − 𝑆̂∗ 𝛼0̂𝐼∗ − 𝑆̂∗ 𝐼
𝜙 (𝐼)
𝜙 (𝐼)
𝑆̂
=
𝑅0
𝐼 + ∗ 𝐼 − ]𝑅𝑑 ∗ [1 − ∗ ]
∗
̂𝐼
̂𝐼
̂𝐼
̂𝐼
𝜙 (𝐼)
∗
∗
∗
∗
𝛼0 (𝑅0 − 1) ̂𝐼 − (𝑅0 𝑆̂ − 𝑆̂ + ̂𝐼 ) 𝐼
= −𝐼 (̂𝐼 − 𝐼)
𝛼0 − 𝑆 + 𝐼
× [1 − (𝛼0 − 𝑆 + 𝐼)] + ] (𝛼0 − 𝑆)
−𝐼
𝛼0 (𝑅0 − 1)
(𝛼0 − 𝑆) 𝐼
∗
(𝛼0 − 𝑆) 𝐼 𝑆̂∗
𝑆̂
= (1 − ∗ ) 𝑅0
+ ∗ 𝐼 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼)
̂𝐼
𝛼0 − 𝑆 + 𝐼 ̂𝐼
As for the 𝐼 component, we have
𝐼 − 𝑐𝐼 + 𝑅0
+ ] (𝛼0 − 𝑆)
× [1 − (𝛼0 − 𝑆 + 𝐼)] + 𝑅0
Proof. On the boundary,
𝑆
0
( ) (−∞) = ( ) ,
0
𝐼
𝛼0 − 𝑆 + 𝐼
∗
󸀠󸀠
󸀠
𝑆̂
∗ (𝐼 − 𝑐𝐼 ) − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
̂𝐼
+ 𝑅0
then we can get the following result.
(𝛼0 − 𝑆) 𝐼
𝛼0 (𝑅0 − 1) ̂𝐼 − 𝛼0 (𝑅0 − 1) 𝐼
𝐼 (−∞) = 0,
𝜙 (𝐼)
∗
= 0.
𝛼0 (𝑅0 − 1)
= 0,
∗
∗
∗
̂
𝛼0 𝐼 + (̂𝐼 − 𝑆̂ ) 𝐼
̂𝐼∗
𝐼 (+∞) =
.
1+𝜀
∗
(37)
∗
Define 𝑔(𝐼) = 𝐼[̂𝐼 − (1 + 𝜀)𝐼](𝛼0 (𝑅0 − 1)/(𝛼0̂𝐼 + (̂𝐼 −
̂𝑆 )𝐼)), 𝐼 ∈ [0, ̂𝐼∗ /(1 + 𝜀)]. One can easily verify that all of the
following conditions hold:
∗
∗
∗
As for the 𝑆 component, since ] > 1, then ̂𝐼 − 𝑆̂ = (] −
1)(𝑅0 − 1)/]𝑅0 𝑅𝑑 > 0. And
∗
∗
(i) if 𝜂0 < ̂𝐼 /2, then max𝜉∈[0,̂𝐼∗ ] 𝜓(𝐼) = 𝜓(̂𝐼 ) = 𝜓(𝐵);
∗
(ii) if 𝜂0 ≥ ̂𝐼 /2, then max𝜉∈[0,̂𝐼∗ ] 𝜓(𝐼) = 𝜓(0) = 𝜓(𝐵).
∗
∗
∗
∗
(i) 𝑔(𝐼) = 𝐼[̂𝐼 −(1+𝜀)𝐼](𝛼0 (𝑅0 −1)/(𝛼0̂𝐼 +(̂𝐼 − 𝑆̂ )𝐼)) ∈
∗
𝐶2 ([0, ̂𝐼 /(1 + 𝜀)]);
∗
∗
(ii) 𝑔(𝐼) > 0, for all 𝐼 ∈ (0, ̂𝐼 /(1+𝜀)) and 𝑔(0) = 𝑔(̂𝐼 /(1+
𝜀)) = 0;
6
Abstract and Applied Analysis
∗
(iii) 𝑔󸀠 (0) = 𝑅0 − 1 > 0, 𝑔󸀠 (̂𝐼 /(1 + 𝜀)) = −(1 + 𝜀)𝛼0 (𝑅0 −
∗
1)/(𝜀𝛼0 + (𝑅0 /(𝑅0 − 1))̂𝐼 ) < 0.
From Lemma 7, we know that, for each fixed 𝑐 ≥
2√𝑅0 − 1, model (37) has a unique traveling wave solution
𝐼(𝜉) (up to a translation of the origin), satisfying the given
boundary value conditions (28).
Define
∗
𝑆̂
𝐼 (𝜉)
𝑆 (𝜉)
) = ( ̂𝐼∗
),
(
𝐼 (𝜉)
𝐼 (𝜉)
− ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)] + 𝑅0
𝜉 ∈ 𝑅,
(41)
Thus (𝑆, 𝐼) forms a smooth lower solution for model (24).
Proof. On the boundary,
∗
𝑆̂
∗
𝑆̂
𝑆
1+𝜀
( ) (+∞) = ( ∗ ) ≤ ( ∗ ) .
̂𝐼
𝐼
̂𝐼
1+𝜀
(39)
Next, we show that, by shifting the upper solution far
enough to the left, then the upper-lower solution in Lemmas
8 and 9 are ordered.
Lemma 10. Let 𝑐 ≥ 2√𝑅0 − 1, (𝑆, 𝐼)𝑇 and (𝑆, 𝐼)𝑇 be the upper
solution and the lower solution defined in (33) and (38), then
there exists a positive number 𝑟, such that (𝑆, 𝐼)𝑇 (𝜉 + 𝑟) ≥
(𝑆, 𝐼)𝑇 (𝜉) for all 𝜉 ∈ 𝑅.
Proof. Our proof is only for 𝑐 > 2√𝑅0 − 1, and the proof for
the case of 𝑐 = 2√𝑅0 − 1 is similar to it.
First, we derive the asymptotic behaviors of the upper
solution and the lower solution at infinities.
According to Lemma 7, when 𝜉 → −∞, we can obtain:
∗
𝑆̂
𝐴
𝑆
√ 2
( ) (𝜉) = ( ̂𝐼∗ 1 ) 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝐼
𝐴1
As for the 𝐼 component, we have
(𝛼0 − 𝑆) 𝐼
−𝐼
𝛼0 − 𝑆 + 𝐼
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
(𝛼 − 𝑆) 𝐼
𝛼 (𝑅 − 1)
= −𝐼 [̂𝐼∗ − (1 + 𝜀) 𝐼] 0 0
−𝐼
+ 𝑅0 0
𝛼0 − 𝑆 + 𝐼
𝜙 (𝐼)
= −𝐼 [̂𝐼∗ − (1 + 𝜀) 𝐼]
2 𝛼0
= 𝜀(𝐼)
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
(𝑅0 − 1)
≥ 0.
𝜙 (𝐼)
As for the 𝑆 component, we have
𝑆󸀠󸀠 − 𝑐𝑆󸀠 − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆)
𝛼0 − 𝑆 + 𝐼
∗
𝑆̂
= ∗ (𝐼󸀠󸀠 − 𝑐𝐼󸀠 ) − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
̂𝐼
+ 𝑅0
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆)
𝛼0 − 𝑆 + 𝐼
),
(42)
∗
𝑆̂
𝐵
𝑆
√ 2
( ) (𝜉) = ( ̂𝐼∗ 1 ) 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝐼
𝐵1
𝛼0 (𝑅0 − 1)
𝛼 (𝑅 − 1)
∗
+ 𝐼 (̂𝐼 − 𝐼) 0 0
𝜙 (𝐼)
𝜙 (𝐼)
(40)
+ 𝑅0
∗
∗
∗
(̂𝐼 − 𝑆̂ ) 𝐼𝜓 (𝐼)
𝑆̂
2 𝛼0 (𝑅0 − 1)
(𝐼)
+
≥ 0.
∗ 2
̂𝐼∗
𝜙 (𝐼)
(̂𝐼 ) 𝜙 (𝐼)
(38)
Lemma 9. For each fixed 𝑐 ≥ 2√𝑅0 − 1, (38) is a lower
solution of model (24).
𝐼󸀠󸀠 − 𝑐𝐼󸀠 + 𝑅0
(𝛼0 − 𝑆) 𝐼
𝛼0 − 𝑆 + 𝐼
+ ] (𝛼0 − 𝑆)
=𝜀
then we have the following result:
𝑆
0
( ) (−∞) = ( ) ,
𝐼
0
∗
(𝛼 − 𝑆) 𝐼
𝑆̂
2 𝛼 (𝑅 − 1)
{[𝜀 (𝐼) 0 0
− 𝐼]}
] − [𝑅0 0
̂𝐼∗
𝛼0 − 𝑆 + 𝐼
𝜙 (𝐼)
=
And
√𝑐2
𝜎0
let
∗
+ 4(𝛼0 (𝑅0 − 1) /𝑅0̂𝐼 )
2
).
=
<
0, 𝛿0
(1/2)(𝑐
=
(1/2)(𝑐 −
∗
√𝑐2 + 4((1 + 𝜀)𝛼0 (𝑅0 − 1)/(𝜀𝛼0 + (𝑅0 /(𝑅0 − 1))̂𝐼 ))
when 𝜉 → +∞, we can get
∗
𝑆̂
∗
𝑆̂
𝑆
∗ 𝐴2
( ) (𝜉) = ( ∗ ) − ( ̂𝐼
) 𝑒𝜎0 𝜉 + 𝑜 (𝑒𝜎0 𝜉 ) ,
̂𝐼
𝐼
𝐴2
∗
𝑆̂
̂
𝐵
1
𝑆
𝑆
( ) (𝜉) =
( ∗ ) − ( ̂𝐼∗ 2 ) 𝑒𝛿0 𝜉 + 𝑜 (𝑒𝛿0 𝜉 ) ,
̂
𝐼
1+𝜀 𝐼
𝐵2
∗
where, 𝐴 1 , 𝐴 2 , 𝐵1 , 𝐵2 are all positive constants.
−
<
0,
(43)
Abstract and Applied Analysis
7
̃𝑟
Since for any ̃𝑟 > 0, 𝐼 (𝜉) ≡ 𝐼(𝜉 + ̃𝑟) is also a solution of
̃𝑟 𝑇
̃𝑟
model (32). Thus, (𝑆 , 𝐼 ) (𝜉) is an upper solution of model
(24). So, according to Lemma 7, when 𝜉 → −∞, we can get:
∗
𝑆̂
̃𝑟
𝐴
𝑆
√ 2
√ 2
( ̃𝑟 ) (𝜉) = ( ̂𝐼∗ 1 ) 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)̃𝑟𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝐼
𝐴1
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒
branches of the upper solution touches its counterpart of
the lower solution at some point 𝜉1 in the interval (−𝑁1 +
(𝑟1 − 𝑟2 ), 𝑁2 − (𝑟1 − ̃𝑟)). On the endpoints of the interval
𝑟
𝑟
(−𝑁1 +(𝑟1 −𝑟2 ), 𝑁2 −(𝑟1 −̃𝑟)), we still have (𝑆 2 , 𝐼 2 )𝑇 > (𝑆, 𝐼)𝑇 .
𝑟2 𝑟2 𝑇
⃗
Let 𝑊(𝜉)
= (𝑆 , 𝐼 ) − (𝑆, 𝐼)𝑇 and 𝐹⃗ = (𝐹1 , 𝐹2 )𝑇 , where
𝐹1 = −]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼) [1 − (𝛼0 − 𝑆 + 𝐼)]
+ 𝑅0
).
𝐹2 = 𝑅0
(44)
Since (𝑐 − √𝑐2 − 4(𝑅0 − 1))/2 > 0, we can choose a large
enough number ̃𝑟 ≫ 0, such that
∗
∗
𝑆̂
𝑆̂
𝐴
𝐵
√ 2
( ̂𝐼∗ 1 ) 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)̃𝑟 > ( ̂𝐼∗ 1 ) ,
𝐴1
𝐵1
(51)
(𝛼0 − 𝑆) 𝐼
− 𝐼.
𝛼0 − 𝑆 + 𝐼
For 𝜉 ∈ (−𝑁1 + (𝑟1 − 𝑟2 ), 𝑁2 − (𝑟1 − ̃𝑟)), we get that
𝜕𝐹1 𝑟2
(𝑆 , 𝐼 + 𝜁2 𝜔2 )
𝜕𝐼
) 𝑊⃗
𝜕𝐹2 𝑟2
(𝑆 , 𝐼 + 𝜁4 𝜔2 )
𝜕𝐼
𝜕𝐹1
𝑟
(𝑆 + 𝜁1 𝜔1 , 𝐼 2 )
𝜕𝑆
𝑊⃗ − 𝑐𝑊⃗ + (
𝜕𝐹2
𝑟
(𝑆 + 𝜁3 𝜔1 , 𝐼 2 )
𝜕𝑆
󸀠󸀠
(45)
(𝛼0 − 𝑆) 𝐼
+ ] (𝛼0 − 𝑆) ,
𝛼0 − 𝑆 + 𝐼
󸀠
= 0,
hence, there exists a large number 𝑁1 ≫ 1, such that
̃𝑟
(
𝑆 (𝜉)
𝑆 (𝜉)
),
)>(
̃𝑟
𝐼 (𝜉)
𝐼 (𝜉)
𝜉 ∈ (−∞, −𝑁1 ] .
(52)
(46)
By using a similar argument as above, there exists a large
enough number 𝑁2 ≫ 1, such that
̃𝑟
(
𝑆 (𝜉)
𝑆 (𝜉)
),
)>(
̃𝑟
𝐼 (𝜉)
𝐼 (𝜉)
𝜉 ∈ [𝑁2 , +∞) .
(47)
where 𝜁𝑖 ∈ [0, 1], 𝑖 = 1, 2, 3, 4. Since the above model is
∗ ∗
monotone and the cube [(0, 0), (𝑆̂ , ̂𝐼 )] is convex, thus we
can deduce by Maximum Principle that 𝑊⃗ > 0 for 𝜉 ∈
[−𝑁1 + (𝑟1 − 𝑟2 ), 𝑁2 − (𝑟1 − ̃𝑟)]. So 𝜉1 does not exist and we can
decrease 𝑟2 further to ̃𝑟. It is calculated that the point 𝜉0 does
not exist either. The proof of this lemma is completed.
To ease the burden of notations, we still use (𝑆, 𝐼)𝑇 to
denote the shifted upper solution as given in Lemma 8. Let
Second, we show that
̃𝑟
𝑆 (𝜉)
𝑆 (𝜉)
),
( ̃𝑟 ) > (
𝐼 (𝜉)
𝐼 (𝜉)
𝜉 ∈ [−𝑁1 , 𝑁2 ] .
(48)
𝐷11 = −
We deal with such two possible cases:
𝑅02 + ]𝑅0 𝑅𝑑 + ]𝑅0 − 4𝑅0 − 2] + 3
,
𝑅0
𝐷12 =
Case 1. If
]𝑅0 𝑅𝑑 − 2𝑅0 − 2] + 3
,
𝑅0
2
̃𝑟
𝑆 (𝜉)
𝑆 (𝜉)
),
( ̃𝑟 ) > (
𝐼 (𝜉)
𝐼 (𝜉)
𝜉 ∈ [−𝑁1 , 𝑁2 ] ,
𝐷21
(49)
then, the proof is completed.
𝐷22
Case 2. If there exists a point 𝜉0 ∈ (−𝑁1 , 𝑁2 ), such that
̃𝑟
(50)
̃𝑟
satisfying 𝑆 (𝜉0 ) < 𝑆(𝜉0 ) or 𝐼 (𝜉0 ) < 𝐼(𝜉0 ).
In this case, we use the Sliding Domain method [15].
̃𝑟
̃𝑟
Step 1. we shift (𝑆 , 𝐼 )𝑇 to the left by increasing the number
𝑟
𝑟
̃𝑟 until finding a new number 𝑟1 > ̃𝑟 such that (𝑆 1 , 𝐼 1 )𝑇 >
𝑇
(𝑆, 𝐼) on the smaller interval [−𝑁1 , 𝑁2 − (𝑟1 − ̃𝑟)].
𝑟
𝑟
(53)
𝑅 −1
=− 0
,
𝑅0
2
̃𝑟
𝑆 (𝜉 )
𝑆 (𝜉 )
( ̃𝑟 0 ) ≤ ( 0 )
𝐼 (𝜉0 )
𝐼 (𝜉0 )
(𝑅 − 1)
=− 0
,
𝑅0
Step 2. we shift (𝑆 1 , 𝐼 1 )𝑇 back to the right by decreasing 𝑟1
to a smaller number ̃𝑟 < 𝑟2 < 𝑟1 such that one of the
𝜇1 =
− (𝐷11 + 𝐷22 ) + √(𝐷11 − 𝐷22 ) + 4𝐷12 𝐷21
2
,
2
𝜇2 =
− (𝐷11 + 𝐷22 ) − √(𝐷11 − 𝐷22 ) + 4𝐷12 𝐷21
2
.
With such constructed ordered upper-lower solution pair,
we can get the following.
Theorem 11. For 𝑐 ≥ 2√𝑅0 − 1, model (24) has a unique
(up to a translation of the origin) traveling wave solution.
The traveling wave solution is strictly increasing and has the
following asymptotic properties:
8
Abstract and Applied Analysis
(i) if 𝑐 > 2√𝑅0 − 1, when 𝜉 → −∞,
𝑆
𝐴
√ 2
( ) (𝜉) = ( 1 ) 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝐴2
𝐼
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
the first two equations in model (24) for every 𝑐 ≥ 2√𝑅0 − 1,
which satisfies
(54)
).
((𝑐−√𝑐2 +4𝜇)/2)𝜉
+ 𝑜 (𝑒
𝑈 (𝜉) = (𝑆 (𝜉) , 𝐼 (𝜉))𝑇 ,
(55)
),
while 𝜇1 = 𝜇2 :
∗
𝐴 11 + 𝐴 12 𝜉
2
𝑆
𝑆̂
( ) (𝜉) = ( ∗ ) − (
) 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
̂𝐼
𝐼
𝐴 21 + 𝐴 22 𝜉
((𝑐−√𝑐2 +4𝜇)/2)𝜉
+ 𝑜 (𝑒
𝜒2󸀠󸀠 − 𝑐𝜒2󸀠 + 𝐶21 (𝑆, 𝐼) 𝜒1 + 𝐶22 (𝑆, 𝐼) 𝜒2 = 0,
(56)
𝐶11 (𝑆, 𝐼) = ]𝑅𝑑 [1 − (𝛼0 − 𝑆 + 𝐼)] − ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼)
−
(57)
𝐶21 (𝑆, 𝐼) = −
𝐶22 (𝑆, 𝐼) =
(58)
+ 𝑜 (𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 ) ,
𝑅 (𝛼 − 𝑆) 𝐼
𝑅0 𝐼
,
+ 0 0
𝛼0 − 𝑆 + 𝐼 (𝛼0 − 𝑆 + 𝐼)2
𝑅0 (𝛼0 − 𝑆) 𝑅0 (𝛼0 − 𝑆) 𝐼
− 1.
−
2
𝛼0 − 𝑆 + 𝐼
(𝛼0 − 𝑆 + 𝐼)
(63)
Now, we study the exponential decay rate of the traveling
wave solution as 𝜉 → −∞. The asymptotic model of model
(62) as 𝜉 → −∞ is
while 𝜇1 = 𝜇2 ,
𝜆󸀠󸀠 − 𝑐𝜆󸀠 + 𝐸11 𝜆 + 𝐸12 𝜇 = 0,
𝐵11 + 𝐵12 𝜉
+ 𝑜 (𝑒
𝑅0 (𝛼0 − 𝑆) 𝑅0 (𝛼0 − 𝑆) 𝐼
,
−
2
𝛼0 − 𝑆 + 𝐼
(𝛼0 − 𝑆 + 𝐼)
+
when 𝜉 → +∞, and if 𝜇1 ≠ 𝜇2 , then
(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉
𝑅 (𝛼 − 𝑆) 𝐼
𝑅0 𝐼
− ],
+ 0 0
𝛼0 − 𝑆 + 𝐼 (𝛼0 − 𝑆 + 𝐼)2
𝐶12 (𝑆, 𝐼) = −]𝑅𝑑 [1 − (𝛼0 − 𝑆 + 𝐼)] + ]𝑅𝑑 (𝛼0 − 𝑆 + 𝐼)
(ii) if 𝑐 = 2√𝑅0 − 1, when 𝜉 → −∞,
𝑆
𝑆̂
( ) (𝜉) = ( ∗ ) − (
) 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉
̂𝐼
𝐼
𝐵21 + 𝐵22 𝜉
(62)
where
𝐴 12 and 𝐴 22 are all positive constants.
∗
(61)
be the traveling wave solution of model (24) generated form
the monotone iteration, since the case of (ii) 𝑐 = 2√𝑅0 − 1 is
similar to it.
We differentiate model (24) with respect to 𝜉, and note
that 𝑈󸀠 (𝜉) = (𝜒1 , 𝜒2 )𝑇 (𝜉) satisfies
where, 𝜇 = min{𝜇1 , 𝜇2 } > 0, 𝐴 11 , 𝐴 21 ∈ R, 𝐴 1 , 𝐴 2 , 𝐴 1 , 𝐴 2 ,
∗
𝐵11
𝑆
𝑆̂
( ) (𝜉) = ( ∗ ) − ( ) 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉
̂𝐼
𝐼
𝐵22
−∞ < 𝜉 < +∞
𝜒1󸀠󸀠 − 𝑐𝜒1󸀠 + 𝐶11 (𝑆, 𝐼) 𝜒1 + 𝐶12 (𝑆, 𝐼) 𝜒2 = 0,
),
𝑆
𝐵 + 𝐵12 𝜉 √𝑅0 −1𝜉
( ) (𝜉) = ( 11
+ 𝑜 (𝜉𝑒√𝑅0 −1𝜉 ) ,
)𝑒
𝐼
𝐵21 + 𝐵22 𝜉
(60)
According to the above inequality, we can get that, on
the boundary, the solution tends to (0, 0)𝑇 as 𝜉 → −∞ and
∗ ∗
(𝑆̂ , ̂𝐼 )𝑇 as 𝜉 → +∞.
To derive the asymptotic decay rate of the traveling wave
solutions as 𝜉 → ±∞, we just let 𝑐 > 2√𝑅0 − 1 and
when 𝜉 → +∞, and if 𝜇1 ≠ 𝜇2 , then
∗
𝐴1
2
𝑆
𝑆̂
( ) (𝜉) = ( ∗ ) − ( ) 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
̂𝐼
𝐼
𝐴2
𝑆 (𝜉)
𝑆 (𝜉)
𝑆 (𝜉)
)≤(
)≤(
).
(
𝐼 (𝜉)
𝐼 (𝜉)
𝐼 (𝜉)
𝜇󸀠󸀠 − 𝑐𝜇󸀠 + 𝐸21 𝜆 + 𝐸22 𝜇 = 0,
(59)
(64)
where
),
𝐸11 = −] (𝑅𝑑 − 1) ,
where 𝜇 = min{𝜇1 , 𝜇2 } > 0, 𝐵12 , 𝐵22 < 0, 𝐵11 , 𝐵21 , 𝐵11 , 𝐵21 ∈
R, and 𝐵11 , 𝐵22 , 𝐵12 , 𝐵22 are all positive constants.
Proof. From Lemma 3 and Remark 4, we know that model
(24) is a quasi-monotone nonincreasing system in (𝑆, 𝐼) ∈
̂1 , 𝐸
̂2 ], and by using the monotone iteration scheme given
[𝐸
𝑇
in [3, 13], we can obtain the existence of the solution (𝑆, 𝐼) to
𝐸12 = ]𝑅𝑑 + 𝑅0 − 2],
𝐸21 = 0,
𝐸22 = 𝑅0 − 1.
(65)
The second equation of model (64) has two independent
solutions with the following form:
√𝑐2 −4(𝑅0 −1))/2)𝜉
𝜇(1) (𝜉) = 𝑒((𝑐−
(2)
𝜇
((𝑐+√𝑐2 −4(𝑅0 −1))/2)𝜉
(𝜉) = 𝑒
,
.
(66)
Abstract and Applied Analysis
9
Relating the second equation of model (62) with the
second equation of model (64), we can deduce that 𝜒2 (𝜉) has
the following property as 𝜉 → −∞:
√
√𝑐2 −4(𝑅0 −1))/2)𝜉
𝜒2 (𝜉) = 𝛼 [1 + 𝑜 (1)] 𝑒((𝑐−
(67)
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝛽 [1 + 𝑜 (1)] 𝑒((𝑐+
√𝑐2 −4(𝑅0 −1))/2)𝜉
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝛽𝑒((𝑐+
(68)
𝜓1󸀠󸀠 − 𝑐𝜓1󸀠 + 𝐷11 𝜓1 + 𝐷12 𝜓2 = 0,
+ Υ1 (𝜉) + Υ2 (𝜉) ,
Υ1 (𝜉)
= 0,
𝜉 → −∞ ((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
𝑒
(69)
Υ2 (𝜉)
lim
= 0.
𝜉 → −∞ ((𝑐+√𝑐2 −4(𝑅0 −1))/2)𝜉
𝑒
̃ 𝑖 , 𝑖 = 1, 2, we rewrite model (76)
By setting (𝜓𝑖 )󸀠 (𝜉) = 𝜓
as a first order model of ordinary differential equation in the
̃ 1 , 𝜓2 , 𝜓
̃ 2 )𝑇 :
four components (𝜓1 , 𝜓
̃ 1,
𝜓1󸀠 = 𝜓
So we obtain that
̃ 󸀠1 = 𝑐𝜓
̃ 1 − 𝐷11 𝜓1 − 𝐷12 𝜓2 ,
𝜓
√𝑐2 −4(𝑅0 −1))/2)𝜉
lim
𝜒2 (𝜉) − 𝛼𝑒((𝑐−
𝜉 → −∞
= lim
̃ 󸀠2 = 𝑐𝜓
̃ 2 − 𝐷21 𝜓1 − 𝐷22 𝜓2 .
𝜓
√𝑐2 −4(𝑅0 −1))/2)𝜉
Υ1 (𝜉) + Υ2 (𝜉) + 𝛽𝑒((𝑐+
𝜉 → −∞
√𝑐2 −4(𝑅0 −1))/2)𝜉
𝑒((𝑐−
Υ1 (𝜉)
𝑒
+ lim
(70)
√𝑐2 −4(𝑅0 −1)𝜉
𝜉 → −∞ ((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝛽 lim 𝑒
In the case of (i) 𝜇1 ≠ 𝜇2 , we can obtain that the solution
of model (77) has the following form:
𝜉 → −∞
Υ2 (𝜉)
√𝑐2 −4(𝑅0 −1)𝜉
lim 𝑒
𝑒
̃ 1 , 𝜓2 , 𝜓
̃ 2 ) = ∑𝑐𝑖 ℎ𝑖 𝑒𝜆 𝑖 𝜉 ,
(𝜓1 , 𝜓
= 0.
where
Thus, 𝜒2 (𝜉) = 𝛼[1 + 𝑜(1)]𝑒
.
Now, we consider the first equation of model (64). We
rewrite it as
𝜆󸀠󸀠 − 𝑐𝜆󸀠 − ] (𝑅𝑑 − 1) 𝜆 = − (]𝑅𝑑 + 𝑅0 − 2]) 𝜇.
𝜆󸀠󸀠 − 𝑐𝜆󸀠 − ] (𝑅𝑑 − 1) 𝜆 = 0.
(72)
The above equation has two independent solutions of the
following form:
(2)
𝜆
(𝜉) = 𝑒
((𝑐+√𝑐2 +4](𝑅𝑑 −1))/2)𝜉
(𝜉) = 𝑒
,
(73)
Thus, when 𝜉 → −∞, 𝜒1 (𝜉) has the following property:
+ 𝛾 [1 + 𝑜 (1)] 𝑒
𝑐 + √𝑐2 + 4𝜇2
2
,
,
𝜆2 =
2
(79)
𝑐 − √𝑐2 + 4𝜇2
𝜆4 =
2
,
,
and ℎ𝑖 (𝑖 = 1, 2, 3, 4) are the eigenvectors of the constant
matrix with 𝜆 𝑖 (𝑖 = 1, 2, 3, 4) as the corresponding eigenvalues, 𝑐𝑖 (𝑖 = 1, 2, 3, 4) are arbitrary constants. Since
̃ 1 , 𝜓2 , 𝜓
̃ 2 )𝑇 = 0,
lim (𝜓1 , 𝜓
(80)
̃ 1 , 𝜓2 , 𝜓
̃ 2 )𝑇 = 𝑐2 ℎ2 𝑒𝜆 2 𝜉 + 𝑐4 ℎ4 𝑒𝜆 4 𝜉 , so when 𝜉 →
thus (𝜓1 , 𝜓
+∞, we can get that
2
𝜒1 (𝜉) = 𝛼 [1 + 𝑜 (1)] 𝑒
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
2
𝑐 − √𝑐2 + 4𝜇1
𝜅 [Λ + 𝑜 (1)] 𝑒((𝑐−√𝑐 +4𝜇1 )/2)𝜉
𝜒 (𝜉)
)
( 1 )=( 1 1
2
𝜒2 (𝜉)
𝜅1 [Γ1 + 𝑜 (1)] 𝑒((𝑐−√𝑐 +4𝜇1 )/2)𝜉
((𝑐+√𝑐2 +4](𝑅𝑑 −1))/2)𝜉
√𝑐2 +4](𝑅𝑑 −1))/2)𝜉
𝜆3 =
𝑐 + √𝑐2 + 4𝜇1
𝜉 → +∞
.
+ 𝛽 [1 + 𝑜 (1)] 𝑒((𝑐−
𝜆1 =
(71)
One can verify that (𝑐 − √𝑐2 − 4(𝑅0 − 1))/2 is not a
characteristic of
𝜆
(78)
𝑖=1
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
((𝑐−√𝑐2 +4](𝑅𝑑 −1))/2)𝜉
4
𝑇
𝜉 → −∞ ((𝑐+√𝑐2 −4(𝑅0 −1))/2)𝜉 𝜉 → −∞
(1)
(77)
̃ 2,
𝜓2󸀠 = 𝜓
√𝑐2 −4(𝑅0 −1))/2)𝜉
𝑒((𝑐−
= lim
(75)
(76)
𝜓2󸀠󸀠 − 𝑐𝜓2󸀠 + 𝐷21 𝜓1 + 𝐷22 𝜓2 = 0.
where
lim
2
𝛾 [1 + 𝑜 (1)] 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝜒 (𝜉)
).
( 1 )=(
𝜒2 (𝜉)
√ 2
𝛼 [1 + 𝑜 (1)] 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
Then, we study the exponential decay rate of the traveling
wave solution as 𝜉 → +∞. The asymptotic model of model
(62) as 𝜉 → +∞ is
for some constants 𝛼 and 𝛽. Thus, we can obtain that
𝜒2 (𝜉) = 𝛼𝑒((𝑐−
for some constants 𝛼, 𝛽; 𝛾 ≠ 0. Since 𝜒1 (−∞) = 0, thus 𝛽 = 0.
So, when 𝜉 → −∞, we have the following formula:
(74)
+(
((𝑐−√𝑐2 +4𝜇2 )/2)𝜉
𝜅2 [Λ 2 + 𝑜 (1)] 𝑒
2
𝜅2 [Γ2 + 𝑜 (1)] 𝑒((𝑐−√𝑐 +4𝜇2 )/2)𝜉
(81)
).
10
Abstract and Applied Analysis
Furthermore, we can obtain that
where
2
𝜅Λ ,
Δ 1 (𝜅1 , 𝜅2 , Λ 1 , Λ 2 ) = { 1 1
𝜅2 Λ 2 ,
2
𝜒1 (𝜉) = 𝜅1 Λ 1 𝑒((𝑐−√𝑐 +4𝜇1 )/2)𝜉 + 𝜅2 Λ 2 𝑒((𝑐−√𝑐 +4𝜇2 )/2)𝜉
+ Ω11 (𝜉) + Ω12 (𝜉) ,
((𝑐−√𝑐2 +4𝜇1 )/2)𝜉
𝜒2 (𝜉) = 𝜅1 Γ1 𝑒
𝜇1 < 𝜇2 ,
𝜇1 > 𝜇2 ,
𝜅 Γ , 𝜇1 < 𝜇2 ,
Δ 2 (𝜅1 , 𝜅2 , Γ1 , Γ2 ) = { 1 1
𝜅2 Γ2 , 𝜇1 > 𝜇2 ;
(82)
((𝑐−√𝑐2 +4𝜇2 )/2)𝜉
+ 𝜅2 Γ2 𝑒
(85)
thus, when 𝜉 → +∞, we can get that
+ Ω21 (𝜉) + Ω22 (𝜉) ,
𝜒 (𝜉)
( 1 )
𝜒2 (𝜉)
where
Ω11 (𝜉)
lim
= 0,
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇1 )/2)𝜉
𝑒
Ω21 (𝜉)
lim
lim
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇2 )/2)𝜉
𝑒
= 0,
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇1 )/2)𝜉
𝑒
2
Ω12 (𝜉)
Ω22 (𝜉)
lim
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇2 )/2)𝜉
𝑒
= 0,
= 0.
In the case of (ii) 𝜇1 = 𝜇2 , we can obtain that the solution
of model (77) has the following form:
(83)
𝑇
̃ 1 , 𝜓2 , 𝜓
̃ 2 ) = (𝐺1 + 𝐺2 𝜉) 𝐻1,2 𝑒𝜆 1 𝜉
(𝜓1 , 𝜓
𝜅1 , 𝜅2 , Λ 1 , Λ 2 , Γ1 , and Γ2 are all constants.
Let 𝜇 = min{𝜇1 , 𝜇2 }, then
+ (𝐺3 + 𝐺4 𝜉) 𝐻3,4 𝑒𝜆 3 𝜉 ,
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇)/2)𝜉
𝑒
2
2
= 𝜅1 Λ 1 lim 𝑒((√𝑐 +4𝜇−√𝑐 +4𝜇1 )/2)𝜉
𝜉 → +∞
2
𝑇
̃ 1 , 𝜓2 , 𝜓
̃ 2 ) = (𝐺3 + 𝐺4 𝜉) 𝐻3,4 𝑒𝜆 3 𝜉 .
(𝜓1 , 𝜓
2
+ 𝜅2 Λ 2 lim 𝑒((√𝑐 +4𝜇−√𝑐 +4𝜇2 )/2)𝜉
𝜉 → +∞
+ lim
((√𝑐2 +4𝜇−√𝑐2 +4𝜇1 )/2)𝜉
lim 𝑒
𝑒
Ω12 (𝜉)
((√𝑐2 +4𝜇−√𝑐2 +4𝜇2 )/2)𝜉
lim 𝑒
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇2 )/2)𝜉 𝜉 → +∞
𝑒
𝜒2 (𝜉)
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇)/2)𝜉
𝑒
2
(𝐺1,3 + 𝐺1,4 𝜉) 𝑒𝜆 3 𝜉 + 𝑜 (𝑒𝜆 3 𝜉 )
𝜒 (𝜉)
).
( 1 )=(
𝜒2 (𝜉)
(𝐺 + 𝐺 𝜉) 𝑒𝜆 4 𝜉 + 𝑜 (𝑒𝜆 4 𝜉 )
2,3
2
= 𝜅1 Γ1 lim 𝑒((√𝑐 +4𝜇−√𝑐 +4𝜇1 )/2)𝜉
𝜉 → +∞
𝜒1󸀠󸀠 − 𝑐𝜒1󸀠 +
((√𝑐2 +4𝜇−√𝑐2 +4𝜇2 )/2)𝜉
+ 𝜅2 Γ2 lim 𝑒
𝜉 → +∞
+ lim
Ω21 (𝜉)
2
+ lim
Ω22 (𝜉)
2
−
𝑐𝜒2󸀠
𝜕𝐹1
𝜕𝐹
𝜒 + 1 𝜒 = 0,
𝜕𝑆 1 𝜕𝐼 2
𝜕𝐹
𝜕𝐹
+ 2 𝜒1 + 2 𝜒2 = 0,
𝜕𝑆
𝜕𝐼
(90)
satisfying
2
2
lim 𝑒((√𝑐 +4𝜇−√𝑐 +4𝜇2 )/2)𝜉
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇2 )/2)𝜉 𝜉 → +∞
𝑒
𝜒2󸀠󸀠
lim 𝑒((√𝑐 +4𝜇−√𝑐 +4𝜇1 )/2)𝜉
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇1 )/2)𝜉 𝜉 → +∞
𝑒
(89)
2,4
By comparing the upper solution and roughness of the
exponential dichotomy [24], we obtain the asymptotic decay
rate of the traveling wave solutions at +∞ given in Theorem
11.
According to the monotone iteration process [3], the
traveling wave solution 𝑈(𝜉) is increasing; thus 𝑈󸀠 (𝜉) ≥ 0 and
𝑈󸀠 (𝜉) = (𝜒1 , 𝜒2 )𝑇 (𝜉) hold
= Δ 1 (𝜅1 , 𝜅2 , Λ 1 , Λ 2 ) ,
lim
(88)
So, when 𝜉 → +∞, we can get that
Ω11 (𝜉)
𝜉 → +∞ ((𝑐−√𝑐2 +4𝜇1 )/2)𝜉 𝜉 → +∞
+ lim
(87)
where 𝐻1,2 is the eigenvector of the constant matrix with 𝜆 1 =
𝜆 2 as the corresponding eigenvalues, 𝐻3,4 is the eigenvector
of the constant matrix with 𝜆 3 = 𝜆 4 as the corresponding
eigenvalues, 𝐺𝑖 (𝑖 = 1, 2, 3, 4) are arbitrary constants.
̃ 1 , 𝜓2 , 𝜓
̃ 2 )𝑇 = 0, thus
Since lim𝜉 → +∞ (𝜓1 , 𝜓
𝜒1 (𝜉)
lim
(86)
Δ (𝜅 , 𝜅 , Λ , Λ ) (1 + 𝑜 (1)) 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
=( 1 1 2 1 2
).
2
Δ 2 (𝜅1 , 𝜅2 , Γ1 , Γ2 ) (1 + 𝑜 (1)) 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
= Δ 2 (𝜅1 , 𝜅2 , Γ1 , Γ2 ) ,
(84)
𝑇
(𝜒1 , 𝜒2 ) (𝜉) ≥ 0,
𝑇
(𝜒1 , 𝜒2 ) (±∞) = 0.
(91)
The strong Maximum Principle implies that (𝜒1 , 𝜒2 )𝑇 (𝜉)
> 0. So the strict monotonicity of the traveling wave solutions
is concluded.
Abstract and Applied Analysis
11
Now, we use the Sliding domain method to prove the
uniqueness of the traveling wave solution. Let 𝑈1 (𝜉) =
(𝑆1 , 𝐼1 )𝑇 (𝜉) and 𝑈2 (𝜉) = (𝑆2 , 𝐼2 )𝑇 (𝜉) be the traveling wave
solution of model (24), with 𝑐 > 2√𝑅0 − 1. Thus, there are
some positive numbers 𝐴 𝑖 , 𝐵𝑖 (𝑖 = 1, 2, 3, 4), such that for a
big enough number 𝑁 ≫ 1, when 𝜉 < −𝑁, we have
√
If 𝜃 is large enough, then we can obtain the following inequalities:
2
+ 𝑜 (𝑒
2
),
Thus, if 𝜃 is large enough, then 𝑈1𝜃 (𝜉) > 𝑈2 (𝜉), for all 𝜉 ∈
𝑅 \ [−𝑁, 𝑁].
Now, we consider model (24) on the interval [−𝑁, 𝑁].
First, suppose that
𝑊 (𝜉) ≡ 𝑈1𝜃 (𝜉) − 𝑈2 (𝜉) ≥ 0,
),
𝜉 ∈ [−𝑁, 𝑁] ,
(97)
then
when 𝜉 > 𝑁,
𝑊󸀠󸀠 − 𝑐𝑊󸀠
2
∗
𝑆̂ − 𝐵1 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
𝑆1 (𝜉)
(
)=( ∗
)
𝐼1 (𝜉)
̂𝐼 − 𝐵2 𝑒((𝑐−√𝑐2 +4𝜇)/2)𝜉
𝜕𝐹1
(𝑆 + 𝜁 𝜔 , 𝐼 )
𝜕𝑆 2 1 1 1
+(
𝜕𝐹2
(𝑆 + 𝜁 𝜔 , 𝐼 )
𝜕𝑆 2 3 1 1
2
+ 𝑜 (𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉 ) ,
(93)
2
∗
𝑆̂ − 𝐵3 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
𝑆 (𝜉)
( 2 )=( ∗
)
𝐼2 (𝜉)
̂𝐼 − 𝐵4 𝑒((𝑐−√𝑐2 +4𝜇)/2)𝜉
𝑊 (−𝑁) > 0,
𝜕𝐹1
(𝑆 , 𝐼 + 𝜁 𝜔 )
𝜕𝐼 1 2 2 2
) 𝑊 = 0,
𝜕𝐹2
(𝑆1 , 𝐼2 + 𝜁4 𝜔2 )
𝜕𝐼
𝑊 (𝑁) > 0,
(98)
2
+ 𝑜 (𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉 ) .
Since the traveling wave solutions of model (24) are translation-invariant, then for any 𝜃 ∈ 𝑅, 𝑈1𝜃 (𝜉) ≡ 𝑈1 (𝜉 + 𝜃) ≡
(𝑆1 (𝜉+𝜃), 𝐼1 (𝜉 + 𝜃))𝑇 is also a traveling wave solution of model
(24). Thus, by using the same method as above, when 𝜉 < −𝑁,
we can get
where, 𝜁𝑖 ∈ (0, 1) (𝑖 = 1, 2, 3, 4), 𝜉 ∈ (−𝑁, 𝑁). Since the
above model is monotone, by the Maximum Principle, we can
deduce that 𝑊(𝜉) > 0, 𝜉 ∈ [−𝑁, 𝑁]. Consequently, we get
that 𝑈1𝜃 (𝜉) > 𝑈2 (𝜉), 𝜉 ∈ 𝑅.
Second, we suppose that there exists a point 𝜉∗ ∈ (−𝑁, 𝑁)
such that
𝑆1𝜃 (𝜉∗ ) < 𝑆2 (𝜉∗ )
(99)
𝐼1𝜃 (𝜉∗ ) < 𝐼2 (𝜉∗ ) .
(100)
or
𝑆 (𝜉 + 𝜃)
)
( 1
𝐼1 (𝜉 + 𝜃)
=(
(96)
𝐵2 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜃 < 𝐵4 .
2
√𝑐2 −4(𝑅0 −1))/2)𝜉
> 𝐴 4,
2
𝑆 (𝜉)
𝐴 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
( 2 )=( 3
)
𝐼2 (𝜉)
√ 2
𝐴 4 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
√𝑐2 −4(𝑅0 −1))/2)𝜃
𝐵1 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜃 < 𝐵3 ,
(92)
√
> 𝐴 3,
𝐴 2 𝑒((𝑐+
𝐴 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝑆 (𝜉)
)
( 1 )=( 1
𝐼1 (𝜉)
√ 2
𝐴 2 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
√𝑐2 −4(𝑅0 −1))/2)𝜃
𝐴 1 𝑒((𝑐+
√𝑐2 −4(𝑅0 −1))/2)𝜃 ((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
𝐴 1 𝑒((𝑐−
𝑒
𝐴 2𝑒
𝑒
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜃
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
((𝑐−√𝑐2 −4(𝑅0 −1))/2)𝜉
)
(94)
),
when 𝜉 > 𝑁,
In this case, we increase 𝜃, that is shifting 𝑈1𝜃 to the left,
so that 𝑈1𝜃 (−𝑁) > 𝑈2 (−𝑁) and 𝑈1𝜃 (𝑁) > 𝑈2 (𝑁). According
to the monotonicity of 𝑈1𝜃 and 𝑈2 , we can find a number 𝜃 ∈
(0, 2𝑁) such that 𝑈1𝜃 (𝜉 + 𝜃) > 𝑈2 (𝜉), 𝜉 ∈ (−𝑁, 𝑁). Shifting
𝑈1𝜃 (𝜉 + 𝜃) back until one component of 𝑈1𝜃 (𝜉 + 𝜃) touches its
counterpart of 𝑈2 (𝜉) at some point 𝜉 ∈ (−𝑁, 𝑁). Since 𝑈1𝜃 (𝜉 +
𝑆 (𝜉 + 𝜃)
( 1
)
𝐼1 (𝜉 + 𝜃)
2
2
∗
𝑆̂ − 𝐵1 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜃 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
=( ∗
)
̂𝐼 − 𝐵2 𝑒((𝑐−√𝑐2 +4𝜇)/2)𝜃 𝑒((𝑐−√𝑐2 +4𝜇)/2)𝜉
2
+ 𝑜 (𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉 ) .
(95)
𝜃) and 𝑈2 (𝜉) are strictly increasing, 𝜉 ∈ (−𝑁, 𝑁), thus, we get
that 𝑈1𝜃 (𝜉 + 𝜃) > 𝑈2 (𝜉), 𝜉 = ±𝑁. However, by the Maximum
Principle for that component again, we find that components
of 𝑈1𝜃 and 𝑈2 are identically equal for all 𝜉 ∈ [−𝑁, 𝑁] for a
larger number 𝜃. This is a contradiction, thus 𝑈1𝜃 (𝜉) > 𝑈2 (𝜉),
𝜉 ∈ 𝑅. Here, 𝜃 is a new number which is chosen by the above
mean.
Now, decrease the 𝜃 until one of the following happens.
12
Abstract and Applied Analysis
Case (a). There is a 𝜃 ≥ 0, such that 𝑈1𝜃 = 𝑈2 (𝜉), 𝜉 ∈ 𝑅. In this
case, we have finished the proof.
Case (b). There are a 𝜃 and a point 𝜉1 ∈ 𝑅, such that one of the
components of 𝑈1𝜃 and 𝑈2 are equal. And 𝑈1𝜃 ≥ 𝑈2 , 𝜉 ∈ 𝑅. On
𝑅 for that component, according to the Maximum Principle,
we find that 𝑈1𝜃 and 𝑈2 must be identical on that component.
We can return to Case (a).
Consequently, in either situation, their exists a number
𝜃 ≥ 0 such that
𝑈1𝜃
(𝜉) = 𝑈2 (𝜉) ,
𝜉 ∈ (−∞, +∞) .
(101)
(ii) 𝑐 = 2√𝑅0 − 1: when 𝜉 → −∞,
𝑆 (𝜉) =
𝐼 (𝜉) = (𝐴 11 + 𝐴 12 𝜉) 𝑒√𝑅0 −1𝜉 + 𝑜 (𝜉𝑒√𝑅0 −1𝜉 ) .
(105)
when 𝜉 → +∞, and if 𝜇1 ≠ 𝜇2 , then
𝑆 (𝜉) = 𝑆∗ + 𝐵11 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 + 𝑜 (𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 ) ,
𝐼 (𝜉) = 𝐼∗ − 𝐵22 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 + 𝑜 (𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 ) ,
(106)
This ends of the proof.
By Theorem 11, we can get the following theorem:
Theorem 12. For each 𝑐 ≥ 2√𝑅0 − 1, model (3) has a unique
(up to a translation of the origin) traveling wave solution.
The traveling wave solution is strictly increasing and has the
following asymptotic properties:
if 𝜇1 = 𝜇2 , then
𝑆 (𝜉) = 𝑆∗ + (𝐵11 + 𝐵12 𝜉) 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉
+ 𝑜 (𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 ) ,
(i) 𝑐 > 2√𝑅0 − 1: when 𝜉 → −∞,
𝑆 (𝜉) =
𝐼 (𝜉) = 𝐼∗ − (𝐵21 + 𝐵22 𝜉) 𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉
𝑅𝑑 − 1
√ 2
− 𝐴 1 𝑒((𝑐− 𝑐 −4(𝑅0 −1))/2)𝜉
𝑅𝑑
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
√𝑐2 −4(𝑅0 −1))/2)𝜉
+ 𝑜 (𝑒((𝑐−
where 𝜇 = min{𝜇1 , 𝜇2 } > 0, 𝐵12 , 𝐵22 < 0, 𝐵11 , 𝐵21 , 𝐵11 , 𝐵21 ∈
).
(102)
when 𝜉 → +∞, and if 𝜇1 ≠ 𝜇2 , then
2
𝑆 (𝜉) = 𝑆∗ + 𝐴 1 𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉
2
+ 𝑜 (𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉 ) ,
(103)
((𝑐−√𝑐2 +4𝜇)/2)𝜉
∗
𝐼 (𝜉) = 𝐼 − 𝐴 2 𝑒
+ 𝑜 (𝑒
),
Proof. Suppose there is a monotone traveling wave solution
L(𝜉) = (𝑙1 (𝜉), 𝑙2 (𝜉))𝑇 of model (24) with the wave speed 𝑐0 ,
where 𝑐0 ∈ (0, 2√𝑅0 − 1).
The asymptotic model of L(𝜉) = (𝑙1 (𝜉), 𝑙2 (𝜉))𝑇 as 𝜉 →
−∞ is
󸀠
𝜇󸀠󸀠 − 𝑐0 𝜇󸀠 + (𝑅0 − 1) 𝜇 = 0.
𝑆 (𝜉) = 𝑆 + (𝐴 11 + 𝐴 12 𝜉) 𝑒
√4(𝑅0 − 1) − 𝑐02 𝑖)/2. Thus it has two independent solutions of
the following form:
),
(104)
((𝑐−√𝑐2 +4𝜇)/2)𝜉
∗
𝐼 (𝜉) = 𝐼 − (𝐴 21 + 𝐴 22 𝜉) 𝑒
𝜇11 = 𝑒(𝑐0 /2)𝜉 cos (
2
√4 (𝑅0 − 1) − 𝑐02
2
𝜉) ,
+ 𝑜 (𝑒((𝑐−√𝑐 +4𝜇)/2)𝜉 ) ,
where 𝜇 = min{𝜇1 , 𝜇2 } > 0, 𝐴 11 , 𝐴 21 ∈ R 𝐴 1 , 𝐴 2 , 𝐴 1 , 𝐴 2 ,
𝐴 12 and 𝐴 22 are all positive constants.
(108)
The second function of (108) has two characteristics
as the following ones: (𝑐0 + √4(𝑅0 − 1) − 𝑐02 𝑖)/2, (𝑐0 −
((𝑐−√𝑐2 +4𝜇)/2)𝜉
∗
+ 𝑜 (𝑒
Theorem 13. There is no monotone traveling wave solution of
model (24) for any 0 < 𝑐 < 2√𝑅0 − 1. In other words, there
is no monotone traveling wave solution of model (3) for any
0 < 𝑐 < 2√𝑅0 − 1.
𝜆 − 𝑐0 𝜆 − ] (𝑅𝑑 − 1) 𝜆 + (]𝑅𝑑 + 𝑅0 − 2]) 𝜇 = 0,
if 𝜇1 = 𝜇2 ,
((𝑐−√𝑐2 +4𝜇)/2)𝜉
R, 𝐵11 , 𝐵22 , 𝐵12 , 𝐵22 are all positive constants.
󸀠󸀠
((𝑐−√𝑐2 +4𝜇)/2)𝜉
(107)
+ 𝑜 (𝑒(√𝑅0 −1−√𝑅0 −1+𝜇)𝜉 ) ,
),
√𝑐2 −4(𝑅0 −1))/2)𝜉
𝐼 (𝜉) = 𝐴 2 𝑒((𝑐−
𝑅𝑑 − 1
− (𝐴 11 + 𝐴 12 𝜉) 𝑒√𝑅0 −1𝜉 + 𝑜 (𝜉𝑒√𝑅0 −1𝜉 ) ,
𝑅𝑑
(109)
𝜇22 = 𝑒(𝑐0 /2)𝜉 sin (
√4 (𝑅0 − 1) −
2
𝑐02
𝜉) .
Abstract and Applied Analysis
13
Similar to the proof of Theorem 11, we can get that, when
𝜉 → −∞, 𝜒2 (𝜉) can be described as the following equation:
𝜒2 (𝜉) = 𝐾1 𝑒(𝑐0 /2)𝜉 cos (
√4 (𝑅0 − 1) − 𝑐02
+ 𝐾2 𝑒(𝑐0 /2)𝜉 sin (
2
𝜉)
√4 (𝑅0 − 1) − 𝑐02
= √𝐾12 + 𝐾22 𝑒(𝑐0 /2)𝜉 sin (
2
𝜉) + h.o.t,
√4 (𝑅0 − 1) − 𝑐02
2
𝜉 + 𝜏 (𝜉))
+ h.o.t,
(110)
where tan(𝜏(𝜉)) = 𝐾1 /𝐾2 , and h.o.t is the short notation for
the higher order terms.
That is to say, 𝑙2 (𝜉) is oscillating. Thus, any solution of
model (24) with 0 < 𝑐 < 2√𝑅0 − 1 is not strictly monotone.
Theorems 12 and 13 indicate that 𝑐 = 2√𝑅0 − 1 is the critical minimal wave speed.
Acknowledgments
The authors would like to thank the anonymous referee for
the very helpful suggestions and comments which led to
improvements of our original paper. This research was supported by the Natural Science Foundation of Zhejiang Province (LY12A01014, R1110261, and LQ12A01009), the National
Science Foundation of China (61272018), and the National
Basic Research Program of China (2012CB426510).
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