# Stochastic Predator-Prey System Subject to Lévy Jumps

Authors: Xinzhu Meng; Xiaohong Wang
Journal: Discrete Dynamics in Nature and Society
 (2016)

---

## Abstract

This paper investigates a new nonautonomous impulsive stochastic predator-prey system with the omnivorous predator. First, we show that the system has a unique global positive solution for any given initial positive value. Second, the extinction of the system under some appropriate conditions is explored. In addition, we obtain the sufficient conditions for almost sure permanence in mean and stochastic permanence of the system by using the theory of impulsive stochastic differential equations. Finally, we discuss the biological implications of the main results and show that the large noise can make the system go extinct. Simulations are also carried out to illustrate our theoretical analysis conclusions.


---

## Body

## 1. Introduction

Omnivory is considered as a common ecological phenomenon in the natural world. Omnivorous predator feeds on both animal prey and plant, so the intrinsic growth rate for predator should be positive. For example, the giant panda is omnivorous animal, since it can eat both meat and plant such as bamboo. With the development of the economy, pollution is becoming more and more serious. Thus pollution models have widely attracted the focus of the people [1–5]. A deterministic predator-prey system with omnivorous predator in an impulsive polluted environment takes the following form:(1)
d
x
t
=
x
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
d
t
,
d
y
t
=
y
t
r
2
t
-
d
2
c
0
t
-
a
2
t
y
t
+
β
t
x
t
d
t
,
c
˙
0
t
=
k
c
e
t
-
g
c
0
t
-
m
c
0
t
,
c
˙
e
t
=
-
h
c
e
t
,
t
≠
τ
k
,
Δ
x
τ
k
=
Δ
y
τ
k
=
Δ
c
0
τ
k
=
0
,
Δ
c
e
τ
k
=
u
,
τ
k
=
k
τ
,
k
=
0,1
,
…
,where Δ
x
(
τ
k
)
=
x
(
τ
k
+
)
-
x
(
τ
k
), Δ
y
(
τ
k
)
=
y
(
τ
k
+
)
-
y
(
τ
k
), Δ
c
0
(
τ
k
)
=
c
0
(
τ
k
+
)
-
c
0
(
τ
k
), c
e
(
τ
k
)
=
c
e
(
τ
k
+
)
-
c
e
(
τ
k
), δ
i
(
i
=
1,2
), k, g, m, h, and u are positive constants, x
(
t
) and y
(
t
) denote prey and omnivorous predator densities, and c
0
(
t
) and c
e
(
t
) denote the concentrations of the toxicant in the organism and in the environment, respectively. r
i
(
t
), a
i
(
t
)
(
i
=
1,2
), and β
(
t
) are all positive bounded continuous functions on R
+
≔
[
0
,
+
∞
). r
1
(
t
) and r
2
(
t
) represent the prey intrinsic growth rate and the predator intrinsic growth rate, respectively, a
i
(
t
)
(
i
=
1,2
) are the density-dependent coefficients of the prey and the predator, β
(
t
) is the capturing rate of the predator, d
1 and d
2 are damage rates of the prey and predator by the toxicant, respectively, k represents environmental toxicant uptake rate per unit mass organism, g and m are organismal net ingestion and depuration rates of toxicant, respectively, h denotes the loss rate of toxicant from the environment itself by volatilization, and u is the amount of pulsed input concentration of the toxicant at each τ.System (1) is a deterministic model, where all parameters in the model are deterministic. However, there are some limitations in mathematical model from a biological viewpoint. Therefore, it is significant to study the effects of noises on population systems [6–9]. There are many kinds of environmental noises. First, we assume that the toxicant uptake rates are disturbed by white noise. If we still let d
i
c
0
(
t
)
(
i
=
1,2
) represent the toxicant uptake rates, then d
1
c
0
(
t
) and d
2
c
0
(
t
) can be replaced by(2)
d
1
c
0
t
⟶
d
1
c
0
t
+
σ
1
t
B
˙
1
t
,
d
2
c
0
t
⟶
d
2
c
0
t
+
σ
2
t
B
˙
2
t
,where B
˙
i
(
t
)
(
i
=
1,2
) are white noises and σ
i
(
t
)
(
i
=
1,2
) are the intensities of the white noises, which are bounded continuous functions on [
0
,
+
∞
). Then we obtain the following stochastic system with impulsive toxicant input in a polluted environment: (3)
d
x
t
=
x
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
d
t
-
σ
1
t
x
t
d
B
1
t
,
d
y
t
=
y
t
r
2
t
-
d
2
c
0
t
-
a
2
t
y
t
+
β
t
x
t
d
t
-
σ
2
t
y
t
d
B
2
t
,
c
˙
0
t
=
k
c
e
t
-
g
c
0
t
-
m
c
0
t
,
c
˙
e
t
=
-
h
c
e
t
,
t
≠
τ
k
,
Δ
x
τ
k
=
Δ
y
τ
k
=
Δ
c
0
τ
k
=
0
,
Δ
c
e
τ
k
=
u
,
τ
k
=
k
τ
,
k
=
0,1
,
…
,where B
i
(
t
) are mutually independent standard Brownian motions defined on a complete probability space (
Ω
,
F
,
{
F
}
t
≥
0
,
P
).On the other hand, populations may be affected by sudden environmental fluctuations, such as severe weather, earthquakes, floods, and epidemics. Brownian motion cannot describe these phenomena better, so it is very important to introduce Lévy noise into the population system [10]. There are many researches about autonomous stochastic predator-prey system with Lévy jumps [11, 12].Inspired by these, we focus on nonautonomous impulsive stochastic predator-prey system with white noises and Lévy jumps(4)
d
x
t
=
x
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
d
t
-
σ
1
t
x
t
d
B
1
t
+
∫
Y
x
t
-
γ
1
t
,
u
N
~
d
t
,
d
u
,
d
y
t
=
y
t
r
2
t
-
d
2
c
0
t
-
a
2
t
y
t
+
β
t
x
t
d
t
-
σ
2
t
y
t
d
B
2
t
+
∫
Y
y
t
-
γ
2
t
,
u
N
~
d
t
,
d
u
,
c
˙
0
t
=
k
c
e
t
-
g
c
0
t
-
m
c
0
t
,
c
˙
e
t
=
-
h
c
e
t
,
t
≠
τ
k
,
Δ
x
τ
k
=
Δ
y
τ
k
=
Δ
c
0
τ
k
=
0
,
Δ
c
e
τ
k
=
u
,
τ
k
=
k
τ
,
k
=
0,1
,
…
,where x
(
t
-
) and y
(
t
-
) represent the left limit of x
(
t
) and y
(
t
), respectively, N
(
d
t
,
d
u
) is a Poisson counting measure with characteristic measure ν on a measurable bounded subset Y of (
0
,
∞
) with ν
(
Y
)
<
∞, and B
i
(
i
=
1,2
) are independent of N. The Poisson counting measure is represented by N
~
(
d
t
,
d
u
)
≔
N
(
d
t
,
d
u
)
-
ν
(
d
u
)
d
t; γ
i
>
-
1
,
γ
i
:
R
+
×
Y
→
R
(
i
=
1,2
) are continuous functions on [
0
,
+
∞
), which are assumed to be periodic with period τ
>
0. Other parameters are defined as in system (1).The paper is arranged as follows. In Section2, we prove that system (4) has a global positive solution. Section 3 shows the main result; in Section 3.1 we prove the extinction of system (4). We also examine almost sure permanence in mean and the stochastic permanence of the system in Sections 3.2 and 3.3. Finally we present some simulations and conclusions to close the paper in Section 4.


## 2. Notations and Global Positive Solution

For the purpose of convenience, we introduce some notions and some lemmas which will be used for our main results. We throughout this paper assume thatx
(
t
), y
(
t
), and c
0
(
t
) are continuous at t
=
k
τ, and c
e
(
t
) is left continuous at t
=
k
τ and c
e
(
k
τ
+
)
=
l
i
m
t
→
k
τ
+
c
e
(
t
) and let (
Ω
,
F
,
{
F
}
t
≥
0
,
P
) be a complete probability space with a filtration {
F
t
}
t
≥
0 satisfying the usual conditions (i.e., it is increasing and right continuous while F
0 contains all P-null sets). Further assume that B
(
t
) is a scalar Brownian motion defined on the complete probability space Ω.We denoteR
+
2 as the positive cone in R
2; that is, R
+
2
=
{
z
=
(
z
1
,
z
2
)
∈
R
2
∣
z
i
>
0
,
i
=
1,2
}. If f
(
t
) is a bounded continuous function on [
0
,
+
∞
), we define (5)
f
t
=
1
t
∫
0
t
f
t
d
t
,
f
M
=
max
t
∈
0
,
+
∞
⁡
f
t
,
f
L
=
min
t
∈
0
,
+
∞
⁡
f
t
,
f
∗
=
lim
⁡
sup
t
→
+
∞
⁡
f
t
,
f
∗
=
lim
⁡
inf
t
→
+
∞
⁡
f
t
. Now we give some basic properties of the following subsystem of systems (1) and (4):(6)
d
c
0
t
=
k
c
e
t
-
g
c
0
t
-
m
c
0
t
d
t
,
d
c
e
t
=
-
h
c
e
t
d
t
,
t
≠
k
τ
,
k
∈
Z
+
,
Δ
c
0
t
=
0
,
Δ
c
e
t
=
u
,
t
=
k
τ
,
k
∈
Z
+
.Lemma 1 (see [3]).
System (6) has a unique positive τ-periodic solution (
c
0
∗
(
t
)
,
c
e
∗
(
t
)
)
T which is globally asymptotically stable. Moreover, c
0
(
t
)
>
c
0
∗
(
t
)
,
c
e
(
t
)
>
c
e
∗
(
t
) for all t
≥
0 if c
0
(
0
)
>
c
0
∗
(
0
)
,
c
e
(
0
)
>
c
e
∗
(
0
), where(7)
c
0
∗
t
=
c
0
∗
0
e
-
g
+
m
t
-
n
τ
+
k
u
e
-
g
+
m
t
-
n
τ
-
e
-
h
t
-
n
τ
h
-
g
-
m
1
-
e
-
h
τ
,
c
e
∗
t
=
u
e
-
h
t
-
n
τ
1
-
e
-
h
τ
,
c
0
∗
0
=
k
u
e
-
g
+
m
τ
-
e
-
h
τ
h
-
g
-
m
1
-
e
-
g
+
m
τ
1
-
e
-
h
τ
,
c
e
∗
0
=
u
1
-
e
-
h
τ
,for t
∈
(
k
τ
,
(
k
+
1
)
τ
] and k
∈
Z
+.Lemma 2.
For any positive solution(
c
0
(
t
)
,
c
e
(
t
)
) of system (1) or (4) with initial value (
c
0
0
,
c
e
0
+
)
∈
R
+
2, one has(8)
lim
t
→
+
∞
⁡
c
0
t
=
k
u
h
g
+
m
τ
≜
c
0
¯
.Proof.
Through a simple calculation, we can get(9)
∫
n
τ
n
+
1
τ
c
0
∗
t
d
t
=
k
u
h
g
+
m
. Moreover, since c
0
∗
(
t
) is a periodic function, we have (10)
lim
t
→
+
∞
⁡
c
0
t
∗
≤
lim
n
→
+
∞
⁡
1
n
τ
∫
0
n
+
1
τ
c
0
∗
t
d
t
=
lim
n
→
+
∞
⁡
n
+
1
n
τ
∫
n
τ
n
+
1
τ
c
0
∗
t
d
t
=
k
u
h
g
+
m
τ
,
lim
t
→
+
∞
⁡
c
0
t
∗
≥
lim
n
→
+
∞
⁡
1
n
+
1
τ
∫
0
n
τ
c
0
∗
t
d
t
=
lim
n
→
+
∞
⁡
n
n
+
1
τ
∫
n
-
1
τ
n
τ
c
0
∗
t
d
t
=
k
u
h
g
+
m
τ
.Hence one can observe that (11)
lim
t
→
+
∞
⁡
c
0
t
=
k
u
h
g
+
m
τ
≜
c
0
¯
.Then we show an assumption which will be used in the following proof.Assumption 3.
There exists a bounded continuous functionc
i
(
t
) such that (12)
∫
Y
γ
i
-
ln
⁡
1
+
γ
i
ν
d
u
≤
c
i
t
,
i
=
1,2
.Theorem 4.
For any given initial value(
x
(
0
)
,
y
(
0
)
)
∈
R
+
2, there is a unique solution X
(
t
)
=
(
x
(
t
)
,
y
(
t
)
) of (4) and the solution will remain in R
+
2 with probability 1; that is, X
(
t
)
∈
R
+
2 for all t
≥
0 almost surely.Proof.
The coefficient of (4) is locally Lipschitz continuous, and so, for any given initial value (
x
(
0
)
,
y
(
0
)
)
∈
R
+
2, there is a unique local solution (
x
(
t
)
,
y
(
t
)
) for t
∈
[
0
,
τ
e
], where τ
e is the explosion time. To demonstrate that this solution is global, we need to show that τ
e
=
∞ a.s. Let k
0
>
0 be sufficiently large for x
0
∈
[
1
/
k
0
,
k
0
] and y
0
∈
[
1
/
k
0
,
k
0
]. For each integer k
≥
k
0, define the stopping time(13)
τ
k
=
inf
⁡
t
∈
0
,
τ
e
:
x
t
⋷
1
k
,
k
or
y
t
⋷
1
k
,
k
,where we set inf
⁡
∅
=
∞ (∅ denotes the empty set). Obviously, τ
k is increasing as k
→
+
∞. Set τ
∞
=
l
i
m
k
→
+
∞
τ
k; thus τ
∞
≤
τ
e a.s., so we just need to demonstrate that τ
∞
=
∞. If τ
∞
=
∞ is not true, then there exist two constants T
>
0 and ε
∈
(
0,1
) such that P
{
τ
∞
≤
T
}
>
ε. Thus there is an integer k
1
≥
k
0 such that P
{
τ
k
≤
T
}
≥
ε for all k
≥
k
1.
Define a functionV
:
R
+
2
→
R
+ as follows: (14)
V
x
,
y
=
x
-
1
-
ln
⁡
x
+
y
-
1
-
ln
⁡
y
. Let T
>
0 be arbitrary. Applying Itô’s formula and Assumption 3 leads to (15)
d
V
x
,
y
=
L
V
d
t
-
x
-
1
σ
1
t
d
B
1
t
-
y
-
1
σ
2
t
d
B
2
t
+
∫
Y
x
t
-
γ
1
t
,
u
-
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
+
∫
Y
y
t
-
γ
2
t
,
u
-
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
,where (16)
L
V
=
x
-
1
r
1
t
-
d
1
c
0
t
-
a
1
t
x
-
β
t
y
+
1
2
σ
1
2
t
+
y
-
1
r
2
t
-
d
2
c
0
t
-
a
1
t
y
+
β
t
x
+
1
2
σ
2
2
t
+
∫
Y
γ
1
t
,
u
-
ln
⁡
1
+
γ
1
t
,
u
ν
d
u
+
∫
Y
γ
2
t
,
u
-
ln
⁡
1
+
γ
2
t
,
u
ν
d
u
≤
-
a
1
t
x
2
+
a
1
t
+
r
1
t
-
d
1
c
0
t
-
β
t
x
-
r
1
t
+
d
1
c
0
t
+
1
2
σ
1
2
t
+
c
1
t
+
-
a
2
t
y
2
+
a
2
t
+
r
2
t
-
d
2
c
0
t
+
β
t
y
-
r
2
t
+
d
2
c
0
t
+
1
2
σ
2
2
t
+
c
2
t
≤
K
1
,where (17)
K
1
=
max
0
≤
t
<
+
∞
⁡
∑
i
=
1
2
4
a
i
t
-
r
i
t
+
d
i
c
0
t
+
1
/
2
σ
i
2
t
+
c
i
t
+
a
i
t
+
r
i
t
-
d
i
c
0
t
+
β
t
2
4
a
i
t
.So(18)
d
V
x
,
y
≤
K
1
d
t
-
x
-
1
σ
1
t
d
B
1
t
-
y
-
1
σ
2
t
d
B
2
t
+
∫
Y
x
t
-
γ
1
t
,
u
-
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
+
∫
Y
y
t
-
γ
2
t
,
u
-
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
.Integrating (18) from 0 to τ
k
∧
T and taking the expectations for both sides result in(19)
E
V
x
τ
k
∧
T
,
y
τ
k
∧
T
≤
V
x
0
,
y
0
+
K
1
T
.Let Ω
k
=
{
τ
k
≤
T
} for k
≥
k
1; we have P
(
Ω
k
)
≥
ε. Then(20)
V
x
τ
k
∧
T
,
ω
,
y
τ
k
∧
T
,
ω
≥
k
-
1
-
ln
⁡
k
∧
1
k
-
1
+
ln
⁡
k
.It is inferred from (19) and (20) that (21)
V
x
0
,
y
0
+
K
1
T
≥
E
1
Ω
V
x
τ
k
∧
T
,
y
τ
k
∧
T
≥
ε
k
-
1
-
ln
⁡
k
∧
1
k
-
1
+
ln
⁡
k
, where 1
Ω is the indicator function of Ω
k. Let k
→
+
∞; we have that (22)
+
∞
>
V
x
0
,
y
0
+
K
1
T
≥
+
∞is a contradiction; then we have τ
∞
=
∞.
This completes the proof of Theorem4.


## 3. Main Results

### 3.1. Extinction

For convenience, we prepare the following lemma.Lemma 5.
Suppose thatη
(
t
)
∈
C
(
Ω
×
[
0
,
∞
)
,
R
+
) and let Assumption 3 hold.
(I) If there exist two positive constantsT and δ
0 such that (23)
ln
⁡
η
t
≤
∫
0
t
δ
s
d
s
-
δ
0
∫
0
t
η
s
d
s
+
α
B
t
+
∑
i
=
1
2
δ
i
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
u
N
~
d
s
,
d
u
a
.
s
. for all t
≥
T, where α, δ
1, and δ
2 are constants, then (24)
η
∗
≤
δ
∗
δ
0
a
.
s
.
,
i
f
δ
∗
≥
0
;
lim
t
→
∞
⁡
η
t
=
0
a
.
s
.
,
i
f
δ
∗
<
0
.
(II) If there exist two positive constantsT and δ
0 such that (25)
ln
⁡
η
t
≥
∫
0
t
δ
s
d
s
-
δ
0
∫
0
t
η
s
d
s
+
α
B
t
+
∑
i
=
1
2
δ
i
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
u
N
~
d
s
,
d
u
a.s. for all t
≥
T, then η
∗
≥
δ
∗
/
δ
0
a.s. provided that δ
∗
≥
0.Lemma5 is proved in Appendix by using a similar method in [13].Now, we will prove the extinction of system (4).Define(26)
b
1
t
=
r
1
t
-
1
2
σ
1
2
t
+
∫
Y
ln
⁡
1
+
γ
1
t
,
u
-
γ
1
t
,
u
ν
d
u
,
b
2
t
=
r
2
t
-
1
2
σ
2
2
t
+
∫
Y
ln
⁡
1
+
γ
2
t
,
u
-
γ
2
t
,
u
ν
d
u
,
c
0
¯
=
k
u
h
g
+
m
τ
.Theorem 6.
If(27)
b
1
∗
<
d
1
c
0
¯
,
(28)
b
2
∗
<
d
2
c
0
¯
,then(29)
lim
t
→
∞
⁡
x
t
=
0
,
lim
t
→
∞
⁡
y
t
=
0
a.s.Proof.
By Assumption3 and the strong law of large numbers for local martingales, one has (30)
lim
t
→
+
∞
⁡
1
t
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
s
,
u
N
~
d
s
,
d
u
=
0
,
lim
t
→
+
∞
⁡
∫
0
t
σ
i
s
d
B
i
s
t
=
0
a.s.
,
i
=
1,2
.Define(31)
V
1
x
=
ln
⁡
x
t
,
V
2
x
=
l
n
y
t
.Applying Itô’s formula yields(32)
d
ln
⁡
x
t
=
b
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
d
t
-
σ
1
t
d
B
1
t
+
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
,
(33)
d
ln
⁡
y
t
=
b
2
t
-
d
2
c
0
t
-
a
2
t
y
t
+
β
t
x
t
d
t
-
σ
2
t
d
B
2
t
+
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
.Integrating both sides of (32) and (33) from 0 to t, respectively, we have(34)
ln
⁡
x
t
-
ln
⁡
x
0
=
∫
0
t
b
1
s
-
d
1
c
0
s
-
a
1
s
x
s
-
β
s
y
s
d
t
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
,
(35)
ln
⁡
y
t
-
ln
⁡
y
0
=
∫
0
t
b
2
s
-
d
2
c
0
s
-
a
2
s
y
s
+
β
s
x
s
d
s
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
.From (34), we can obtain that(36)
ln
⁡
x
t
-
ln
⁡
x
0
t
≤
∫
0
t
b
1
s
d
s
t
-
d
1
∫
0
t
c
0
s
d
s
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.Taking the limit superior results in(37)
ln
⁡
x
t
t
∗
≤
lim
⁡
sup
t
→
∞
⁡
∫
0
t
b
1
s
d
s
t
-
d
1
∫
0
t
c
0
s
d
s
t
a.s.From (27) we can know that ln
⁡
x
(
t
)
/
t
∗
<
0
. Thus we have(38)
lim
t
→
∞
⁡
x
t
=
0
a.s.Applying (35) leads to(39)
ln
⁡
y
t
-
ln
⁡
y
0
t
≤
∫
0
t
b
2
s
d
s
t
+
∫
0
t
β
s
x
s
d
s
t
-
d
2
∫
0
t
c
0
s
d
s
t
-
∫
0
t
σ
2
s
d
B
2
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.Taking the limit superior, together with (28) and (38), we can see that (40)
ln
⁡
y
t
t
∗
<
0
a.s. Therefore, (41)
lim
t
→
∞
⁡
y
t
=
0
a.s. This completes the proof of Theorem 6.For simplicity, we define(42)
R
∗
=
b
2
∗
+
β
M
/
a
1
L
·
b
1
∗
d
2
+
d
1
β
M
/
a
1
L
c
0
¯
.Theorem 7.
Ifb
1
∗
>
d
1
c
0
¯ and R
∗
<
1, then (43)
b
1
∗
-
d
1
c
0
¯
a
1
M
≤
x
t
∗
≤
x
t
∗
≤
b
1
∗
-
d
1
c
0
¯
a
1
L
,
lim
t
→
∞
⁡
y
t
=
0
a.s.Proof.
According to (34), we can obtain that (44)
ln
⁡
x
t
-
ln
⁡
x
0
≤
∫
0
t
b
1
s
d
s
-
d
1
∫
0
t
c
0
s
d
s
-
∫
0
t
a
1
s
x
s
d
s
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
≤
∫
0
t
b
1
s
d
s
-
d
1
∫
0
t
c
0
s
d
s
-
a
1
L
∫
0
t
x
s
d
s
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
. It follows from Lemma 5 that (45)
x
t
∗
≤
b
1
∗
-
d
1
c
0
¯
a
1
L
a.s
. By (35), we have (46)
ln
⁡
y
t
-
ln
⁡
y
0
≤
∫
0
t
b
2
s
d
s
+
∫
0
t
β
s
x
s
d
s
-
d
2
∫
0
t
c
0
s
d
s
-
∫
0
t
a
2
s
y
s
d
s
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
≤
∫
0
t
b
2
s
d
s
t
+
β
M
b
1
∗
-
d
1
c
0
¯
a
1
L
-
d
2
∫
0
t
c
0
s
d
s
t
t
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
,and then (47)
y
t
≤
y
0
exp
⁡
∫
0
t
b
2
t
d
t
t
+
β
M
b
1
∗
-
d
1
c
0
¯
a
1
L
-
d
2
∫
0
t
c
0
s
d
s
t
t
-
∫
0
t
σ
2
t
d
B
2
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
. Since R
∗
<
1, then we know that (48)
lim
t
→
∞
⁡
y
t
=
0
a.s.
,which combined with (34) and the property of the limit superior shows that, for any ε
>
0, there exists a random number T
>
0 for t
>
T such that (49)
ln
⁡
x
t
x
0
≥
∫
0
t
b
1
s
-
d
1
c
0
s
d
s
-
β
M
ε
-
a
1
M
∫
0
t
x
t
d
t
-
∫
0
t
σ
1
t
d
B
1
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
. By Lemma 5, we have that(50)
x
t
∗
≥
b
1
∗
-
d
1
c
0
¯
a
1
M
a.s.This completes the proof of Theorem 7.


### 3.2. Permanence in Mean

In this section, we need to show the permanence in mean. Note thatb
1
∗
<
d
1
c
0
¯ implies that productiveness of the prey is less than its death loss rate; then the prey can go extinct. Naturally, we assume b
1
∗
>
d
1
c
0
¯ in the rest of this paper.Define(51)
Δ
1
=
β
L
b
1
∗
+
a
1
M
b
2
∗
-
β
L
d
1
+
a
1
M
d
2
c
0
¯
,
Δ
2
=
β
L
β
M
+
a
1
M
a
2
M
,
Δ
3
=
b
1
∗
-
d
1
c
0
¯
-
β
L
Δ
1
Δ
2
,
Δ
4
=
b
1
∗
-
d
1
c
0
¯
-
β
M
Δ
5
a
2
L
,
Δ
5
=
b
2
∗
-
d
2
c
0
¯
+
β
M
Δ
3
a
1
L
,
R
∗
=
b
2
∗
+
β
L
/
a
1
M
·
b
1
∗
d
2
+
d
1
β
L
/
a
1
M
c
0
¯
.Theorem 8.
If(52)
b
2
∗
>
d
2
c
0
¯
,
R
∗
>
1
,
Δ
3
>
0
,
Δ
4
>
0
,then(53)
Δ
4
a
1
M
≤
x
t
∗
≤
x
t
∗
≤
Δ
3
a
1
L
,
Δ
1
Δ
2
≤
y
t
∗
≤
y
t
∗
≤
Δ
5
a
2
L
.Proof.
Defineu
(
t
) such that x
(
t
)
≤
u
(
t
), where u
(
t
) satisfies(54)
d
u
t
=
u
t
r
1
t
-
a
1
t
u
t
d
t
-
σ
1
t
u
t
d
B
1
t
+
∫
Y
u
t
-
γ
1
t
,
u
N
~
d
t
,
d
u
.In [9], we know that (55)
lim
t
→
∞
⁡
ln
⁡
u
t
t
=
0
,
a.s. It can be inferred from comparison theorem that(56)
ln
⁡
x
t
t
∗
≤
lim
t
→
∞
⁡
ln
⁡
u
t
t
=
0
.From (34) and (35), we know that(57)
ln
⁡
x
t
-
ln
⁡
x
0
t
=
∫
0
t
b
1
s
d
s
t
-
∫
0
t
d
1
c
0
s
d
s
t
-
∫
0
t
a
1
s
x
s
d
s
t
-
∫
0
t
β
s
y
s
d
s
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
,
(58)
ln
⁡
y
t
-
ln
⁡
y
0
t
=
∫
0
t
b
2
s
d
s
t
-
∫
0
t
d
2
c
0
s
d
s
t
-
∫
0
t
a
2
s
y
s
d
s
t
+
∫
0
t
β
s
x
s
d
s
t
-
∫
0
t
σ
2
s
d
B
2
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.By β
L × (57) + a
1
M × (58), we obtain that(59)
β
L
×
ln
⁡
x
t
-
ln
⁡
x
0
t
+
a
1
M
×
ln
⁡
y
t
-
ln
⁡
y
0
t
≥
Δ
1
-
Δ
2
y
t
-
β
L
∫
0
t
σ
1
s
d
B
1
s
t
+
β
L
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
-
a
1
M
∫
0
t
σ
2
s
d
B
2
s
t
+
a
1
M
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.When R
∗
>
1 is used in (59), applying Lemma 5 and (56) leads to(60)
y
t
∗
≥
Δ
1
Δ
2
.According to (57) and (60), we can have(61)
ln
⁡
x
t
-
ln
⁡
x
0
t
≤
b
1
∗
-
d
1
c
0
¯
-
β
L
Δ
1
Δ
2
-
a
1
L
x
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.From the conditions of Theorem 8, we know that Δ
3
>
0 and applying Lemma 5 results in(62)
x
t
∗
≤
Δ
3
a
1
L
.By (58) and (62), we obtain that(63)
ln
⁡
y
t
-
ln
⁡
y
0
t
≤
b
2
∗
-
d
2
c
0
¯
+
β
M
Δ
3
a
1
L
-
a
2
L
y
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.From the conditions of Theorem 8, we see that Δ
5
>
0 and applying Lemma 5 leads to(64)
y
t
∗
≤
Δ
5
a
2
L
.By (58) and (64), we can have(65)
ln
⁡
x
t
-
ln
⁡
x
0
t
≥
b
1
∗
-
d
1
c
0
¯
-
β
M
Δ
5
a
2
L
-
a
1
M
x
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.According to the conditions of Theorem 8, we see that Δ
4
>
0 and applying Lemma 5 yields(66)
x
t
∗
≥
Δ
4
a
1
M
.This completes the proof of Theorem 8.


### 3.3. Stochastic Permanence

For the sake of proving stochastic permanence, we should examine thepth moment boundedness firstly.Define(67)
λ
1
=
max
t
∈
0
,
+
∞
⁡
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
,
λ
1
′
=
max
t
∈
0
,
+
∞
⁡
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
,
λ
1
″
=
max
⁡
λ
1
,
λ
1
′
,
λ
2
=
min
t
∈
0
,
+
∞
⁡
a
1
t
p
-
p
β
t
p
+
1
,
λ
2
′
=
min
t
∈
0
,
+
∞
⁡
a
2
t
p
-
p
2
β
t
p
+
1
,
λ
2
″
=
min
⁡
λ
2
,
λ
2
′
.Theorem 9.
If there exists a constantp such that(68)
∫
Y
1
+
γ
i
t
,
u
p
-
1
-
p
γ
i
t
,
u
ν
d
u
<
∞
,
(69)
λ
1
′
′
>
0
,
λ
2
′
′
>
0
,then (70)
lim
⁡
sup
t
→
∞
⁡
E
x
2
t
+
y
2
t
p
/
2
≤
K
p
a.s
.Proof.
Define(71)
V
x
,
y
=
x
p
t
+
y
p
t
.Applying Itô’s formula and (68) yields(72)
d
V
x
,
y
=
L
V
d
t
-
p
σ
1
t
x
p
t
d
B
1
t
-
p
σ
2
t
y
p
t
d
B
2
t
+
∫
Y
x
p
t
-
1
+
γ
1
t
,
u
p
-
1
N
~
d
t
,
d
u
+
∫
Y
y
p
t
-
1
+
γ
2
t
,
u
p
-
1
N
~
d
t
,
d
u
,where (73)
L
V
=
p
x
p
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
+
p
y
p
t
r
2
t
-
d
2
c
0
t
-
a
2
t
x
t
+
β
t
x
t
+
p
p
-
1
2
σ
1
2
t
x
p
t
+
σ
2
2
t
y
p
t
+
∫
Y
x
p
t
-
1
+
γ
1
t
,
u
p
-
1
-
p
γ
1
t
,
u
ν
d
u
+
∫
Y
y
p
t
-
1
+
γ
1
t
,
u
p
-
1
-
p
γ
1
t
,
u
ν
d
u
≤
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
x
p
t
+
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
y
p
t
-
a
1
t
p
x
p
+
1
t
-
β
t
p
x
p
t
y
t
-
a
2
t
p
y
p
+
1
t
+
p
β
t
x
t
y
p
t
≤
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
x
p
t
-
a
1
t
p
-
p
β
t
p
+
1
x
p
+
1
t
+
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
y
p
t
-
a
2
t
p
-
p
2
β
t
p
+
1
y
p
+
1
t
, where c
i
(
t
)
(
i
=
1,2
) is a bounded continuous function.
Integrating (72) from 0 to t and taking the expectations for both sides yield (74)
E
V
x
,
y
=
V
x
0
,
y
0
+
E
∫
0
t
L
V
d
t
, and then (75)
d
E
V
x
,
y
d
t
≤
λ
1
E
x
p
-
λ
2
E
x
p
+
1
+
λ
1
′
E
y
p
-
λ
2
′
E
y
p
+
1
≤
λ
1
′
′
E
x
p
+
y
p
-
λ
2
′
′
E
x
p
+
1
+
y
p
+
1
≤
λ
1
′
′
E
x
p
+
y
p
-
λ
2
′
′
·
2
-
1
/
p
E
x
p
+
y
p
p
+
1
/
p
=
E
x
p
+
y
p
λ
1
′
′
-
λ
2
′
′
·
2
-
1
/
p
E
x
p
+
y
p
1
/
p
. We know that the auxiliary equation(76)
z
˙
t
=
z
t
λ
1
-
λ
2
z
t
1
/
phas a globally asymptotically stable positive equilibrium z
¯
=
(
λ
1
/
λ
2
)
p. Let z
(
t
) be the solution of (76) with z
(
0
)
=
E
[
(
x
(
0
)
)
p
+
(
y
(
0
)
)
p
], and according to the comparison theorem we have E
[
x
p
+
y
p
]
≤
z
(
t
), t
≥
0
. Hence, we obtain(77)
lim
⁡
sup
t
→
+
∞
⁡
E
V
x
,
y
≤
2
1
/
p
·
λ
1
′
′
λ
2
′
′
≔
K
1
p
a.s.We find that(78)
x
2
+
y
2
p
/
2
≤
2
p
/
2
·
max
⁡
x
p
,
y
p
≤
2
p
/
2
·
V
x
,
y
,so we have(79)
lim
⁡
sup
t
→
+
∞
⁡
E
x
2
+
y
2
p
/
2
≤
2
·
λ
1
′
′
λ
2
′
′
≔
K
p
a.s.This completes the proof of Theorem 9.Define(80)
r
t
=
min
⁡
r
1
t
-
d
1
c
0
t
,
r
2
t
-
d
2
c
0
t
,
σ
t
=
max
⁡
σ
1
t
,
σ
2
t
,
H
t
=
r
t
-
σ
2
t
-
∑
i
=
1
2
∫
Y
γ
i
2
1
+
γ
i
ν
d
u
.Theorem 10.
If(81)
H
t
∗
>
0
,then system (4) is stochastically permanent.Proof.
Define(82)
V
x
,
y
=
1
x
t
+
y
t
.By applying Itô’s formula, we have(83)
d
V
x
,
y
=
L
V
d
t
+
1
x
+
y
2
σ
1
t
x
t
d
B
1
t
+
σ
2
t
y
t
d
B
2
t
+
∫
Y
1
x
t
-
+
y
t
-
+
x
t
-
γ
1
t
,
u
+
y
t
-
γ
2
t
,
u
-
1
x
t
-
+
y
t
-
N
~
d
t
,
d
u
,where (84)
L
V
=
-
1
x
+
y
2
x
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
2
t
+
r
2
t
-
d
2
c
0
t
y
t
-
a
2
t
y
2
t
+
σ
1
2
t
x
2
t
+
σ
2
2
t
y
2
t
+
2
σ
1
t
σ
2
t
x
t
y
t
x
t
+
y
t
3
+
∫
Y
1
x
+
y
+
x
γ
1
t
,
u
+
y
γ
2
t
,
u
-
1
x
+
y
+
1
x
+
y
2
x
γ
1
t
,
u
+
y
γ
2
t
,
u
ν
d
u
≤
-
1
x
t
+
y
t
r
t
-
σ
2
t
-
∑
i
=
1
2
∫
Y
γ
i
2
t
,
u
1
+
γ
i
t
,
u
ν
d
u
+
a
t
=
-
V
·
H
t
+
a
t
, where a
(
t
)
=
m
a
x
{
a
1
(
t
)
,
a
2
(
t
)
}
.
Integrating (83) from 0 to t and taking the expectation on both sides yield (85)
E
V
t
=
V
0
+
E
∫
0
t
L
V
s
d
s
. Therefore, (86)
d
E
V
t
=
E
L
V
t
d
t
≤
-
H
t
E
V
t
+
a
t
d
t
. By the variation of constants method, we can derive that (87)
E
V
t
≤
∫
0
t
a
s
exp
⁡
-
∫
s
t
H
τ
d
τ
d
s
+
V
0
exp
⁡
-
∫
0
t
H
s
d
s
, which together with (81) results in (88)
E
V
x
,
y
≤
K
,where K is a positive constant. Then, for any given ε
>
0 and constant δ
(
ε
)
=
ε
/
K, according to the Chebyshev inequality, we can know that (89)
P
1
V
x
,
y
<
δ
=
P
V
x
,
y
>
1
δ
≤
E
V
x
,
y
1
/
δ
=
δ
E
V
x
,
y
≤
ε
, and then(90)
P
1
V
x
,
y
≥
δ
≥
1
-
ε
.Therefore,(91)
lim
⁡
inf
t
→
∞
⁡
P
x
t
+
y
t
≥
δ
≥
1
-
ε
a.s.Then, we have (92)
lim
⁡
inf
t
→
∞
⁡
P
x
2
t
+
y
2
t
≥
2
2
δ
≥
1
-
ε
a.s
. Using Chebyshev inequality and Theorem 9, we can obtain (93)
lim
⁡
sup
t
→
∞
⁡
P
x
2
t
+
y
2
t
≤
χ
≥
1
-
ε
a.s.
, where χ is a constant. This completes the proof of Theorem 10.


## 3.1. Extinction

For convenience, we prepare the following lemma.Lemma 5.
Suppose thatη
(
t
)
∈
C
(
Ω
×
[
0
,
∞
)
,
R
+
) and let Assumption 3 hold.
(I) If there exist two positive constantsT and δ
0 such that (23)
ln
⁡
η
t
≤
∫
0
t
δ
s
d
s
-
δ
0
∫
0
t
η
s
d
s
+
α
B
t
+
∑
i
=
1
2
δ
i
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
u
N
~
d
s
,
d
u
a
.
s
. for all t
≥
T, where α, δ
1, and δ
2 are constants, then (24)
η
∗
≤
δ
∗
δ
0
a
.
s
.
,
i
f
δ
∗
≥
0
;
lim
t
→
∞
⁡
η
t
=
0
a
.
s
.
,
i
f
δ
∗
<
0
.
(II) If there exist two positive constantsT and δ
0 such that (25)
ln
⁡
η
t
≥
∫
0
t
δ
s
d
s
-
δ
0
∫
0
t
η
s
d
s
+
α
B
t
+
∑
i
=
1
2
δ
i
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
u
N
~
d
s
,
d
u
a.s. for all t
≥
T, then η
∗
≥
δ
∗
/
δ
0
a.s. provided that δ
∗
≥
0.Lemma5 is proved in Appendix by using a similar method in [13].Now, we will prove the extinction of system (4).Define(26)
b
1
t
=
r
1
t
-
1
2
σ
1
2
t
+
∫
Y
ln
⁡
1
+
γ
1
t
,
u
-
γ
1
t
,
u
ν
d
u
,
b
2
t
=
r
2
t
-
1
2
σ
2
2
t
+
∫
Y
ln
⁡
1
+
γ
2
t
,
u
-
γ
2
t
,
u
ν
d
u
,
c
0
¯
=
k
u
h
g
+
m
τ
.Theorem 6.
If(27)
b
1
∗
<
d
1
c
0
¯
,
(28)
b
2
∗
<
d
2
c
0
¯
,then(29)
lim
t
→
∞
⁡
x
t
=
0
,
lim
t
→
∞
⁡
y
t
=
0
a.s.Proof.
By Assumption3 and the strong law of large numbers for local martingales, one has (30)
lim
t
→
+
∞
⁡
1
t
∫
0
t
∫
Y
ln
⁡
1
+
γ
i
s
,
u
N
~
d
s
,
d
u
=
0
,
lim
t
→
+
∞
⁡
∫
0
t
σ
i
s
d
B
i
s
t
=
0
a.s.
,
i
=
1,2
.Define(31)
V
1
x
=
ln
⁡
x
t
,
V
2
x
=
l
n
y
t
.Applying Itô’s formula yields(32)
d
ln
⁡
x
t
=
b
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
d
t
-
σ
1
t
d
B
1
t
+
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
,
(33)
d
ln
⁡
y
t
=
b
2
t
-
d
2
c
0
t
-
a
2
t
y
t
+
β
t
x
t
d
t
-
σ
2
t
d
B
2
t
+
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
.Integrating both sides of (32) and (33) from 0 to t, respectively, we have(34)
ln
⁡
x
t
-
ln
⁡
x
0
=
∫
0
t
b
1
s
-
d
1
c
0
s
-
a
1
s
x
s
-
β
s
y
s
d
t
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
,
(35)
ln
⁡
y
t
-
ln
⁡
y
0
=
∫
0
t
b
2
s
-
d
2
c
0
s
-
a
2
s
y
s
+
β
s
x
s
d
s
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
.From (34), we can obtain that(36)
ln
⁡
x
t
-
ln
⁡
x
0
t
≤
∫
0
t
b
1
s
d
s
t
-
d
1
∫
0
t
c
0
s
d
s
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.Taking the limit superior results in(37)
ln
⁡
x
t
t
∗
≤
lim
⁡
sup
t
→
∞
⁡
∫
0
t
b
1
s
d
s
t
-
d
1
∫
0
t
c
0
s
d
s
t
a.s.From (27) we can know that ln
⁡
x
(
t
)
/
t
∗
<
0
. Thus we have(38)
lim
t
→
∞
⁡
x
t
=
0
a.s.Applying (35) leads to(39)
ln
⁡
y
t
-
ln
⁡
y
0
t
≤
∫
0
t
b
2
s
d
s
t
+
∫
0
t
β
s
x
s
d
s
t
-
d
2
∫
0
t
c
0
s
d
s
t
-
∫
0
t
σ
2
s
d
B
2
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.Taking the limit superior, together with (28) and (38), we can see that (40)
ln
⁡
y
t
t
∗
<
0
a.s. Therefore, (41)
lim
t
→
∞
⁡
y
t
=
0
a.s. This completes the proof of Theorem 6.For simplicity, we define(42)
R
∗
=
b
2
∗
+
β
M
/
a
1
L
·
b
1
∗
d
2
+
d
1
β
M
/
a
1
L
c
0
¯
.Theorem 7.
Ifb
1
∗
>
d
1
c
0
¯ and R
∗
<
1, then (43)
b
1
∗
-
d
1
c
0
¯
a
1
M
≤
x
t
∗
≤
x
t
∗
≤
b
1
∗
-
d
1
c
0
¯
a
1
L
,
lim
t
→
∞
⁡
y
t
=
0
a.s.Proof.
According to (34), we can obtain that (44)
ln
⁡
x
t
-
ln
⁡
x
0
≤
∫
0
t
b
1
s
d
s
-
d
1
∫
0
t
c
0
s
d
s
-
∫
0
t
a
1
s
x
s
d
s
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
≤
∫
0
t
b
1
s
d
s
-
d
1
∫
0
t
c
0
s
d
s
-
a
1
L
∫
0
t
x
s
d
s
-
∫
0
t
σ
1
s
d
B
1
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
. It follows from Lemma 5 that (45)
x
t
∗
≤
b
1
∗
-
d
1
c
0
¯
a
1
L
a.s
. By (35), we have (46)
ln
⁡
y
t
-
ln
⁡
y
0
≤
∫
0
t
b
2
s
d
s
+
∫
0
t
β
s
x
s
d
s
-
d
2
∫
0
t
c
0
s
d
s
-
∫
0
t
a
2
s
y
s
d
s
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
≤
∫
0
t
b
2
s
d
s
t
+
β
M
b
1
∗
-
d
1
c
0
¯
a
1
L
-
d
2
∫
0
t
c
0
s
d
s
t
t
-
∫
0
t
σ
2
s
d
B
2
s
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
,and then (47)
y
t
≤
y
0
exp
⁡
∫
0
t
b
2
t
d
t
t
+
β
M
b
1
∗
-
d
1
c
0
¯
a
1
L
-
d
2
∫
0
t
c
0
s
d
s
t
t
-
∫
0
t
σ
2
t
d
B
2
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
. Since R
∗
<
1, then we know that (48)
lim
t
→
∞
⁡
y
t
=
0
a.s.
,which combined with (34) and the property of the limit superior shows that, for any ε
>
0, there exists a random number T
>
0 for t
>
T such that (49)
ln
⁡
x
t
x
0
≥
∫
0
t
b
1
s
-
d
1
c
0
s
d
s
-
β
M
ε
-
a
1
M
∫
0
t
x
t
d
t
-
∫
0
t
σ
1
t
d
B
1
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
. By Lemma 5, we have that(50)
x
t
∗
≥
b
1
∗
-
d
1
c
0
¯
a
1
M
a.s.This completes the proof of Theorem 7.


## 3.2. Permanence in Mean

In this section, we need to show the permanence in mean. Note thatb
1
∗
<
d
1
c
0
¯ implies that productiveness of the prey is less than its death loss rate; then the prey can go extinct. Naturally, we assume b
1
∗
>
d
1
c
0
¯ in the rest of this paper.Define(51)
Δ
1
=
β
L
b
1
∗
+
a
1
M
b
2
∗
-
β
L
d
1
+
a
1
M
d
2
c
0
¯
,
Δ
2
=
β
L
β
M
+
a
1
M
a
2
M
,
Δ
3
=
b
1
∗
-
d
1
c
0
¯
-
β
L
Δ
1
Δ
2
,
Δ
4
=
b
1
∗
-
d
1
c
0
¯
-
β
M
Δ
5
a
2
L
,
Δ
5
=
b
2
∗
-
d
2
c
0
¯
+
β
M
Δ
3
a
1
L
,
R
∗
=
b
2
∗
+
β
L
/
a
1
M
·
b
1
∗
d
2
+
d
1
β
L
/
a
1
M
c
0
¯
.Theorem 8.
If(52)
b
2
∗
>
d
2
c
0
¯
,
R
∗
>
1
,
Δ
3
>
0
,
Δ
4
>
0
,then(53)
Δ
4
a
1
M
≤
x
t
∗
≤
x
t
∗
≤
Δ
3
a
1
L
,
Δ
1
Δ
2
≤
y
t
∗
≤
y
t
∗
≤
Δ
5
a
2
L
.Proof.
Defineu
(
t
) such that x
(
t
)
≤
u
(
t
), where u
(
t
) satisfies(54)
d
u
t
=
u
t
r
1
t
-
a
1
t
u
t
d
t
-
σ
1
t
u
t
d
B
1
t
+
∫
Y
u
t
-
γ
1
t
,
u
N
~
d
t
,
d
u
.In [9], we know that (55)
lim
t
→
∞
⁡
ln
⁡
u
t
t
=
0
,
a.s. It can be inferred from comparison theorem that(56)
ln
⁡
x
t
t
∗
≤
lim
t
→
∞
⁡
ln
⁡
u
t
t
=
0
.From (34) and (35), we know that(57)
ln
⁡
x
t
-
ln
⁡
x
0
t
=
∫
0
t
b
1
s
d
s
t
-
∫
0
t
d
1
c
0
s
d
s
t
-
∫
0
t
a
1
s
x
s
d
s
t
-
∫
0
t
β
s
y
s
d
s
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
,
(58)
ln
⁡
y
t
-
ln
⁡
y
0
t
=
∫
0
t
b
2
s
d
s
t
-
∫
0
t
d
2
c
0
s
d
s
t
-
∫
0
t
a
2
s
y
s
d
s
t
+
∫
0
t
β
s
x
s
d
s
t
-
∫
0
t
σ
2
s
d
B
2
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.By β
L × (57) + a
1
M × (58), we obtain that(59)
β
L
×
ln
⁡
x
t
-
ln
⁡
x
0
t
+
a
1
M
×
ln
⁡
y
t
-
ln
⁡
y
0
t
≥
Δ
1
-
Δ
2
y
t
-
β
L
∫
0
t
σ
1
s
d
B
1
s
t
+
β
L
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
-
a
1
M
∫
0
t
σ
2
s
d
B
2
s
t
+
a
1
M
∫
0
t
∫
Y
ln
⁡
1
+
γ
2
t
,
u
N
~
d
t
,
d
u
t
.When R
∗
>
1 is used in (59), applying Lemma 5 and (56) leads to(60)
y
t
∗
≥
Δ
1
Δ
2
.According to (57) and (60), we can have(61)
ln
⁡
x
t
-
ln
⁡
x
0
t
≤
b
1
∗
-
d
1
c
0
¯
-
β
L
Δ
1
Δ
2
-
a
1
L
x
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.From the conditions of Theorem 8, we know that Δ
3
>
0 and applying Lemma 5 results in(62)
x
t
∗
≤
Δ
3
a
1
L
.By (58) and (62), we obtain that(63)
ln
⁡
y
t
-
ln
⁡
y
0
t
≤
b
2
∗
-
d
2
c
0
¯
+
β
M
Δ
3
a
1
L
-
a
2
L
y
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.From the conditions of Theorem 8, we see that Δ
5
>
0 and applying Lemma 5 leads to(64)
y
t
∗
≤
Δ
5
a
2
L
.By (58) and (64), we can have(65)
ln
⁡
x
t
-
ln
⁡
x
0
t
≥
b
1
∗
-
d
1
c
0
¯
-
β
M
Δ
5
a
2
L
-
a
1
M
x
t
-
∫
0
t
σ
1
s
d
B
1
s
t
+
∫
0
t
∫
Y
ln
⁡
1
+
γ
1
t
,
u
N
~
d
t
,
d
u
t
.According to the conditions of Theorem 8, we see that Δ
4
>
0 and applying Lemma 5 yields(66)
x
t
∗
≥
Δ
4
a
1
M
.This completes the proof of Theorem 8.


## 3.3. Stochastic Permanence

For the sake of proving stochastic permanence, we should examine thepth moment boundedness firstly.Define(67)
λ
1
=
max
t
∈
0
,
+
∞
⁡
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
,
λ
1
′
=
max
t
∈
0
,
+
∞
⁡
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
,
λ
1
″
=
max
⁡
λ
1
,
λ
1
′
,
λ
2
=
min
t
∈
0
,
+
∞
⁡
a
1
t
p
-
p
β
t
p
+
1
,
λ
2
′
=
min
t
∈
0
,
+
∞
⁡
a
2
t
p
-
p
2
β
t
p
+
1
,
λ
2
″
=
min
⁡
λ
2
,
λ
2
′
.Theorem 9.
If there exists a constantp such that(68)
∫
Y
1
+
γ
i
t
,
u
p
-
1
-
p
γ
i
t
,
u
ν
d
u
<
∞
,
(69)
λ
1
′
′
>
0
,
λ
2
′
′
>
0
,then (70)
lim
⁡
sup
t
→
∞
⁡
E
x
2
t
+
y
2
t
p
/
2
≤
K
p
a.s
.Proof.
Define(71)
V
x
,
y
=
x
p
t
+
y
p
t
.Applying Itô’s formula and (68) yields(72)
d
V
x
,
y
=
L
V
d
t
-
p
σ
1
t
x
p
t
d
B
1
t
-
p
σ
2
t
y
p
t
d
B
2
t
+
∫
Y
x
p
t
-
1
+
γ
1
t
,
u
p
-
1
N
~
d
t
,
d
u
+
∫
Y
y
p
t
-
1
+
γ
2
t
,
u
p
-
1
N
~
d
t
,
d
u
,where (73)
L
V
=
p
x
p
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
t
-
β
t
y
t
+
p
y
p
t
r
2
t
-
d
2
c
0
t
-
a
2
t
x
t
+
β
t
x
t
+
p
p
-
1
2
σ
1
2
t
x
p
t
+
σ
2
2
t
y
p
t
+
∫
Y
x
p
t
-
1
+
γ
1
t
,
u
p
-
1
-
p
γ
1
t
,
u
ν
d
u
+
∫
Y
y
p
t
-
1
+
γ
1
t
,
u
p
-
1
-
p
γ
1
t
,
u
ν
d
u
≤
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
x
p
t
+
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
y
p
t
-
a
1
t
p
x
p
+
1
t
-
β
t
p
x
p
t
y
t
-
a
2
t
p
y
p
+
1
t
+
p
β
t
x
t
y
p
t
≤
p
r
1
t
-
p
d
1
c
0
t
+
p
p
-
1
2
σ
1
2
t
+
c
1
t
x
p
t
-
a
1
t
p
-
p
β
t
p
+
1
x
p
+
1
t
+
p
r
2
t
-
p
d
2
c
0
t
+
p
p
-
1
2
σ
2
2
t
+
c
2
t
y
p
t
-
a
2
t
p
-
p
2
β
t
p
+
1
y
p
+
1
t
, where c
i
(
t
)
(
i
=
1,2
) is a bounded continuous function.
Integrating (72) from 0 to t and taking the expectations for both sides yield (74)
E
V
x
,
y
=
V
x
0
,
y
0
+
E
∫
0
t
L
V
d
t
, and then (75)
d
E
V
x
,
y
d
t
≤
λ
1
E
x
p
-
λ
2
E
x
p
+
1
+
λ
1
′
E
y
p
-
λ
2
′
E
y
p
+
1
≤
λ
1
′
′
E
x
p
+
y
p
-
λ
2
′
′
E
x
p
+
1
+
y
p
+
1
≤
λ
1
′
′
E
x
p
+
y
p
-
λ
2
′
′
·
2
-
1
/
p
E
x
p
+
y
p
p
+
1
/
p
=
E
x
p
+
y
p
λ
1
′
′
-
λ
2
′
′
·
2
-
1
/
p
E
x
p
+
y
p
1
/
p
. We know that the auxiliary equation(76)
z
˙
t
=
z
t
λ
1
-
λ
2
z
t
1
/
phas a globally asymptotically stable positive equilibrium z
¯
=
(
λ
1
/
λ
2
)
p. Let z
(
t
) be the solution of (76) with z
(
0
)
=
E
[
(
x
(
0
)
)
p
+
(
y
(
0
)
)
p
], and according to the comparison theorem we have E
[
x
p
+
y
p
]
≤
z
(
t
), t
≥
0
. Hence, we obtain(77)
lim
⁡
sup
t
→
+
∞
⁡
E
V
x
,
y
≤
2
1
/
p
·
λ
1
′
′
λ
2
′
′
≔
K
1
p
a.s.We find that(78)
x
2
+
y
2
p
/
2
≤
2
p
/
2
·
max
⁡
x
p
,
y
p
≤
2
p
/
2
·
V
x
,
y
,so we have(79)
lim
⁡
sup
t
→
+
∞
⁡
E
x
2
+
y
2
p
/
2
≤
2
·
λ
1
′
′
λ
2
′
′
≔
K
p
a.s.This completes the proof of Theorem 9.Define(80)
r
t
=
min
⁡
r
1
t
-
d
1
c
0
t
,
r
2
t
-
d
2
c
0
t
,
σ
t
=
max
⁡
σ
1
t
,
σ
2
t
,
H
t
=
r
t
-
σ
2
t
-
∑
i
=
1
2
∫
Y
γ
i
2
1
+
γ
i
ν
d
u
.Theorem 10.
If(81)
H
t
∗
>
0
,then system (4) is stochastically permanent.Proof.
Define(82)
V
x
,
y
=
1
x
t
+
y
t
.By applying Itô’s formula, we have(83)
d
V
x
,
y
=
L
V
d
t
+
1
x
+
y
2
σ
1
t
x
t
d
B
1
t
+
σ
2
t
y
t
d
B
2
t
+
∫
Y
1
x
t
-
+
y
t
-
+
x
t
-
γ
1
t
,
u
+
y
t
-
γ
2
t
,
u
-
1
x
t
-
+
y
t
-
N
~
d
t
,
d
u
,where (84)
L
V
=
-
1
x
+
y
2
x
t
r
1
t
-
d
1
c
0
t
-
a
1
t
x
2
t
+
r
2
t
-
d
2
c
0
t
y
t
-
a
2
t
y
2
t
+
σ
1
2
t
x
2
t
+
σ
2
2
t
y
2
t
+
2
σ
1
t
σ
2
t
x
t
y
t
x
t
+
y
t
3
+
∫
Y
1
x
+
y
+
x
γ
1
t
,
u
+
y
γ
2
t
,
u
-
1
x
+
y
+
1
x
+
y
2
x
γ
1
t
,
u
+
y
γ
2
t
,
u
ν
d
u
≤
-
1
x
t
+
y
t
r
t
-
σ
2
t
-
∑
i
=
1
2
∫
Y
γ
i
2
t
,
u
1
+
γ
i
t
,
u
ν
d
u
+
a
t
=
-
V
·
H
t
+
a
t
, where a
(
t
)
=
m
a
x
{
a
1
(
t
)
,
a
2
(
t
)
}
.
Integrating (83) from 0 to t and taking the expectation on both sides yield (85)
E
V
t
=
V
0
+
E
∫
0
t
L
V
s
d
s
. Therefore, (86)
d
E
V
t
=
E
L
V
t
d
t
≤
-
H
t
E
V
t
+
a
t
d
t
. By the variation of constants method, we can derive that (87)
E
V
t
≤
∫
0
t
a
s
exp
⁡
-
∫
s
t
H
τ
d
τ
d
s
+
V
0
exp
⁡
-
∫
0
t
H
s
d
s
, which together with (81) results in (88)
E
V
x
,
y
≤
K
,where K is a positive constant. Then, for any given ε
>
0 and constant δ
(
ε
)
=
ε
/
K, according to the Chebyshev inequality, we can know that (89)
P
1
V
x
,
y
<
δ
=
P
V
x
,
y
>
1
δ
≤
E
V
x
,
y
1
/
δ
=
δ
E
V
x
,
y
≤
ε
, and then(90)
P
1
V
x
,
y
≥
δ
≥
1
-
ε
.Therefore,(91)
lim
⁡
inf
t
→
∞
⁡
P
x
t
+
y
t
≥
δ
≥
1
-
ε
a.s.Then, we have (92)
lim
⁡
inf
t
→
∞
⁡
P
x
2
t
+
y
2
t
≥
2
2
δ
≥
1
-
ε
a.s
. Using Chebyshev inequality and Theorem 9, we can obtain (93)
lim
⁡
sup
t
→
∞
⁡
P
x
2
t
+
y
2
t
≤
χ
≥
1
-
ε
a.s.
, where χ is a constant. This completes the proof of Theorem 10.


## 4. Simulations and Conclusions

In this section, we show some numerical examples, which will demonstrate our results.Choose the parameters in (4) as follows: (94)
r
1
t
=
1.8
,
r
2
t
=
1.3
,
d
1
=
0.35
,
d
2
=
0.4
,
a
1
t
=
0.3
,
a
2
t
=
0.4
,
β
t
=
0.1
,
k
=
0.5
,
g
=
0.2
,
h
=
0.2
,
m
=
0.3
,
u
=
1
,
τ
=
1
.In Figure1, we choose σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.4 and the values of γ
1
(
t
,
u
) and γ
2
(
t
,
u
) are different between (a) and (b).Figure 1
The red line and the black line representx
(
t
) and y
(
t
), respectively. The initial value is x
(
0
)
=
5, y
(
0
)
=
10, c
0
(
0
)
=
0.5, and c
e
(
0
)
=
1. In (a), x
(
t
) is permanent and y
(
t
) is extinct; in (b), both of x
(
t
) and y
(
t
) are extinct. In (c), x
(
t
) is permanent and y
(
t
) is extinct; in (d), both of x
(
t
) and y
(
t
) are extinct.
(a)
σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.4; γ
1
(
t
,
u
)
=
0.1, γ
2
(
t
,
u
)
=
0.3
(b)
σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.4; γ
1
(
t
,
u
)
=
0.3, γ
2
(
t
,
u
)
=
0.3
(c)
σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.9; γ
1
(
t
,
u
)
=
0.1, γ
2
(
t
,
u
)
=
0.1
(d)
σ
1
(
t
)
=
0.8, σ
2
(
t
)
=
0.9; γ
1
(
t
,
u
)
=
0.1, γ
2
(
t
,
u
)
=
0.1In Figure1(a), we choose γ
1
(
t
,
u
)
=
0.1, γ
2
(
t
,
u
)
=
0.3; then the conditions in Theorem 7 are satisfied. According to Theorem 7, we know that x
(
t
) is permanent and y
(
t
) is extinct.In Figure1(b), we choose γ
1
(
t
,
u
)
=
0.3, γ
2
(
t
,
u
)
=
0.3; by Theorem 6, we can obtain that both of x
(
t
) and y
(
t
) are extinct.Then, we chooseγ
1
(
t
,
u
)
=
0.1, γ
2
(
t
,
u
)
=
0.1, the values of σ
1
(
t
) and σ
2
(
t
) are different between Figures 1(c) and 1(d).In Figure1(c), we choose σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.9; then the conditions in Theorem 7 are satisfied. According to Theorem 7, we know that x
(
t
) is permanent and y
(
t
) is extinct.In Figure1(d), we choose σ
1
(
t
)
=
0.8, σ
2
(
t
)
=
0.9; by Theorem 6, we get that both of x
(
t
) and y
(
t
) are extinct.In Figure2, we choose γ
1
=
2.8, γ
2
=
2.3, σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.4, γ
1
(
t
,
u
)
=
0.1, and γ
2
(
t
,
u
)
=
0.1; other parameters are the same as in Figure 1; according to Theorems 8 and 10, we obtain that system (4) is permanent in mean and stochastically permanent.Figure 2
The red line and the black line representx
(
t
) and y
(
t
), respectively. The initial value is x
(
0
)
=
5, y
(
0
)
=
10, c
0
(
0
)
=
0.5, and c
e
(
0
)
=
1. Both x
(
t
) and y
(
t
) are stochastically permanent when σ
1
(
t
)
=
0.3, σ
2
(
t
)
=
0.4, γ
1
(
t
,
u
)
=
0.1, and γ
2
(
t
,
u
)
=
0.1.In conclusion, permanence and extinction of the predator and the prey for the intensity of white noise and Lévy noise are given in Table1.Table 1
Permanence and extinction for different noise parameters.
White noiseσ
1
White noiseσ
2
Lévy noiseγ
1
Lévy noiseγ
2
Permanence and extinction
0.3
0.4
0.1
0.3
Permanence forx and extinction for y
0.3
0.4
0.3
0.3
Extinction forx and y
0.3
0.9
0.1
0.1
Permanence forx and extinction for y
0.8
0.9
0.1
0.1
Extinction forx and y
0.3
0.4
0.1
0.1
Permanence forx and yIn this paper, we consider a nonautonomous impulsive stochastic predator-prey system with Lévy jumps, which considers the predator is omnivorous. We prove that the system has a unique global positive solution. From Theorem6 and Figure 1, we can know that if the intensities of white noise and Lévy noise are sufficiently large, then the species will be extinct (see Figures 1(b) and 1(d)). According to Theorems 8 and 10, we know that system (4) is permanent in mean and stochastically permanent under some conditions in Figure 2. The change of permanent conditions shows an important property; that is, the permanence of species has a close relationship with the intensity of white noise and Lévy noise. Obviously, white noises and Lévy noises are harmful to the permanence of populations.There are some investigations for stochastic epidemic models [14, 15]. It is very significant for the natural world. Some interesting questions deserve further investigation. One could study more realistic but more complex models. Also it is interesting to investigate evolutionary dynamics of stochastic evolutionary model, and we leave these for future work.



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