Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 256249, 13 pages
http://dx.doi.org/10.1155/2013/256249
Research Article
Persistence and Nonpersistence of a Nonautonomous Stochastic
Mutualism System
Peiyan Xia,1,2 Xiaokun Zheng,2 and Daqing Jiang2
1
2
School of Basic Sciences, Changchun University of Technology, Changchun, Jilin 130021, China
School of Mathematics and Statistics, Northeast Normal University, Changchun, Jilin 130024, China
Correspondence should be addressed to Daqing Jiang; daqingjiang2010@hotmail.com
Received 22 October 2012; Accepted 29 November 2012
Academic Editor: Jifeng Chu
Copyright © 2013 Peiyan Xia 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.
In this paper, a two-species nonautonomous stochastic mutualism system is investigated. The intrinsic growth rates of the two
species at time t are estimated by 𝑟𝑖 (𝑡) + 𝜎𝑖 (𝑡)𝐵̇𝑖 (𝑡), 𝑖 = 1, 2, respectively. Viewing the different intensities of the noises 𝜎𝑖 (𝑡), 𝑖 = 1, 2
as two parameters at time t, we conclude that there exists a global positive solution and the pth moment of the solution is bounded.
We also show that the system is permanent, including stochastic permanence, persistence in mean, and asymptotic boundedness in
time average. Besides, we show that the large white noise will make the system nonpersistent. Finally, we establish sufficient criteria
for the global attractivity of the system.
1. Introduction
For more than three decades, mutualism of multispecies has
attracted the attention of both mathematicians and ecologists.
By definition, in a mutualism of multispecies, the interaction is beneficial for the growth of other species. LotkaVolterra mutualism systems have long been used as standard
models to mathematically address questions related to this
interaction. Among these, nonautonomous Lotka-Volterra
mutualism models are studied by many authors, see [1–7]
and references therein. The classical nonautonomous LotkaVolterra mutualism system can be expressed as follows:
𝑛
[
]
𝑎𝑖𝑗 (𝑡) 𝑥𝑗 (𝑡)]
𝑟
−
𝑎
𝑥
+
𝑥𝑖̇ (𝑡) = 𝑥𝑖 (𝑡) [
∑
(𝑡)
(𝑡)
(𝑡)
𝑖𝑖
𝑖
[𝑖
],
[
𝑗=1
𝑗 ≠ 𝑖
]
(1)
𝑖 = 1, 2, . . . , 𝑛,
where 𝑥𝑖 (𝑡), 𝑖 = 1, 2, . . . , 𝑛 is the density of the 𝑖th population
at time 𝑡, 𝑟𝑖 (𝑡) > 0, 𝑖 = 1, 2, . . . , 𝑛 is the intrinsic growth
rate of the 𝑖th population at time 𝑡, 𝑟𝑖 (𝑡)/𝑎𝑖𝑖 (𝑡) > 0, 𝑖 =
1, 2, . . . , 𝑛 is the carrying capacity at time 𝑡, and coefficient
𝑎𝑖𝑗 (𝑡) > 0, 𝑖, 𝑗 = 1, 2, . . . , 𝑛 describes the influence of the 𝑗th
population upon the 𝑖th population at time 𝑡.
It is shown in [1] that if different conditions hold (see
conditions (a)–(e) in [1]), then the solution of system (1) is
bounded, permanent, extinct, and global attractive, respectively. However, when the intrinsic growth rate and coefficient
𝑎𝑖𝑗 (𝑡) are periodic, it is shown in [3] that there exists positive
periodic solution and almost periodic solutions are obtained.
From another point of view, environmental noise always
exists in real life. It is an interesting problem, both mathematically and biologically, to determine how the structure of the
model changes under the effect of a fluctuating environment.
Many authors studied the biological models with stochastic
perturbation, see [8–12] and references therein. In [8] Ji et
al. discussed the following two-species stochastic mutualism
system
𝑑𝑥1 (𝑡) = 𝑥1 (𝑡) [(𝑟1 − 𝑎11 𝑥1 (𝑡) + 𝑎12 𝑥2 (𝑡)) 𝑑𝑡 + 𝜎1 𝑑𝐵1 (𝑡)] ,
𝑑𝑥2 (𝑡) = 𝑥2 (𝑡) [(𝑟2 + 𝑎21 𝑥1 (𝑡) − 𝑎22 𝑥2 (𝑡)) 𝑑𝑡 + 𝜎2 𝑑𝐵2 (𝑡)] ,
(2)
where 𝐵𝑖 (𝑡), 𝑖 = 1, 2 are mutually independent one dimensional standard Brownian motions with 𝐵𝑖 (0) = 0, 𝑖 = 1, 2,
and 𝜎𝑖 , 𝑖 = 1, 2 are the intensities of white noise. It is shown in
[8] that if 𝑎11 𝑎22 > 𝑎12 𝑎21 then there is a unique nonnegative
solution of system (2). For small white noise there is a stationary distribution of (2) and it has ergodic property.
2
Abstract and Applied Analysis
Biologically, this implies that with small perturbation of
environment, the stability of the two species varies with the
intensity of white noise, and both species will survive.
However, almost all known stochastic models assume that
the growth rate and the carrying capacity of the population
are independent of time 𝑡. In contrast, the natural growth
rates of many populations vary with 𝑡 in real situation, for
example, due to the seasonality. As a matter of fact, nonautonomous stochastic population systems have recently been
studied by many authors, for example, [13–17].
In this paper we consider the system
𝑑𝑥1 (𝑡) = 𝑥1 (𝑡) [(𝑟1 (𝑡) − 𝑎11 (𝑡) 𝑥1 (𝑡) + 𝑎12 (𝑡) 𝑥2 (𝑡)) 𝑑𝑡
+𝜎1 (𝑡) 𝑑𝐵1 (𝑡) ] ,
𝑑𝑥2 (𝑡) = 𝑥2 (𝑡) [(𝑟2 (𝑡) + 𝑎21 (𝑡) 𝑥1 (𝑡) − 𝑎22 (𝑡) 𝑥2 (𝑡)) 𝑑𝑡
+𝜎2 (𝑡) 𝑑𝐵2 (𝑡) ] ,
(3)
where 𝑟𝑖 (𝑡), 𝑎𝑖𝑗 (𝑡), 𝜎𝑖 (𝑡), 𝑖, 𝑗 = 1, 2 are all continuous bounded
nonnegative functions on [0, +∞). The objective of our study
is to investigate the long-time behavior of system (3). As in
[8], we mainly discuss when the system is persistent and
when it is not under a fewer conditions. More specifically,
we show that there is a positive solution of system (3) and its
𝑝th moment bounded in Section 2. In Section 3, we deduce
the persistence of the system. If the white noise is not large
such that 𝑟𝑖𝑙 − ((𝜎𝑖𝑢 )2 /2) > 0, 𝑖 = 1, 2, we will prove that the
solution of system (3) is a stochastic persistence. In addition,
we show that every component of the solution is persistent
in mean. We further deduce that every component of the
solution of system (3) is an asymptotic boundedness in mean.
In Section 4, we show that larger white noise will make system
(3) nonpersistent. Finally, we study the global attractivity of
system (3).
Throughout this paper, unless otherwise specified, let
(Ω, {F𝑡 }𝑡≥0 , 𝑃) be a complete probability space with a filtration {F𝑡 }𝑡≥0 satisfying the usual conditions (i.e., it is right
continuous and F0 contains all 𝑃-null sets). Let 𝑅+2 be the
positive cone of 𝑅2 , namely, 𝑅+2 = {𝑥 ∈ 𝑅2 : 𝑥𝑖 > 0, 𝑖 = 1, 2}.
(∑𝑛𝑖=1
𝑛
1/2
𝑥𝑖2 ) .
satisfy the linear growth condition, though they are locally
Lipschitz continuous, so the solution of system (3) may
explode at a finite time. Following the way developed by Mao
et al. [19], we show that there is a unique positive solution of
(3).
𝑙 𝑙
𝑢 𝑢
𝑎22 > 𝑎12
𝑎21 . Then, there is a
Theorem 1. Assume that 𝑎11
unique positive solution 𝑥(𝑡) = (𝑥1 (𝑡), 𝑥2 (𝑡)) of system (3) on
𝑡 ≥ 0 for any given initial value 𝑥(0) ∈ 𝑅+2 , and the solution
will remain in 𝑅+2 with probability 1, namely, 𝑥(𝑡) ∈ 𝑅+2 for all
𝑡 ≥ 0 almost surely.
The proof of Theorem 1 is similar to [8]. But it is skilled in
taking the value of 𝜖. We show it here.
Proof. Since the coefficients of the equation are locally Lipschitz continuous, for any given initial value 𝑥(0) ∈ 𝑅+2 there
is an unique local solution 𝑥(𝑡) = (𝑥1 (𝑡), 𝑥2 (𝑡)) on 𝑡 ∈ [0, 𝜏𝑒 ),
where 𝜏𝑒 is the explosion time. To show that this solution is
global, we need to show that 𝜏𝑒 = ∞ a.s. Let 𝑚0 > 1 be
sufficiently large for every component of 𝑥(0) lying within
the interval [1/𝑚0 , 𝑚0 ]. For each integer 𝑚 ≥ 𝑚0 , define the
stopping time
𝜏𝑚 = inf {𝑡 ∈ [0, 𝜏𝑒 ) : min {𝑥1 (𝑡) , 𝑥2 (𝑡)}
1
≤
or max {𝑥1 (𝑡) , 𝑥2 (𝑡)} ≥ 𝑚},
𝑚
where throughout this paper we set inf 0 = ∞ (as usual 0
denotes the empty set). Clearly, 𝜏𝑚 is increasing as 𝑚 → ∞.
Set 𝜏∞ = lim𝑚 → ∞ 𝜏𝑚 , whence 𝜏∞ ≤ 𝜏𝑒 a.s. If we can show
that 𝜏∞ = ∞ a.s., then 𝜏𝑒 = ∞ a.s. and 𝑥(𝑡) ∈ 𝑅+2 a.s. for all
𝑡 ≥ 0. In other words, to complete the proof, all we need to
show is that 𝜏∞ = ∞ a.s. If this statement is false, there is a
pair of constant 𝑇 > 0 and 𝜀 ∈ (0, 1) such that
𝑃 {𝜏∞ ≤ 𝑇} > 𝜀.
𝑓 = sup 𝑓 (𝑡) ,
𝑡∈[0,+∞)
(6)
Hence, there is an integer 𝑚1 ≥ 𝑚0 such that
If 𝑥 ∈ 𝑅 , its norm is denoted by |𝑥| =
If 𝑓(𝑡)
is a continuous bounded function on [0, +∞), we use the
notation sup
𝑢
(5)
𝑃 {𝜏𝑚 ≤ 𝑇} ≥ 𝜀
∀𝑚 ≥ 𝑚1 .
(7)
We define
𝑙
𝑓 = min 𝑓 (𝑡) .
𝑡∈[0,+∞)
(4)
2. Existence and Uniqueness of
the Positive Solution
In population dynamics, the first concern is that the solution
should be nonnegative. In order to do that a stochastic differential equation can have a unique global (i.e., no explosion
at any finite time) solution for any given initial value, the
coefficients of the equation are generally required to satisfy
the linear growth condition and local Lipschitz condition
(Mao [18]). However, the coefficients of system (3) do not
𝑢
𝑢
(𝑥1 − 1 − log 𝑥1 ) + 𝑎12
(𝑥2 − 1 − log 𝑥2 ) .
𝑉 (𝑥) = 𝑎21
(8)
By Itô’s formula, we have
𝑢
𝑑𝑉 (𝑥) = {𝑎21
(1−
1
) 𝑥1 [𝑟1 (𝑡)−𝑎11 (𝑡) 𝑥1 +𝑎12 (𝑡) 𝑥2 ]
𝑥1
𝑢
+𝑎12
(1−
1
) 𝑥2 [𝑟2 (𝑡)+𝑎21 (𝑡) 𝑥1 −𝑎22 (𝑡) 𝑥2 ]
𝑥2
1 𝑢 2
𝑢 2
𝜎1 (𝑡) + 𝑎12
𝜎2 (𝑡)] } 𝑑𝑡
+ [𝑎21
2
Abstract and Applied Analysis
3
𝑢
+ 𝑎21
𝜎1 (𝑡) (𝑥1 − 1) 𝑑𝐵1 (𝑡)
𝑢
𝑢
𝑢 𝑙
+ [𝑎12
(𝑟2𝑢 + 𝑎22
) − 𝑎21
𝑎12 ] 𝑥2
𝑢
+ 𝑎12
𝜎2 (𝑡) (𝑥2 − 1) 𝑑𝐵2 (𝑡)
1 𝑢 𝑢 2
𝑢 𝑙
− 𝑎12
𝑟2 + 𝑎12
(𝜎2 )
2
𝑢
:= 𝐿𝑉𝑑𝑡 + 𝑎21
𝜎1 (𝑡) (𝑥1 − 1) 𝑑𝐵1 (𝑡)
≤ 𝐾.
𝑢
+ 𝑎12
𝜎2 (𝑡) (𝑥2 − 1) 𝑑𝐵2 (𝑡) ,
(11)
(9)
where
𝑢
𝑙
𝑙
𝑢
𝑢 𝑙
𝑢
/2𝑎22
< 𝜖 < 𝑎11
/2𝑎12
, we obtain −(𝑎21
𝑎11 − 2𝜖𝑎12
)<
Since 𝑎21
𝑢 𝑙
𝑢 𝑢
0 and −(𝑎12 𝑎22 − (1/2𝜖)𝑎21 𝑎12 ) < 0. Hence, 𝐾 is a positive
constant. Integrating both sides of (9) from 0 to 𝜏𝑚 ∧ 𝑇, we
therefore obtain
𝑉 (𝑥 (𝜏𝑚 ∧ 𝑇)) − 𝑉 (𝑥 (0))
𝑢
(1 −
𝐿𝑉 = 𝑎21
1
) 𝑥1 [𝑟1 (𝑡) − 𝑎11 (𝑡) 𝑥1 + 𝑎12 (𝑡) 𝑥2 ]
𝑥1
𝜏𝑚 ∧𝑇
≤∫
1
𝑢
(1 − ) 𝑥2 [𝑟2 (𝑡) + 𝑎21 (𝑡) 𝑥1 − 𝑎22 (𝑡) 𝑥2 ]
+ 𝑎12
𝑥2
+
1 𝑢 2
𝑢 2
𝜎2 (𝑡)]
[𝑎 𝜎 (𝑡) + 𝑎12
2 21 1
0
𝐾𝑑𝑡 + ∫
𝜏𝑚 ∧𝑇
0
+∫
𝜏𝑚 ∧𝑇
0
𝑢
𝑎21
𝜎1 (𝑡) (𝑥1 (𝑡) − 1) 𝑑𝐵1 (𝑡)
(12)
𝑢
𝑎12
𝜎2 (𝑡) (𝑥2 (𝑡) − 1) 𝑑𝐵2 (𝑡) .
Whence, taking expectations yields
𝐸 [𝑉 (𝑥 (𝜏𝑚 ∧ 𝑇))] ≤ 𝑉 (𝑥 (0)) + 𝐾𝐸 (𝜏𝑚 ∧ 𝑇)
𝑢
𝑢
𝑙
𝑙
≤ 𝑎21
[ (𝑟1𝑢 + 𝑎11
) 𝑥1 − 𝑎12
𝑥2 − 𝑎11
𝑥12
≤ 𝑉 (𝑥 (0)) + 𝐾𝑇.
1
2
𝑢
+𝑎12
𝑥1 𝑥2 − 𝑟1𝑙 + (𝜎1𝑢 ) ]
2
(13)
Set Ω𝑚 = {𝜏𝑚 ≤ 𝑇} for 𝑚 ≥ 𝑚1 and by (7), 𝑃(Ω𝑚 ) ≥ 𝜀. Note
that for every 𝜔 ∈ Ω𝑚 , there is 𝑥1 (𝜏𝑚 , 𝜔) or 𝑥2 (𝜏𝑚 , 𝜔) equals
either 𝑚 or 1/𝑚, and therefore
𝑢
𝑢
𝑙
𝑙
+ 𝑎12
[ (𝑟2𝑢 + 𝑎22
) 𝑥2 − 𝑎21
𝑥1 − 𝑎22
𝑥22
𝑉 (𝑥 (𝜏𝑚 , 𝜔))
1
2
𝑢
+𝑎21
𝑥1 𝑥2 − 𝑟2𝑙 + (𝜎2𝑢 ) ] .
2
(10)
𝑢
𝑢
≥ min {𝑎21
, 𝑎12
} (𝑚 − 1 − log 𝑚) ∧ (
1
1
− 1 − log )
𝑚
𝑚
:= ℎ (𝑚) ,
According to Young inequality, note that 𝑥1 𝑥2 ≤ 𝜖𝑥12 +
𝑢
𝑙
𝑙
𝑢
/2𝑎22
< 𝜖 < 𝑎11
/2𝑎12
, then,
(1/4𝜖)𝑥22 , where 𝑎21
(14)
where lim𝑚 → ∞ ℎ(𝑚) = ∞. It then follows from (13) that
𝐸 [𝑉 (𝑥 (0))] + 𝐾𝑇 ≥ 𝐸 [1Ω𝑚 ⋅ 𝑉 (𝑥 (𝜏𝑚 , 𝜔))] ≥ 𝜖ℎ (𝑚) ,
(15)
𝑢
𝑢
𝑙
𝑙
[ (𝑟1𝑢 + 𝑎11
) 𝑥1 − 𝑎12
𝑥2 − 𝑎11
𝑥12
𝐿𝑉 ≤ 𝑎21
𝑢
+𝑎12
(𝜖𝑥12 +
+
𝑢
𝑎12
[ (𝑟2𝑢
+
1 2
1
2
𝑥 ) − 𝑟1𝑙 + (𝜎1𝑢 ) ]
4𝜖 2
2
𝑢
𝑎22
) 𝑥2
𝑢
+𝑎21
(𝜖𝑥12 +
−
𝑙
𝑎21
𝑥1
−
1 2
1
2
𝑥2 ) − 𝑟2𝑙 + (𝜎2𝑢 ) ]
4𝜖
2
𝑢 𝑙
𝑢 𝑢
= − (𝑎21
𝑎11 − 2𝜖𝑎21
𝑎12 ) 𝑥12
𝑢
𝑢
𝑢 𝑙
+ [𝑎21
(𝑟1𝑢 + 𝑎11
) − 𝑎12
𝑎21 ] 𝑥1
1 𝑢 𝑢 2
𝑢 𝑙
− 𝑎21
𝑟1 + 𝑎21
(𝜎1 )
2
𝑢 𝑙
− (𝑎12
𝑎22 −
𝑙
𝑎22
𝑥22
1 𝑢 𝑢
𝑎 𝑎 ) 𝑥2
2𝜖 21 12 2
where 1Ω𝑚 is the indicator function of Ω𝑚 . Letting 𝑚 → ∞
leads to the contradiction
∞ > 𝑉 (𝑥 (0)) + 𝐾𝑇 = ∞,
(16)
so we must have 𝜏∞ = ∞ a.s. This completes the proof of
Theorem 1.
Remark 2. By Theorem 1, we observe that for any given
initial value 𝑥(0) ∈ 𝑅+2 , there is a unique solution 𝑥(𝑡) =
(𝑥1 (𝑡), 𝑥2 (𝑡)) of system (3) on 𝑡 ≥ 0 and the solution will
remain in 𝑅+2 with probability 1, no matter how large the
intensities of white noise are. So, under the same assumption
there is an global unique positive solution of the corresponding deterministic system of system (3).
Next, we show that the 𝑝th moment of the solution of
system (3) is bounded in time average.
4
Abstract and Applied Analysis
𝑙 𝑙
𝑢 𝑢
Theorem 3. Assume that 𝑎11
𝑎22 > 𝑎12
𝑎21 . Then there exists a
positive constant 𝐾(𝑝) such that the solution 𝑥(𝑡) of system (3)
has the following property:
𝑝
𝐸 [𝑐1 𝑥1
(𝑡) +
𝑝
𝑐2 𝑥2
(𝑡)] ≤ 𝐾 (𝑝) ,
where 𝛼2 (𝑡) = 𝑟2 (𝑡) + ((𝑝 − 1)/2)𝜎22 (𝑡). According to Young
inequality, we obtain
𝑝
𝑝+1
𝑥1 (𝑡) 𝑥2 (𝑡) ≤ 𝜖1 𝑥1
∀𝑡 ∈ [0, ∞) , 𝑝 > 1,
(𝑡) +
𝑝 𝑝 1 𝑝 𝑝+1
1
(
) ( ) 𝑥2 (𝑡) ,
𝑝+1 𝑝+1
𝜖1
(17)
𝜖1 =
where 𝑐1 , 𝑐2 satisfy
𝑝
𝑙
𝑙
𝑐1 𝑎22 (𝑎11 )
<
<
.
𝑙
𝑙 𝑝
𝑢 𝑝+1
𝑐2
𝑎11
(𝑎22
)
(𝑎12
)
𝑝+1
𝑢
(𝑎21
)
(18)
𝑝
𝑝+1
𝑥2 (𝑡) 𝑥1 (𝑡) ≤ 𝜖2 𝑥2
(𝑡) +
𝑝 𝑝 1 𝑝 𝑝+1
1
(
) ( ) 𝑥1 (𝑡) ,
𝑝+1 𝑝+1
𝜖2
Proof. By Itô’s formula, we have
𝑝
𝜖2 =
𝑝
𝑑𝑥1 (𝑡) = 𝑝𝑥1 (𝑡) [(𝑟1 (𝑡) − 𝑎11 (𝑡) 𝑥1 (𝑡) + 𝑎12 (𝑡) 𝑥2 (𝑡)) 𝑑𝑡
+𝜎1 (𝑡) 𝑑𝐵1 (𝑡) ]
𝑝
𝑝−1 2
𝑝
𝑝+1
𝜎 (𝑡)) 𝑥1 (𝑡) − 𝑎11 (𝑡) 𝑥1 (𝑡)
2 1
𝑝
𝑢
+𝑎12
𝑝
+
𝑝
𝑝+1
𝑝
𝑑𝑥2
(𝑡)
(𝑡) ≤
𝑢
+𝑎21
𝑝
+ 𝜎1 (𝑡) 𝑝𝑥1 (𝑡) 𝑑𝐵1 (𝑡)
𝑝+1
𝑝
𝑢
𝑥1 (𝑡) 𝑥2 (𝑡)] 𝑑𝑡
(𝑡) + 𝑎12
(19)
𝑝
𝑝
+ 𝜎2 (𝑡) 𝑝𝑥2 (𝑡) 𝑑𝐵2 (𝑡)
2
𝑝+1
𝑎22 (𝑡) 𝑥2
(𝑡)
𝑝
𝑝+1 𝑝
𝜖2
𝑢 𝑝
𝑎21
𝑥2
𝑝+1
) 𝑥1 (𝑡)
𝑝𝑝
𝑝+1 𝑝
(𝑝+1) 𝜖1
𝑝+1
) 𝑥2
(𝑡)
𝑝
𝑖=1
(24)
(𝑡) 𝑥1 (𝑡)] 𝑑𝑡
From (23) and the values of 𝜖1 , 𝜖2 , we obtain
𝑢
𝑎21
(𝑝𝑝 /(𝑝 + 1)
𝑝+1 𝑝
𝜖2 )
𝑙
𝑢
𝑎11
− 𝑎12
𝜖1
𝑝
+ 𝑝𝜎2𝑢 𝑥2 (𝑡) 𝑑𝐵2 (𝑡) ,
(𝑝+1)
2
𝑝
𝑖=1
𝑝
(𝑡) +
𝑝𝑝
−∑𝑐𝑖 𝛼𝑖𝑢 𝑥𝑖 (𝑡)] 𝑑𝑡 + ∑𝑐𝑖 𝑝𝜎𝑖𝑢 𝑥𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡) .
+𝑎21 (𝑡) 𝑥2 (𝑡) 𝑥1 (𝑡)] 𝑑𝑡 + 𝜎2 (𝑡) 𝑝𝑥2 (𝑡) 𝑑𝐵2 (𝑡)
(𝑡) −
𝑙
𝑢
𝑢
−𝑐1 𝑎12
𝜖1 −𝑐2 𝑎21
≤ −𝑝 [(𝑐1 𝑎11
𝑙
𝑢
𝑢
+(𝑐2 𝑎22
−𝑐1 𝑎21
𝜖2 −𝑐1 𝑎12
𝑝
+𝑎21 (𝑡) 𝑥2 (𝑡) 𝑥1 (𝑡) ] 𝑑𝑡
≤
𝑝
𝑑 (𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡))
𝑝−1 2
𝑝
𝑝+1
𝜎 (𝑡)) 𝑥2 (𝑡) − 𝑎22 (𝑡) 𝑥2 (𝑡)
2 2
𝑝+1
𝑙
𝑎22
𝑥2
(23)
Therefore,
1
𝑝
+ 𝑝 (𝑝 − 1) 𝑥2 (𝑡) 𝜎22 (𝑡) 𝑑𝑡
2
𝑝
𝑝 [𝛼2𝑢 𝑥2
𝑝
𝑙
𝑙
𝑐1 𝑎22 (𝑎11 )
<
<
.
𝑙
𝑙 𝑝
𝑢 𝑝+1
𝑐2
𝑎11
(𝑎22
)
(𝑎12
)
𝑝+1
+𝜎2 (𝑡) 𝑑𝐵2 (𝑡) ]
(𝑡) −
(𝑡)
𝑝 𝑝 1 𝑝 𝑝+1
1
(
) ( ) 𝑥1 (𝑡)] 𝑑𝑡
𝑝+1 𝑝+1
𝜖2
𝑢
)
(𝑎21
𝑝
=
(𝑡) +
𝑙 𝑙
𝑢 𝑢
𝑎22 > 𝑎12
𝑎21 , there exist two positive constants 𝑐1 , 𝑐2
Since 𝑎11
which satisfy
𝑑𝑥2 (𝑡) = 𝑝𝑥2 (𝑡) [(𝑟2 (𝑡) − 𝑎22 (𝑡) 𝑥2 (𝑡) + 𝑎21 (𝑡) 𝑥1 (𝑡)) 𝑑𝑡
𝑝
𝑝 [𝛼2 (𝑡) 𝑥2
(22)
𝑝+1
𝑢
𝜖2 𝑥2
𝑎21
𝑝
where 𝛼1 (𝑡) = 𝑟1 (𝑡) + ((𝑝 − 1)/2)𝜎12 (𝑡), and
= 𝑝 [(𝑟2 (𝑡) +
(𝑡) −
𝑝+1
𝑙
𝑥2
𝑎22
+ 𝑝𝜎2𝑢 𝑥2 (𝑡) 𝑑𝐵2 (𝑡) .
𝑝
+ 𝑝𝜎1𝑢 𝑥1 (𝑡) 𝑑𝐵1 (𝑡) ,
𝑝
𝑝 𝑝 1 𝑝 𝑝+1
1
(
) ( ) 𝑥2 (𝑡)] 𝑑𝑡
𝑝+1 𝑝+1
𝜖1
𝑝
𝑝 [𝛼2𝑢 𝑥2
𝑝
𝑝
(𝑡)
𝑝
+𝑎12 (𝑡) 𝑥1 (𝑡) 𝑥2 (𝑡)] 𝑑𝑡
𝑙
≤ 𝑝 [𝛼1𝑢 𝑥1 (𝑡) − 𝑎11
𝑥1
𝑝+1
𝑢
𝜖1 𝑥1
(𝑡) + 𝑎12
+ 𝑝𝜎1𝑢 𝑥1 (𝑡) 𝑑𝐵1 (𝑡) ,
(𝑡) 𝑑𝐵1 (𝑡)
= 𝑝 [𝛼1 (𝑡) 𝑥1 (𝑡) − 𝑎11 (𝑡) 𝑥1
𝑝+1
𝑙
𝑑𝑥1 (𝑡) ≤ 𝑝 [𝛼1𝑢 𝑥1 (𝑡) − 𝑎11
𝑥1
+𝑎12 (𝑡) 𝑥1 (𝑡) 𝑥2 (𝑡) ] 𝑑𝑡
𝑝
𝜎1 (𝑡) 𝑝𝑥1
𝑙
𝑝𝑎22
.
𝑢
(𝑝 + 1) 𝑎21
(21)
Thus, we have
1
𝑝
+ 𝑝 (𝑝 − 1) 𝑥1 (𝑡) 𝜎12 (𝑡) 𝑑𝑡
2
= 𝑝 [(𝑟1 (𝑡) +
𝑙
𝑝𝑎11
,
𝑢
(𝑝 + 1) 𝑎12
(20)
<
𝑙
𝑢
𝑎22
− 𝑎21
𝜖2
𝑐1
,
<
𝑐2 𝑎𝑢 (𝑝𝑝 /(𝑝 + 1)𝑝+1 𝜖𝑝 )
12
1
(25)
Abstract and Applied Analysis
5
𝑝
𝑙
𝑢
𝑢
which implies that 𝑐1 𝑎11
− 𝑐1 𝑎12
𝜖1 − 𝑐2 𝑎21
(𝑝𝑝 /(𝑝 + 1)𝑝+1 𝜖2 ) > 0
𝑝+1 𝑝
𝑙
𝑢
𝑢
𝑝
and 𝑐2 𝑎22 − 𝑐1 𝑎21 𝜖2 − 𝑐1 𝑎12 (𝑝 /((𝑝 + 1) 𝜖1 )) > 0. Let
𝛼=
max {𝛼1𝑢 , 𝛼2𝑢 } ,
−(𝑝+1)/𝑝
𝑙
𝑢
𝑢
[𝑐1 𝑎11
− 𝑐1 𝑎12
𝜖1 − 𝑐2 𝑎21
𝛽 = min {𝑐1
−(𝑝+1)/𝑝
𝑐2
𝑙
[𝑐2 𝑎22
−
𝑢
𝑐1 𝑎21
𝜖2
𝑝𝑝
𝑢
𝑐1 𝑎12
−
𝑝+1 𝑝
𝜖2
(𝑝 + 1)
],
𝑝𝑝
𝑝+1 𝑝
𝜖1
(𝑝 + 1)
]} ,
(26)
then we have
𝑝
𝑝
𝑑 (𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡))
2
2
𝑝
1+(1/𝑝) 𝑝+1
𝑥𝑖 )] 𝑑𝑡
≤ 𝑝 [𝛼 (∑𝑐𝑖 𝑥𝑖 (𝑡)) − 𝛽 (∑𝑐𝑖
𝑖=1
2
𝑖=1
̃
Let 𝐾(𝑝) = max{2𝑀(𝑝), 𝑀(𝑝)},
then
𝑝
𝑝
𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)] ≤ 𝐾 (𝑝) ,
∀𝑡 ∈ [0, ∞) .
3. Persistence
Theorem 1 shows that the solution of system (3) will remain in
𝑙 𝑙
𝑢 𝑢
the positive cone 𝑅+2 if 𝑎11
𝑎22 > 𝑎12
𝑎21 . Studying a population
system, we pay more attention on whether the system is
persistent. In this section, we first show that the solution
is a stochastic permanence. Next we show that the solution
is persistent in time average. Moreover, we show that the
solution 𝑥(𝑡) of system (3) is an asymptotic boundedness in
time average.
3.1. Stochastic Permanence. Let 𝑦(𝑡) be the solution of a
randomized nonautonomous competitive equation:
𝑛
𝑝
+∑𝑐𝑖 𝑝𝜎𝑖𝑢 𝑥𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡) .
𝑖=1
(27)
𝑑𝑦𝑖 (𝑡) = 𝑦𝑖 (𝑡) [(𝑏𝑖 (𝑡)− ∑ 𝑎𝑖𝑗 (𝑡) 𝑦𝑗 (𝑡)) 𝑑𝑡 + 𝜎𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡)] ,
𝑗=1
[
]
𝑖 = 1, 2, . . . , 𝑛,
(33)
Hence, we get
𝑝
𝑝
𝑑𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)]
where 𝐵𝑖 (𝑡), 𝑖 = 1, 2, . . . , 𝑛, are independent standard
Brownian motions, 𝑦(0) = 𝑦0 > 0 while 𝑦0 is independent
of 𝐵(𝑡), and 𝑏𝑖 (𝑡), 𝑎𝑖𝑗 (𝑡), 𝜎𝑖 (𝑡) are all continuous bounded
nonnegative functions on [0, +∞).
𝑑𝑡
𝑝
𝑝
≤ 𝑝𝛼𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)]
1+(1/𝑝) 𝑝+1
𝑥1
− 𝑝𝛽𝐸 [𝑐1
𝑝
1+(1/𝑝) 𝑝+1
𝑥2
(𝑡) + 𝑐2
(𝑡)]
2
𝑝
≤ 𝑝𝛼𝐸 [𝑐1 𝑥1 (𝑡)+𝑐2 𝑥2 (𝑡)]
(28)
1+(1/𝑝)
𝑝
− 𝑝𝛽 {[𝐸 (𝑐1 𝑥1 (𝑡))]
𝑝
+[𝐸 (𝑐2 𝑥2
𝑝
1+(1/𝑝)
(𝑡))]
}
𝑝
1+(1/𝑝)
𝑝
− 𝑝𝛽 ⋅ 2−1/𝑝 [𝐸 (𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡))]
.
2
𝜃 max(𝜎𝑖𝑢 ) < 2 min (𝑏𝑖 (𝑡) −
1≤𝑖≤𝑛
By the comparison theorem, we get
𝛼 𝑝
𝑝
𝑝
lim sup 𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)] ≤ 2( ) := 𝑀 (𝑝) ,
𝛽
𝑡→∞
(29)
which implies that there is a 𝑇0 > 0, such that
𝑝
𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)] ≤ 2𝑀 (𝑝) ,
Besides, note that
+
̃
there is a 𝑀(𝑝)
> 0 such that
𝑝
𝐸 [𝑐1 𝑥1
(𝑡) +
𝑝
𝑐2 𝑥2
∀𝑡 > 𝑇0 .
𝑝
𝑐2 𝑥2 (𝑡)]
̃ (𝑝) ,
(𝑡)] ≤ 𝑀
(34)
where 𝐻 is a constant, 𝜃 is an arbitrary positive constant
satisfying
𝑝
𝑝
𝐸[𝑐1 𝑥1 (𝑡)
Lemma 4 (see [15]). Assume that 𝑏𝑖𝑙 − ((𝜎𝑖𝑢 ) /2) > 0, 𝑖 =
1, 2, . . . , 𝑛, then for any given initial value 𝑦(0) ∈ 𝑅+𝑛 , the
solution 𝑦(𝑡) of (36) has the properties
1
) ≤ 𝐻,
lim sup 𝐸 ( 󵄨
󵄨󵄨𝑦 (𝑡)󵄨󵄨󵄨𝜃
𝑡→∞
󵄨
󵄨
≤ 𝑝𝛼𝐸 [𝑐1 𝑥1 (𝑡) + 𝑐2 𝑥2 (𝑡)]
𝑝
(32)
(35)
Let 𝑁(𝑡) be the solution of a randomized nonautonomous
logistic equation
𝑑𝑁 (𝑡) = 𝑁 (𝑡) [(𝑎 (𝑡) − 𝑏 (𝑡) 𝑁 (𝑡)) 𝑑𝑡 + 𝛼 (𝑡) 𝑑𝐵 (𝑡)] , (36)
(30)
is continuous, then
∀𝑡 ∈ [0, 𝑇0 ] .
1≤𝑖≤𝑛
𝑙
𝜎𝑖2 (𝑡)
).
2
(31)
where 𝐵(𝑡) is a 1-dimensional standard Brownian motion,
𝑁(0) = 𝑁0 > 0, and 𝑁0 is independent of 𝐵(𝑡).
Lemma 5 (see [13]). Assume that 𝑎(𝑡), 𝑏(𝑡), and 𝛼(𝑡) are
bounded continuous functions defined on [0, ∞), 𝑎(𝑡) > 0
and 𝑏(𝑡) > 0. Then there exists a unique continuous positive
6
Abstract and Applied Analysis
solution of (36) for any initial value 𝑁(0) = 𝑁0 > 0, which is
global and represented by
𝑡
𝑁 (𝑡) = exp {∫ [𝑎 (𝑠) − (
0
× ((
where 𝐻𝑖 , 𝑖 = 1, 2 are two constants, 𝜃 is positive constant
satisfying
𝛼2 (𝑠)
)] 𝑑𝑠 + 𝛼 (𝑠) 𝑑𝐵 (𝑠)}
2
2
𝜃(𝜎𝑖𝑢 ) < 2𝑟𝑙𝑖 ,
𝑡
𝑠
𝛼2 (𝜏)
1
)+∫ 𝑏 (𝑠) exp {∫ [𝑎 (𝜏)−(
)] 𝑑𝜏
𝑁0
2
0
0
−1
+𝛼 (𝜏) 𝑑𝐵 (𝜏) } 𝑑𝑠) ,
𝑡 ≥ 0.
(37)
𝑖 = 1, 2.
(45)
Proof. Equation (43) follows directly from the classical comparison theorem of stochastic differential equations (see
[20]). Thus, we obtain
lim sup 𝐸 (
𝑡→∞
1
1
) ≤ lim sup 𝐸 ( 𝜃 ) ≤ 𝐻𝑖 ,
𝑥𝑖𝜃 (𝑡)
𝜙𝑖 (𝑡)
𝑡→∞
𝑖 = 1, 2.
(46)
From Lemma 4 we have the following.
Lemma 6. Assume that 𝑎𝑙 − ((𝛼𝑢 )2 /2) > 0, then for any given
initial value 𝑁(0) ∈ 𝑅+ , the solution 𝑁(𝑡) of (36) has the
properties
lim sup 𝐸 (
𝑡→∞
1
) ≤ 𝐻,
(𝑡)
(38)
𝑁𝜃
lim inf 𝑃 {𝑥𝑖 (𝑡) ≤ 𝑀 (𝜖)} ≥ 1 − 𝜖,
where 𝐻 is a constant, 𝜃 is positive constant satisfying
2
𝜃(𝛼𝑢 ) < 2 [𝑎𝑙 −
𝑡→∞
lim inf 𝑃 {𝑥𝑖 (𝑡) ≥ 𝛿 (𝜖)} ≥ 1 − 𝜖,
𝑢 2
(𝛼 )
].
2
(39)
𝑑𝜙𝑖 (𝑡) = 𝜙𝑖 (𝑡) [(𝑟𝑖 (𝑡) − 𝑎𝑖𝑖 (𝑡) 𝜙𝑖 (𝑡)) 𝑑𝑡 + 𝜎𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡)] ,
𝑖 = 1, 2,
(40)
where 𝐵𝑖 (𝑡), 𝑖 = 1, 2, are independent standard Brownian
motions, 𝜙(0) = 𝜙0 > 0, and 𝜙0 ∈ 𝑅+2 , 𝑟𝑖 (𝑡), 𝑎𝑖𝑖 (𝑡), 𝜎𝑖 (𝑡), 𝑖 =
1, 2 are all continuous bounded nonnegative functions on
[0, +∞). From Lemma 4 it is easy to know the following.
2
Lemma 7. Assume that 𝑟𝑙𝑖 = 𝑟𝑖𝑙 − ((𝜎𝑖𝑢 ) /2) > 0, 𝑖 = 1, 2, then
for any given initial value 𝜙(0) ∈ 𝑅+2 , the solution 𝜙(𝑡) of (40)
has the properties
𝑡→∞
1
) ≤ 𝐻𝑖 ,
(𝑡)
𝜙𝑖𝜃
𝑖 = 1, 2,
(41)
where 𝐻𝑖 , 𝑖 = 1, 2 are two constants, 𝜃 is positive constant
satisfying
2
𝜃(𝜎𝑖𝑢 ) < 2𝑟𝑙𝑖 ,
𝑖 = 1, 2.
lim sup 𝐸 (
𝑡→∞
𝑖 = 1, 2,
1
) ≤ 𝐻𝑖 ,
𝜃
𝑥𝑖 (𝑡)
𝑖 = 1, 2,
(47)
The proof is a simple application of the Chebyshev
inequality, we omit it.
3.2. Persistence in Time Average. Theorem 10 shows that if the
white noise is not large, the solution of system (3) is survive
with large probability. In this part, we show 𝑥(𝑡) is persistence
in mean.
Lemma 11. Assume that 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2, then for any given
initial value 𝜙(0) ∈ 𝑅+2 , the solution 𝜙(𝑡) of (40) has the
properties
𝑠
𝑡
𝑧𝑖 (𝑡) 𝑒−[max0≤𝑠≤𝑡 ∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)−∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)]
𝑠
(43)
(44)
𝑡
≤ 𝜙𝑖 (𝑡) ≤ 𝑧𝑖 (𝑡) 𝑒−[min0≤𝑠≤𝑡 ∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)−∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)] ,
𝑖 = 1, 2,
(42)
Lemma 8. Assume that 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2, then for any given
initial value 𝑥0 ∈ 𝑅+2 , the solution 𝑥(𝑡) of system (3) has the
properties
𝑥𝑖 (𝑡) ≥ 𝜙𝑖 (𝑡) ,
𝑡→∞
𝑖 = 1, 2.
𝑙 𝑙
𝑢 𝑢
𝑎22 > 𝑎12
𝑎21 , 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2,
Theorem 10. Assume that 𝑎11
then system (3) is stochastically permanent.
Let 𝜙(𝑡) = (𝜙1 (𝑡), 𝜙2 (𝑡))𝑇 be the solution of
lim sup 𝐸 (
Definition 9. System (3) is said to be stochastically permanent
if for any 𝜀 ∈ (0, 1), there exists a pair of positive constants 𝛿 =
𝛿(𝜖) and 𝑀 = 𝑀(𝜖) such that for any initial value 𝑥0 ∈ 𝑅+2 ,
the solution obeys
(48)
where 𝑧(𝑡) = (𝑧1 (𝑡), 𝑧2 (𝑡)) is the solution of
𝑑𝑧𝑖 (𝑡) = 𝑧𝑖 (𝑡) [𝑟𝑖 (𝑡) −
𝜎𝑖2 (𝑡)
− 𝑎𝑖𝑖𝑢 𝑧𝑖 (𝑡)] 𝑑𝑡,
2
𝑧𝑖 (0) = 𝜙𝑖 (0) , 𝑖 = 1, 2.
(49)
Abstract and Applied Analysis
7
where 𝑧̃(𝑡) = (̃𝑧1 (𝑡), 𝑧̃2 (𝑡)), 𝑧̂(𝑡) = (̂𝑧1 (𝑡), 𝑧̂2 (𝑡)) are the
solutions of the two equations, respectively,
Proof. From Lemma 5, we know
1
1 − ∫0𝑡 [𝑟𝑖 (𝑠)−(𝜎𝑖2 (𝑠)/2)]𝑑𝑠+𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠)
=
𝑒
𝜙𝑖 (𝑡) 𝜙𝑖 (0)
𝑡
𝑡
2
+ 𝑎𝑖𝑖𝑢 ∫ 𝑒− ∫0 [𝑟𝑖 (𝑠)−(𝜎𝑖 (𝑠)/2)]𝑑𝑠+𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠)
𝑑̃𝑧𝑖 (𝑡) = 𝑧̃𝑖 (𝑡) [𝑟𝑙𝑖 − 𝑎𝑖𝑖𝑢 𝑧̃𝑖 (𝑡)] 𝑑𝑡,
𝑧̃𝑖 (0) = 𝑧𝑖 (0) , 𝑖 = 1, 2,
(53)
𝑑̂𝑧𝑖 (𝑡) = 𝑧̂𝑖 (𝑡) [𝑟𝑢𝑖 − 𝑎𝑖𝑖𝑢 𝑧̂𝑖 (𝑡)] 𝑑𝑡,
𝑧̂𝑖 (0) = 𝑧𝑖 (0) , 𝑖 = 1, 2.
(54)
0
𝑠
Proof. Let 𝑧̃(𝑡) = (̃𝑧1 (𝑡), 𝑧̃2 (𝑡)), 𝑧̂(𝑡) = (̂𝑧1 (𝑡), 𝑧̂2 (𝑡)) are the
solutions of SDE (53) and (54), respectively, with the positive
initial value 𝑧(0). By Lemma 5, we know
2
× 𝑒+ ∫0 [𝑟𝑖 (𝜏)−(𝜎𝑖 (𝜏)/2)]𝑑𝜏+𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏) 𝑑𝑠
𝑡
= 𝑒− ∫0 𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠) [
1 − ∫0𝑡 [𝑟𝑖 (𝑠)−(𝜎𝑖2 (𝑠)/2)]𝑑𝑠
𝑒
𝜙𝑖 (0)
𝑡
𝑡
𝑙
𝑧̃𝑖 (𝑡) =
2
+ 𝑎𝑖𝑖𝑢 ∫ 𝑒− ∫𝑠 [𝑟𝑖 (𝜏)−(𝜎𝑖 (𝜏)/2)]𝑑𝜏
0
𝑠
𝑡
𝑧̂𝑖 (𝑡) =
1 − ∫0𝑡 [𝑟𝑖 (𝑠)−(𝜎𝑖2 (𝑠)/2)]𝑑𝑠
𝑒
𝜙𝑖 (0)
𝑠
max0≤𝑠≤𝑡 (∫0
+ 𝑎𝑖𝑖𝑢 𝑒
𝑡
𝑒𝑟𝑖 𝑡
𝑢
1/̂𝑧𝑖 (0) + (𝑎𝑖𝑖𝑢 /𝑟𝑢𝑖 ) (𝑒𝑟𝑖 𝑡 − 1)
lim 𝑧̃𝑖 (𝑡) =
𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏))
𝑡
𝑡→∞
(50)
Similarly, we have
lim
𝑡
1
1 − ∫0𝑡 [𝑟𝑖 (𝑠)−(𝜎𝑖2 (𝑠)/2)]𝑑𝑠
≥ 𝑒− ∫0 𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠) [
𝑒
𝜙𝑖 (𝑡)
𝜙𝑖 (0)
𝑠
𝑡→∞
𝑟𝑢𝑖
.
𝑎𝑖𝑖𝑢
𝑡
− ∫𝑠 [𝑟𝑖 (𝜏)−(𝜎𝑖2 (𝜏)/2)]𝑑𝜏
(56)
(57)
log 𝜙𝑖 (𝑡)
= 0,
𝑡
𝑎.𝑠.
(58)
𝑠
𝑡
𝑒−[max0≤𝑠≤𝑡 ∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)−∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)]
𝑠
0
lim 𝑧̂𝑖 (𝑡) =
Proof. By Lemma 12, we know
𝑎𝑖𝑖𝑢 𝑒min0≤𝑠≤𝑡 (∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏))
×∫ 𝑒
𝑟𝑙𝑖
,
𝑎𝑖𝑖𝑢
Lemma 13. Assume that 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2, then for any given
initial value 𝜙(0) ∈ 𝑅+2 , the solution 𝜙(𝑡) of (40) has the
properties
𝑡→∞
𝑡
.
𝑧̃𝑖 (𝑡) ≤ 𝑧𝑖 (𝑡) ≤ 𝑧̂𝑖 (𝑡) .
𝑡
𝑒max0≤𝑠≤𝑡 [∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)]−∫0 𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠)
.
𝑧𝑖 (𝑡)
+
(55)
By the classical comparison theorem of ordinary differential
equations, we know
2
0
𝑠
,
Thus,
× ∫ 𝑒− ∫𝑠 [𝑟𝑖 (𝜏)−(𝜎𝑖 (𝜏)/2)]𝑑𝜏 𝑑𝑠]
≤
𝑙
1/̃𝑧𝑖 (0) + (𝑎𝑖𝑖𝑢 /𝑟𝑙𝑖 ) (𝑒𝑟𝑖 𝑡 − 1)
𝑢
× 𝑒∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏) 𝑑𝑠]
≤ 𝑒− ∫0 𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠) [
𝑒𝑟𝑖 𝑡
≤
𝑑𝑠]
𝑠
𝑡
𝜙𝑖 (𝑡)
≤ 𝑒−[min0≤𝑠≤𝑡 ∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)−∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)] .
𝑧𝑖 (𝑡)
(59)
So, we have
𝑡
𝑒min0≤𝑠≤𝑡 [∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏)]−∫0 𝜎𝑖 (𝑠)𝑑𝐵𝑖 (𝑠)
.
≥
𝑧𝑖 (𝑡)
(51)
𝑡
𝑠
0
0≤𝑠≤𝑡 0
∫ 𝜎𝑖 (𝜏) 𝑑𝐵𝑖 (𝜏)−max ∫ 𝜎𝑖 (𝜏) 𝑑𝐵𝑖 (𝜏)
≤ log 𝜙𝑖 (𝑡)−log 𝑧𝑖 (𝑡)
𝑡
Lemma 12. Assume that 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2, then for any given
initial value 𝑧(0) ∈ 𝑅+2 , the solution 𝑧(𝑡) of (49) has the
following properties
≤ ∫ 𝜎𝑖 (𝜏) 𝑑𝐵𝑖 (𝜏)
(60)
0
𝑠
−min ∫ 𝜎𝑖 (𝜏) 𝑑𝐵𝑖 (𝜏) .
0≤𝑠≤𝑡 0
𝑡
𝑧̃𝑖 (𝑡) ≤ 𝑧𝑖 (𝑡) ≤ 𝑧̂𝑖 (𝑡) ,
lim 𝑧̃𝑖 (𝑡) =
𝑡→∞
𝑟𝑙𝑖
,
𝑎𝑖𝑖𝑢
lim 𝑧̂𝑖 (𝑡) =
𝑡→∞
Let 𝑀𝑖 (𝑡) = ∫0 𝜎𝑖 (𝜏)𝑑𝐵𝑖 (𝜏), then
𝑟𝑢𝑖
,
𝑎𝑖𝑖𝑢
(52)
𝑡
⟨𝑀𝑖 , 𝑀𝑖 ⟩𝑡 = ∫ 𝜎𝑖2 (𝜏) 𝑑𝜏.
0
(61)
8
Abstract and Applied Analysis
Since 𝜎𝑖 (𝑡), 𝑖 = 1, 2 are bounded, then
⟨𝑀𝑖 , 𝑀𝑖 ⟩𝑡
1 𝑡
= lim ∫ 𝜎𝑖2 (𝜏) 𝑑𝜏 < ∞,
𝑡→∞
𝑡→∞ 𝑡 0
𝑡
lim
Definition 15. System (3) is said to be persistent in time
average if
a.s.
By the strong law of large numbers, we know
𝑡
∫ 𝜎𝑖 (𝜏) 𝑑𝐵𝑖 (𝜏)
𝑀𝑖 (𝑡)
= lim 0
= 0,
𝑡→∞
𝑡→∞
𝑡
𝑡
lim
a.s.
1 𝑡
lim inf ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 > 0,
𝑡→∞ 𝑡 0
(62)
(63)
𝑟𝑙
1 𝑡
lim inf ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 ≥ 𝑢𝑖 ,
𝑡→∞ 𝑡 0
𝑎𝑖𝑖
lim
log 𝑥𝑖 (𝑡)
lim inf
≥ 0,
𝑡→∞
𝑡
(64)
a.s.
log 𝜙𝑖 (𝑡)
= 0,
𝑡
(65)
a.s.
𝑡
1
lim inf ∫ 𝜙𝑖 (𝑠) 𝑑𝑠 ≥
𝑡→∞ 𝑡 0
𝑟𝑙𝑖
,
𝑎𝑖𝑖𝑢
𝑎.𝑠.
𝑟𝑙
1 𝑡
1 𝑡
lim inf ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 ≥ lim inf ∫ 𝜙𝑖 (𝑠) 𝑑𝑠 ≥ 𝑢𝑖 ,
𝑡→∞ 𝑡 0
𝑡→∞ 𝑡 0
𝑎𝑖𝑖
Integrating both sides of this equation from 0 to 𝑡 yields
𝑡
2
log 𝜙𝑖 (𝑡) log 𝜙𝑖 (0) ∫0 [𝑟𝑖 (𝑠) − (𝜎𝑖 (𝑠) /2)] 𝑑𝑠
−
=
𝑡
𝑡
𝑡
𝑡
𝑡
𝑡
+
∫0 𝜎𝑖 (𝑠) 𝑑𝐵𝑖 (𝑠)
𝑡
.
(68)
lim
𝑡→∞
𝑡
log 𝜙𝑖 (𝑡)
= lim
= 0,
𝑡→∞
𝑡
a.s.
(69)
𝜎2 (𝑠)
1
1 𝑡
1 𝑡
lim inf ∫ 𝜙𝑖 (𝑠) 𝑑𝑠 = 𝑢 lim inf ∫ [𝑟𝑖 (𝑠) − 𝑖
] 𝑑𝑠
𝑡→∞ 𝑡 0
𝑎𝑖𝑖 𝑡 → ∞ 𝑡 0
2
𝑟𝑙𝑖
,
𝑎𝑖𝑖𝑢
𝑡→∞
log 𝜙𝑖 (𝑡)
log 𝑥𝑖 (𝑡)
≥ lim inf
= 0,
𝑡→∞
𝑡
𝑡
(74)
a.s.
(75)
3.3. Asymptotic Boundedness of Integral Average. Theorem 16
shows that every component of the solution 𝑥(𝑡) of system (3)
will survive forever in time average, if the white noise is not
large. In this part, we further deduce that every component of
𝑥(𝑡) of system (3) will be an asymptotic boundedness in time
average. Before we give the result, we do some preparation
work.
Lemma 17. Let 𝑓 ∈ 𝐶[[0, ∞) × Ω, (0, ∞)], 𝐹(𝑡) ∈ ((0, ∞) ×
Ω, 𝑅). If there exist positive constants 𝜆 0 and 𝜆 such that
𝑡
0
Hence,
≥
lim inf
log 𝑓 (𝑡) ≥ 𝜆𝑡 − 𝜆 0 ∫ 𝑓 (𝑠) 𝑑𝑠 + 𝐹 (𝑡) ,
By Lemma 13, we know that
𝑡
a.s.
Hence, by Lemma 13 we know that
(𝑡)
− 𝑎𝑖𝑖𝑢 𝜙𝑖 (𝑡)] 𝑑𝑡 + 𝜎𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡) .
2
(67)
−
(73)
(66)
𝜎𝑖2
𝑎𝑖𝑖𝑢 ∫0 𝜙𝑖 (𝑠) 𝑑𝑠
𝑖 = 1, 2,
where 𝜙(𝑡) = (𝜙1 (𝑡), 𝜙2 (𝑡)) is the solution of system (40).
Moreover, by Lemma 14 we know that
Proof. By Itô’s formula, we have
∫0 𝜎𝑖 (𝑠) 𝑑𝐵𝑖 (𝑠)
𝑎.𝑠.,
Proof. By Lemma 8, we know that
𝑥𝑖 (𝑡) ≥ 𝜙𝑖 (𝑡)
Lemma 14. Assume that 𝑟𝑙𝑖 > 0, 𝑖 = 1, 2, then for any given
initial value 𝜙(0) ∈ 𝑅+2 , the solution 𝜙(𝑡) of (40) has the
properties
𝑑 log 𝜙𝑖 (𝑡) = [𝑟𝑖 (𝑡) −
(72)
and so system (3) is persistent in time average.
Then from (60) we obtain
𝑡→∞
(71)
𝑙 𝑙
𝑢 𝑢
𝑎22 > 𝑎12
𝑎21 and 𝑟𝑙𝑖 > 0, 𝑖 =
Theorem 16. Assume that 𝑎11
1, 2, then the solution 𝑥(𝑡) of system (3) with any initial value
𝑥(0) ∈ 𝑅+2 has the following property:
Thus,
󵄨󵄨 𝑀 (𝑠) 󵄨󵄨
󵄨
󵄨
lim max 󵄨󵄨󵄨 𝑖 󵄨󵄨󵄨 = 0,
𝑡 → ∞ 0≤𝑠≤𝑡 󵄨󵄨
𝑡 󵄨󵄨
𝑖 = 1, 2.
𝑡 ≥ 0, 𝑎.𝑠.,
(76)
and lim𝑡 → ∞ (𝐹(𝑡)/𝑡) = 0 a.s., then
1 𝑡
𝜆
lim inf ∫ 𝑓 (𝑠) 𝑑𝑠 ≥
, 𝑎.𝑠.
𝑡→∞ 𝑡 0
𝜆0
(77)
Proof. The proof is similar to the proof of Lemma in [21]. Let
𝑡
𝜑 (𝑡) = ∫ 𝑓 (𝑠) 𝑑𝑠.
0
(78)
Since 𝑓 ∈ 𝐶[[0, ∞) × Ω, (0, ∞)], 𝜑(𝑡) is differentiable on
[0, ∞) and
a.s.
(70)
𝑑𝜑 (𝑡)
= 𝑓 (𝑡) > 0,
𝑑𝑡
for 𝑡 ≥ 0.
(79)
Abstract and Applied Analysis
9
Substituting 𝑑𝜑(𝑡)/𝑑𝑡 and 𝜑(𝑡) into (76), we obtain the
following:
log
𝑑𝜑 (𝑡)
≥ 𝜆𝑡 − 𝜆 0 𝜑 (𝑡) + 𝐹 (𝑡) ,
𝑑𝑡
(80)
thus
𝑑𝜑 (𝑡)
𝑒𝜆 0 𝜑(𝑡)
≥ 𝑒𝜆𝑡+𝐹(𝑡) ,
𝑑𝑡
for 𝑡 ≥ 0.
𝑑𝜑 (𝑡)
≥ 𝑒(𝜆−𝜀)𝑡 ,
𝑑𝑡
for 𝑡 ≥ 𝑇, 𝜔 ∈ Ω𝜀 .
(82)
Integrating inequality (82) from 0 to 𝑡 results in the following:
𝜆 0 𝜑(𝑡)
− 𝑒𝜆 0 𝜑(𝑇) ] ≥ (𝜆 − 𝜀)−1 [𝑒(𝜆−𝜀)𝑡 − 𝑒(𝜆−𝜀)𝑇 ] .
𝜆−1
0 [𝑒
(83)
This inequality can be rewritten into
𝑒𝜆 0 𝜑(𝑡) ≥ 𝑒𝜆 0 𝜑(𝑇) + 𝜆 0 (𝜆 − 𝜀)−1 [𝑒(𝜆−𝜀)𝑡 − 𝑒(𝜆−𝜀)𝑇 ] .
(84)
Taking the logarithm of both sides and dividing both sides by
𝑡(> 0) yields
𝜆 0 𝜑(𝑇)
log {𝑒
𝜑 (𝑡)
≥ 𝜆−1
0
𝑡
+ 𝜆 0 (𝜆 − 𝜀)−1 [𝑒(𝜆−𝜀)𝑡−𝑒
(𝜆−𝜀)𝑇
1 𝑡
lim inf ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 ≥ 𝑥̃𝑖∗ ,
𝑡→∞ 𝑡 0
𝑖 = 1, 2, a.s.
(90)
1 𝑡
lim sup ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 ≤ 𝑥̂𝑖∗ ,
𝑡→∞ 𝑡 0
𝑖 = 1, 2, a.s.
(91)
(81)
Note that lim𝑡 → ∞ (𝐹(𝑡)/𝑡) = 0 a.s., then for 0 < 𝜀 <
min{1, 𝜆}, ∃𝑇 = 𝑇(𝜔) > 0 and Ω𝜀 ⊂ Ω such that 𝑃(Ω𝜀 ) > 1−𝜀
and 𝐹(𝑡) ≥ −𝜀𝑡, 𝑡 ≥ 𝑇, 𝜔 ∈ Ω𝜀 . Then we have
𝑒𝜆 0 𝜑(𝑡)
Proof. To prove the results, we only need to prove
]}
𝑡
By Itô’s formula, we have
𝑑 log 𝑥1 (𝑡)
1
= [𝑟1 (𝑡) − 𝜎12 (𝑡) − 𝑎11 (𝑡) 𝑥1 (𝑡) + 𝑎12 (𝑡) 𝑥2 (𝑡)] 𝑑𝑡
2
+ 𝜎1 (𝑡) 𝑑𝐵1 (𝑡) ,
𝑑 log 𝑥2 (𝑡)
1
= [𝑟2 (𝑡) − 𝜎22 (𝑡) + 𝑎21 (𝑡) 𝑥1 (𝑡) − 𝑎22 (𝑡) 𝑥2 (𝑡)] 𝑑𝑡
2
+ 𝜎2 (𝑡) 𝑑𝐵2 (𝑡) .
(92)
First, we prove (91). Integrating both sides of (92) from 0 to 𝑡
yields
. (85)
𝑡
𝑡
log 𝑥1 (𝑡) = log 𝑥1 (0) + ∫ 𝑟1 (𝑠) 𝑑𝑠 − ∫ 𝑎11 (𝑠) 𝑥1 (𝑠) 𝑑𝑠
Then,
0
lim inf
𝑡→∞
𝜑 (𝑡) 𝜆 − 𝜀
,
≥
𝑡
𝜆0
0
𝑡
𝜔 ∈ Ω𝜀 .
(86)
Letting 𝜀 → ∞ yields
𝑡
+ ∫ 𝑎12 (𝑠) 𝑥2 (𝑠) 𝑑𝑠 + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠) ,
0
0
𝑡
𝑡
0
0
log 𝑥2 (𝑡) = log 𝑥2 (0) + ∫ 𝑟2 (𝑠) 𝑑𝑠 − ∫ 𝑎22 (𝑠) 𝑥2 (𝑠) 𝑑𝑠
𝑡
1
𝜆
lim inf ∫ 𝑓 (𝑠) 𝑑𝑠 ≥
,
𝑡→∞ 𝑡 0
𝜆0
a.s.
(87)
𝑡
𝑡
0
0
+ ∫ 𝑎21 (𝑠) 𝑥1 (𝑠) 𝑑𝑠 + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠) ,
This finishes the proof of the Lemma.
𝑙 𝑙
𝑢 𝑢
Theorem 18. Assume that 𝑎11
𝑎22 > 𝑎12
𝑎21 and 𝑟𝑙𝑖 > 0, 𝑖 =
1, 2, then the solution 𝑥(𝑡) of system (3) with any initial value
𝑥(0) ∈ 𝑅+2 has the property
1 𝑡
∫ 𝑥 (𝑠) 𝑑𝑠 ≤ 𝑥̂𝑖∗ ,
𝑡→∞ 𝑡 0 𝑖
𝑥̃𝑖∗ ≤ lim
𝑖 = 1, 2, 𝑎.𝑠.,
(88)
where
𝑥̃1∗
(93)
where 𝑟𝑖 (𝑠) = 𝑟𝑖 (𝑠) − (1/2)𝜎𝑖2 (𝑠), 𝑖 = 1, 2. Since 𝑥𝑖 (𝑡) > 0, 𝑖 =
1, 2, hence
𝑡
𝑙
log 𝑥1 (𝑡) ≤ log 𝑥1 (0) + 𝑟𝑢1 𝑡 − 𝑎11
∫ 𝑥1 (𝑠) 𝑑𝑠
0
𝑡
𝑡
0
0
𝑢
+ 𝑎12
∫ 𝑥2 (𝑠) 𝑑𝑠 + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠) ,
𝑙 𝑙
𝑟2
𝑎𝑢 𝑟𝑙 + 𝑎12
= 𝑢22 𝑢1
,
𝑙 𝑙
𝑎11 𝑎22 − 𝑎12 𝑎21
𝑢 𝑢
𝑎𝑙 𝑟𝑢 + 𝑎12
𝑟2
,
𝑥̂1∗ = 𝑙 22 𝑙 1
𝑢 𝑢
𝑎11 𝑎22 − 𝑎12 𝑎21
𝑥̃2∗
𝑙 𝑙
𝑟1
𝑎𝑢 𝑟𝑙 + 𝑎21
= 𝑢11 𝑢2
,
𝑙 𝑙
𝑎11 𝑎22 − 𝑎12 𝑎21
𝑢 𝑢
𝑎𝑙 𝑟𝑢 + 𝑎21
𝑟1
𝑥̂2∗ = 𝑙 11 𝑙 2
.
𝑢 𝑢
𝑎11 𝑎22 − 𝑎12 𝑎21
(89)
log 𝑥2 (𝑡) ≤ log 𝑥2 (0) +
𝑟𝑢2 𝑡
−
𝑙
𝑎22
𝑡
∫ 𝑥2 (𝑠) 𝑑𝑠
0
𝑡
𝑡
0
0
𝑢
+ 𝑎21
∫ 𝑥1 (𝑠) 𝑑𝑠 + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠) .
(94)
10
Abstract and Applied Analysis
for 𝑡 > 𝑇(𝜔). It follows from (100) that, for 𝑡 > 𝑇(𝜔),
So we have
𝑡
𝑙
𝑢
𝑎22
log 𝑥1 (𝑡) + 𝑎12
log 𝑥2 (𝑡)
≤
𝑙
𝑎22
𝑢
log 𝑥1 (𝑡) ≥ log 𝑥1 (0) + 𝑟𝑙1 𝑡 − 𝑎11
∫ 𝑥1 (𝑠) 𝑑𝑠
0
𝑡
[log 𝑥1 (0) + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠)] +
0
𝑙 𝑢
𝑎22
𝑟1 𝑡
𝑡
+
𝑢
𝑎12
−
𝑙 𝑙
𝑎22
(𝑎11
[log 𝑥2 (0) + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠)] +
0
−
𝑡
𝑢 𝑢
𝑎21
𝑎12 ) ∫
0
𝑢 𝑢
𝑟2 𝑡
𝑎12
𝑡
(95)
𝑙
+ 𝑎12
(𝑁1 − 𝜀) 𝑡 + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠)
0
𝑡
𝑢
= log 𝑥1 (0) − 𝑎11
∫ 𝑥1 (𝑠) 𝑑𝑠
0
𝑡
𝑥1 (𝑠) 𝑑𝑠.
𝑙
+ [𝑟𝑙1 + 𝑎12
(𝑁1 − 𝜀)] 𝑡 + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠) .
0
By Theorem 16, we know that
lim inf
𝑡→∞
log 𝑥𝑖 (𝑡)
≥ 0,
𝑡
From Lemma 17, we have
𝑖 = 1, 2, a.s.
(96)
Obviously,
𝑡
lim
log 𝑥𝑖 (0) + ∫0 𝜎𝑖 (𝑠) 𝑑𝐵𝑖 (𝑠)
𝑡
𝑡→∞
= 0,
𝑖 = 1, 2, a.s.
(97)
𝑙
𝑟𝑙 + 𝑎12
(𝑁1 − 𝜀)
1 𝑡
lim inf ∫ 𝑥1 (𝑠) 𝑑𝑠 ≥ 1
:= 𝑀2 > 𝑀1 .
𝑢
𝑡→∞ 𝑡 0
𝑎11
(104)
Similarly, we have
𝑙
𝑟𝑙 + 𝑎21
(𝑀1 − 𝜀)
1 𝑡
lim inf ∫ 𝑥2 (𝑠) 𝑑𝑠 ≥ 2
:= 𝑁2 > 𝑁1 .
𝑢
𝑡→∞ 𝑡 0
𝑎22
(105)
Hence, we have
𝑢 𝑢
𝑎𝑙 𝑟𝑢 + 𝑎12
𝑟2
1 𝑡
≜ 𝑥̂1∗ ,
lim sup ∫ 𝑥1 (𝑠) 𝑑𝑠 ≤ 𝑙 22 𝑙 1
𝑢 𝑢
𝑡
𝑎11 𝑎22 − 𝑎21 𝑎12
0
𝑡→∞
a.s. (98)
Continuing this process, we obtain two sequences 𝑀𝑛 ,
𝑁𝑛 (𝑛 = 1, 2, . . .) such that
Similarly, we have
𝑢 𝑢
𝑟1
𝑎𝑙 𝑟𝑢 + 𝑎12
1 𝑡
lim sup ∫ 𝑥2 (𝑠) 𝑑𝑠 ≤ 𝑙 11 𝑙 2
≜ 𝑥̂2∗ ,
𝑢 𝑢
𝑡
𝑎
𝑎
−
𝑎
𝑎
0
𝑡→∞
11 22
21 12
𝑡
𝑢
log 𝑥1 (𝑡) ≥ log 𝑥1 (0) + 𝑟𝑙1 𝑡 − 𝑎11
∫ 𝑥1 (𝑠) 𝑑𝑠
0
𝑡
0
0
𝑙
+ 𝑎12
∫ 𝑥2 (𝑠) 𝑑𝑠 + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠) ,
log 𝑥2 (𝑡) ≥ log 𝑥2 (0) +
𝑟𝑙2 𝑡
−
𝑢
𝑎22
(100)
𝑡
0
0
𝑙
+ 𝑎21
∫ 𝑥1 (𝑠) 𝑑𝑠 + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠) .
𝑁𝑛 =
𝑙
𝑟𝑙2 + 𝑎21
(𝑀𝑛−1 − 𝜀)
.
𝑢
𝑎22
(107)
By induction, we can easily show that 𝑀𝑛+1 > 𝑀𝑛 , 𝑁𝑛+1 >
𝑁𝑛 , 𝑛 = 1, 2, . . ., that is, sequences {𝑀𝑛 , 𝑛 = 1, 2, . . .} and
{𝑁𝑛 , 𝑛 = 1, 2, . . .} are nondecreasing. Moreover, note that (98)
and (99), then the sequences {𝑀𝑛 , 𝑛 = 1, 2, . . .} and {𝑁𝑛 , 𝑛 =
1, 2, . . .}, have upper bounds. Therefore, there are two positive
𝑀, 𝑁 such that
lim 𝑀𝑛 = 𝑀,
lim 𝑁𝑛 = 𝑁,
𝑛→∞
1 𝑡
lim inf ∫ 𝑥2 (𝑠) 𝑑𝑠 ≥ 𝑁,
𝑡→∞ 𝑡 0
(108)
which together with (106) implies
By Theorem 16 we know that
𝑢
𝑙
𝑙
𝑎11
𝑀 − 𝑎12
𝑁 = 𝑟𝑙1 − 𝜀𝑎12
,
𝑟𝑙
1 𝑡
lim inf ∫ 𝑥1 (𝑠) 𝑑𝑠 ≥ 𝑢1 ≜ 𝑀1 ,
𝑡→∞ 𝑡 0
𝑎11
a.s.,
𝑟𝑙
1 𝑡
lim inf ∫ 𝑥2 (𝑠) 𝑑𝑠 ≥ 𝑢2 ≜ 𝑁1 ,
𝑡→∞ 𝑡 0
𝑎22
a.s.,
(101)
𝑢
𝑙
𝑙
𝑁 − 𝑎21
𝑀 = 𝑟𝑙2 − 𝜀𝑎21
.
𝑎22
𝑡
(102)
(109)
Letting 𝜀 → 0 yields
𝑀=
then for any 𝜀 > 0, there is a 𝑇(𝜔) > 0 such that
1
∫ 𝑥 (𝑠) 𝑑𝑠 ≥ 𝑁1 − 𝜀,
𝑡 0 2
(106)
1 𝑡
lim inf ∫ 𝑥1 (𝑠) 𝑑𝑠 ≥ 𝑀,
𝑡→∞ 𝑡 0
0
𝑡
𝑙
𝑟𝑙1 + 𝑎12
(𝑁𝑛−1 − 𝜀)
,
𝑢
𝑎11
𝑛→∞
∫ 𝑥2 (𝑠) 𝑑𝑠
𝑡
𝑀𝑛 =
a.s. (99)
Next, we prove that (90) is true. Taking integration both
sides of (92) from 0 to 𝑡, we have
𝑡
(103)
𝑢 𝑙
𝑙 𝑙
𝑎22
𝑟1 + 𝑎12
𝑟2
≜ 𝑥̃1∗ ,
𝑢 𝑢
𝑙 𝑙
𝑎11 𝑎22 − 𝑎12 𝑎21
𝑙 𝑙
𝑎𝑢 𝑟𝑙 + 𝑎21
𝑟2
𝑁 = 𝑢11 𝑢1
≜ 𝑥̃1∗ .
𝑙 𝑙
𝑎11 𝑎22 − 𝑎12 𝑎21
(110)
Abstract and Applied Analysis
11
If 𝐾1 < 0, then there must be
Hence,
1 𝑡
lim inf ∫ 𝑥𝑖 (𝑠) 𝑑𝑠 ≥ 𝑥̃𝑖∗ ,
𝑡→∞ 𝑡 0
𝑖 = 1, 2, a.s.,
𝑎𝑙
𝑎𝑢
lim 𝑥122 (𝑡) 𝑥212 (𝑡) = 0,
(111)
𝑡→∞
a.s.
(116)
Hence, system (3) is nonpersistent.
which is as required.
4. Nonpersistence
In this section, we discuss the dynamics of system (3) as the
white noise is getting larger. We show that system (3) will
be nonpersistent if the white noise is large, which does not
happen in the deterministic system.
Definition 19. System (3) is said to be nonpersistent, if there
are positive constants 𝑞1 , 𝑞2 such that
2
𝑞
lim ∏𝑥𝑖 𝑖 (𝑡) = 0
𝑡→∞
𝑙 𝑙
𝑢 𝑢
𝑢 𝑢
Theorem 21. Assume that 𝑎11
𝑎22 > 𝑎12
𝑎21 and (𝑎21
𝑟1 +
𝑙 𝑢
𝑟2 ) < 0, then system (3) is nonpersistent, where 𝑟𝑖 (𝑠) =
𝑎11
𝑟𝑖 (𝑠) − (𝜎𝑖2 (𝑠)/2), 𝑖 = 1, 2.
Here we omit the proof of Theorem 21 which is similar to
the proof of Theorem 20.
Remark 22. If (𝜎𝑖𝑙 )2 > 2𝑟𝑖𝑢 , 𝑖 = 1, 2, then the conditions in
Theorems 20 and 21 are obviously satisfied, respectively. That
is to say, the large white noise will lead to the population
system being non-persistent.
(112)
a.s.
𝑖=1
5. Global Attractivity
𝑙 𝑙
𝑎22
Assume that 𝑎11
𝑢 𝑢
𝑎12
𝑎21
𝑙 𝑢
𝑢 𝑢
and 𝑎22
𝑟1 +𝑎12
𝑟2
>
<
Theorem 20.
0, then system (3) is nonpersistent, where 𝑟𝑖 (𝑠) = 𝑟𝑖 (𝑠) −
(𝜎𝑖2 (𝑠)/2), 𝑖 = 1, 2.
𝑙 𝑙
𝑢 𝑢
Proof. Since 𝑥𝑖 (𝑡) > 0, 𝑖 = 1, 2 and 𝑎11
𝑎22 > 𝑎12
𝑎21 , from (93)
we have
In this section, we turn to establishing sufficient criteria for
the global attractivity of stochastic system (3).
Definition 23. Let 𝑥(𝑡), 𝑦(𝑡) be two arbitrary solutions of
system (3) with initial values 𝑥(0), 𝑦(0) ∈ 𝑅+2 , respectively.
If
󵄨
󵄨
lim 󵄨󵄨𝑥 (𝑡) − 𝑦 (𝑡)󵄨󵄨󵄨 = 0,
𝑙
𝑢
log 𝑥1 (𝑡) + 𝑎12
log 𝑥2 (𝑡)
𝑎22
𝑡→∞ 󵄨
a.s.,
(117)
𝑡
𝑙 𝑢
𝑢 𝑢
𝑙 𝑙
𝑢 𝑢
≤ {𝑎22
𝑟1 + 𝑎12
𝑟2 } 𝑡 − (𝑎11
𝑎22 − 𝑎21
𝑎12 ) ∫ 𝑥1 (𝑠) 𝑑𝑠
0
then we say system (3) is globally attractive.
𝑙 𝑙
𝑢 𝑢
Theorem 24. Assume that 𝑎11
𝑎22 > 𝑎12
𝑎21 , then system (3) is
globally attractive.
𝑡
𝑙
+ 𝑎22
[log 𝑥1 (0) + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠)]
0
Proof. Let 𝑥(𝑡), 𝑦(𝑡) be two arbitrary solutions of system (3)
with initial values 𝑥(0), 𝑦(0) ∈ 𝑅+2 . By the Itô’s formula, we
have
𝑡
𝑢
+ 𝑎12
[log 𝑥2 (0) + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠)]
0
𝑡
𝑙
≤ 𝐾1 𝑡 + 𝑎22
[log 𝑥1 (0) + ∫ 𝜎1 (𝑠) 𝑑𝐵1 (𝑠)]
1
𝑑 log 𝑥𝑖 (𝑡) = [𝑟𝑖 (𝑡) − 𝜎𝑖2 (𝑡) − 𝑎𝑖𝑖 (𝑡) 𝑥𝑖 (𝑡) + 𝑎𝑖𝑗 (𝑡) 𝑥𝑗 (𝑡)] 𝑑𝑡
2
0
𝑡
𝑢
[log 𝑥2 (0) + ∫ 𝜎2 (𝑠) 𝑑𝐵2 (𝑠)] ,
+ 𝑎12
+ 𝜎𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡) ,
0
where 𝐾1 =
𝑙 𝑢
𝑎22
𝑟1
+
𝑢 𝑢
𝑎12
𝑟2
(113)
which together with
1
𝑑 log 𝑦𝑖 (𝑡) = [𝑟𝑖 (𝑡) − 𝜎𝑖2 (𝑡) − 𝑎𝑖𝑖 (𝑡) 𝑦𝑖 (𝑡) + 𝑎𝑖𝑗 (𝑡) 𝑦𝑗 (𝑡)] 𝑑𝑡
2
+ 𝜎𝑖 (𝑡) 𝑑𝐵𝑖 (𝑡) ,
𝑡
lim
𝑙
𝑎22
[log 𝑥1 (0) + ∫0 𝜎1 (𝑠) 𝑑𝐵1 (𝑠)]
Then,
𝑡
= lim
𝑡→∞
𝑢
𝑎12
[log 𝑥2 (0) + ∫0 𝜎2 (𝑠) 𝑑𝐵2 (𝑠)]
𝑡
= 0,
a.s.,
(114)
𝑑 (log 𝑥𝑖 (𝑡) − log 𝑦𝑖 (𝑡))
= {−𝑎𝑖𝑖 (𝑡) [𝑥𝑖 (𝑡) − 𝑦𝑖 (𝑡)] + 𝑎𝑖𝑗 (𝑡) [𝑥𝑖 (𝑡) − 𝑦𝑖 (𝑡)]} 𝑑𝑡,
implies
lim
𝑖, 𝑗 = 1, 2, 𝑗 ≠ 𝑖.
(118)
𝑡
𝑡→∞
𝑖, 𝑗 = 1, 2, 𝑗 ≠ 𝑖,
1
𝑡→∞ 𝑡
𝑙
𝑢
log 𝑥1 (𝑡) + 𝑎12
log 𝑥2 (𝑡)] ≤ 𝐾1 ,
[𝑎22
a.s.
(115)
𝑖, 𝑗 = 1, 2, 𝑗 ≠ 𝑖.
(119)
12
Abstract and Applied Analysis
𝑙 𝑙
𝑢 𝑢
Since 𝑎11
𝑎22 > 𝑎12
𝑎21 , there exist two positive constants 𝑐1 , 𝑐2
which satisfy
Note that 𝑢(𝑡) = 𝑥(𝑡) − 𝑦(𝑡). Clearly, 𝑢(𝑡) ∈ 𝐶(𝑅+ , 𝑅2 ) a.s. It
is straightforward to see from (124) that
𝑢
𝑙
𝑎21
𝑐1 𝑎22
<
<
𝑢 .
𝑙
𝑐2 𝑎12
𝑎11
lim inf |𝑢 (𝑡)| = 0 a.s.
𝑙
𝑐1 𝑎11
𝑢
𝑐2 𝑎21
𝑙
𝑐2 𝑎22
󵄨
󵄨
+ 𝑐2 󵄨󵄨󵄨log 𝑥2 (𝑡) − log 𝑦2 (𝑡)󵄨󵄨󵄨 ,
lim |𝑢 (𝑡)| = 0 a.s.
𝑡→∞
(121)
𝑡 ≥ 0.
A direct calculation of the right differential 𝑑+ 𝑉(𝑡) of 𝑉(𝑡)
along the ordinary differential equation (119) leads to
𝑑+ 𝑉 (𝑡) = 𝑐1 sgn (𝑥1 (𝑡) − 𝑦1 (𝑡)) 𝑑 [log 𝑥1 (𝑡) − log 𝑦1 (𝑡)]
+ 𝑐2 sgn (𝑥2 (𝑡) − 𝑦2 (𝑡)) 𝑑 [log 𝑥2 (𝑡) − log 𝑦2 (𝑡)]
= 𝑐1 sgn (𝑥1 (𝑡) − 𝑦1 (𝑡))
+𝑎12 (𝑡) (𝑥2 (𝑡) − 𝑦2 (𝑡)) 𝑑𝑡]
+ 𝑐2 sgn (𝑥2 (𝑡) − 𝑦2 (𝑡))
× [𝑎21 (𝑡) (𝑥1 (𝑡) − 𝑦1 (𝑡)) 𝑑𝑡
−𝑎22 (𝑡) (𝑥2 (𝑡) − 𝑦2 (𝑡)) 𝑑𝑡]
󵄨
𝑙 󵄨󵄨
≤ − 𝑐1 𝑎11
󵄨󵄨𝑥1 (𝑡) − 𝑦1 (𝑡)󵄨󵄨󵄨 𝑑𝑡
󵄨
𝑢 󵄨󵄨
+ 𝑐1 𝑎12
󵄨󵄨𝑥2 (𝑡) − 𝑦2 (𝑡)󵄨󵄨󵄨 𝑑𝑡
󵄨
𝑙 󵄨󵄨
− 𝑐2 𝑎22
󵄨󵄨𝑥2 (𝑡) − 𝑦2 (𝑡)󵄨󵄨󵄨 𝑑𝑡
󵄨
𝑢 󵄨󵄨
+ 𝑐2 𝑎21
󵄨󵄨𝑥1 (𝑡) − 𝑦1 (𝑡)󵄨󵄨󵄨 𝑑𝑡
󵄨
𝑙
𝑢 󵄨󵄨
= − (𝑐1 𝑎11
− 𝑐2 𝑎21
) 󵄨󵄨𝑥1 (𝑡) − 𝑦1 (𝑡)󵄨󵄨󵄨 𝑑𝑡
󵄨
𝑙
𝑢 󵄨󵄨
− (𝑐2 𝑎22
− 𝑐1 𝑎12
) 󵄨󵄨𝑥2 (𝑡) − 𝑦2 (𝑡)󵄨󵄨󵄨 𝑑𝑡
2
󵄨
󵄨
≤ − 𝛾∑ 󵄨󵄨󵄨𝑥𝑖 (𝑡) − 𝑦𝑖 (𝑡)󵄨󵄨󵄨 𝑑𝑡,
𝑖=1
(122)
𝑙
𝑢
𝑙
𝑢
where 𝛾 = min{𝑐1 𝑎11
− 𝑐2 𝑎21
, 𝑐2 𝑎22
− 𝑐1 𝑎12
}. Integrating both
sides of (122) form 0 to 𝑡, we have
𝑡 2
󵄨
󵄨
𝑉 (𝑡) + 𝛾 ∫ ∑ 󵄨󵄨󵄨𝑥𝑖 (𝑠) − 𝑦𝑖 (𝑠)󵄨󵄨󵄨 𝑑𝑠 ≤ 𝑉 (0) < ∞.
0
(123)
𝑖=1
Let 𝑡 → ∞, we obtain
∞ 2
󵄨
󵄨
󵄨
󵄨
∫ 󵄨󵄨󵄨𝑥 (𝑠) − 𝑦 (𝑠)󵄨󵄨󵄨 𝑑𝑠 ≤ ∫ ∑ 󵄨󵄨󵄨𝑥𝑖 (𝑠) − 𝑦𝑖 (𝑠)󵄨󵄨󵄨 𝑑𝑠
0
0
𝑉 (0)
≤
< ∞ a.s.
𝛾
(126)
By Theorem 3 we obtain that the 𝑝th moment of the solution
of system (3) is bounded, the following proof is similar to the
proof of Theorem 6.2 in [15] and hence is omitted.
Acknowledgments
The work was supported by the Ph.D. Programs Foundation
of Ministry of China (no. 200918), NSFC of China (no.
10971021), and Program for Changjiang Scholars and Innovative Research Team in University.
References
× [−𝑎11 (𝑡) (𝑥1 (𝑡) − 𝑦1 (𝑡)) 𝑑𝑡
𝑖=1
(125)
Next, we prove that
𝑢
𝑐1 𝑎12
Thus,
−
> 0,
−
> 0.
Consider a Lyapunov function 𝑉(𝑡) defined by
󵄨
󵄨
𝑉 (𝑡) = 𝑐1 󵄨󵄨󵄨log 𝑥1 (𝑡) − log 𝑦1 (𝑡)󵄨󵄨󵄨
∞
𝑡→∞
(120)
(124)
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