# Global Attractivity for Lasota–Wazewska-Type System with Patch Structure and Multiple Time-Varying Delays

Authors: Zhiwen Long; Yanxiang Tan
Journal: Complexity
 (2020)

---

## Abstract

This paper aims to study the asymptotic behavior of Lasota–Wazewska-type system with patch structure and multiple time-varying delays. Based on the fluctuation lemma and some differential inequality techniques, we prove that the positive equilibrium is a global attractor of the addressed system with small time delay. Finally, we provide an example to illustrate the feasibility of the theoretical results.


---

## Body

## 1. Introduction

In 1988, in order to describe the survival of red blood cells in animals, Wazewska–Czyzewska and Lasota in [1] presented the following delayed differential equation model(1)
x
′
t
=
−
δ
x
t
+
∑
j
=
1
m
p
j
e
−
a
j
x
t
−
τ
j
t
,wherex
t represents the number of red blood cells at time t, δ denotes the death rate of red blood cells, p
j and a
j are related to the production of red blood cells per unit time, τ
j
t represents the time required to produce a red blood cell. Since the model was proposed, there have been a large number of results about the dynamical behaviors for (1) and its modifications (see [2–7] and the references therein) due to their comprehensive practical application background.As pointed out by Yao in [8], populations usually spread between different patches for survival and development. Recently, many scholars have paid attention to the population models with patch structure and time delays (see [9–16]). As far as we know, fewer works have been done concerning with the effect of time delay on dynamical behaviors of Lasota–Wazewska-type model with patch structure. The purpose of the present paper is to establish some sufficient conditions to guarantee the global attractivity of the following Lasota–Wazewska-type delay system with patch structure(2)
x
i
′
t
=
−
δ
i
x
i
t
+
∑
j
=
1
,
j
≠
i
n
d
i
j
x
j
t
+
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
t
−
τ
i
j
t
,
i
∈
J
=
1
,
2
,
…
,
n
,wherex
i denotes the number of species in the patch i, generation delay function τ
i
j
:
R
→
0
,
+
∞ is bounded and continuous, d
i
j represents the dispersal coefficient of the species from patch j to patch i, δ
i, p
i
j, and a
i
j are all positive. In what follows, we always assume that(3)
a
i
=
max
1
≤
j
≤
m
a
i
j
>
0
,
r
i
=
max
1
≤
j
≤
m
sup
t
∈
R
τ
i
j
t
>
0
,
τ
=
max
1
≤
i
≤
n
r
i
,
0
<
δ
i
−
∑
j
=
1
,
j
≠
i
n
d
i
j
<
∑
j
=
1
m
p
i
j
for
all
i
∈
J
.For convenience, letC
=
∏
i
=
1
n
C
−
r
i
,
0
,
R and C
+
=
∏
i
=
1
n
C
−
r
i
,
0
,
0
,
+
∞. If x
i
t is defined on t
0
−
r
i
,
σ with t
0
,
σ
∈
R and i
∈
J, then we write x
t
∈
C as x
t
=
x
t
1
,
x
t
2
,
…
,
x
t
n where x
t
i
θ
=
x
i
t
+
θ for all θ
∈
−
r
i
,
0 and t
0
≤
t
<
σ and i
∈
J. Denote x
t
t
0
,
φ
x
t
;
t
0
,
φ as an admissible solution of (2) with the following admissible initial condition:(4)
x
t
0
=
φ
,
φ
∈
C
+
,
φ
i
0
>
0
,
i
∈
J
.Also, lett
0
,
η
φ be the maximal right-interval of existence of x
t
t
0
,
φ.In the following, we further assume that there exists at least one positive constantx
∗
=
x
1
∗
,
x
2
∗
,
…
,
x
n
∗ such that x
∗ is the positive equilibrium point of (2) satisfying(5)
−
δ
i
x
i
∗
+
∑
j
=
1
,
j
≠
i
n
d
i
j
x
j
∗
+
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
=
0
,
for
all
i
∈
J
.


## 2. Global Attractivity of the Positive Equilibrium Pointx
1
∗
,
x
2
∗
,
…
,
x
n
∗

First, we will discuss the properties of the solutionx
t
;
t
0
,
φ of the system (2) with (4).Lemma 1.
x
t
;
t
0
,
φ is positive and bounded on t
0
,
η
φ
, and
η
φ
=
+
∞. Moreover, lim
inf
t
→
+
∞
x
j
t
;
t
0
,
φ
>
0
,
j
∈
J
.Proof.
For simplicity, we denotex
t
;
t
0
,
φ by x
t. We first prove that(6)
x
i
t
>
0
for
all
t
∈
t
0
,
η
φ
,
i
=
1
,
2
,
…
,
n
.Assume by contradiction that there existt
∗
∈
t
0
,
η
φ and k
∈
J such that(7)
x
k
t
∗
=
0
,
x
i
t
>
0
for
all
t
∈
t
0
,
t
∗
,
i
=
1
,
2
,
…
,
n
.It follows from (2) that(8)
0
≥
D
−
x
k
t
∗
=
−
δ
k
x
k
t
∗
+
∑
j
=
1
,
j
≠
k
n
d
k
j
x
j
t
∗
+
∑
j
=
1
m
p
k
j
e
−
a
k
j
x
k
t
∗
−
τ
k
j
t
∗
>
0
.This contradiction implies that (6) holds. Furthermore, define
(9)
M
t
=
max
ξ
:
ξ
≤
t
∣
there
exists
j
0
∈
J
such
that
x
j
0
ξ
=
max
t
0
≤
s
≤
t
x
j
0
s
=
max
j
∈
J
max
t
0
≤
s
≤
t
x
j
s
.We claim thatx
i
t is bounded on t
0
,
η
φ
,
i
=
1
,
2
,
…
,
n. Otherwise, we have M
t
→
η
φ as t
→
η
φ. Moreover, we can choose k
∈
J and l
p
p
≥
1 with lim
p
→
+
∞
l
p
=
η
φ such that(10)
x
k
M
l
p
=
max
t
0
≤
s
≤
l
p
x
k
s
=
max
j
∈
J
max
t
0
≤
s
≤
l
p
x
j
sand(11)
lim
p
→
+
∞
x
k
M
l
p
=
+
∞
.According to the definition ofM
l
p,(12)
0
≤
D
−
x
k
M
l
p
=
−
δ
k
x
k
M
l
p
+
∑
j
=
1
,
j
≠
k
n
d
k
j
x
j
M
l
p
+
∑
j
=
1
m
p
k
j
e
−
a
k
j
x
k
M
l
p
−
τ
k
j
M
l
p
.Furthermore, we have(13)
δ
k
≤
∑
j
=
1
,
j
≠
k
n
d
k
j
x
j
M
l
p
x
k
M
l
p
+
∑
j
=
1
m
p
k
j
x
k
M
l
p
e
−
a
k
j
x
k
M
l
p
−
τ
k
j
M
l
p
.Lettingp
→
+
∞ gives us that(14)
δ
k
≤
∑
j
=
1
,
j
≠
k
n
d
k
j
,which contradicts with the inequality in (3). This shows that x
i
t is positive and bounded for all t
∈
t
0
,
η
φ
,
i
∈
J
. From Theorem 2.3.1 in [17], we easily obtain η
φ
=
+
∞
.Next we prove that any positive solution of (2) with (4) satisfies(15)
lim
inf
t
→
+
∞
x
j
t
>
0
,
j
∈
J
.Denotel
=
min
j
∈
J
lim
inf
t
→
+
∞
x
j
t. We claim that l
>
0. Suppose on the contrary that l
=
0. Define (16)
m
t
=
max
ξ
:
ξ
≤
t
there
exists
j
0
∈
J
such
that
x
j
o
ξ
=
min
t
0
≤
s
≤
t
x
j
0
s
=
min
j
∈
J
min
t
0
≤
s
≤
t
x
j
s
.Thenm
t
→
+
∞ as t
→
+
∞. Moreover, for a sequence t
p
p
≥
1 with lim
p
→
+
∞
t
p
=
+
∞, we can choose i
0
∈
J and a subsequence t
p
k
k
≥
1
⊆
t
p
p
≥
1 such that(17)
x
i
0
m
t
p
k
=
min
t
0
≤
s
≤
t
p
k
x
i
0
s
=
min
j
∈
J
min
t
0
≤
s
≤
t
p
k
x
j
s
,and(18)
lim
k
→
+
∞
x
i
0
m
t
p
k
=
0
.By virtue of the definition ofm
t
p
k, we obtain that D
−
x
i
0
m
t
p
k
≤
0 or(19)
δ
i
0
x
i
0
m
t
p
k
≥
∑
j
=
1
,
j
≠
i
0
n
d
i
0
j
x
j
m
t
p
k
+
∑
j
=
1
m
p
i
0
j
exp
−
a
i
0
j
x
i
0
m
t
p
k
−
τ
i
0
j
m
t
p
k
≥
∑
j
=
1
m
p
i
0
j
exp
−
a
i
0
j
sup
t
≥
t
0
−
r
i
0
x
i
0
t
,
m
t
p
k
>
t
0
+
τ
.Lettingk
→
+
∞ leads to(20)
0
≥
∑
j
=
1
m
p
i
0
j
exp
−
a
i
0
j
sup
t
≥
t
0
−
r
i
0
x
i
0
t
>
0
.This is a contradiction, and the claim holds. The proof of Lemma1 is completed.Now, we show the global attractivity ofx
1
∗
,
x
2
∗
,
…
,
x
n
∗ by the following two propositions:Proposition 2.
Ify
i
t
=
x
i
t
−
x
i
∗ is eventually nonoscillating about zero, then lim
t
→
+
∞
x
i
t
=
x
i
∗
,
f
o
r
a
l
l
i
∈
J
.Proof.
We only give the proofs for the case thaty
i
t
=
x
i
t
−
x
i
∗ is eventually nonnegative for all i
∈
J, since the eventually nonpositive case can be proved by a similar argument. In this case, we can choose T
>
t
0 such that.(21)
y
i
t
=
x
i
t
−
x
i
∗
≥
0
for
all
i
∈
J
,
t
≥
T
.Leti
∗
∈
J such that lim
sup
t
→
+
∞
y
i
∗
t
=
max
j
∈
J
lim
sup
t
→
+
∞
y
j
t. We claim that(22)
lim
sup
t
→
+
∞
y
i
∗
t
=
0
.Assume the contrary thatlim
sup
t
→
+
∞
y
i
∗
t
>
0
. In view of the fluctuation lemma [18, Lemma A.1], there exists a sequence t
k
k
≥
1 such that(23)
t
k
→
+
∞
,
y
i
∗
t
k
→
lim
sup
t
→
+
∞
y
i
∗
t
,
y
i
∗
′
t
k
→
0
as
k
→
+
∞
.It follows from (2) that(24)
y
i
∗
′
t
k
=
−
δ
i
∗
x
i
∗
t
k
+
∑
j
=
1
,
j
≠
i
∗
n
d
i
∗
j
x
j
t
k
+
∑
j
=
1
m
p
i
∗
j
e
−
a
i
∗
j
x
i
∗
t
k
−
τ
i
∗
j
t
k
,
t
k
>
T
+
τ
.Without loss of generality, we can pick a subsequence oft
k (not relabelled) such that lim
k
→
+
∞
x
i
t
k
,
lim
k
→
+
∞
x
i
∗
t
k and lim
k
→
+
∞
x
i
∗
t
k
−
τ
i
∗
j
t
k exist for all i
∈
J
,
j
∈
I
=
1
,
2
,
…
,
m. Then,(25)
x
i
∗
∗
≤
lim
k
→
+
∞
x
i
∗
t
k
−
τ
i
∗
j
t
k
≤
x
i
∗
∗
+
lim
sup
t
→
+
∞
y
i
∗
t
=
lim
k
→
+
∞
x
i
∗
t
k
,
j
∈
I
.It follows from (24) that (taking limits)(26)
0
≤
−
δ
i
∗
lim
sup
t
→
+
∞
y
i
∗
t
+
x
i
∗
∗
+
∑
j
=
1
,
j
≠
i
∗
n
d
i
∗
j
lim
sup
t
→
+
∞
y
i
∗
t
+
x
j
∗
+
∑
j
=
1
m
p
i
∗
j
e
−
a
i
∗
j
x
i
∗
∗
<
−
δ
i
∗
+
∑
j
=
1
,
j
≠
i
∗
m
d
i
∗
j
lim
sup
t
→
+
∞
y
i
∗
t
<
0
,which leads to a contradiction. Hence,lim
sup
t
→
+
∞
y
i
∗
t
=
0
. This completes the proof.Remark 3.
It is worth noting that, from Proposition2 the nonoscillating solutions of system (2) converge to the positive equilibrium point which does not depend on the delays.Inspired by Theorem 4.1 in [19], we can obtain the following more general conclusion.Lemma 4.
Letv
≤
0
≤
u be such that u
≤
e
−
v
−
1 and k
v
≥
e
−
k
u
−
1 for some k
≥
1. Then u
=
v
=
0.Proof.
Sinceu
≤
e
−
v
−
1 and k
v
≥
e
−
k
u
−
1, we have(27)
1
+
u
≤
e
−
v
≤
e
1
/
k
−
1
/
k
e
−
k
u
.Letf
x
=
e
1
/
k
−
1
/
k
e
−
k
x
−
x
−
1. Then,(28)
f
′
x
=
e
1
/
k
−
k
x
−
1
/
k
e
−
k
x
−
1and(29)
f
″
x
=
e
1
/
k
−
k
x
−
1
/
k
e
−
k
x
−
k
+
e
−
k
x
≤
0
,
for
all
x
>
0
.Due to the fact thatf
′
0
=
0, it follows that(30)
f
′
x
<
0
,
for
all
x
>
0
.On the other hand, sincef
0
=
0, we get f
x
<
0 for all x
>
0. This implies that u
=
0. According to k
v
≥
e
−
k
u
−
1, we have v
≥
0. As v
≤
0, we must have v
=
0. This finishes the proof.Next, we consider the attractivity of (2) on the premise that the conditions in Proposition 5 are not satisfied.Let(31)
y
i
t
=
x
i
t
x
i
∗
−
1
,
i
∈
J
.Then, from (2), we get(32)
y
i
′
t
=
−
δ
i
y
i
t
+
1
+
∑
j
=
1
,
j
≠
i
n
d
i
j
x
j
∗
x
i
∗
y
j
t
+
1
+
∑
j
=
1
m
p
i
j
1
x
i
∗
e
−
a
i
j
x
i
∗
y
i
t
−
τ
i
j
t
+
1
=
−
δ
i
y
i
t
+
1
+
∑
j
=
1
,
j
≠
i
n
d
i
j
∗
y
j
t
+
1
+
∑
j
=
1
m
p
i
j
∗
e
−
a
i
j
x
i
∗
y
i
t
−
τ
i
j
t
+
1
,whered
i
j
∗
=
d
i
j
x
j
∗
/
x
i
∗
,
p
i
j
∗
=
p
i
j
1
/
x
i
∗. It is easy to see that the global attractivity of the equilibrium x
∗
=
x
1
∗
,
x
2
∗
,
…
,
x
n
∗ for (2) is equivalent to the global attractivity of the trivial solution for (35).Set(33)
λ
=
min
i
∈
J
lim
inf
t
→
+
∞
y
i
t
,
μ
=
max
i
∈
J
lim
sup
t
→
+
∞
y
i
t
.By the fluctuation lemma [14, Lemma A.1], we can take i
1
∗
∈
J and a strictly monotonically increasing sequence l
q
∗
q
≥
1 such that(34)
l
q
∗
→
+
∞
,
y
i
1
∗
l
q
∗
→
μ
,
as
q
→
+
∞
.Furthermore, choosei
1
∈
J and a subsequence l
q
∗
∗
q
≥
1
⊆
l
q
∗
q
≥
1 satisfying(35)
l
q
∗
∗
>
t
0
,
y
i
1
l
¯
q
=
max
t
∈
l
q
∗
∗
,
l
q
+
1
∗
∗
y
i
1
t
=
max
i
∈
J
max
t
∈
l
q
∗
∗
,
l
q
+
1
∗
∗
y
i
t
>
0
,
l
¯
q
∈
l
q
∗
∗
,
l
q
+
1
∗
∗
.By the boundedness ofy
i
1
′
l
¯
q, we pick a strictly monotonically increasing sequence l
q
q
≥
1
⊆
l
¯
q
q
≥
1 such that lim
q
→
+
∞
y
i
1
′
l
q exists. It follows from (35) that(36)
y
i
1
l
q
>
0
,
lim
q
→
+
∞
l
q
=
+
∞
,
lim
q
→
+
∞
y
i
1
l
q
=
μ
,
lim
q
→
+
∞
y
i
1
′
l
q
≥
0
.Adopting the same procedure as in the proof of (12), there exist i
2
∈
J and a strictly monotonically increasing sequence s
q
q
≥
1 such that(37)
y
i
2
s
q
<
0
,
lim
q
→
+
∞
s
q
=
+
∞
,
lim
q
→
+
∞
y
i
2
s
q
=
λ
,
lim
q
→
+
∞
y
i
2
′
s
q
≤
0
.Subsequently, we prove that there is a positive integerQ
1 such that, for any q
≥
Q
1, there exists L
q
∈
l
q
−
r
i
1
,
l
q such that(38)
y
i
1
L
q
=
0
,
y
i
1
t
>
0
,
for
all
t
∈
L
q
,
l
q
.In the contrary case, there exists a subsequence ofl
q (for convenience, we still denote by l
q) such that(39)
y
i
1
t
>
0
,
for
all
t
∈
l
q
−
r
i
1
,
l
q
,
q
=
1
,
2
,
…
.According to the definition of thel
q, we conclude from (32) that(40)
y
i
1
′
l
q
=
−
δ
i
1
y
i
1
l
q
+
1
+
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
y
j
l
q
+
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
j
x
i
1
∗
y
i
1
l
q
−
τ
i
1
j
l
q
+
1
<
−
δ
i
1
y
i
1
l
q
+
1
+
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
y
j
l
q
+
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
j
x
i
1
∗
,
l
q
>
t
0
∗
+
r
i
1
.Assume thatlim
q
→
+
∞
y
j
l
q exists for all j
∈
J
\
i
1, from the fact that lim
sup
t
→
+
∞
y
i
t
≤
μ
i
∈
J, (40) implies that(41)
0
≤
lim
q
→
+
∞
y
i
1
′
l
q
≤
−
δ
i
1
lim
q
→
+
∞
y
i
1
l
q
+
1
+
∑
j
=
1
,
j
≠
i
m
d
i
1
j
∗
lim
q
→
+
∞
y
j
l
q
+
1
+
∑
j
=
1
m
p
i
1
j
e
−
a
i
1
j
x
i
1
∗
≤
−
δ
i
1
μ
+
1
+
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
μ
+
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
j
x
i
1
∗
<
μ
+
1
−
δ
i
1
+
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
j
x
i
1
∗
=
0
,which is a contradiction and (38) is true.Based on the above discussions, the following conclusion can be drawn.Proposition 5.
Suppose that the assumption mentioned in Proposition2 does not hold, and(42)
a
i
e
δ
i
r
i
−
1
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
δ
i
+
e
δ
i
r
i
−
1
/
x
i
∗
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
≤
1
,
f
o
r
a
l
l
i
∈
J
.(43)
a
i
1
x
i
1
∗
=
min
1
≤
i
≤
n
a
i
x
i
∗
,Thenlim
t
→
+
∞
x
i
t
=
x
i
∗
,
i
∈
J
.Proof.
Observe that, from Theorem6, −
1
<
λ
≤
0
≤
μ
. Now, we show that λ
=
μ
=
0
. Otherwise, either μ
>
0 or λ
<
0 holds. We only consider the case that μ
>
0 holds (the situation is analogous for λ
<
0).For any givenε
>
0 such that λ
−
ε
>
−
1, by (33) there exists a positive integer q
∗
>
Q
2
+
t
0
∗ such that(44)
λ
−
ε
<
y
i
t
<
μ
+
ε
for
all
t
>
min
l
q
∗
,
s
q
∗
−
2
τ
,
i
∈
J
.Furthermore, from the fact that(45)
y
i
′
t
=
−
δ
i
y
i
t
+
1
+
∑
j
=
1
,
j
≠
i
n
d
i
j
∗
y
j
t
+
1
+
∑
j
=
1
m
p
i
j
∗
e
−
a
i
1
j
x
i
∗
y
i
t
−
τ
i
j
t
+
1
,
t
∈
L
q
,
l
q
,we have(46)
y
i
1
l
q
e
δ
i
1
l
q
≤
−
e
δ
i
1
l
q
−
e
δ
i
1
L
q
+
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
∫
L
q
l
q
y
i
t
+
1
e
δ
i
1
t
d
t
+
∑
j
=
1
m
p
i
1
j
∗
∫
L
q
l
q
e
−
a
i
1
j
x
i
∗
y
i
1
t
q
−
τ
i
1
j
t
q
+
1
e
δ
i
1
t
d
t
≤
−
δ
i
1
+
μ
+
ε
+
1
∑
j
=
1
,
j
≠
i
1
n
d
i
1
j
∗
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
j
x
i
1
∗
λ
−
ε
e
δ
i
1
l
q
−
e
δ
i
1
L
q
δ
i
1
=
−
δ
i
1
+
μ
+
ε
+
1
δ
i
1
−
∑
j
=
1
m
p
i
1
j
∗
e
−
x
i
1
∗
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
1
j
x
i
1
∗
λ
−
ε
e
δ
i
1
l
q
−
e
δ
i
1
L
q
δ
i
1
=
μ
+
ε
δ
i
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
1
j
x
i
1
∗
λ
−
ε
−
1
−
μ
+
ε
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
δ
i
1
l
q
−
e
δ
i
1
L
q
δ
i
1
,
q
>
q
∗
,that is,(47)
y
i
1
l
q
≤
μ
+
ε
δ
i
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
1
j
x
i
1
∗
λ
−
ε
−
1
−
μ
+
ε
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
1
−
e
−
δ
i
1
r
i
1
δ
i
1
,
q
>
q
∗
,Lettingq
→
+
∞ and ε
→
0
+, (36) and (47) give us that(48)
μ
≤
μ
1
−
e
−
δ
i
1
r
i
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
1
j
x
i
1
∗
λ
−
1
−
μ
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
1
−
e
−
δ
i
1
r
i
1
δ
i
1
,which implies(49)
μ
≤
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
1
−
e
−
δ
i
1
r
i
1
/
δ
i
1
e
−
δ
i
1
r
i
1
+
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
1
−
e
−
δ
i
1
r
i
1
/
δ
i
1
e
−
a
i
1
x
i
1
∗
λ
−
1
=
e
δ
i
1
r
i
1
−
1
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
δ
i
1
+
e
δ
i
1
r
i
1
−
1
∑
j
=
1
m
p
i
1
j
∗
e
−
a
i
1
j
x
i
1
∗
e
−
a
i
1
x
i
1
∗
λ
−
1
.Let’s assume thatλ
<
0 (if λ
=
0, from (49) and μ
≥
0 we have μ
=
0). Using the same arguments in the proof of (38), we can obtain that there is a positive integer Q
2
>
Q
1 such that, for any q
≥
Q
2, there exists S
q
∈
s
q
−
r
i
2
,
s
q such that(50)
y
i
2
S
q
=
0
,
and
y
i
2
t
<
0
,
for
all
t
∈
S
q
,
s
q
.Again from (32), we have(51)
y
i
2
s
q
e
δ
i
2
s
q
≥
−
e
δ
i
2
s
q
−
e
δ
i
2
S
q
+
∑
j
=
1
,
j
≠
i
2
m
d
i
2
j
∗
∫
S
q
s
q
y
i
t
+
1
e
δ
i
2
t
d
t
+
∑
j
=
1
m
p
i
2
j
∗
∫
S
q
s
q
e
−
a
i
2
j
x
i
2
∗
y
i
s
q
−
τ
i
2
j
s
q
+
1
e
δ
i
2
t
d
t
≥
−
δ
i
2
+
λ
−
ε
+
1
∑
j
=
1
,
j
≠
i
2
m
d
i
2
j
∗
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
j
x
i
2
∗
μ
+
ε
e
δ
i
2
s
q
−
e
δ
i
2
S
q
δ
i
2
=
−
δ
i
2
+
λ
−
ε
+
1
δ
i
2
−
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
j
x
i
2
∗
μ
+
ε
e
δ
i
2
s
q
−
e
δ
i
2
S
q
δ
i
2
≥
λ
−
ε
δ
i
2
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
x
i
2
∗
μ
+
ε
−
1
−
λ
−
ε
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
δ
i
2
s
q
−
e
δ
i
2
S
q
δ
i
2
,
q
>
q
∗
,that is,(52)
y
i
2
s
q
≥
λ
−
ε
δ
i
2
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
x
i
2
∗
μ
+
ε
−
1
−
λ
−
ε
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
1
−
e
−
δ
i
2
r
i
2
δ
i
2
,
q
>
q
∗
.Lettingq
→
+
∞ and ε
→
0
+, (37) and (52) give us that(53)
λ
≥
λ
1
−
e
−
δ
i
2
r
i
2
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
x
i
2
∗
μ
−
1
−
λ
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
1
−
e
−
δ
i
2
r
i
2
δ
i
2
.which implies(54)
λ
≥
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
1
−
e
−
δ
i
2
r
i
2
/
δ
i
2
e
−
δ
i
2
r
i
2
+
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
1
−
e
−
δ
i
2
r
i
2
/
δ
i
2
e
−
a
i
2
x
i
2
∗
μ
−
1
=
e
δ
i
2
r
i
2
−
1
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
δ
i
2
+
e
δ
i
2
r
i
2
−
1
∑
j
=
1
m
p
i
2
j
∗
e
−
a
i
2
j
x
i
2
∗
e
−
a
i
2
x
i
2
∗
μ
−
1
.This, combined with (42) and (49) imply that(55)
a
i
1
x
i
1
∗
μ
≤
e
−
a
i
1
x
i
1
∗
λ
−
1
,
a
i
2
x
i
2
∗
λ
≥
e
−
a
i
2
x
i
2
∗
μ
−
1
.Denote(56)
μ
∗
=
a
i
1
x
i
1
∗
μ
,
λ
∗
=
a
i
1
x
i
1
∗
λ
,
k
=
a
i
2
x
i
2
∗
a
i
1
x
i
1
∗
.From (43), one can find that k
≥
1. By virtue of (55) and (56), we have(57)
μ
∗
≤
e
−
λ
∗
−
1
,
k
λ
∗
≥
e
−
k
μ
∗
−
1
.According to Lemma4, it is easy to see that λ
∗
=
μ
∗
=
0. Furthermore, we get λ
=
μ
=
0, which is a contradiction, and μ
>
0 is not true. This completes the proof.Combining Propositions2 and 5, we have the following delay-dependent criterion of global attraction.Theorem 6.
Assume that conditions (3), (42) and (43) are satisfied. Then the positive equilibrium x
1
∗
,
x
2
∗
,
…
,
x
n
∗ is a global attractor of (2).Remark 7.
Let us note that, from(58)
lim
r
i
→
0
+
a
i
e
δ
i
r
i
−
1
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
δ
i
+
e
δ
i
r
i
−
1
/
x
i
∗
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
=
0
for
all
i
∈
J
,
one can find that condition (42) naturally holds under the sufficiently small delay, and the positive equilibrium point x
1
∗
,
x
2
∗
,
…
,
x
n
∗ is a global attractor of (2) with small delays. Moreover,(59)
lim
r
i
→
+
∞
a
i
e
δ
i
r
i
−
1
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
δ
i
+
e
δ
i
r
i
−
1
/
x
i
∗
∑
j
=
1
m
p
i
j
e
−
a
i
j
x
i
∗
=
a
i
x
i
∗
for
all
i
∈
J
,
implies that condition (42) is not satisfied when the delays in (2) is sufficiently large and a
i
x
i
∗
>
1 for i
∈
J.


## 3. A Numerical Example

Example 1.
Consider the following Lasota–Wazewska-type delay system with two groups and patch structures:(60)
x
1
′
t
=
−
8
x
1
t
+
3
x
2
t
+
4
e
5
e
−
x
1
t
−
τ
11
t
+
6
e
e
−
1
/
5
−
x
1
t
−
τ
12
t
,
x
2
′
t
=
−
5
x
2
t
+
4
x
1
t
+
10
e
5
e
−
1
/
2
x
2
t
−
τ
21
t
+
20
e
5
/
2
e
−
1
/
4
−
x
2
t
−
τ
22
t
.
It is easy to check that(61)
a
1
=
1
,
a
2
=
1
2
,
0
<
8
−
3
<
4
e
5
+
6
e
,
0
<
5
−
4
<
10
e
5
+
20
e
5
/
2
.
Therefore, (3) is true. Obviously, x
1
∗
,
x
2
∗
=
5
,
10 is the positive equilibrium point of (60). In addition, a
1
x
1
∗
=
a
2
x
2
∗
=
5 implies (43) holds. Now, we choose(62)
τ
i
j
t
=
0.05
sin
i
+
j
t
,
i
,
j
=
1
,
2
,
such that (42) holds. By Theorem 6, we conclude that the positive equilibrium point 5
,
10 is a global attractor of (60) with delays (62). This implies that small delays are harmless on the asymptotic behavior of system (60). Numerical runs with Matlab illustrate convergence of positive solutions to x
1
∗
,
x
2
∗ (see Figure 1).Figure 1
Numerical solutionx
1
t
,
x
2
t of Eq. (60) with different initial values x
1
s
,
x
2
s
≡
4
,
9
,
6
,
10
,
7
,
11
,
s
∈
−
0.05
,
0
.Remark 8.
Observe that the methods used in [13, 14] are not suitable for (60) with (62) since the system (60) with time-varying delays (62) does not generate a semiflow. In addition, it is also worth pointing out that the components of the positive equilibrium point in this paper are not required to be equal, which is also different from the literature [12].



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