# Permanence and Extinction of Stochastic Logistic System with Feedback Control under Regime Switching

Authors: Hongxiao Hu; Ling Zhu
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

We study a stochastic logistic system with feedback control under regime switching. Sufficient conditions for extinction, non-persistence in the mean, weak persistence, and persistence in the mean are established. A very important fact is found in our results; that is, the feedback control is harmless to the permanence of species even under the regime switching and stochastic perturbation environments. Finally, some examples are introduced to illustrate the main results.


---

## Body

## 1. Introduction

The classical logistic equation with feedback control is(1)
x
˙
t
=
x
t
r
t
-
a
t
x
t
-
c
t
u
t
,
u
˙
t
=
-
e
t
u
t
+
f
t
x
t
,where x
(
t
) denotes the population size at time t and u
(
t
) is an “indirect control” variable (see [1]). It has been studied extensively and many important results on the global dynamics of solutions have been founded (see [2–5] and references therein). On the other hand, population systems in the real world are often affected by environmental noise and there are various types of environmental noise, for example, white or color noise (see [6–14] and references therein) and it has been shown that the presence of such noise affects population systems significantly. But the white and color noise are unobservable, and we can only observe the species x
(
t
). Hence, we can use the same feedback control with system (1) to regulate the species which is affected by the environmental noises. To the best of the authors’ knowledge, few scholars still consider the stochastic perturbation logistic system with feedback controls under regime switching. And we have known very little about how the feedback control affects the survival of species which is under the random factors and switching environment.In this paper, motivated by the above analysis, we will study the following stochastic logistic system with feedback control under regime switching:(2)
d
x
t
=
x
t
r
γ
t
-
a
γ
t
x
t
-
c
γ
t
u
t
d
t
+
σ
γ
t
x
t
d
B
t
,
u
˙
t
=
-
e
t
u
t
+
f
t
x
t
,where γ
(
t
) is a Markov chain on the state space S as defined in the next section, for each i
∈
S, r
(
i
) represents the intrinsic growth rate, r
(
i
)
/
a
(
i
) is the carrying capacity in regime i, a
(
i
), c
(
i
), and σ
(
i
) are nonnegative constants, and e
(
t
) and f
(
t
) are nonnegative continuous bounded functions on [
0
,
+
∞
).In this work, our purpose is to design feedback controls such that the system becomes permanent or extinct. We will establish the sufficient conditions for extinction, nonpersistence in the mean, weak persistence, and persistence in the mean of system (2).


## 2. Main Results

Throughout this paper, unless otherwise specified, let(
Ω
,
F
,
{
F
t
}
t
≥
0
,
P
) be a complete probability space with filtration {
F
t
}
t
≥
0 satisfying the usual conditions (i.e., it is right-continuous and F
0 contains all P-null set). Let B
(
t
), t
≥
0, be 1-dimension standard Brownian motion defined on this probability space. We also denote by R
+ the interval [
0
,
+
∞
) and denote by R
+
2 the set {
x
,
y
∣
x
>
0
,
y
>
0
}. Let γ
(
t
) be a right-continuous Markov chain on the probability space, taking values in a finite-state space S
=
{
1,2
,
…
,
n
} for all t
≥
0, with the generator Γ
=
(
γ
i
j
)
n
×
n given by (3)
P
γ
t
+
Δ
t
=
j
∣
γ
t
=
i
=
γ
i
j
Δ
t
+
o
Δ
t
,
i
f
j
≠
i
,
1
+
γ
i
i
Δ
t
+
o
Δ
t
,
i
f
j
=
i
,where γ
i
j
≥
0 for i
,
j
=
1,2
,
…
,
n with j
≠
i and ∑
j
=
1
n
γ
i
j
=
0 for i
=
1,2
,
…
,
n. We assume that the Markov chain γ
(
·
) is independent of the Brownian motion B
(
·
). As a standing hypothesis, we assume in this paper that the Markov chain γ
(
t
) is irreducible; that is, for any i
,
j
∈
S, one can find finite numbers i
1
,
i
2
,
…
,
i
k
∈
S such that γ
i
,
i
1
γ
i
1
,
i
2
γ
i
k
,
j
>
0. Under this condition, the Markov chain has a unique stationary distribution π
=
(
π
1
,
π
2
,
…
,
π
n
) which can be obtained by solving the following linear equation π
Γ
=
0 subject to ∑
i
=
1
n
π
i
=
1 and π
i
>
0 for all i
∈
S. For convenience and simplicity in the following discussion, define (4)
v
u
=
max
i
∈
S
⁡
v
i
,
v
l
=
min
i
∈
S
⁡
v
i
,and we also define (5)
f
u
=
sup
s
∈
R
+
⁡
f
s
,
f
l
=
inf
s
∈
R
+
⁡
f
s
,
f
∗
=
limsup
t
→
∞
⁡
1
t
∫
0
t
f
s
d
s
,
f
∗
=
liminf
t
→
∞
⁡
1
t
∫
0
t
f
s
d
s
,where f
(
s
) is a bounded function on R
+.In system (2), x
(
t
) is the size of the species and u
(
t
) is the regulator; thus, we are only interested in the positive solutions. Moreover, in order for a stochastic differential equation to have a unique global (i.e., no explosion in a 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 (cf. Mao [15]). However, the coefficients of system (2) do not satisfy the linear growth condition, though they are locally Lipschitz continuous. In this section, using the comparison theorem of stochastic equations (see [16]), we will show there is a unique positive solution with positive initial value of system (2).Theorem 1.
For any given initial value(
x
0
,
u
0
)
∈
R
+
2, there is a unique continuous positive solution (
x
(
t
)
,
u
(
t
)
) to system (2) on t
≥
0 a.s.Proof.
Since the coefficients of the equation are locally Lipschitz continuous, it is known that for any given initial value(
x
0
,
u
0
)
∈
R
+
2 there is a unique maximal local solution (
x
(
t
)
,
u
(
t
)
) for all t
∈
[
0
,
τ
e
), where τ
e is the explosion time. Furthermore, by Theorem 2.1 in [17], we have (6)
x
t
=
exp
⁡
∫
0
t
b
γ
s
-
c
γ
s
u
s
d
s
+
∫
0
t
σ
γ
s
d
B
s
1
/
x
0
+
∫
0
t
a
γ
s
exp
⁡
∫
0
s
b
γ
τ
-
c
γ
s
u
s
d
τ
+
∫
0
s
σ
γ
τ
d
B
τ
d
s
>
0
,
u
t
≥
u
0
exp
⁡
-
∫
0
t
e
s
d
s
>
0
,
∀
t
∈
0
,
τ
e
,where b
(
i
)
=
r
(
i
)
-
0.5
σ
2
(
i
) for each i
∈
S. Hence, to show this solution is global positive, we only show that τ
e
=
∞ a.s. By the first equation of (2), we have (7)
d
x
t
≤
x
t
r
γ
t
-
a
γ
t
x
t
d
t
+
σ
γ
t
x
t
d
B
t
.Consider the following auxiliary equation: (8)
d
y
t
=
y
t
r
γ
t
-
a
γ
t
y
t
d
t
+
σ
γ
t
y
t
d
B
t
,
y
0
=
x
0
.From Theorem 2.1 in [17], we known that there exists a unique continuous positive solution y
(
t
) of system (8) for any positive initial value x
0, which will remain in R
+ with probability one. Consequently, by the comparison theorem of stochastic differential equation, we have (9)
x
t
≤
y
t
∀
t
≥
0
a.s.Therefore, x
(
t
)
<
∞ for all t
>
0 a.s. By the second equation of (2), we can represent u
(
t
) by (10)
u
t
=
∫
0
t
f
s
x
s
exp
⁡
∫
0
s
e
τ
d
τ
+
u
0
exp
⁡
-
∫
0
t
e
s
d
s
.From this, we can find that if x
(
t
) is global, then u
(
t
) also is a global solution; that is, τ
e
=
∞ a.s. This completes the proof of the theorem.Now, we will discuss extinction and persistence of system (2). For convenience and simplicity in the following discussion, we denote b
(
i
)
=
r
(
i
)
-
0.5
σ
2
(
i
) and b
~
=
∑
i
=
1
n
‍
π
i
b
(
i
) for i
∈
S and write (
x
(
t
)
,
u
(
t
)
)
=
(
x
(
t
,
0
,
x
0
,
u
0
)
,
u
(
t
,
0
,
x
0
,
u
0
)
) simply for any (
x
0
,
u
0
)
∈
R
+
2. Applying Itô’s formula to ln
⁡
x
(
t
), we have that (11)
d
ln
⁡
x
t
=
b
γ
t
-
a
γ
t
x
t
-
c
γ
t
u
t
d
t
+
σ
γ
t
d
B
t
.Then, we have (12)
ln
⁡
x
t
=
ln
⁡
x
0
+
∫
0
t
b
γ
s
-
a
γ
s
x
s
-
c
γ
s
u
s
d
s
+
M
t
,where M
(
t
)
=
∫
0
t
‍
σ
(
γ
(
s
)
)
d
B
(
s
). By the second equation of system (2), we have (13)
u
t
-
u
0
=
-
∫
0
t
e
s
u
s
d
s
+
∫
0
t
f
s
x
s
d
s
.Note that M
(
t
) is a local martingale. Making use of the strong law of large numbers for local martingales (see Mao [15]), we have (14)
lim
t
→
∞
⁡
M
t
t
=
0
a.s.We denote Ω
0
=
{
l
i
m
t
→
∞
M
(
t
)
/
t
=
0
}; obviously, P
(
Ω
0
)
=
1.Theorem 2.
Ifb
~
<
0 and there is a positive constant λ such that (15)
liminf
t
→
∞
⁡
∫
t
t
+
λ
e
s
d
s
>
0
,then system (2) will go to extinction almost surely; that is, l
i
m
t
→
∞
x
(
t
)
=
0 and l
i
m
t
→
∞
u
(
t
)
=
0 a.s.Proof.
For anyω
∈
Ω
0, from (12), we have (16)
ln
⁡
x
t
,
ω
t
≤
ln
⁡
x
0
t
+
1
t
∫
0
t
b
γ
s
d
s
+
M
t
,
ω
t
.Making use of (14) and the ergodicity, we obtain l
i
m
s
u
p
t
→
∞
ln
⁡
x
(
t
,
ω
)
/
t
≤
b
~
<
0
. That is, we say l
i
m
t
→
∞
x
(
t
,
ω
)
=
0. Now, we will prove l
i
m
t
→
∞
u
(
t
,
ω
)
=
0. Since l
i
m
t
→
∞
x
(
t
,
ω
)
=
0, then, for any α
0
>
0, there is a positive constant T
0 such that x
t
,
ω
<
α
0
/
f
u for all t
≥
T
0. Consequently, from (13), we have that (17)
d
u
t
,
ω
d
t
≤
-
e
t
u
t
,
ω
+
α
0
,
∀
t
≥
T
0
.We consider the comparison equation (18)
d
v
t
d
t
=
-
e
t
v
t
+
α
0
.By Lemma 2.1 of [5] and (15), we have that for any positive constant ε there are constants δ
=
δ
(
ε
) and T
1
=
T
1
(
u
(
T
0
)
)
>
T
0 such that when α
0
<
δ, we have v
t
<
ε for all t
≥
T
1, where v
(
t
) is the solution of system (18) with initial condition v
(
T
0
)
=
u
(
T
0
,
ω
). Therefore, by the comparison theorem, we obtain u
(
t
,
ω
)
<
ε for all t
≥
T
1. Since ε is arbitrary, we have l
i
m
t
→
∞
u
(
t
,
ω
)
=
0. Since P
(
Ω
0
)
=
1, this completes the proof of the theorem.Theorem 3.
Ifa
l
>
0, e
l
>
0 and b
~
=
0, then system (2) will be nonpersistent in the mean a.s.; that is, l
i
m
t
→
∞
∫
0
t
‍
x
(
s
)
d
s
/
t
=
0 and l
i
m
t
→
∞
∫
0
t
‍
u
(
s
)
d
s
/
t
=
0 a.s.Proof.
b
~
=
0 and (14) imply that, for any ε
>
0 and ω
∈
Ω
0, there is a positive constant T
0 such that (19)
1
t
∫
0
t
b
γ
s
d
s
≤
lim
t
→
∞
⁡
1
t
∫
0
t
b
γ
s
d
s
+
ε
2
=
b
~
+
ε
2
=
ε
2
,
M
t
,
ω
≤
ε
t
2
,
∀
t
≥
T
0
.Then, it follows from (12) that ln
⁡
x
(
t
,
ω
)
≤
ln
⁡
x
0
+
ε
t
-
a
l
∫
0
t
‍
x
(
s
,
ω
)
d
s. Let h
(
t
)
=
∫
0
t
‍
x
(
s
,
ω
)
d
s; then, we have deduced that e
a
l
h
(
t
)
d
h
(
t
)
≤
x
0
e
ε
t
d
t. Integrating this inequality from T
>
T
0 to t results in e
a
l
h
(
t
)
≤
e
a
l
h
(
T
)
+
x
0
a
l
(
e
ε
t
-
e
ε
T
)
/
ε. It follows that (20)
h
t
≤
1
a
l
ln
⁡
e
a
l
h
T
+
x
0
a
l
ε
e
ε
t
-
e
ε
T
.Using the L’Hospital’s rule, we get l
i
m
s
u
p
t
→
∞
h
(
t
)
/
t
≤
ε
/
a
l. Since ε is arbitrary and x
(
t
,
ω
)
>
0
(
t
>
0
), we can obtain that l
i
m
t
→
∞
∫
0
t
‍
x
(
s
,
ω
)
d
s
/
t
=
0.
Now, we will provel
i
m
t
→
∞
∫
0
t
‍
u
(
s
,
ω
)
d
s
/
t
=
0. Dividing both sides of (13) by t, we get (21)
1
t
∫
0
t
u
s
,
ω
d
s
≤
f
u
e
l
t
∫
0
t
x
s
,
ω
d
s
-
u
t
,
ω
e
l
t
+
u
0
e
l
t
≤
f
u
e
l
t
∫
0
t
x
s
,
ω
d
s
+
u
0
e
l
t
.From l
i
m
t
→
∞
∫
0
t
‍
x
(
s
,
ω
)
d
t
/
t
=
0, letting t
→
∞ we obtain l
i
m
t
→
∞
∫
0
t
‍
u
(
s
,
ω
)
d
t
/
t
=
0. Since P
(
Ω
0
)
=
1, this completes the proof of the theorem.Theorem 4.
Ife
l
>
0 and b
~
>
0, then species x will be weakly persistent in the mean a.s.; that is, 〈
x
〉
∗
>
0 a.s.Proof.
We claim thatΩ
0
⊂
{
〈
x
〉
∗
>
0
}
. If the claim is not true, then {
〈
x
〉
∗
=
0
}
∩
Ω
0
≠
⌀. By the proof of Theorem 3, if e
l
>
0, we have 〈
u
(
t
,
ω
)
〉
∗
=
0 for any ω
∈
{
〈
x
〉
∗
=
0
}
∩
Ω
0. It is easy to see that (22)
a
γ
t
x
t
,
ω
∗
≤
a
u
x
t
,
ω
∗
=
0
,
c
γ
t
u
t
,
ω
∗
≤
c
u
u
t
,
ω
∗
=
0
.From (12), we get (23)
ln
⁡
x
t
,
ω
t
=
ln
⁡
x
0
t
+
1
t
∫
0
t
b
γ
s
d
s
-
1
t
∫
0
t
a
γ
s
x
s
,
ω
d
s
-
1
t
∫
0
t
c
γ
s
u
s
,
ω
d
s
+
M
t
,
ω
t
.Combining this equation with (14) and (22), we have (24)
limsup
t
→
∞
⁡
ln
⁡
x
t
,
ω
t
=
limsup
t
→
∞
⁡
1
t
∫
0
t
b
γ
s
d
s
=
b
~
.Hence, there are positive constant T
0 and time sequence {
t
n
} with t
n
≥
T
0 and t
n
+
1
-
t
n
≥
1 for all n
≥
1 such that (25)
ln
⁡
x
t
n
,
ω
t
n
>
b
~
2
,
M
t
,
ω
t
<
b
~
8
,
∀
t
>
T
0
.Let b
-
=
m
a
x
1
≤
i
≤
n
{
b
i
}. Any positive constant Δ
t
<
m
i
n
{
1
,
t
1
b
~
/
(
8
b
-
)
} from (12) leads to (26)
ln
⁡
x
t
n
,
ω
-
ln
⁡
x
t
,
ω
≤
∫
t
t
n
b
γ
s
d
s
+
M
t
n
,
ω
-
M
t
,
ω
≤
b
-
Δ
t
+
b
~
4
t
n
,for all t
∈
[
t
n
-
Δ
t
,
t
n
]. Combining with (25), we obtain (27)
ln
⁡
x
t
,
ω
≥
ln
⁡
x
t
n
,
ω
-
b
-
Δ
t
-
b
~
4
t
n
≥
b
~
4
t
n
-
b
-
Δ
t
≥
b
~
8
t
n
,
∀
t
∈
t
n
-
Δ
t
,
t
n
,
n
≥
1
.Consequently, (28)
1
t
n
∫
t
1
t
n
x
s
,
ω
d
s
≥
1
t
n
∑
t
1
<
t
m
≤
t
n
∫
t
m
-
Δ
t
t
m
x
s
,
ω
d
s
≥
1
t
n
∑
t
1
<
t
m
≤
t
n
Δ
t
exp
⁡
b
~
8
t
m
≥
Δ
t
t
n
exp
⁡
b
~
8
t
n
.Since b
~
>
0, l
i
m
s
u
p
t
→
∞
∫
t
1
t
n
‍
x
(
s
,
ω
)
d
s
/
t
n
=
+
∞, which contradicts with ω
∈
{
〈
x
〉
∗
=
0
}
∩
Ω
0. Therefore, Ω
0
⊂
{
〈
x
〉
∗
>
0
}; that is, 〈
x
〉
∗
>
0 a.s.Theorem 5.
Ifa
l
>
0, e
l
>
0, f
l
>
0 and b
~
>
0, then system (2) will be uniform permanent in the mean a.s. Moreover, (29)
x
_
≤
x
∗
≤
x
∗
≤
x
-
,
u
_
≤
u
∗
≤
u
∗
≤
u
-
a
.
s
.
,where x
_
=
b
~
e
l
/
(
a
u
e
l
+
c
u
f
u
), u
_
=
f
l
e
l
b
~
/
e
u
(
a
u
e
l
+
c
u
f
u
), x
-
=
b
~
(
e
u
(
a
u
e
l
+
c
u
f
u
)
-
c
l
f
l
e
l
)
/
a
l
e
u
(
a
u
e
l
+
c
u
f
u
), and u
-
=
f
u
b
~
(
e
u
(
a
u
e
l
+
c
u
f
u
)
-
c
l
f
l
e
l
)
/
a
l
e
l
e
u
(
a
u
e
l
+
c
u
f
u
).Proof.
From (13), we have (30)
∫
0
t
e
s
u
s
d
s
=
∫
0
t
f
s
x
s
d
s
-
u
t
+
u
0
≤
f
u
∫
0
t
x
s
d
s
+
u
0
.Consequently, we have (31)
∫
0
t
u
s
d
s
≤
f
u
e
l
∫
0
t
x
s
d
s
+
u
0
e
l
.For any ε
>
0 and ω
∈
Ω
0, there is T such that (32)
1
t
∫
0
t
b
γ
s
d
s
>
b
~
-
ε
2
,
M
t
t
>
-
ε
2
,
∀
t
>
T
.Substituting these inequalities and (31) into (12), we get (33)
ln
⁡
x
t
-
ln
⁡
x
0
≥
ν
t
-
a
u
+
c
u
f
u
e
l
∫
0
t
x
s
d
s
-
c
u
u
0
e
l
∀
t
≥
T
,where ν
=
b
~
-
ε. Let g
(
t
)
=
∫
0
t
‍
x
(
s
)
d
s; then, we have (34)
ln
⁡
d
g
t
d
t
-
ln
⁡
x
0
≥
ν
t
-
c
u
u
0
e
l
-
a
u
+
c
u
f
u
e
l
g
t
.In a similar discussion with Theorem 3, we can obtain 〈
x
〉
∗
≥
ν
e
l
/
(
a
u
e
l
+
c
u
f
u
). Since ε is arbitrary, we obtain 〈
x
〉
∗
≥
b
~
e
l
/
(
a
u
e
l
+
c
u
f
u
)
:
=
x
_ for all ω
∈
Ω
0.
Now, we will prove〈
u
〉
∗ also has a lower bound. From the above proof, we can imply for any ε
>
0 and ω
∈
Ω
0 that there is a positive constant T such that ∫
0
t
‍
x
(
s
)
d
s
≥
(
x
_
-
ε
)
t for all t
≥
T
. Substituting this inequality into (13), we have u
(
t
)
≥
-
e
u
∫
0
t
‍
u
(
s
)
d
s
+
f
l
(
x
_
-
ε
)
t for all t
≥
T
. Let h
(
t
)
=
∫
0
t
‍
u
(
s
)
d
s; then, we have d
h
(
t
)
/
d
t
≥
-
e
u
h
(
t
)
+
f
l
(
x
_
-
ε
)
t for all t
≥
T. Consider the following comparison equation d
y
(
t
)
/
d
t
=
-
e
u
y
(
t
)
+
f
l
(
x
_
-
ε
)
t, with initial value y
(
T
)
=
h
(
T
). By the well-known variation-of-constants formula, we have (35)
y
t
=
f
l
x
_
-
ε
∫
T
t
s
exp
⁡
-
e
u
t
-
s
d
s
+
h
T
exp
⁡
-
e
u
t
-
T
=
f
l
e
u
x
_
-
ε
t
-
1
e
u
-
T
-
1
e
u
exp
⁡
-
e
u
t
-
T
+
h
T
exp
⁡
-
e
u
t
-
T
.By the comparison theorem, we have that l
i
m
i
n
f
t
→
∞
h
(
t
)
/
t
≥
l
i
m
i
n
f
t
→
∞
y
(
t
)
/
t
=
f
l
(
x
_
-
ε
)
/
e
u. Since ε is arbitrary, we obtain that (36)
u
∗
=
liminf
t
→
∞
⁡
h
t
t
≥
f
l
e
u
x
_
≔
u
_
.
In the following, we will prove the upper bound of〈
x
〉
∗ and 〈
u
〉
∗. From (14) and (36), for any ε
>
0 and ω
∈
Ω
0, there exists a positive constant T
0 such that (37)
∫
0
t
b
s
d
s
≤
b
~
+
c
l
ε
3
t
,
∫
0
t
u
s
d
s
≥
u
_
-
ε
3
t
,
M
t
≤
c
l
ε
3
t
,for all t
≥
T
0. Substituting (37) into (12), we have (38)
ln
⁡
x
t
-
ln
⁡
x
0
≤
b
~
-
c
l
u
_
-
ε
t
-
a
l
∫
0
t
x
s
d
s
,
∀
t
≥
T
0
.Let k
(
t
)
=
∫
0
t
‍
x
(
s
)
d
s; then, we have ln
⁡
(
d
k
(
t
)
/
d
t
)
-
ln
⁡
x
0
≤
ρ
t
-
a
l
k
(
t
) for all t
≥
T
0, where ρ
=
b
~
-
c
l
(
u
_
-
ε
). In a similar discussion with Theorem 3, we consequently have 〈
x
〉
∗
≤
ρ
/
a
l. Since ε is arbitrary, we obtain (39)
x
∗
≤
b
~
-
c
l
u
_
a
l
≔
x
-
.Rewriting (13), we have (40)
∫
0
t
u
s
d
s
≤
f
u
e
l
∫
0
t
x
s
d
s
-
u
t
e
l
+
u
0
e
l
≤
f
u
e
l
∫
0
t
x
s
d
s
+
u
0
e
l
.Combine this inequality with (39), we have 〈
u
〉
∗
≤
f
u
x
-
/
e
l
:
=
u
-. This completes the proof.


## 3. Remarks

In Theorem4, we used a new method to study the weakly persistent in the mean of species x. If c
(
i
)
≡
0 for all i
∈
S, system (2) becomes a stochastic logistic system without feedback control under the regime switching which has been studied by [10]. By Theorem 4, species x is weakly persistent in the mean if b
~
>
0, and the condition a
l
>
0 is not necessary. Hence, Theorem 3 in [10] is improved by Theorem 4 in this paper.In [10], the authors studied system (2) without the feedback control and obtained the critical value between weak persistence and extinction; that is, species x is weakly persistent if b
~
>
0 and goes to extinction if b
~
<
0. Hence, from Theorems 2–5, we can find that if e
l and f
l are positive in system (2), species x has the same extinction and permanence property as the system without feedback control under the same conditions. Therefore, the feedback control is harmless to the permanence of species even under the white and color noise perturbation environments.


## 4. Numerical Simulation

In this section, we will introduce some examples to illustrate our main results. Considering system (2), we can choose that the γ
(
t
) is a Markov chain with state space S
=
{
1,2
} and generator (41)
Q
=
-
q
12
q
12
q
21
-
q
21
,
r
(
1
)
=
0.4, r
(
2
)
=
0.3, a
(
γ
(
t
)
)
≡
0.01, c
(
γ
(
t
)
)
≡
0.01, e
(
t
)
≡
0.1, f
(
t
)
≡
0.1, σ
2
(
1
)
=
0.5, and σ
2
(
2
)
=
1; then, b
(
1
)
=
r
(
1
)
-
σ
2
(
1
)
/
2
=
0.15
>
0 and b
(
2
)
=
r
(
2
)
-
σ
2
(
2
)
/
2
=
-
0.2
<
0.Example 1.
We chooseq
12
=
1 and q
21
=
0. Obviously, we have π
1
=
0, π
2
=
1, and b
~
=
b
(
2
)
=
-
0.2
<
0. By Theorem 2, we have that system (2) is almost surely extinct.Example 2.
Letq
12
=
0.4 and q
21
=
0.6. Then, (42)
π
1
=
q
21
q
12
+
q
21
=
0.6
,
π
2
=
q
12
q
12
+
q
21
=
0.4
,and b
~
=
π
1
b
(
1
)
+
π
2
b
(
2
)
=
0.01
>
0. By Theorem 5, system (2) will be uniform permanent in the mean a.s.From the two examples, we can find that the species is in the good environment when statei
=
1, and it is in the bad environment when state i
=
2. If the species stay more time in state i
=
1, it may be persistent even though b
(
2
)
<
0. On the contrary, the species may go to extinction almost surely. Therefore, we can change the dynamic property of system (2) through changing the generator Q, but the feedback control is harmless to the permanence of species.



---

*Source: 862507-2015-09-29.xml*