Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 578345, 20 pages
http://dx.doi.org/10.1155/2013/578345
Research Article
Existence and Asymptotic Behavior of Traveling Wave Fronts for
a Time-Delayed Degenerate Diffusion Equation
Weifang Yan and Rui Liu
Department of Mathematics, South China University of Technology, Guangzhou 510640, China
Correspondence should be addressed to Rui Liu; scliurui@scut.edu.cn
Received 15 November 2012; Revised 15 February 2013; Accepted 18 February 2013
Academic Editor: Peixuan Weng
Copyright © 2013 W. Yan and R. Liu. 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.
This paper is concerned with traveling wave fronts for a degenerate diffusion equation with time delay. We first establish the necessary and sufficient conditions to the existence of monotone increasing and decreasing traveling wave fronts, respectively. Moreover,
special attention is paid to the asymptotic behavior of traveling wave fronts connecting two uniform steady states. Some previous
results are extended.
1. Introduction
In this paper, we consider the traveling wave fronts for the following reaction diffusion equation with Hodgkin-Huxley
source:
𝜕𝑢 𝜕2 𝑢𝑚
+ 𝑓 (𝑢, 𝑢𝜏 ) ,
=
𝜕𝑡
𝜕𝑥2
𝑡 ≥ 0, 𝑥 ∈ R,
(1)
where 𝑚 > 0, 𝑓(𝑢, V) = 𝑢𝑝 (1 − 𝑢)𝑞 (V − 𝑎), 𝑝 > 0, 𝑞 > 0,
𝑚+𝑝 > 1, 𝑎 ∈ (0, 1) is a constant, and 𝑢𝜏 (𝑥, 𝑡) = 𝑢(𝑥, 𝑡−𝜏) for
𝜏 > 0.
In 1952, Hodgkin and Huxley [1] proposed the HodgkinHuxley (H-H) equation
𝑑𝑢
= 𝑢 (𝑡) (1 − 𝑢 (𝑡)) (𝑢 (𝑡 − 𝜏) − 𝑎) ,
𝑑𝑡
𝑎 ∈ (0, 1) ,
(2)
which describes the propagation of a voltage pulse through
the nerve axon of a squid. Recently, more and more attention
has been paid to the linear and semilinear parabolic equations
with and without time delay; see, for example, [2–7]. A
natural extension of the H-H model is the following linear diffusion equation:
𝜕𝑢 𝜕2 𝑢
+ 𝑢 (1 − 𝑢) (𝑢𝜏 − 𝑎) .
=
𝜕𝑡 𝜕𝑥2
(3)
For this equation, there have been many interesting results
on the existence and stability of the traveling wave solutions,
for instance, [8–10]. By a traveling wave solution, we mean a
solution 𝑢(𝑥, 𝑡) of (3) of the form 𝑢(𝑥, 𝑡) = 𝜑(𝑥 + 𝑐𝑡) with the
wave speed 𝑐.
On the other hand, the classical research of traveling
waves for the standard linear diffusion equations with various
sources has been extended to some degenerate or singular
diffusion equations. For example, Aronson [11] considered
the following equation:
𝜕𝑢 𝜕2 𝑢𝑚
(4)
+ 𝑢 (1 − 𝑢) (𝑢 − 𝑎) ,
𝑎 ∈ (0, 1) .
=
𝜕𝑡
𝜕𝑥2
When 𝑚 > 1, the equation degenerates at 𝑢 = 0. Hence, it has
a different feature from the case 𝑚 = 1; that is, if the initial
distribution of 𝑢(𝑥, 𝑡) has compact support, then 𝑢(𝑥, 𝑡) also
has compact support for each 𝑡 > 0. When 𝑚 > 1, 𝑎 ∈ (0,
1/2), Aronson [11] showed that (4) possesses a unique sharp
traveling wave solution with positive wave speed. Hosono
[12] solved the existence problem of traveling wave solutions
for (4) especially with nonpositive wave speed and discussed
the shape of the solutions. Sánchez-Garduño and Maini [13]
considered the following degenerate diffusion equation:
𝜕
𝜕𝑢
𝜕𝑢
=
(𝐷 (𝑢) ) + 𝑢 (1 − 𝑢) (𝑢 − 𝑎) ,
𝜕𝑡 𝜕𝑥
𝜕𝑥
(5)
𝑎 ∈ (0, 1) , 𝐷 (0) = 0,
and obtained the existence of traveling wave solutions of
smooth or sharp (oscillatory and monotone) type.
2
Abstract and Applied Analysis
For other papers concerning the traveling wave solutions
for degenerate diffusion equations without time delay, see
[14–22]. From these results, we see that an obvious difference
between the linear diffusion equations and the degenerate
diffusion equations is that, in the degenerate diffusion case,
there may exist traveling wave fronts of sharp type; that is, the
support of the solution is bounded above or below, and at the
boundary of the support, the derivative of the traveling wave
solution is discontinuous. However, in the linear diffusion
case, all traveling wave fronts are of smooth type; that is, the
solutions are classical solutions, which approach the steady
states at infinity.
As far as we know, there are only two articles dealing with
the traveling wave solutions for degenerate diffusion equations with time delay. In [23, 24], Jin et al. considered the following time-delayed Newtonian filtration and non-Newtonian filtration equations:
𝜕𝑢 𝜕2 𝑢𝑚
+ 𝑢 (1 − 𝑢) (𝑢𝜏 − 𝑎) ,
=
𝜕𝑡
𝜕𝑥2
𝜕𝑢
𝜕 󵄨󵄨󵄨󵄨 𝜕𝑢 󵄨󵄨󵄨󵄨𝑝−2 𝜕𝑢
=
(󵄨 󵄨
) + 𝑢𝑞 (1 − 𝑢) (𝑢𝜏 − 𝑎) .
𝜕𝑡 𝜕𝑥 󵄨󵄨󵄨 𝜕𝑥 󵄨󵄨󵄨 𝜕𝑥
󸀠󸀠
𝑞
Definition 1. A function 𝜑(𝜉) is called a traveling wave front
with wave speed 𝑐 > 0 if there exist 𝜉𝑎 , 𝜉𝑏 with −∞ ≤ 𝜉𝑎 <
𝜉𝑏 ≤ +∞ such that 𝜑 ∈ 𝐶2 (𝜉𝑎 , 𝜉𝑏 ) is monotonic increasing
and
𝑞
𝑐𝜑󸀠 (𝜉) = (𝜑 ) (𝜉) + 𝜑(𝜉)𝑝 (1 − 𝜑 (𝜉)) (𝜑𝑐𝜏 (𝜉) − 𝑎) ,
(8)
𝑚 󸀠
(𝜑 ) (𝜉𝑎 ) = (𝜑 ) (𝜉𝑏 ) = 0,
𝜑 (𝜉) = 1 for 𝜉̃𝑎 − 𝑐𝜏 < 𝜉 ≤ 𝜉̃𝑎 ,
𝜑 (𝜉̃𝑏 ) = 0,
(11)
(𝜑𝑚 ) (𝜉̃𝑎 ) = (𝜑𝑚 ) (𝜉̃𝑏 ) = 0.
󸀠
󸀠
(i) If 𝜉𝑎 > −∞ and 𝜑󸀠 (𝜉𝑎+ ) ≠ 0, then 𝜑(𝜉) is called
an increasing sharp-type traveling wave front (see
Figure 1(a)).
(ii) If 𝜉𝑎 = −∞ or 𝜑󸀠 (𝜉𝑎+ ) = 0, then 𝜑(𝜉) is called an
increasing smooth-type traveling wave front (see
Figure 2(a)).
Similarly, we have the following.
(iii) If 𝜉̃𝑏 < +∞ and 𝜑󸀠 (𝜉̃𝑏− ) ≠ 0, then 𝜑(𝜉) is called
a decreasing sharp-type traveling wave front (see
Figure 1(b)).
(iv) If 𝜉̃𝑏 = +∞ or 𝜑󸀠 (𝜉̃𝑏− ) = 0, then 𝜑(𝜉) is called
a decreasing smooth-type traveling wave front (see
Figure 2(b)).
󸀠
Let 𝜓(𝜉) = (𝜑𝑚 ) (𝜉). Then, (8) or (10) is transformed into
𝜑󸀠 (𝜉) =
𝜓󸀠 (𝜉) =
1 1−𝑚
𝜑
(𝜉) 𝜓 (𝜉) ,
𝑚
𝑐 1−𝑚
𝑞
𝑝
𝜑
(𝜉) 𝜓 (𝜉) − 𝜑(𝜉) (1 − 𝜑 (𝜉)) (𝜑𝑐𝜏 (𝜉) − 𝑎) .
𝑚
(12)
Furthermore, by (9) or (11), we give the asymptotic boundary
conditions for traveling wave fronts as follows:
𝜑 (𝜉) = 0 for 𝜉𝑎 − 𝑐𝜏 < 𝜉 ≤ 𝜉𝑎 ,
𝜑 (𝜉𝑏 ) = 1,
(13)
𝜓 (𝜉𝑎 ) = 𝜓 (𝜉𝑏 ) = 0,
or
𝜑 (𝜉) = 1 for 𝜉̃𝑎 − 𝑐𝜏 < 𝜉 ≤ 𝜉̃𝑎 ,
(14)
𝜓 (𝜉̃𝑎 ) = 𝜓 (𝜉̃𝑏 ) = 0.
If 𝜓(𝜉) is strictly positive or negative for 0 < 𝜑 < 1, then (12)
is equivalent to
𝜑 (𝜉) = 0 for 𝜉𝑎 − 𝑐𝜏 < 𝜉 ≤ 𝜉𝑎 ,
𝑚 󸀠
(10)
𝜑 (𝜉̃𝑏 ) = 0,
𝜉 ∈ (𝜉𝑎 , 𝜉𝑏 ) ,
𝜑 (𝜉𝑏 ) = 1,
𝑞
𝜉 ∈ (𝜉̃𝑎 , 𝜉̃𝑏 ) ,
(7)
where 𝜑𝑐𝜏 (𝜉) = 𝜑(𝜉 − 𝑐𝜏).
Before going further, we first give the definition of sharpand smooth-type traveling wave fronts.
𝑚 󸀠󸀠
󸀠󸀠
𝑐𝜑󸀠 (𝜉) = (𝜑𝑚 ) (𝜉) + 𝜑(𝜉)𝑝 (1 − 𝜑 (𝜉)) (𝜑𝑐𝜏 (𝜉) − 𝑎) ,
(6)
By using the shooting method together with the comparison
technique, they first obtained the necessary and sufficient
conditions to the existence of monotone increasing and decreasing traveling wave solutions, respectively, and then gave
an accurate estimation on the convergent rate for the semifinite or infinite traveling waves.
Motivated by [23, 24], in this paper, we discuss the existence and asymptotic behavior of traveling wave fronts for (1).
Let 𝑢(𝑥, 𝑡) = 𝜑(𝜉) with 𝜉 = 𝑥 + 𝑐𝑡. Then, (1) is transformed
into the following form:
𝑐𝜑󸀠 (𝜉) = (𝜑𝑚 ) (𝜉) + 𝜑(𝜉)𝑝 (1 − 𝜑 (𝜉)) (𝜑𝑐𝜏 (𝜉) − 𝑎) ,
or there exist 𝜉̃𝑎 , 𝜉̃𝑏 with −∞ ≤ 𝜉̃𝑎 < 𝜉̃𝑏 ≤ +∞ such that
𝜑 ∈ 𝐶2 (𝜉̃𝑎 , 𝜉̃𝑏 ) is monotonic decreasing and
(9)
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜓
=𝑐−
.
𝑑𝜑
𝜓
(15)
Abstract and Applied Analysis
3
(a)
(b)
Figure 1: Sharp-type traveling wave fronts. (a) Monotonic increasing. (b) Monotonic decreasing.
(a)
(b)
Figure 2: Smooth-type traveling wave fronts. (a) Monotonic increasing. (b) Monotonic decreasing.
Clearly, for any given 𝜑 > 0, if
∫
𝜑
0
where 𝜓+ is a solution of the following problem:
𝑚𝜎𝑚−1
𝑑𝜎 = +∞,
𝜓 (𝜎)
𝑞
(16)
𝑚𝑎𝜑𝑚+𝑝−1 (1 − 𝜑)
𝑑𝜓
=𝑐+
,
𝑑𝜑
𝜓
𝑐𝜏 = ∫
𝜑
𝜑𝑐𝜏
𝑚𝜎𝑚−1
𝑑𝜎.
𝜓 (𝜎)
𝜓 (𝜑) > 0,
(17)
0
𝑚𝜎𝑚−1
𝑑𝜎 < +∞,
𝜓 (𝜎)
𝑐𝜏 = ∫
𝜑
𝑐𝜏 = ∫
𝜑𝑐𝜏
𝑚𝜎𝑚−1
𝑑𝜎,
𝜓 (𝜎)
𝜑𝑐𝜏 = 0,
𝑚𝜎𝑚−1
if ∫
𝑑𝜎 ≥ 𝑐𝜏,
0 𝜓+ (𝜎)
𝜑
𝑚𝜎𝑚−1
𝑑𝜎 < 𝑐𝜏,
if ∫
0 𝜓+ (𝜎)
𝜑
𝑚𝜎𝑚−1
𝑑𝜎,
−𝜓 (𝜎)
𝜑𝑐𝜏 = 1,
then ∫0 (𝑚𝜎𝑚−1 /𝜓(𝜎))𝑑𝜎 may be less than 𝑐𝜏 when 𝜑 is near
0. Therefore, the previous definition is not reasonable. In what
follows, we give the definition of 𝜑𝑐𝜏 .
(i) If 𝜓 is positive, define 𝜑𝑐𝜏 by
𝜑
𝜑𝑐𝜏
(18)
𝜑
(19)
for 𝜑 ∈ (0, 1) .
(ii) If 𝜓 is negative, define 𝜑𝑐𝜏 by
However, if for some 𝜑 > 0,
𝜑
(20)
𝜓 (0+ ) = 0,
then 𝜑𝑐𝜏 can be defined by
∫
for 𝜑 ∈ (0, 1) ,
if ∫
1
𝜑
𝑚𝜎𝑚−1
𝑑𝜎 ≥ 𝑐𝜏,
−𝜓− (𝜎)
𝑚𝜎𝑚−1
if ∫
𝑑𝜎 < 𝑐𝜏,
𝜑 −𝜓− (𝜎)
1
(21)
where 𝜓− is a solution of the following problem:
𝑞
𝑚 (1 − 𝑎) 𝜑𝑚+𝑝−1 (1 − 𝜑)
𝑑𝜓
=𝑐−
,
𝑑𝜑
𝜓
for 𝜑 ∈ (0, 1) ,
𝜓 (1− ) = 0,
𝜓 (𝜑) < 0,
for 𝜑 ∈ (0, 1) .
(22)
4
Abstract and Applied Analysis
where 𝜑𝑖𝑠 = 𝜑𝑖 (𝜉𝑖 −𝑠), 𝑖 = 1, 2. The desired conclusion follows
immediately.
Consider the following problem:
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜓
=𝑐−
,
𝑑𝜑
𝜓
+
(23)
−
𝜓 (0 ) = 0,
𝜓 (1 ) = 0.
In Sections 2 and 3, we will verify the following two conclusions are equivalent, that is, (1) 𝜑 is a monotonic solution of
the problem (8)-(9) (or (10)-(11)); (2) 𝜓(𝜑) > 0 (or 𝜓(𝜑) < 0)
is a solution of the problem (23).
2. Existence of Increasing Traveling Waves
In this section, we aim to find a solution 𝜓(𝜑) of the problem
(23) with 𝜓(𝜑) > 0 for 𝜑 ∈ (0, 1).
Since 𝜓 > 0, we see that 𝜑(𝜉) is increasing in 𝜉, and so
𝜑𝑐𝜏 (𝜉) ≤ 𝜑(𝜉). Thus, to investigate the behavior of the trajectories 𝜓𝑐 (𝜑) of (23), we have to study the trajectories starting
from (0, 0), since the property of 𝜓𝑐 at 𝜑 depends on the
behavior of 𝜓𝑐 at 𝜑𝑐𝜏 closely. Consider the following problem:
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜓
=𝑐−
𝑑𝜑
𝜓
for 𝜑 ∈ (0, 𝛽) ,
Lemma 3. For any given 𝜏 > 0, and 𝑐1 > 𝑐2 ≥ 0, let 𝜓1 (𝜑),
𝜓2 (𝜑) be solutions of (24) corresponding to 𝑐1 , 𝑐2 , respectively.
Then, 𝜓1 (𝜑) > 𝜓2 (𝜑) for any 𝜑 ∈ (0, 𝛽2 ), where (0, 𝛽2 ) is the
maximal existence interval of the solution 𝜓2 (𝜑) > 0. In addition, 𝜓1 (𝛽2 ) > 0. (See Figure 3(a).)
Proof. We first show that 𝜓1 (𝜑) > 𝜓2 (𝜑) for sufficiently small
𝜑 > 0. The argument consists of three cases, 𝑚 + 𝑝 > 2, 𝑚 +
𝑝 = 2, and 1 < 𝑚 + 𝑝 < 2.
(i) Consider the case 𝑚 + 𝑝 > 2. According to (24), we
have
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑑𝜓
=𝑐+
.
𝑑𝜑
𝜓
Noticing that 𝜑𝑐𝜏 ≤ 𝜑, we have for 𝜑 ≤ 𝑎 that
𝑑𝜓
≥ 𝑐.
𝑑𝜑
Integrating from 0 to 𝜑 yields
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑑𝜓
=𝑐+
𝑑𝜑
𝜓
(24)
where (0, 𝛽) with 𝛽 ≤ 1 is the maximal existence interval of
the solution 𝜓(𝜑) > 0. By (24),
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑐𝜑
𝑎𝑚 𝑚+𝑝−2
≤𝑐+
.
𝜑
𝑐
≤𝑐+
𝜑
𝜑
1 2
𝜓 (𝜑) = ∫ 𝑐𝜓 (𝑠) 𝑑𝑠 − 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠𝑐𝜏 − 𝑎) 𝑑𝑠
2
0
0
𝜑
0
0
≥ ∫ 𝑐𝜓 (𝑠) 𝑑𝑠 − 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠.
1
This excludes 𝜓(1− ) = 0 if ∫0 𝑠𝑚+𝑝−1 (1−𝑠)𝑞 (𝑠−𝑎)𝑑𝑠 ≤ 0. Thus,
we only need to find the increasing traveling wave fronts for
1
the case ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 > 0. We first prove the
following two lemmas.
Lemma 2. Assume that 0 < 𝜑0 < 1 and 𝜓1 (𝜑), 𝜓2 (𝜑) are solutions of (24) corresponding to different wave speeds 𝑐1 , 𝑐2 ,
respectively, where 𝜓1 (𝜑0 ) ≥ 𝜓2 (𝜑0 ) > 0, 𝜓1 (𝜑) > 𝜓2 (𝜑) > 0
for 0 < 𝜑 < 𝜑0 , and 𝜑1 (𝜉1 ) = 𝜑2 (𝜉2 ) = 𝜑0 . Then, 𝜑1 (𝜉1 − 𝑠) <
𝜑
𝜑2 (𝜉2 − 𝑠) if 0 < 𝑠 < ∫0 0 (𝑚𝜎𝑚−1 /𝜓2 (𝜎))𝑑𝜎 and 𝜑1 (𝜉1 − 𝑠) =
𝜑0
𝜑2 (𝜉2 − 𝑠) = 0 if 𝑠 ≥ ∫0 (𝑚𝜎𝑚−1 /𝜓2 (𝜎))𝑑𝜎.
𝜓 (𝜑) ≤ 𝑐𝜑 +
𝜑𝑖𝑠
𝑚𝜎𝑚−1
𝑑𝜎
𝜓𝑖 (𝜎)
𝜑0
0
𝜑𝑖𝑠 = 0 if ∫
𝜑0
0
𝑚𝜎𝑚−1
𝑑𝜎 > 𝑠,
𝜓𝑖 (𝜎)
𝑚𝜎𝑚−1
𝑑𝜎 ≤ 𝑠,
𝜓𝑖 (𝜎)
(31)
as 𝜑 󳨀→ 0+ .
(32)
That is,
𝜓 (𝜑) ∼ 𝑐𝜑
Therefore, 𝜓1 (𝜑) > 𝜓2 (𝜑) for sufficiently small 𝜑 > 0.
(ii) Consider the case 𝑚 + 𝑝 = 2. Similar to (i), we have
𝜓 (𝜑) ≥ 𝑐𝜑,
(33)
and, hence,
𝜓 (𝜑) ≤ (𝑐 +
𝑎𝑚
) 𝜑.
𝑐
𝑎𝑚
,
𝑐
(1 − 𝜀)𝑞 (𝑎 − 𝜀) 𝑚
,
𝐴𝑛 = 𝑐 +
𝐵𝑛−1
𝑎𝑚
𝐵𝑛 = 𝑐 +
, 𝑛 = 2, 3, . . . ,
𝐴 𝑛−1
𝐴 1 = 𝑐,
if ∫
𝑎𝑚
𝜑𝑚+𝑝−1 .
𝑐 (𝑚 + 𝑝 − 1)
(34)
Consider the sequences {𝐴 𝑛 } and {𝐵𝑛 }, where
Proof. Recalling (19), we see that
𝜑0
(30)
Integrating from 0 to 𝜑 gives
(25)
𝑠=∫
(29)
We further have
𝜓 (𝜑) > 0 for 𝜑 ∈ (0, 𝛽) ,
𝜑
(28)
𝜓 (𝜑) ≥ 𝑐𝜑.
𝜓 (0+ ) = 0,
(27)
(26)
𝐵1 = 𝑐 +
(35)
Abstract and Applied Analysis
5
𝜓
𝜓
𝐿 𝑘1
𝑐1 > 𝑐2
𝜓𝑐 (𝜑)
𝜓𝑐1 (𝜑)
𝜓𝑐2 (𝜑)
0
𝛽2
1
𝜑
0
1
(a)
𝜑
(b)
𝜓
𝜓∗ (𝜑)
𝜓𝑐 (𝜑)
0
1
𝜑
(c)
Figure 3: The properties of the trajectory 𝜓𝑐 (𝜑). (a) The monotonicity of 𝜓𝑐 (𝜑) on 𝑐. (b) The trajectory 𝜓𝑐 (𝜑) wanders through 𝐿 𝑘 . (c) The
trajectory 𝜓𝑐 (𝜑) intersects with 𝜓∗ (𝜑) for large 𝑐.
and 𝜀 > 0 is sufficiently small. Noticing that 𝐴 1 < 𝐵1 and
𝐴 1 < 𝐴 2 , by induction, we obtain
𝐴1 < 𝐴2 = 𝐴3 < 𝐴4 = 𝐴5 < ⋅ ⋅ ⋅ < 𝐴𝑛
< ⋅ ⋅ ⋅ < 𝐵𝑛 < ⋅ ⋅ ⋅ < 𝐵4 = 𝐵3 < 𝐵2 = 𝐵1 .
For 0 < 𝜑 < 𝜀, if
𝐴 𝑛−1 𝜑 ≤ 𝜓 (𝜑) ≤ 𝐵𝑛−1 𝜑,
(36)
≤𝑐+
𝑞
𝑚𝜑(1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑑𝜓
=𝑐+
𝑑𝜑
𝜓
𝑞
𝑚𝜑(1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
≤𝑐+
𝐴 𝑛−1 𝜑
𝑞
𝑚𝜑(1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑑𝜓
=𝑐+
𝑑𝜑
𝜓
𝑞
(37)
𝑚𝜑(1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
≥𝑐+
𝐵𝑛−1 𝜑
≥𝑐+
then
𝑎𝑚
,
𝐴 𝑛−1
(1 − 𝜀)𝑞 (𝑎 − 𝜀) 𝑚
.
𝐵𝑛−1
(38)
Integrating from 0 to 𝜑 yields
𝐴 𝑛 𝜑 ≤ 𝜓 (𝜑) ≤ 𝐵𝑛 𝜑.
(39)
6
Abstract and Applied Analysis
Thus, we have
Recalling (46), we see that
∗
lim 𝐴 𝑛 = 𝐴 (𝜀) ,
lim 𝐵
𝑛→∞ 𝑛
𝑛→∞
∗
= 𝐵 (𝜀) ,
(40)
𝐴∗ (𝜀) 𝜑 ≤ 𝜓 (𝜑) ≤ 𝐵∗ (𝜀) 𝜑,
with 𝐴∗ (𝜀) and 𝐵∗ (𝜀) satisfying
𝐴∗ (𝜀) = 𝑐 +
2𝑚𝑎 (𝑚+𝑝)/2
𝑞
+ 2𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑)
𝜑
𝑚+𝑝
≥ 2𝑐√
(1 − 𝜀)𝑞 (𝑎 − 𝜀) 𝑚
,
𝐵∗ (𝜀)
𝐵∗ (𝜀) = 𝑐 +
𝑎𝑚
.
𝐴∗ (𝜀)
(41)
Letting 𝜀 → 0, we obtain
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) ,
(50)
which implies that
lim 𝐴∗ (𝜀) = lim 𝐵∗ (𝜀) =
𝜀→0
𝑑𝜓2
𝑞
= 2𝑐𝜓 + 2𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑𝑐𝜏 )
𝑑𝜑
√𝑐2
𝜀→0
+ 4𝑚𝑎 + 𝑐
.
2
(42)
𝜓2 (𝜑) ≥
That is,
𝜓 (𝜑) ∼
√𝑐2 + 4𝑚𝑎 + 𝑐
𝜑 as 𝜑 󳨀→ 0+ .
2
2𝑚𝑎 𝑚+𝑝
+
+ 𝑜 (𝜑(𝑚+𝑝+2)/2 ) .
𝜑
𝑚+𝑝
(43)
𝑑𝜓2
≤ 2𝑐𝜓 + 2𝑚𝑎𝜑𝑚+𝑝−1
𝑑𝜑
𝑞
(44)
= 2𝑐√
which means that
𝜑
𝜓2 (𝜑) ≥ ∫ 2𝑚𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑎 − 𝑠) 𝑑𝑠
0
2𝑚𝑎 𝑚+𝑝
=
+ 𝑜 (𝜑𝑚+𝑝 ) ;
𝜑
𝑚+𝑝
(45)
2𝑚𝑎 (𝑚+𝑝)/2
+ 2𝑚𝑎𝜑𝑚+𝑝−1 + 𝑜 (𝜑(𝑚+𝑝)/2 ) ,
𝜑
𝑚+𝑝
(52)
and, hence,
𝜓2 (𝜑) ≤
that is,
𝜓 (𝜑) ≥ √
2𝑚𝑎 (𝑚+𝑝)/2
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) .
𝜑
𝑚+𝑝
(46)
𝜓2 (𝜑) =
2𝑚𝑎 (𝑚+𝑝+2)/2
4𝑐
2𝑚𝑎 𝑚+𝑝
√
+
𝜑
𝜑
𝑚+𝑝+2 𝑚+𝑝
𝑚+𝑝
+ 𝑜 (𝜑(𝑚+𝑝+2)/2 ) ,
𝑚+𝑝−1
𝑚𝑎𝜑
as 𝜑 󳨀→ 0+ ,
(54)
√2𝑚𝑎/(𝑚 + 𝑝)𝜑(𝑚+𝑝)/2 + 𝑜 (𝜑(𝑚+𝑝)/2 )
𝑚𝑎 (𝑚 + 𝑝) (𝑚+𝑝−2)/2
+ 𝑜 (𝜑(𝑚+𝑝−2)/2 ) .
𝜑
2
(47)
which implies that 𝜓1 (𝜑) > 𝜓2 (𝜑) for sufficiently small 𝜑 > 0.
We claim that 𝜓1 (𝜑) > 𝜓2 (𝜑) for any 𝜑 ∈ (0, 𝛽2 ). Suppose
for contradiction that there exists 𝜑0 ∈ (0, 𝛽2 ) such that
𝜓1 (𝜑0 ) = 𝜓2 (𝜑0 ) = 𝜓0 and 𝜓1 (𝜑) > 𝜓2 (𝜑) for 𝜑 ∈ (0, 𝜑0 ).
Then,
𝜓1󸀠 (𝜑0 ) ≤ 𝜓2󸀠 (𝜑0 ) ,
That is,
𝜓 (𝜑) ≤ 𝑐𝜑 + √
2𝑚𝑎 (𝑚+𝑝)/2
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) .
𝜑
𝑚+𝑝
(48)
𝑚+𝑝−1
𝑚𝜑0
𝑞
(1 − 𝜑0 ) (𝑎 − 𝜑0𝑐1 𝜏 )
𝜓0
𝑚+𝑝−1
(49)
(55)
which means that
𝑐1 +
Thus, we have
2𝑚𝑎 (𝑚+𝑝)/2
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) .
𝜑
𝜓 (𝜑) = √
𝑚+𝑝
(53)
Summing up, we arrive at
𝑚𝑎𝜑𝑚+𝑝−1
𝑑𝜓
≤𝑐+
𝑑𝜑
𝜓
=𝑐+√
2𝑚𝑎 (𝑚+𝑝+2)/2
4𝑐
√
𝜑
𝑚+𝑝+2 𝑚+𝑝
2𝑚𝑎 𝑚+𝑝
+
+ 𝑜 (𝜑(𝑚+𝑝+2)/2 ) .
𝜑
𝑚+𝑝
Consequently,
≤𝑐+
(51)
On the other hand, by (49), we have
Therefore, 𝜓1 (𝜑) > 𝜓2 (𝜑) for sufficiently small 𝜑 > 0.
(iii) Consider the case 1 < 𝑚 + 𝑝 < 2. Notice that
𝑑𝜓 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑)
≥
,
𝑑𝜑
𝜓
2𝑚𝑎 (𝑚+𝑝+2)/2
4𝑐
√
𝜑
𝑚+𝑝+2 𝑚+𝑝
≤ 𝑐2 +
𝑚𝜑0
(56)
𝑞
(1 − 𝜑0 ) (𝑎 − 𝜑0𝑐2 𝜏 )
𝜓0
;
Abstract and Applied Analysis
7
that is,
0 < 𝑐1 − 𝑐2 ≤
𝑚+𝑝−1
(1
𝑚𝜑0
𝑞
− 𝜑0 )
𝜓0
(𝜑0𝑐1 𝜏 − 𝜑0𝑐2 𝜏 ) .
(57)
since 𝜑(𝜉) is increasing in 𝜉. This implies that (𝜑, 𝜓) wanders through increasing level sets with increasing 𝜉. See
Figure 3(b). Let
1
𝑘1 = 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠;
Thus,
0
𝜑0𝑐1 𝜏 > 𝜑0𝑐2 𝜏 .
(58)
Denote that 𝜑1 (𝜉1 ) = 𝜑2 (𝜉2 ) = 𝜑0 . By Lemma 2, we have
𝜑1 (𝜉1 − 𝑐1 𝜏) ≤ 𝜑2 (𝜉2 − 𝑐1 𝜏) ≤ 𝜑2 (𝜉2 − 𝑐2 𝜏) ,
2
1 𝑑𝜓𝑖
𝑞
= 𝑐𝑖 𝜓𝑖 − 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝑖 𝜏 − 𝑎) ,
2 𝑑𝜑
𝑖 = 1, 2.
(60)
By Lemma 2, for any 𝜑1 (𝜉1 ) = 𝜑2 (𝜉2 ) < 𝛽2 , we have
𝜑1 (𝜉1 − 𝑐1 𝜏) ≤ 𝜑2 (𝜉2 − 𝑐1 𝜏) ≤ 𝜑2 (𝜉2 − 𝑐2 𝜏) ,
In what follows, we will see that 𝜓∗ (𝜑) plays a special role for
the proof of the main result. We first need a lemma as follows.
Lemma 4. The trajectory 𝜓𝑐 (𝜑) of the problem (24) must intersect with 𝜓∗ (𝜑) for sufficiently large 𝑐 > 0. (See Figure 3(c).)
(62)
𝜓𝑐󸀠 (𝜑) ≥ 𝑐 > 0.
𝑚+𝑝−1
(63)
(64)
𝜓𝑐 (𝜑0 ) =
𝑚+𝑝−1
for any 𝑘 ≥ 0. Clearly, for 𝑐 = 0, the level sets {𝐿 𝑘 } are exactly
the trajectories of solutions to (12) or (15). Now, we define
Notice that if (𝜑, 𝜓) ∈
have
𝐿+𝑘
(66)
is a solution of the system (12), we
𝑑𝑘 (𝜉)
𝑞
= 𝜓𝜓󸀠 + 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎) 𝜑󸀠
𝑑𝜉
𝑐 1−𝑚 2
𝑞
𝜑 𝜓 + 𝜓𝜑𝑝 (1 − 𝜑) (𝜑 − 𝜑𝑐𝜏 )
𝑚
𝑐
≥ 𝜑1−𝑚 𝜓2
𝑚
=
> 0,
(67)
(71)
(72)
we have
𝑚𝜑0
𝜑
1
= {(𝜑, 𝜓) ∈ R2 ; 𝜓2 + 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠 = 𝑘}
2
0
(65)
𝑞
(1 − 𝜑0 ) (𝜑0𝑐𝜏 − 𝑎)
.
𝑐
𝜓𝑐 (𝜑0 ) > 𝜓𝑐 (𝑎) ≥ 𝑎𝑐,
𝑞
(1 − 𝜑0 ) (𝜑0𝑐𝜏 − 𝑎) > 𝑎𝑐2 .
(73)
Thus,
𝑚+𝑝−1
𝑚𝜑0
𝐿𝑘
𝐿+𝑘 = 𝐿 𝑘 ∩ {𝜓 > 0} .
𝑚𝜑0
Since
namely, 𝜓1 (𝛽2 ) > 0. The proof is completed.
To deal with the behavior of the trajectories 𝜓𝑐 (𝜑) of the
problem (24), we introduce the level set
(70)
Let 𝜑0 ∈ (𝑎, 1) be the first point such that 𝜓𝑐󸀠 (𝜑0 ) = 0. Then,
we have
Letting 𝜑 → 𝛽2 gives
𝜓12 (𝛽2 ) ≥ 𝜓12 (𝜑0 ) − 𝜓22 (𝜑0 ) > 0;
(69)
𝜑
(61)
for any 𝜑 ∈ (0, 𝛽2 ). Integrating the previous inequality from
𝜑0 to 𝜑 for any 0 < 𝜑0 < 𝜑 < 𝛽2 yields
𝜓12 (𝜑) > 𝜓22 (𝜑) + 𝜓12 (𝜑0 ) − 𝜓22 (𝜑0 ) .
1
𝜓∗ (𝜑) = √ 2𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠.
Proof. For any 𝜑 ∈ (0, 𝑎], we have 𝜑𝑐𝜏 ∈ (0, 𝑎], and
since 𝜓1 > 𝜓2 . Thus, we obtain
𝑑𝜓12 𝑑𝜓22
>
𝑑𝜑
𝑑𝜑
we know that 𝐿 𝑘1 passes through the critical point (1, 0).
Denote that
(59)
which means that 𝜑0𝑐1 𝜏 ≤ 𝜑0𝑐2 𝜏 , a contradiction.
In what follows, we will show that 𝜓1 (𝛽2 ) > 0. Recalling
the first equation of (24), we infer that
(68)
𝑞
(1 − 𝜑0 ) (𝜑0 − 𝑎) > 𝑎𝑐2 .
(74)
Denote that
𝑞
𝑀 = max {
𝜑∈(𝑎,1)
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎)
}.
𝑎
(75)
Then, for any 𝑐 ≥ √𝑀, (74) does not hold and 𝜓𝑐 (𝜑) is increasing on (𝑎, 1). Therefore, 𝜓𝑐 (𝜑) must intersect with 𝜓∗ (𝜑)
for any 𝑐 ≥ √𝑀. The proof is completed.
1
Theorem 5. (i) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≤ 0, then there is
no nontrivial nonnegative solution for problem (23).
1
(ii) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 > 0, then there exists a
unique 𝑐1∗ > 0, such that the problem (23) admits a nonnegative
solution 𝜓(𝜑), and 𝜓(𝜑) > 0 for any 𝜑 ∈ (0, 1).
1
Proof. (i) The case ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≤ 0 has been
discussed.
(ii) We know that for any fixed 𝑐 > 0, (𝜑, 𝜓𝑐 ) wanders
through increasing level sets {𝐿 𝑘 } strictly. Thus, (𝜑, 𝜓𝑐 )
8
Abstract and Applied Analysis
intersects with a level set 𝐿 𝑘 at most once. Let (𝜎𝑐 , 𝜙𝑐 ) be the
intersection point of (𝜑, 𝜓𝑐 ) with 𝐿 𝑘1 . Then, if 𝜙𝑐 > 0, we have
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎) 󵄨󵄨󵄨󵄨
𝑑𝜓∗ 󵄨󵄨󵄨󵄨
󵄨󵄨
=−
󵄨󵄨
󵄨󵄨
𝑑𝜑 󵄨󵄨𝜑=𝜎𝑐
𝜓∗
󵄨𝜑=𝜎𝑐
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎) 󵄨󵄨󵄨󵄨
󵄨󵄨
≤−
󵄨󵄨
𝜙𝑐
󵄨𝜑=𝜎𝑐
=
Proof. The necessity is clear. Consider the sufficient one. Let
𝜓(𝜑) > 0 for any 𝜑 ∈ (0, 1) be a solution of the problem (23),
and 𝜑(𝜉) solves
1 1−𝑚
(83)
𝜑
(𝜉) 𝜓 (𝜑 (𝜉)) .
𝑚
Without loss of generality, let 𝜑(0) = 1/2, and let (𝛼, 𝛽) be the
maximal existence interval of 𝜑 such that 0 < 𝜑 < 1. Firstly,
we have
𝜑󸀠 (𝜉) =
(76)
𝑑𝜓𝑐 󵄨󵄨󵄨󵄨
− 𝑐.
󵄨
𝑑𝜑 󵄨󵄨󵄨𝜑=𝜎𝑐
󸀠󸀠
(𝜑𝑚 ) (𝜉) = 𝜓󸀠 (𝜑) 𝜑󸀠 (𝜉)
𝑞
= (𝑐 −
Define
𝑐1∗
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
) 𝜑󸀠
𝜓
(84)
𝑞
= 𝑐𝜑󸀠 − 𝜑𝑝 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎) .
= inf {𝑐 > 0; 𝜓𝑐 intersects 𝜓∗ at (𝜎𝑐 , 𝜙𝑐 ) with 𝜎𝑐 ∈ (0, 1]} .
(77)
4, 𝑐1∗
By Lemma
is well defined. In what follows, we will show
that 𝑐1∗ is just the desired wave speed.
We first have 𝑐1∗ > 0. Indeed, when 𝑐 = 0, the first equation
of (24) becomes
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎)
𝑑𝜓0
.
=−
𝑑𝜑
𝜓0
󸀠
(85)
Therefore, if 𝛼 > −∞ and 𝜑󸀠 (𝛼+ ) ≠ 0, 𝜑(𝜉) is a sharp-type
traveling wave front; if 𝛼 = −∞ or 𝜑󸀠 (𝛼+ ) = 0, 𝜑(𝜉) is a
smooth-type traveling wave front.
Theorem 5 and Proposition 6 imply the following result.
(79)
Theorem 7. (i) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≤ 0, then there is
no increasing traveling wave front for the problem (8)-(9).
1
(ii) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 > 0, then there is a unique
wave speed 𝑐1∗ > 0, such that the problem (8)-(9) admits an
increasing traveling wave front.
𝜑
0
Then, there exists 𝜑0 ∈ (𝑎, 1) such that
𝜓0 (𝜑0 ) = 0,
(80)
𝑞
since
− 𝑠) (𝑠 − 𝑎)𝑑𝑠 > 0. The continuous dependence of 𝜓𝑐 on 𝑐 ensures that 𝜓𝑐 goes to zero before reaching
𝜓∗ for sufficiently small 𝑐 > 0. Thus, 𝑐1∗ > 0.
In addition, by Lemma 3, we know that 𝜎𝑐 is decreasing
on 𝑐. Assume that
lim 𝜎𝑐 = 𝜎𝑐1∗ .
(81)
𝑐↘𝑐1∗
If 𝜎𝑐1∗ < 1; namely, 𝜙𝑐1∗ > 0, then by (76), we arrive at
󸀠
󸀠
(𝜑𝑚 ) (𝛼) = (𝜑𝑚 ) (𝛽) = 0.
𝜑 (𝛽) = 1,
1
𝜓02 (𝜑) = −2𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠.
󸀠
𝜓𝑐󸀠∗ (𝜎𝑐1∗ ) ≥ 𝜓∗ (𝜎𝑐1∗ ) + 𝑐1∗ > 𝜓∗ (𝜎𝑐1∗ ) .
1
𝜑 (𝜉) = 0 for 𝛼 − 𝑐𝜏 < 𝜉 ≤ 𝛼,
(78)
Noticing that 𝜓0 (0+ ) = 0, we have
1
∫0 𝑠𝑚+𝑝−1 (1
Moreover,
(82)
So, there exists 𝜎𝑐1∗ < 𝜑1 < 1, such that 𝜓𝑐1∗ (𝜑1 ) > 𝜓∗ (𝜑1 ). By
the continuous dependence of 𝜓𝑐 on 𝑐, there is 𝑐 > 0 with
𝑐 < 𝑐1∗ and close to 𝑐1∗ sufficiently such that 𝜓𝑐 (𝜑1 ) ≥ 𝜓∗ (𝜑1 ),
which implies that 𝜓𝑐 intersects with 𝜓∗ . This contradicts the
definition of 𝑐1∗ . Thus, 𝜎𝑐1∗ = 1, which implies 𝜙𝑐1∗ = 0.
Moreover, by Lemma 3, for 𝑐 > 𝑐1∗ , we have 𝜓𝑐 (1) > 0.
However, if 0 < 𝑐 < 𝑐1∗ , there exists 𝜑𝑐 with 𝑎 < 𝜑𝑐 < 1 such
that 𝜓𝑐 (𝜑) → 0 as 𝜑 ↗ 𝜑𝑐 . The proof is completed.
Proposition 6. 𝜑(𝜉) is a monotone increasing sharp- or
smooth-type traveling wave front of the problem (8)-(9) for
some fixed 𝑐 > 0, if and only if 𝜓(𝜑) with 𝜓(𝜑) > 0 for any
𝜑 ∈ (0, 1) is a solution of the problem (23).
Furthermore, we have the following results.
Theorem 8. Let 𝜑(𝜉) be the traveling wave front of the problem
(8)-(9) corresponding to the wave speed 𝑐1∗ .
(i) If 𝑚 < 2, then 𝜑(𝜉) is a smooth-type traveling wave
front.
(ii) If 𝑚 = 2, then 𝜑󸀠 (𝜉𝑎+ ) = 𝑐1∗ /2 and 𝜑(𝜉) is a sharp-type
traveling wave front.
(iii) If 𝑚 > 2, then 𝜑󸀠 (𝜉𝑎+ ) = +∞ and 𝜑(𝜉) is a sharp-type
traveling wave front.
Proof. (i) If 0 < 𝑚 ≤ 1, it is easy to see that
1 1−𝑚 +
(86)
(𝜉𝑎 ) 𝜓 (𝜉𝑎+ ) = 0.
𝜑
𝑚
If 1 < 𝑚 < 2, from the proof of Lemma 3, we see that when
𝜑 > 0 is sufficiently small,
𝜑󸀠 (𝜉𝑎+ ) =
𝜓 (𝜑) = 𝑐1∗ 𝜑 + 𝑜 (𝜑) ,
𝜓 (𝜑) =
√𝑐1∗2 + 4𝑚𝑎 + 𝑐1∗
𝜓 (𝜑) = √
2
𝜑 + 𝑜 (𝜑) ,
𝑚 + 𝑝 > 2;
𝑚 + 𝑝 = 2;
2𝑚𝑎 (𝑚+𝑝)/2
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) ,
𝜑
𝑚+𝑝
1 < 𝑚 + 𝑝 < 2.
(87)
Abstract and Applied Analysis
9
for 1 < 𝑚 + 𝑝 < 2,
Since
𝜑󸀠 (𝜉) =
1 1−𝑚
𝜑
(𝜉) 𝜓 (𝜉) ,
𝑚
(88)
𝜓𝑐2 (𝜑) =
we have
𝜑󸀠 (𝜉) =
𝑐1∗ 2−𝑚
𝜑
(𝜉) + 𝑜 (𝜑2−𝑚 (𝜉)) ,
𝑚
√𝑐1∗2 + 4𝑚𝑎 + 𝑐1∗
𝜑󸀠 (𝜉) =
2𝑚
2𝑚𝑎 𝑚+𝑝
+ 𝑜 (𝜑(𝑚+𝑝+2)/2 ) ,
𝜑
𝑚+𝑝
+
𝑚 + 𝑝 > 2;
𝜓𝑐̂22
𝜑2−𝑚 (𝜉) + 𝑜 (𝜑2−𝑚 (𝜉)) ,
2𝑚𝑎 𝑚+𝑝
+ 𝑜 (𝜑(𝑚+𝑝+2)/2 ) .
𝜑
𝑚+𝑝
+
2𝑎
𝜑(2−𝑚+𝑝)/2 (𝜉) + 𝑜 (𝜑(2−𝑚+𝑝)/2 (𝜉)) ,
𝑚 (𝑚 + 𝑝)
Since 𝑐 > 𝑐̂2 , we have
1 < 𝑚 + 𝑝 < 2.
(89)
Thus,
𝜑󸀠 (𝜉𝑎+ ) = 0.
𝑐1∗
.
2
𝜑
(𝜉𝑎+ )
(91)
𝜓𝑐󸀠 (𝜑0 ) ≤ 𝜓𝑐̂󸀠2 (𝜑0 ) ;
= +∞.
(92)
𝑚+𝑝−1
0 < 𝑐 − 𝑐̂2 ≤
Theorem 9. 𝑐1∗ (𝜏) is nonincreasing in delay 𝜏; namely, if 𝜏1 >
𝜏2 , then 𝑐1∗ (𝜏1 ) ≤ 𝑐1∗ (𝜏2 ).
Proof. Suppose for contradiction that there exist time delays
𝜏1 and 𝜏2 , with 𝜏1 > 𝜏2 and 𝑐1∗ (𝜏1 ) > 𝑐1∗ (𝜏2 ). Denote that 𝑐̂1 =
𝑐1∗ (𝜏1 ) and 𝑐̂2 = 𝑐1∗ (𝜏2 ) for simplicity. Take 𝑐 such that
𝑐̂2 < 𝑐 < 𝑐̂1 .
(93)
In what follows, denote the solutions of (24) corresponding
to wave speed and time delay (̂𝑐1 , 𝜏1 ), (𝑐, 𝜏1 ), (̂𝑐2 , 𝜏2 ) by 𝜓𝑐̂1 , 𝜓𝑐 ,
𝜓𝑐̂2 , respectively.
Similar to the proof of Lemma 3, we know that when 𝜑 >
0 is sufficiently small, for 𝑚 + 𝑝 > 2,
𝜓𝑐 (𝜑) = 𝑐𝜑 + 𝑜 (𝜑) ,
𝜓𝑐̂2 (𝜑) = 𝑐̂2 𝜑 + 𝑜 (𝜑) ;
(94)
for 𝑚 + 𝑝 = 2,
𝜓𝑐̂2 (𝜑) =
(98)
that is,
The proof is completed.
𝜓𝑐 (𝜑) =
(97)
for sufficiently small 𝜑 > 0. Furthermore, we claim that
𝜓𝑐 (𝜑) > 𝜓𝑐̂2 (𝜑) for 𝜑 ∈ (0, 1). Otherwise, there exists 𝜑0 ∈
(0, 1) such that 𝜓𝑐 (𝜑) > 𝜓𝑐̂2 (𝜑) for 𝜑 ∈ (0, 𝜑0 ), and 𝜓𝑐 (𝜑0 ) =
𝜓𝑐̂2 (𝜑0 ) = 𝜓0 . Then, we have
(iii) If 𝑚 > 2, we have
󸀠
𝜓𝑐 (𝜑) > 𝜓𝑐̂2 (𝜑)
(90)
(ii) If 𝑚 = 2, we have
𝜑󸀠 (𝜉𝑎+ ) =
(96)
4̂𝑐2
2𝑚𝑎 (𝑚+𝑝+2)/2
√
(𝜑) =
𝜑
𝑚+𝑝+2 𝑚+𝑝
𝑚 + 𝑝 = 2;
𝜑󸀠 (𝜉) = √
2𝑚𝑎 (𝑚+𝑝+2)/2
4𝑐
√
𝜑
𝑚+𝑝+2 𝑚+𝑝
√𝑐2 + 4𝑚𝑎 + 𝑐
2
𝑞
(1 − 𝜑0 ) (𝜑0𝑐𝜏1 − 𝜑0̂𝑐2 𝜏2 )
𝜓0
𝜑 + 𝑜 (𝜑) ;
(99)
3. Existence of Decreasing Traveling Waves
In this section, we aim to find a solution 𝜓(𝜑) of the problem
(23) with 𝜓(𝜑) < 0 for 𝜑 ∈ (0, 1). We first introduce a comparison lemma.
Lemma 10. Let 𝑢, V be the solutions of the following problems,
respectively:
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎)
𝑑𝑢
=𝑐−
𝑑𝜑
𝑢
𝑞
(95)
,
which implies 𝜑0𝑐𝜏1 > 𝜑0̂𝑐2 𝜏2 . On the other hand, by Lemma 2,
we have 𝜑0𝑐𝜏1 ≤ 𝜑0̂𝑐2 𝜏2 , a contradiction. Similarly, we can get
𝜓𝑐̂1 (𝜑) > 𝜓𝑐 (𝜑). From the uniqueness of wave speed on any
fixed 𝜏 > 0, this contradicts the definition of 𝑐̂1 .
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (1 − 𝑎)
𝑑V
=𝑐−
𝑑𝜑
V
𝜑 + 𝑜 (𝜑) ,
√̂𝑐22 + 4𝑚𝑎 + 𝑐̂2
2
𝑚𝜑0
for 𝜑 < 1, 𝑢 (1− ) = 0,
for 𝜑 < 1, V (1− ) = 0.
(100)
And let 𝜓 solve the problem (23). Then, 𝜓(𝜑) ≤ 𝑢(𝜑) ≤ V(𝜑)for
10
Abstract and Applied Analysis
𝜑 < 1 if 𝜓, 𝑢, V are positive, while V(𝜑) ≤ 𝜓(𝜑) ≤ 𝑢(𝜑) for
𝜑 < 1 if 𝜓, 𝑢, V are negative.
Proof. Notice that 𝜑𝑐𝜏 ≤ 𝜑 when 𝜓 is positive, and thus
1 𝑑𝜓2
𝑞
− 𝑐𝜓 = −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
2 𝑑𝜑
𝑞
≥ −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎)
=
(101)
1 𝑑𝑢2
− 𝑐𝑢;
2 𝑑𝜑
1
Hence, 𝜓(0+ ) = 0 is impossible if ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≥
0. So, we only need to find the decreasing traveling wave
1
fronts for the case ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 < 0. We first
give the following three lemmas.
Lemma 11. Assume that 0 < 𝜑0 < 1 and 𝜓1 (𝜑), 𝜓2 (𝜑) are
solutions of (106) corresponding to different wave speeds 𝑐1 , 𝑐2 ,
respectively, where 𝜓1 (𝜑0 ) ≤ 𝜓2 (𝜑0 ) < 0, 𝜓1 (𝜑) < 𝜓2 (𝜑) < 0
for 𝜑0 < 𝜑 < 1, and 𝜑1 (𝜉1 ) = 𝜑2 (𝜉2 ) = 𝜑0 . Then, 𝜑1 (𝜉1 − 𝑠) >
1
𝜑2 (𝜉2 − 𝑠) if 0 < 𝑠 < ∫𝜑 (𝑚𝜎𝑚−1 / − 𝜓2 (𝜎))𝑑𝜎, and 𝜑1 (𝜉1 − 𝑠) =
0
1
𝜑2 (𝜉2 − 𝑠) = 1 if 𝑠 ≥ ∫𝜑 (𝑚𝜎𝑚−1 / − 𝜓2 (𝜎))𝑑𝜎.
0
that is,
Proof. The proof is similar to that of Lemma 2.
𝑑 (𝜓2 − 𝑢2 )
𝑑𝜑
− 2𝑐 (𝜓2 − 𝑢2 ) 𝑔 (𝜑) ≥ 0,
(102)
(103)
Lemma 12. For any given 𝜏 > 0, and 𝑐1 > 𝑐2 ≥ 0, let 𝜓1 (𝜑),
𝜓2 (𝜑) be solutions of (106) corresponding to 𝑐1 , 𝑐2 , respectively.
Then, 𝜓1 (𝜑) < 𝜓2 (𝜑) for any 𝜑 ∈ (𝛼2 , 1), where (𝛼2 , 1) is
the maximal existence interval of the solution 𝜓2 (𝜑) < 0. In
addition, 𝜓1 (𝛼2 ) < 0. (See Figure 4(a).)
(104)
Proof. We first show that 𝜓1 (𝜑) < 𝜓2 (𝜑) if 𝜑 is in a sufficiently
small left neighborhood of 1. The argument consists of three
cases, 𝑞 > 1, 𝑞 = 1, and 𝑞 < 1.
(i) Consider the case 𝑞 > 1. According to (106), we have
where
𝑔 (𝜑) =
1
.
𝜓+𝑢
Consequently,
1
2𝑐 ∫𝜑 𝑔(𝑠)𝑑𝑠
𝑑 ((𝜓2 − 𝑢2 ) 𝑒
)
𝑑𝜑
≥ 0.
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜓
=𝑐−
.
𝑑𝜑
𝜓
Integrating from 𝜑 to 1 yields
𝜓2 (𝜑) − 𝑢2 (𝜑) ≤ 0,
(105)
and so 𝜓(𝜑) ≤ 𝑢(𝜑) if 𝜓, 𝑢 > 0. Similarly, 𝑢(𝜑) ≤ V(𝜑) if
𝑢, V > 0.
The proof for 𝜓, 𝑢, V < 0 is similar and omitted here.
Since 𝜓 < 0, we see that 𝜑(𝜉) is decreasing in 𝜉, and so
𝜑𝑐𝜏 (𝜉) ≥ 𝜑(𝜉). To get the behavior of the trajectories 𝜓𝑐 (𝜑)
of (23), we have to study the trajectories starting from (1, 0),
since the property of 𝜓𝑐 at 𝜑 depends on the behavior of 𝜓𝑐 at
𝜑𝑐𝜏 closely. Consider the following problem:
𝑚+𝑝−1
𝑚𝜑
𝑑𝜓
=𝑐−
𝑑𝜑
Noticing that 𝜑𝑐𝜏 ≥ 𝜑, we have for 𝜑 ≥ 𝑎 that
𝑑𝜓
≥ 𝑐.
𝑑𝜑
−𝜓 (𝜑) ≥ 𝑐 (1 − 𝜑) .
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜓
=𝑐−
𝑑𝜑
𝜓
for 𝜑 ∈ (𝛼, 1) ,
𝑞
𝜓 (1 ) = 0,
for 𝜑 ∈ (𝛼, 1) ,
(106)
where (𝛼, 1) with 𝛼 ≥ 0 is the maximal existence interval of
the solution 𝜓(𝜑) < 0. By (106),
1
1
1 2
𝜓 (𝜑) = − ∫ 𝑐𝜓 (𝑠) 𝑑𝑠 + 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠𝑐𝜏 − 𝑎) 𝑑𝑠
2
𝜑
𝜑
1
1
𝜑
𝜑
𝑚+𝑝−1
≥ − ∫ 𝑐𝜓 (𝑠) 𝑑𝑠 + 𝑚 ∫ 𝑠
(110)
We further have
−
𝜓 (𝜑) < 0
(109)
Integrating from 𝜑 to 1 yields
𝑞
(1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝜓
(108)
𝑞
(1 − 𝑠) (𝑠 − 𝑎) 𝑑𝑠.
(107)
≤𝑐+
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑐 (1 − 𝜑)
≤𝑐+
𝑚 (1 − 𝑎)
𝑞−1
(1 − 𝜑) .
𝑐
(111)
Integrating from 𝜑 to 1 gives
−𝜓 (𝜑) ≤ 𝑐 (1 − 𝜑) +
𝑚 (1 − 𝑎)
𝑞
(1 − 𝜑) .
𝑐𝑞
(112)
as 𝜑 󳨀→ 1− .
(113)
Thus,
𝜓 (𝜑) ∼ 𝑐 (𝜑 − 1) ,
Therefore, 𝜓1 (𝜑) < 𝜓2 (𝜑) in a left neighborhood of 𝜑 = 1.
Abstract and Applied Analysis
11
𝜓
𝜓
0
1
𝛼2
0
𝜑
1
𝜑
𝜓𝑐2 (𝜑)
𝜓𝑐 (𝜑)
𝜓𝑐1 (𝜑)
𝑐1 > 𝑐2
𝑅𝑘2
(a)
(b)
𝜓
0
1
𝜑
𝜓𝑐 (𝜑)
𝜓̃ ∗ (𝜑)
(c)
Figure 4: The properties of the trajectory 𝜓𝑐 (𝜑). (a) The monotonicity of 𝜓𝑐 (𝜑) on 𝑐. (b) The trajectory 𝜓𝑐 (𝜑) wanders through 𝑅𝑘 . (c) The
trajectory 𝜓𝑐 (𝜑) intersects with 𝜓̃∗ (𝜑) for large 𝑐.
(ii) When 𝑞 = 1, consider the following two systems:
𝜑󸀠 (𝜉) =
𝑢1󸀠 (𝜉) =
1
𝑚
)
(115)
𝑃=(
𝑐
1−𝑎
𝑚
at (1, 0). By a simple calculation, we get the eigenvalues of the
matrix 𝑃 as
0
𝑐 1−𝑚
𝑝
𝜑
(𝜉) 𝑢1 (𝜉) − 𝜑(𝜉) (1 − 𝜑 (𝜉)) (𝜑 (𝜉) − 𝑎) ,
𝑚
𝜑󸀠 (𝜉) =
𝑢2󸀠 (𝜉) =
1 1−𝑚
𝜑
(𝜉) 𝑢1 (𝜉) ,
𝑚
Notice that the right hand side of the previous two systems
shares the same Jacobian matrix
1 1−𝑚
𝜑
(𝜉) 𝑢2 (𝜉) ,
𝑚
𝑐 1−𝑚
𝑝
𝜑
(𝜉) 𝑢2 (𝜉) − 𝜑(𝜉) (1 − 𝜑 (𝜉)) (1 − 𝑎) .
𝑚
𝜆+ =
(114)
𝑐 + √𝑐2 + 4𝑚 (1 − 𝑎)
,
2𝑚
𝜆− =
𝑐 − √𝑐2 + 4𝑚 (1 − 𝑎)
.
2𝑚
(116)
12
Abstract and Applied Analysis
It is easy to see that (1, 0) is a saddle point. And the eigenvector associated with the eigenvalue 𝜆 + can be
1
𝑄=(
).
𝑚𝜆 +
Thus, we have
−𝜓 (𝜑) = √
(117)
We express the local explicit solutions of the problem (114) in
the left neighborhood of (1, 0) to reach
2
𝜑 (𝜉) = 1 − 𝑙𝑒𝜆 + 𝜉 + O ((1 − 𝜑) ) ,
𝑢𝑖 (𝜉) = −𝑙𝑚𝜆 + 𝑒𝜆 + 𝜉 + O (𝑢𝑖2 ) ,
(118)
𝑖 = 1, 2.
Recalling (123), we see that
−
𝑑𝜓2
𝑞
= −2𝑐𝜓 + 2𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐𝜏 − 𝑎)
𝑑𝜑
Therefore, near (1, 0), we have
𝑢𝑖 ∼ 𝑚𝜆 + (𝜑 − 1) =
≥ 2𝑐√
𝑐 + √𝑐2 + 4𝑚 (1 − 𝑎)
(𝜑 − 1) ,
2
𝑖 = 1, 2.
𝑐 + √𝑐2 + 4𝑚 (1 − 𝑎)
𝜓 (𝜑) ∼
(𝜑 − 1) ,
2
𝑞
as 𝜑 󳨀→ 1− .
(120)
𝜓2 (𝜑) ≥
2𝑚 (1 − 𝑎)
4𝑐
(𝑞+3)/2
√
(1 − 𝜑)
𝑞+3
𝑞+1
2𝑚 (1 − 𝑎)
𝑞+1
𝑞+1
+
(1 − 𝜑) + 𝑜 ((1 − 𝜑) ) .
𝑞+1
𝑞
𝑑𝜓 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑)
≥
,
𝑑𝜑
𝜓
(121)
which means that
1
𝜑
(122)
2𝑚 (𝑎 − 1)
𝑞+1
𝑞+1
(1 − 𝜑) + 𝑜 ((1 − 𝜑) ) ;
𝑞+1
),
(127)
𝑑𝜓2
𝑞
≤ −2𝑐𝜓 + 2𝑚 (1 − 𝑎) (1 − 𝜑)
𝑑𝜑
= 2𝑐√
2𝑚 (1 − 𝑎)
(𝑞+1)/2
𝑞
+ 2𝑚 (1 − 𝑎) (1 − 𝜑)
(1 − 𝜑)
𝑞+1
+ 𝑜 ((1 − 𝜑)
that is,
(128)
On the other hand, by (126), we have
−
−𝜓2 (𝜑) ≤ ∫ 2𝑚𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑎 − 𝑠) 𝑑𝑠
(𝑞+1)/2
),
(129)
2𝑚 (1 − 𝑎)
(𝑞+1)/2
(𝑞+1)/2
−𝜓 (𝜑) ≥ √
+ 𝑜 ((1 − 𝜑)
).
(1 − 𝜑)
𝑞+1
(123)
Consequently,
and, hence,
𝜓2 (𝜑) ≤
𝑞
𝑚 (1 − 𝑎) (1 − 𝜑)
(𝑞+1)/2
√2𝑚 (1−𝑎) / (𝑞+1)(1−𝜑)
= 𝑐+ √
2𝑚 (1 − 𝑎)
4𝑐
(𝑞+3)/2
√
(1 − 𝜑)
𝑞+3
𝑞+1
2𝑚 (1 − 𝑎)
𝑞+1
𝑞+1
+
(1 − 𝜑) + 𝑜 ((1 − 𝜑) ) .
𝑞+1
𝑚 (1 − 𝑎) (1 − 𝜑)
𝑑𝜓
≤𝑐−
𝑑𝜑
𝜓
≤ 𝑐+
(𝑞+1)/2
which implies that
Thus, 𝜓1 (𝜑) < 𝜓2 (𝜑) in a left neighborhood of 𝜑 = 1.
(iii) Consider the case 0 < 𝑞 < 1. Notice that
=
2𝑚 (1 − 𝑎)
(𝑞+1)/2
(1 − 𝜑)
𝑞+1
+ 2𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎) + 𝑜 ((1 − 𝜑)
(119)
By the comparison lemma, 𝑢2 ≤ 𝜓 ≤ 𝑢1 , which implies
2𝑚 (1 − 𝑎)
(𝑞+1)/2
(𝑞+1)/2
+ 𝑜 ((1 − 𝜑)
).
(1 − 𝜑)
𝑞+1
(126)
𝑞
(130)
Summing up, we arrive at
(𝑞+1)/2
+𝑜 ((1−𝜑)
)
𝑚 (1 − 𝑎) (𝑞 + 1)
(𝑞−1)/2
(𝑞−1)/2
+𝑜 ((1−𝜑)
).
(1−𝜑)
2
(124)
𝜓2 (𝜑) =
2𝑚 (1 − 𝑎)
4𝑐
(𝑞+3)/2
√
(1 − 𝜑)
𝑞+3
𝑞+1
+
2𝑚 (1 − 𝑎)
𝑞+1
𝑞+1
(1 − 𝜑) + 𝑜 ((1 − 𝜑) ) ,
𝑞+1
(131)
as 𝜑 󳨀→ 1− ,
That is,
−𝜓 (𝜑) ≤ 𝑐 (1 − 𝜑) + √
2𝑚 (1 − 𝑎)
(𝑞+1)/2
(1 − 𝜑)
𝑞+1
(𝑞+1)/2
+ 𝑜 ((1 − 𝜑)
).
(125)
which implies that 𝜓1 (𝜑) < 𝜓2 (𝜑) in a left neighborhood of
𝜑 = 1.
The proof for the claims that 𝜓1 (𝜑) < 𝜓2 (𝜑) for any 𝜑 ∈
(𝛼2 , 1) and 𝜓1 (𝛼2 ) < 0 is similar to the proof of Lemma 3 and
omitted here.
Abstract and Applied Analysis
13
Similar to Section 2, we denote level set 𝑅𝑘 by
Thus,
𝑚+𝑝−1
𝑚𝜑0
𝑅𝑘
1
1
= {(𝜑, 𝜓) ∈ R2 ; 𝜓2 − 𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠 = 𝑘}
2
𝜑
(132)
for any 𝑘 ≥ 0, and correspondingly, define
𝑅𝑘− = 𝑅𝑘 ∩ {𝜓 < 0} .
(133)
Notice that if (𝜑, 𝜓) ∈ 𝑅𝑘− solves system (12), then
𝑐 1−𝑚 2
𝑞
𝜑 𝜓 + 𝜓𝜑𝑝 (1 − 𝜑) (𝜑 − 𝜑𝑐𝜏 )
𝑚
𝑐
≥ 𝜑1−𝑚 𝜓2
𝑚
(134)
0
(135)
we know that 𝑅𝑘2 passes through the critical point (0, 0).
Denote that
𝜑
𝜓̃∗ (𝜑) = −√ −2𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠.
̃ = max {
𝑀
𝜑∈(0,𝑎)
𝑞
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜑)
}.
1−𝑎
(142)
̃ (141) does not hold, and 𝜓𝑐 (𝜑) is inThen, for any 𝑐 ≥ √𝑀,
creasing on (0, 𝑎). Therefore, 𝜓𝑐 (𝜑) must intersect with 𝜓̃∗ (𝜑)
̃ The proof is completed.
for any 𝑐 ≥ √𝑀.
Theorem 14. (i) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≥ 0, then there
is no nontrivial nonpositive solution for the problem (23).
1
(ii) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 < 0, then there exists a
unique 𝑐2∗ > 0, such that the problem (23) admits a nonpositive
solution 𝜓(𝜑), and 𝜓(𝜑) < 0 for any 𝜑 ∈ (0, 1).
𝑐2∗
since 𝜑(𝜉) is decreasing in 𝜉. This implies that the trajectory
(𝜑, 𝜓) of (106) wanders through increasing level sets with
increasing 𝜉. See Figure 4(b). Letting
1
Denote that
Proof. The proof is similar to that of Theorem 5, and
> 0,
𝑘2 = −𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎) 𝑑𝑠,
(141)
1
𝑑𝑘 (𝜉)
𝑞
= 𝜓𝜓󸀠 + 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎) 𝜑󸀠
𝑑𝜉
=
𝑞
(1 − 𝜑0 ) (𝜑0 − 𝑎) < (𝑎 − 1) 𝑐2 .
= inf {𝑐 > 0; 𝜓𝑐 intersects 𝜓̃∗ at (𝜂𝑐 , 𝜙̃𝑐 ) with 𝜂𝑐 ∈ [0, 1)} .
(143)
Proposition 15. 𝜑(𝜉) is a monotone decreasing sharp- or
smooth-type traveling wave front of the problem (10)-(11) for
some fixed 𝑐 > 0, if and only if 𝜓(𝜑) with 𝜓(𝜑) < 0 for any
𝜑 ∈ (0, 1) is a solution of the problem (23).
Proof. The proof is similar to that of Proposition 6.
(136)
Theorem 14 and Proposition 15 imply the following result.
Lemma 13. The trajectory 𝜓𝑐 (𝜑) of the problem (106) must
intersect with 𝜓̃∗ (𝜑) for sufficiently large 𝑐 > 0. (See
Figure 4(c).)
Theorem 16. (i) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 ≥ 0, then there
is no decreasing traveling wave front for the problem (10)-(11).
1
(ii) If ∫0 𝑠𝑚+𝑝−1 (1 − 𝑠)𝑞 (𝑠 − 𝑎)𝑑𝑠 < 0, then there is a unique
wave speed 𝑐2∗ > 0, such that the problem (10)-(11) admits a
decreasing traveling wave front.
0
1
We introduce the following lemma.
Proof. For any 𝜑 ∈ [𝑎, 1), we have 𝜑𝑐𝜏 ∈ [𝑎, 1) and
𝜓𝑐󸀠 (𝜑) ≥ 𝑐 > 0.
Furthermore, we have the following results.
(137)
Let 𝜑0 ∈ (0, 𝑎) be the first point such that 𝜓𝑐󸀠 (𝜑0 ) = 0. Then,
we have
𝑚+𝑝−1
𝜓𝑐 (𝜑0 ) =
𝑚𝜑0
𝑞
(1 − 𝜑0 ) (𝜑0𝑐𝜏 − 𝑎)
.
𝑐
Theorem 17. The traveling wave front 𝜑(𝜉) of the problem (10)(11) corresponding to the wave speed 𝑐2∗ obtained earlier is of
smooth type.
Proof. It is easy to see that
𝑑𝜓
≤ 0,
𝑑𝜑
(138)
Since
as 𝜑 󳨀→ 0+ .
(144)
Thus,
𝜓𝑐 (𝜑0 ) < 𝜓𝑐 (𝑎) ≤ (𝑎 − 1) 𝑐,
(139)
we have
𝑚+𝑝−1
𝑚𝜑0
𝑞
(1 − 𝜑0 ) (𝜑0𝑐𝜏 − 𝑎) < (𝑎 − 1) 𝑐2 .
(140)
𝑞
𝜓≥
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐2∗ 𝜏 − 𝑎)
𝑚𝑎
≥ − ∗ 𝜑𝑚+𝑝−1 .
𝑐2
𝑐2∗
(145)
14
Abstract and Applied Analysis
Recalling that
𝜑󸀠 (𝜉) =
we know that
−
1 1−𝑚
𝜑
(𝜉) 𝜓 (𝜉) ,
𝑚
(146)
It is easy to prove that 𝜓(𝜑) ≥ 𝑢(𝜑) when 𝜑𝑐1∗ 𝜏 ≥ (1 + 𝑎)/2.
Now, we construct a function 𝑢 satisfying (152). Let
𝜃
𝑢 = 𝐴(1 − 𝜑) .
𝑎 𝑝
𝜑 (𝜉) ≤ 𝜑󸀠 (𝜉) ≤ 0,
𝑐2∗
(147)
Then, 𝑢 satisfies (152) if and only if
𝐴2 𝜃(1 − 𝜑)
which means that
𝜑󸀠 (𝜉̃𝑏− ) = 0.
(148)
2𝜃−1−𝑞
+ 𝑐1∗ 𝐴(1 − 𝜑)
𝜃−𝑞
𝑚 (1 − 𝑎) 1 + 𝑎 𝑚+𝑝−1
≤
.
(
)
2
2
The proof is completed.
Theorem 18. 𝑐2∗ (𝜏) is nonincreasing in delay 𝜏; namely, if 𝜏1 >
𝜏2 , then 𝑐2∗ (𝜏1 ) ≤ 𝑐2∗ (𝜏2 ).
(153)
(154)
Take
𝜃 = max {
Proof. The proof is similar to that of Theorem 9.
1+𝑞
, 𝑞} .
2
(155)
Then, (154) holds when 𝐴 is appropriately small. Thus,
𝜃
𝜓 (𝜑) ≥ 𝐴(1 − 𝜑) ,
4. Asymptotic Behavior of the Traveling
Wave Solutions
In this section, we first pay our attention to the finiteness of
𝜉𝑎 , 𝜉𝑏 , 𝜉̃𝑎 , 𝜉̃𝑏 . On the basis of this, the convergent rates of 𝜑
going to the steady states at far-field are discussed.
with 𝜃 = max{(1 + 𝑞)/2, 𝑞} and 𝐴 appropriately small. Since
𝜑(𝜉) is the solution of the problem (8)-(9), there exists 𝜑0 < 1
with 1 − 𝜑0 small enough such that when 𝜑0 < 𝜑 < 1,
𝜉 (𝜑) − 𝜉 (𝜑0 ) = ∫
Theorem 19. Let 𝜑(𝜉) be the increasing traveling wave solution
of the problem (8)-(9) corresponding to the unique wave speed
𝑐1∗ .
(i) If 0 < 𝑞 < 1, then 𝜉𝑏 < +∞.
(ii) If 𝑞 ≥ 1, then 𝜉𝑏 = +∞. More precisely,
𝜑0
∗2 +4𝑚(1−𝑎)−𝑐∗ )/2𝑚)𝜉
1
𝑑𝜓
≤ 0,
𝑑𝜑
as 𝜉 󳨀→ +∞;
(149)
,
as 𝜑 󳨀→ 1− .
(158)
𝑞
as 𝜉 󳨀→ +∞.
𝜓≤
(150)
Proof. (i) For (1 + 𝑎)/2 ≤ 𝜑𝑐1∗ 𝜏 < 1, we have (1 + 𝑎)/2 ≤ 𝜑𝑐1∗ 𝜏 ≤
𝜑 ≤ 1, and
1 𝑑𝜓2
𝑞
− 𝑐1∗ 𝜓 = −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐1∗ 𝜏 − 𝑎)
2 𝑑𝜑
𝑚 (1 − 𝑎) 1 + 𝑎 𝑚+𝑝−1
𝑞
(1 − 𝜑) .
≤−
(
)
2
2
Consider the following inequality problem:
for 𝜑 ≤ 1,
𝑢 (1− ) = 0,
for 𝜑 < 1.
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐1∗ 𝜏 − 𝑎)
𝑐1∗
𝑞
≤
𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑 − 𝑎)
,
𝑐1∗
𝑚𝑠𝑚−1
𝜉 (𝜑) − 𝜉 (𝜑0 ) = ∫
𝑑𝑠
𝜑0 𝜓 (𝑠)
𝜑
(151)
𝜑
≥∫
𝜑0
𝑚 (1 − 𝑎) 1 + 𝑎 𝑚+𝑝−1
1 𝑑𝑢2
𝑞
(1 − 𝜑) ,
− 𝑐1∗ 𝑢 ≥ −
(
)
2 𝑑𝜑
2
2
𝑢 (𝜑) > 0,
(157)
Thus,
(b) when 𝑞 = 1, one has
1 − 𝜑 (𝜉) ∼ 𝑒−((√𝑐1
𝑚𝑠𝑚−1
𝑑𝑠
𝜓 (𝑠)
𝑚 𝜑 𝑠𝑚−1
𝑑𝑠.
≤ ∫
𝐴 𝜑0 (1 − 𝑠)𝜃
1/(1−𝑞)
,
𝜑
When 0 < 𝑞 < 1, 𝜉(𝜑) < +∞ as 𝜑 → 1− . Thus, 𝜉𝑏 < +∞.
(ii) It is easy to see that
(a) when 𝑞 > 1, one has
(1 − 𝑎) (𝑞 − 1)
𝜉)
1 − 𝜑 (𝜉) ∼ (
𝑐1∗
(156)
(159)
𝑐1∗
𝑑𝑠.
𝑠𝑝 (1 − 𝑠)𝑞 (𝑠 − 𝑎)
Letting 𝜑 → 1, we obtain
lim 𝜉 (𝜑) = +∞;
(160)
𝜉𝑏 = +∞.
(161)
𝜑 → 1−
(152)
that is,
Abstract and Applied Analysis
15
For any 𝜀 > 0, when 𝜑 approaches 1 enough, we have
𝜑𝑐1∗ 𝜏 > 1 − 𝜀. Consider the following two problems:
2
1 𝑑𝑢
𝑞
− 𝑐1∗ 𝑢 ≤ −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (1 − 𝑎) ,
2 𝑑𝜑
for 𝜑 ≤ 1,
(162)
(163)
𝑐1
(𝑚 (1 − 𝑎))2/(𝑞−1) 𝑞2/(𝑞−1)
𝜃
𝑢 = 𝐴(1 − 𝜑) .
(164)
̃=
𝐴
𝜃−𝑞
2
𝑐1∗
𝑞
(1 − 𝜑)
(175)
𝑚 (1 − 𝑎)
𝑞
≤ 𝜓 (𝜑) ≤
(1 − 𝜑) .
𝑐1∗
Noticing that
𝑚 (1 − 𝑎)
𝐴=
.
𝑐1∗
(166)
𝜉 (𝜑) = 𝜉 (𝜑0 ) + ∫
𝜑
𝜑0
When 𝑞 = 1, take
𝑚𝑠𝑚−1
𝑑𝑠,
𝜓 (𝑠)
(176)
we have
𝜃 = 1,
+ 4𝑚 (1 − 𝑎) −
2
𝑐1∗
(167)
.
𝜃̃
̃ − 𝜑) .
V = 𝐴(1
𝑐1∗
(1 − 𝑎) (𝑞 − 1)
𝑞−1
≤ lim− 𝜉 (𝜑) (1 − 𝜑)
Then, 𝑢 is a solution of problem (162).
On the other hand, let
(177)
𝜑→1
(168)
Then, V satisfies (163) if and only if
̃
(174)
.
Then, (169) is ensured.
Summing up, for any 𝜑 < 1 with 𝜑𝑐1∗ 𝜏 > 1 − 𝜀, when 𝑞 > 1,
≥ 𝑚 (1 − 𝑎) 𝜑𝑚+𝑝−1 .
(165)
When 𝑞 > 1, take
𝐴=
(173)
√𝑐1∗2 + 4𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 − 𝑐1∗
𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 (1 − 𝜀(𝑞−1)/2 )
Then, 𝑢 satisfies (162) if and only if
√𝑐1∗2
},
(169) holds. When 𝑞 = 1, take 𝜃̃ = 1, and
It is easy to prove that V(𝜑) ≤ 𝜓(𝜑) ≤ 𝑢(𝜑) when 𝜑 < 1 with
𝜑𝑐1∗ 𝜏 > 1 − 𝜀. In what follows, we construct two functions 𝑢
and V satisfying (162) and (163), respectively. Let
𝜃 = 𝑞,
(172)
∗4/(𝑞−1)
for 𝜑 < 1.
+ 𝑐1∗ 𝐴(1 − 𝜑)
.
𝑐1∗
𝜀 ≤ min {1 − 𝑎,
V (1− ) = 0,
2𝜃−1−𝑞
𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 (1 − 𝜀(𝑞−1)/2 )
Then, when
1 𝑑V2
𝑞
− 𝑐1∗ V ≥ −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (1 − 𝜀 − 𝑎) ,
2 𝑑𝜑
for 𝜑 ≤ 1,
(171)
Take
̃=
𝐴
𝑢 (𝜑) > 0 for 𝜑 < 1,
𝐴2 𝜃(1 − 𝜑)
̃2 𝑞𝜀(𝑞−1)/2 ≤ 𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 ,
𝐴
̃ ≤ 𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 (1 − 𝜀(𝑞−1)/2 ) .
𝑐1∗ 𝐴
𝑢 (1− ) = 0,
V (𝜑) > 0
which holds if
̃
̃ − 𝜑)2𝜃−1−𝑞 + 𝑐∗ 𝐴(1
̃ − 𝜑)𝜃−𝑞 ≤ 𝑚 (1 − 𝜀 − 𝑎) 𝜑𝑚+𝑝−1 .
̃2 𝜃(1
𝐴
1
(169)
When 𝑞 > 1, take 𝜃̃ = 𝑞; then, (169) is ensured by
≤
(1 − 𝜀 − 𝑎) (1 − 𝜀)
(1 − 𝜀(𝑞−1)/2 ) (𝑞 − 1)
;
that is,
1/(1−𝑞)
(1 − 𝑎) (𝑞 − 1)
)
(
𝑐1∗
≤ lim (1 − 𝜑 (𝜉)) 𝜉1/(𝑞−1)
𝜉 → +∞
̃2 𝑞𝜀𝑞−1 + 𝑐∗ 𝐴
̃
𝐴
1
≤ 𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 (1 − 𝜀(𝑞−1)/2 + 𝜀(𝑞−1)/2 ) ,
(170)
𝑐1∗
𝑚+𝑝−1
≤(
(1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 (1 − 𝜀(𝑞−1)/2 ) (𝑞 − 1)
𝑐1∗
1/(1−𝑞)
)
.
(178)
16
Abstract and Applied Analysis
When 𝑞 = 1,
√𝑐1∗2 + 4𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 − 𝑐1∗
2
(c) when 1 < 𝑚 + 𝑝 < 2, noticing that 𝑚 ≤
min{𝑝, 1}, then 𝑚 ≤ 𝑝; if 𝑚 = 𝑝, one has
≤ 𝜓 (𝜑)
≤
𝜑 (𝜉) ∼ 𝑒(√a/𝑚)𝜉 ,
(1 − 𝜑)
2
2/(𝑚−𝑝)
(1 − 𝜑) .
𝑚 − 𝑝 2𝑚𝑎
√
𝜑 (𝜉) ∼ (
𝜉)
2𝑚
𝑚+𝑝
A direct calculation gives
−
≤−
𝜉 (𝜑)
ln (1 − 𝜑)
𝜉 (𝜑) = 𝜉 (𝜑0 ) − ∫
√𝑐1∗2 + 4𝑚 (1 − 𝜀 − 𝑎) (1 − 𝜀)𝑚+𝑝−1 − 𝑐1∗
;
𝜓 (𝜑) = 𝐴𝜑𝛾 + 𝑜 (𝜑𝛾 ) ,
−((√𝑐1∗2 +4𝑚(1−𝑎)−𝑐1∗ )/2𝑚)𝜉
𝑒
≤ 1 − 𝜑 (𝜉)
(181)
√𝑐1∗2 +4𝑚(1−𝜀−𝑎)(1−𝜀)𝑚+𝑝−1 −𝑐1∗ )/2𝑚)𝜉
𝜓 (𝜑) = 𝑐1∗ 𝜑 + 𝑜 (𝜑) .
(i) If 𝑚 > min{𝑝, 1}, then 𝜉𝑎 > −∞.
𝜑→0
(182)
𝜓 (𝜑) =
as 𝜉 󳨀→ −∞;
(183)
(184)
as 𝜉 󳨀→ −∞,
and if 𝑚 < 1, one has
𝜑 (𝜉) ∼ (
(𝑚 − 1) (√𝑐1∗2 + 4𝑚𝑎 + 𝑐1∗ )
2𝑚
1/(𝑚−1)
𝜉)
,
as 𝜉 󳨀→ −∞;
√𝑐1∗2 + 4𝑚𝑎 + 𝑐
2
𝜑 + 𝑜 (𝜑) .
(193)
If 𝑚 = 1, we have
(b) when 𝑚 + 𝑝 = 2, if 𝑚 = 1, one has
,
(192)
When 𝑚 + 𝑝 = 2,
as 𝜉 󳨀→ −∞,
1/(𝑚−1)
∗2 +4𝑎+𝑐∗ )/2)𝜉
1
𝜉 (𝜑)
𝑚
=
.
𝜑𝑚−1 (𝑚 − 1) 𝑐1∗
If 𝑚 < 1,
and if 𝑚 < 1, one has
𝜑 (𝜉) ∼ 𝑒((√𝑐1
(191)
𝜑→0
lim+
(a) when 𝑚 + 𝑝 > 2, if 𝑚 = 1, one has
,
(190)
𝜉 (𝜑)
1
= ∗.
ln 𝜑
𝑐1
lim+
(ii) If 𝑚 ≤ min{𝑝, 1}, then 𝜉𝑎 = −∞. More precisely,
(𝑚 − 1) 𝑐1∗
𝜉)
𝑚
(189)
If 𝑚 = 1, we have
Theorem 20. Let 𝜑(𝜉) be the increasing traveling wave solution of the problem (8)-(9) corresponding to the unique wave
speed 𝑐1∗ .
∗
(188)
where 𝛾 = 1 for 𝑚 + 𝑝 ≥ 2, and 𝛾 = (𝑚 + 𝑝)/2 for 1 < 𝑚 + 𝑝 <
2. It is clear that 𝜉(𝜑) → −∞ as 𝜑 → 0+ if 𝛾 ≥ 𝑚, and
𝜉(𝜑) is finite as 𝜑 → 0+ if 𝛾 < 𝑚. That is, if 𝑚 > min{𝑝, 1},
then 𝜉𝑎 > −∞; however, if 𝑚 ≤ min{𝑝, 1}, then 𝜉𝑎 = −∞.
Furthermore, notice that as 𝜑 → 0+ , when 𝑚 + 𝑝 > 2,
.
By the arbitrariness of 𝜀 > 0, (149)-(150) hold.
𝜑 (𝜉) ∼ 𝑒𝑐1 𝜉 ,
𝑚𝑠𝑚−1
𝑑𝑠.
𝜓 (𝑠)
From the proof of Lemma 3, we see that when 𝜑 > 0 is
sufficiently small,
that is,
≤ 𝑒−((
𝜑0
𝜑
(180)
2𝑚
𝜑 (𝜉) ∼ (
as 𝜉 󳨀→ −∞.
Proof. For any 𝜑0 ∈ (0, 𝑎), we see that
√𝑐1∗2 + 4𝑚 (1 − 𝑎) − 𝑐1∗
𝜑→1
,
(187)
2𝑚
≤ lim−
(186)
and if 𝑚 < 𝑝, one has
(179)
√𝑐1∗2 + 4𝑚 (1 − 𝑎) − 𝑐1∗
as 𝜉 󳨀→ −∞,
∗
∗2
𝜉 (𝜑) √𝑐1 + 4𝑎 − 𝑐1
lim
=
.
𝜑 → 0+ ln 𝜑
2𝑎
(194)
∗
∗2
𝜉 (𝜑) √𝑐1 + 4𝑚𝑎 − 𝑐1
lim
=
.
𝜑 → 0+ 𝜑𝑚−1
2𝑎 (𝑚 − 1)
(195)
If 𝑚 < 1,
When 1 < 𝑚 + 𝑝 < 2,
(185)
𝜓 (𝜑) = √
2𝑚𝑎 (𝑚+𝑝)/2
+ 𝑜 (𝜑(𝑚+𝑝)/2 ) .
𝜑
𝑚+𝑝
(196)
Abstract and Applied Analysis
17
If 𝑚 = 𝑝, we have
lim+
𝜑→0
𝜉 (𝜑)
𝑚
=
.
√𝑎
ln 𝜑
(197)
(i) If 𝑝 < min{𝑚, 1}, then 𝜉̃𝑏 < +∞.
(ii) If 𝑝 ≥ min{𝑚, 1}, then 𝜉̃𝑏 = +∞. More precisely,
If 𝑚 < 𝑝,
𝜉 (𝜑)
lim+
𝜑 → 0 𝜑(𝑚−𝑝)/2
=
Theorem 22. Let 𝜑(𝜉) be the decreasing traveling wave solution of the problem (10)-(11) corresponding to the unique wave
speed 𝑐2∗ .
𝑚+𝑝
2𝑚
√
.
𝑚 − 𝑝 2𝑚𝑎
(198)
(a) when 𝑚 + 𝑝 > 2, if 𝑝 > 1, one has
𝜑 (𝜉) ∼ (
By a simple calculation, (182)–(187) hold.
Theorem 21. Let 𝜑(𝜉) be the decreasing traveling wave solution
of the problem (10)-(11) corresponding to the unique wave speed
𝑐2∗ .
(i) If 0 < 𝑞 < 1, then 𝜉̃𝑎 > −∞.
(ii) If 𝑞 ≥ 1, then 𝜉̃ = −∞. More precisely,
,
as 𝜉 󳨀→ +∞,
(205)
and if 𝑝 = 1, one has
∗
𝜑 (𝜉) ∼ 𝑒−(𝑎/𝑐2 )𝜉 ,
(a) when 𝑞 > 1, one has
∗
1/(1−𝑝)
as 𝜉 󳨀→ +∞;
(206)
(b) when 𝑚 + 𝑝 = 2, if 𝑝 > 1, one has
𝑎
1 − 𝜑 (𝜉) ∼ 𝑒(𝑐2 /𝑚)𝜉 ,
𝑎 (𝑝 − 1)
𝜉)
𝑐2∗
as 𝜉 󳨀→ −∞;
(199)
𝜑 (𝜉) ∼ (
(1 − 𝑚) (√𝑐2∗2 + 4𝑚𝑎 − 𝑐2∗ )
2𝑚
(b) when 𝑞 = 1, one has
1/(𝑚−1)
𝜉)
,
(207)
as 𝜉 󳨀→ +∞,
∗2 +4𝑚(1−𝑎)+𝑐∗ )/2𝑚)𝜉
2
1 − 𝜑 (𝜉) ∼ 𝑒((√𝑐2
,
as 𝜉 󳨀→ −∞.
(200)
∗2 +4𝑎−𝑐∗ )/2)𝜉
2
𝜑 (𝜉) ∼ 𝑒−((√𝑐2
Proof. Let 𝜑0 approach 1 enough. Then, we have
𝜑0
𝜉 (𝜑0 ) − 𝜉 (𝜑) = ∫
𝜑
and if 𝑝 = 1, one has
𝑚𝑠𝑚−1
𝑑𝑠.
𝜓 (𝑠)
(201)
,
as 𝜉 󳨀→ +∞;
(208)
(c) when 1 < 𝑚 + 𝑝 < 2, noticing that 𝑝 ≥
min{𝑚, 1}, then 𝑝 ≥ 𝑚; if 𝑝 > 𝑚, one has
2/(𝑚−𝑝)
From the proof of Lemma 12, we see that
𝜑 (𝜉) ∼ (
𝛾
𝜓 (𝜑) ∼ 𝐴(1 − 𝜑) ,
(202)
𝑝 − 𝑚 2𝑚𝑎
√
𝜉)
2𝑚
𝑚+𝑝
,
as 𝜉 󳨀→ +∞,
(209)
−
as 𝜑 → 1 , where 𝛾 = 1 for 𝑞 ≥ 1, and 𝛾 = (𝑞 + 1)/2 for
0 < 𝑞 < 1. It is clear that 𝜉(𝜑) → −∞ as 𝜑 → 1− if 𝛾 ≥ 1,
and 𝜉(𝜑) is finite as 𝜑 → 1− if 𝛾 < 1. That is, if 0 < 𝑞 < 1, then
𝜉̃𝑎 > −∞; however, if 𝑞 ≥ 1, then 𝜉̃𝑎 = −∞. Furthermore,
notice that as 𝜑 → 1− , when 𝑞 > 1,
𝜑→1
(203)
𝜉 (𝜑)
𝑚
= .
ln (1 − 𝜑) 𝑐2∗
When 𝑞 = 1,
𝜓 (𝜑) =
√𝑐2∗2 + 4𝑚 (1 − 𝑎) + 𝑐2∗
2
as 𝜉 󳨀→ +∞.
1 𝑑𝜓2
𝑞
− 𝑐2∗ 𝜓 = −𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝜑𝑐2∗ 𝜏 − 𝑎)
2 𝑑𝜑
𝑚𝑎
𝑎 𝑞
≥
(1 − ) 𝜑𝑚+𝑝−1 .
2
2
(𝜑 − 1) + 𝑜 (𝜑 − 1) ,
(210)
(204)
(211)
Consider the following inequality problem:
1 𝑑𝑢2
𝑚𝑎
𝑎 𝑞
− 𝑐2∗ 𝑢 ≤
(1 − ) 𝜑𝑚+𝑝−1 ,
2 𝑑𝜑
2
2
𝜉 (𝜑)
2𝑚
lim−
.
=
𝜑 → 1 ln (1 − 𝜑)
√𝑐2∗2 + 4𝑚 (1 − 𝑎) + 𝑐2∗
This yields (199)-(200).
𝜑 (𝜉) ∼ 𝑒−(√𝑎/𝑚)𝜉 ,
Proof. (i) For 0 < 𝜑𝑐2∗ 𝜏 ≤ 𝑎/2, we have 0 ≤ 𝜑 ≤ 𝜑𝑐2∗ 𝜏 ≤ 𝑎/2,
and
𝜓 (𝜑) = 𝑐2∗ (𝜑 − 1) + 𝑜 (𝜑 − 1) ,
lim−
and if 𝑝 = 𝑚, one has
𝑢 (0+ ) = 0,
𝑢 (𝜑) < 0,
for 𝜑 > 0.
for 𝜑 ≥ 0,
(212)
18
Abstract and Applied Analysis
It is easy to prove that 𝜓(𝜑) ≤ 𝑢(𝜑) when 𝜑𝑐2∗ 𝜏 ≤ 𝑎/2. Now, we
construct a function 𝑢 satisfying (212). Let
𝑢 = −𝐵𝜑𝑟 .
For any 0 < 𝜀 < 𝑎, when 𝜑 approaches 0 enough, we have
𝜑𝑐2∗ 𝜏 < 𝜀. Consider the following problem:
1 𝑑𝑢2
𝑞
− 𝑐2∗ 𝑢 ≤ 𝑚𝜑𝑚+𝑝−1 (1 − 𝜑) (𝑎 − 𝜀) ,
2 𝑑𝜑
(213)
Then, 𝑢 satisfies (212) if and only if
𝐵2 𝑟𝜑2𝑟−𝑚−𝑝 + 𝑐2∗ 𝐵𝜑𝑟−𝑚−𝑝+1
𝑚𝑎
𝑎 𝑞
≤
(1 − ) .
2
2
𝑢 (0+ ) = 0,
(214)
𝑢 (𝜑) < 0,
(215)
It is easy to prove that 𝜓(𝜑) ≤ 𝑢(𝜑) when 𝜑 > 0 with 𝜑𝑐2∗ 𝜏 < 𝜀.
In what follows, we construct a function 𝑢 satisfying (223).
Let
𝑢 = −𝐵𝜑𝑟 .
Then, (212) holds when 𝐵 is appropriately small. Thus,
𝜓 (𝜑) ≤ −𝐵𝜑𝑟 ,
(216)
with 𝑟 = max{(𝑚 + 𝑝)/2, 𝑚 + 𝑝 − 1} and 𝐵 appropriately small.
Since 𝜑(𝜉) is the solution of the problem (10)-(11), there exists
𝜑0 > 0 sufficiently small such that
𝜉 (𝜑) − 𝜉 (𝜑0 ) = − ∫
𝜑0
𝜑
𝑚𝑠𝑚−1
𝑑𝑠
𝜓 (𝑠)
𝜑0
1
𝑚
𝑑𝑠.
∫
𝐵 𝜑 𝑠𝜃−𝑚+1
≤
𝜑
𝐵2 𝑟𝜑2𝑟−𝑚−𝑝 + 𝑐2∗ 𝐵𝜑𝑟−𝑚−𝑝+1 ≤ 𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) .
(217)
(𝑚 + 𝑝 − 1) 𝐵2 𝑞𝜀(𝑚+𝑝−2)/2 ≤ 𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) .
𝐵=
𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(𝑚+𝑝−2)/2 )
𝑐2∗
Then, when 𝜀 < (𝑐2∗2 /𝑚(𝑚 + 𝑝 − 1))
Recalling (219), we obtain
𝑚𝑎
− ∗ 𝜑𝑚+𝑝−1
𝑐2
0
2𝑚𝑎 (𝑚+𝑝)/2
.
𝜑
𝑚+𝑝
.
2/(𝑚+𝑝−2)
(227)
, (225) holds.
≤ 𝜓 (𝜑)
On the other hand, from the proof of Theorem 17, we also
note that
(228)
𝑞
(218)
≤−
𝑚(1 − 𝜀) (𝑎 − 𝜀) (1 − 𝜀
(𝑚+𝑝−2)/2
)
𝑐2∗
𝜑𝑚+𝑝−1
when 𝜑𝑐2∗ 𝜏 < 𝜀. When 𝑝 > 1, we have
𝑐2∗
(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(𝑚+𝑝−2)/2 ) (𝑝 − 1)
(219)
≤ lim+ 𝜉 (𝜑) 𝜑𝑝−1
By (218)-(219) and
𝜑
𝜉 (𝜑) − 𝜉 (𝜑0 ) = ∫
𝜑0
(229)
𝜑→0
𝑚−1
𝑚𝑠
𝑑𝑠,
𝜓 (𝑠)
(226)
Take
𝑞
𝑚𝑎
𝜓 (𝜑) ≥ − ∗ 𝜑𝑚+𝑝−1 .
𝑐2
(225)
When 𝑚 + 𝑝 > 2, take 𝑟 = 𝑚 + 𝑝 − 1. Then, (225) is ensured
by the following inequalities:
𝑐2∗ 𝐵 ≤ 𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(𝑚+𝑝−2)/2 ) ,
𝜓 (𝜑) ≥ 𝜓̃ (𝜑) = −√ 2𝑚 ∫ 𝑠𝑚+𝑝−1 (1 − 𝑠) (𝑎 − 𝑠) 𝑑𝑠
≥ −√
(224)
Noticing that 0 < 𝜑 ≤ 𝜑𝑐2∗ 𝜏 < 𝜀, then 𝑢 satisfies (223) if
When 𝑝 < min{𝑚, 1}, 𝜉(𝜑) < +∞ as 𝜑 → 0+ . Thus, 𝜉̃𝑏 <
+∞.
(ii) From the proof of Theorem 14, we know that
∗
for 𝜑 > 0.
(223)
Take
𝑚+𝑝
, 𝑚 + 𝑝 − 1} .
𝑟 = max {
2
for 𝜑 ≥ 0,
(220)
it is easy to see that
≤
𝑐2∗
;
𝑎 (𝑝 − 1)
that is,
lim 𝜉 (𝜑) = +∞,
𝜑 → 0+
(221)
(
(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(𝑚+𝑝−2)/2 ) (𝑝 − 1)
𝑐2∗
if 𝑝 ≥ min{𝑚, 1}. That is,
𝜉̃𝑏 = +∞.
(222)
≤ lim 𝜑 (𝜉) 𝜉1/(𝑝−1)
𝜉 → +∞
1/(1−𝑝)
)
(230)
1/(1−𝑝)
𝑎 (𝑝 − 1)
≤(
)
𝑐2∗
.
Abstract and Applied Analysis
19
When 𝑝 = 1, we have
−
𝑞
𝑐2∗
(1 − 𝜀) (𝑎 − 𝜀) (1 −
𝜀(𝑚+𝑝−2)/2 )
≤ lim+
𝜑→0
𝑐2∗
𝜉 (𝜑)
≤− ;
ln 𝜑
𝑎
(231)
Then, (225) holds. On the other hand, consider the following
problem:
1 𝑑V2
𝑞
− 𝑐2∗ V ≥ 𝑚𝑎𝜑(1 − 𝜑) ,
2 𝑑𝜑
∗
𝑞
(𝑎−𝜀)(1−𝜀(𝑚+𝑝−2)/2 )/𝑐2∗ )𝜉
.
(232)
By the arbitrariness of 𝜀 > 0, (205)-(206) hold.
When 1 < 𝑚 + 𝑝 < 2, take 𝑟 = (𝑚 + 𝑝)/2. Then, (225)
holds if
𝑚+𝑝 2
𝐵 ≤ 𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(2−𝑚−𝑝)/4 ) ,
2
(233)
𝑐2∗ 𝐵𝜀(2−𝑚−𝑝)/4 ≤ 𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) .
V (𝜑) < 0,
V=−
(234)
𝑚 (𝑚 + 𝑝) ∗−2
𝑐2 )
2
√𝑐2∗2 + 4𝑚𝑎 − 𝑐2∗
2
≤−
(241)
𝜑.
𝜑
(242)
√𝑐2∗2 + 4𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) − 𝑐2∗
2
𝜑.
When 𝑚 < 1, that is, 𝑝 > 1,
(235)
2/(2−𝑚−𝑝)
×(
2
≤ 𝜓 (𝜑)
Then, when
𝑎 𝑎
𝑎 𝑞 2/(2−𝑚−𝑝)
𝜀 ≤ min { , ( (1 − ) )
2 2
2
√𝑐2∗2 + 4𝑚𝑎 − 𝑐2∗
Then, V satisfies (240). Thus, we conclude that
−
2𝑚
(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(2−𝑚−𝑝)/4 ).
𝑚+𝑝
for 𝜑 > 0.
We can see that 𝜓(𝜑) ≥ V(𝜑). Let
Take
𝐵=√
(240)
V (0+ ) = 0,
that is,
𝑒−(𝑎/𝑐2 )𝜉 ≤ 𝜑 (𝜉) ≤ 𝑒−((1−𝜀)
for 𝜑 ≥ 0,
},
(
(1 − 𝑚) (√𝑐2∗2 + 4𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) − 𝑐2∗ )
2𝑚
1/(𝑚−1)
)
(225) holds. Combining with (218), we obtain
≤ lim 𝜑 (𝜉) 𝜉1/(1−𝑚)
2𝑚𝑎 (𝑚+𝑝)/2
−√
≤ 𝜓 (𝜑)
𝜑
𝑚+𝑝
≤ −√
𝜉 → +∞
2𝑚
(1 − 𝜀)𝑞 (𝑎 − 𝜀) (1 − 𝜀(2−𝑚−𝑝)/4 )𝜑(𝑚+𝑝)/2
𝑚+𝑝
(236)
when 𝜑𝑐2∗ 𝜏 < 𝜀. When 𝑝 > 𝑚, a simple calculation yields
2/(𝑚−𝑝)
≤ lim 𝜑 (𝜉) 𝜉
𝜉 → +∞
𝐵=
(238)
.
√𝑐2∗2 + 4𝑚(1 − 𝜀)𝑞 (𝑎 − 𝜀) − 𝑐2∗
2
≤ 𝜑 (𝜉)
(244)
.
.
By the arbitrariness of 𝜀 > 0, (209)-(210) hold.
When 𝑚 + 𝑝 = 2, take
𝑟 = 1,
∗2 +4𝑎−𝑐∗ )/2)𝜉
2
𝑒−((√𝑐2
By the arbitrariness of 𝜀 > 0, (207)-(208) hold.
When 𝑝 = 𝑚,
𝑒−(√𝑎/𝑚)𝜉 ≤ 𝜑 (𝜉) ≤ 𝑒
.
(243)
≤𝑒
(237)
−(√(1−𝜀)𝑞 (𝑎−𝜀)(1−𝜀(1−𝑚)/2 )/𝑚)𝜉
)
−((√𝑐2∗2 +4(1−𝜀)𝑞 (𝑎−𝜀)−𝑐2∗ )/2)𝜉
2/(𝑚−𝑝)
𝑝 − 𝑚 2𝑚𝑎
√
≤(
)
2𝑚
𝑚+𝑝
2𝑚
1/(𝑚−1)
When 𝑚 = 1, that is, 𝑝 = 1,
𝑞
(2−𝑚−𝑝)/4
)
𝑝 − 𝑚 √ 2𝑚(1 − 𝜀) (𝑎 − 𝜀) (1 − 𝜀
(
)
2𝑚
𝑚+𝑝
2/(𝑝−𝑚)
≤(
(1 − 𝑚) (√𝑐2∗2 + 4𝑚𝑎 − 𝑐2∗ )
.
(239)
5. Discussion
When 𝑝 = 𝑞 = 1, the outcome in our work is reduced to the
results obtained in [23]. Comparing with [23], our definition
of sharp-type traveling wave fronts and smooth-type traveling wave fronts is more precise. In the proof of Lemma 3, for
the case 𝑚 + 𝑝 = 2, we constructed two sequences to get the
asymptotic expression of 𝜓(𝜑) for 𝜑 > 0 sufficiently small.
This technique is not used in [23]. Moreover, the proof of
Theorem 17 is more concise than the proof of Proposition 2.7
in [23].
20
Acknowledgments
The authors are grateful to the anonymous reviewers for the
careful reading and helpful suggestions which led to an improvement of their original paper. This research is supported
by the Fundamental Research Funds for the Central Universities (no. 2012ZM0057), the National Science Foundation of Guangdong Province (no. S2012040007959), and the
National Natural Science Foundation of China (no. 11171115).
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