# A Model for Influence of Nuclear-Electricity Industry on Area Economy

Authors: Zigen Ouyang; Hui Wang
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

A model for Influence of Nuclear-Electricity Industry on Area Economy is established in this paper; we obtain an existence condition of solutions. Further, the approximation conditions are also constructed; the economic aggregate will tends to the equilibriums point as time tend to infinity.


---

## Body

## 1. Introduction

As a clean energy resource, nuclear electricity is developed more and more quickly in recent years; many nuclear-electricity stations are under construction or planned to construct in China, which will be more beneficial to the development of the economy of the country. On one hand, the development of economy will not be restricted by the absence of electricity. On the other hand, the output of coal electricity will be reduced; then our resource of coal can be saved largely.With the development of nuclear-electricity speedup in China, many regional governments hope that there are some nuclear-electricity stations which could be established in their provinces or counties. This is because if a nuclear-electricity station is established in their provinces or counties, their regional economy will develop faster and faster.By the time, many authors investigated the nuclear-electricity industry [1–11], and more and more authors are interested in area economy [12–20]. However, few authors have investigated how much influence of nuclear-electricity industry on area economy is. Recently, Lixin [21] investigated the economy of source energy of Jiangsu Province. Using the method of dynamics systems, he brought light to the relation of area economy of Jiangsu Province with energy resource.In this paper, we investigate the influence of nuclear-electricity industry on area economy. First, we establish a mathematics model, show the existence of the solutions of the model, investigate the property of solutions, and obtain how much the influence is. For convenience, we consider the influence of nuclear-electricity industry on a county economy. For example, we consider the influence of nuclear-electricity station of Taojiang on the economy of Taojiang county.The paper is arranged as follows. In Section2, we will establish a model about nuclear-electricity economy. We predigest the model in Section 3. In Section 4, We will investigate the existence of the solutions of the above model. We will discuss the approximation of solutions in Section 5. Finally, we will give a conclusion.


## 2. A Model for the Influence of Nuclear-Electricity Industry on Area Economy

As we know, the construction of nuclear-electricity station is a project which is mastered by the country; at the same time, it is also a big project. The construction of nuclear-electricity station is composed of the following four parts mainly: technology, equipment, component, and material.It is easy to know that technology is usually fetched in, the type of Taojiang nuclear-electricity station is AP-1000, and this technology is fetched from America. Equipment and component are domestic; they will be assembled into a group. Therefore, technology, equipment, and component will not affect the economy of Taojiang county. But material is only associate domain, such as cement and rock. Because there is a big industry of cement in Taojiang, so the export of cement and rock can advance the development of economy of Taojiang county, but it is only for a moment; once the nuclear-electricity station has been built, their influence on the economy of Taojiang county will disappear for a long time.Now, we consider the deriving domains which includes real estate, service, and tour. These domains will influence the area economy of Taojiang county for a long time.Through this paper, we suppose thatt
=
0 when we begin to build the nuclear-electricity station.First, we consider the real estate domain. The real estate only influences the economy of Taojiang until Taojiang nuclear-electricity station has been constructed. Once Taojiang nuclear-electricity station building starts, the influence of real estate on Taojiang will become little. Thus, we do not consider the influence of real estate on Taojiang in this paper and only need to consider service and tour domain.Second, we consider the service domain. We suppose that the gross economy of the service domain isx
1
(
t
) at time t. The service domain will be developed when we start building the nuclear-electricity station; that is, x
1
(
t
)
>
0, as t
>
0. But, during the construction of building nuclear-electricity station, the gross of economy of the service domain is small. τ
1 years later, the nuclear-electricity station begins running and the gross of economy of the service domain will stabilize. Thus, the gross of economy of the service domain can be expressed as the following:
(1)
x
1
′
(
t
)
=
r
11
(
t
)
x
1
(
t
)
(
1
-
x
1
(
t
-
τ
1
)
M
1
)
,
t
>
0
,
where r
11
(
t
) is a positive function and M
1 is the upper limit of the gross of economy of the service domain x
1
(
t
); that is, if x
1
(
t
)
>
M
1, then x
1
(
t
) will be decreasing.Finally, we consider the tour domain. Since the type of pile of Taojiang nuclear-electricity station is AP-1000, which is first used in inner-continent nuclear-electricity station in China. Therefore, many tourists will visit Taojiang nuclear-electricity station. We suppose that the gross of economy of the tour domain isx
2
(
t
) at time t; then
(2)
x
2
′
(
t
)
=
r
22
(
t
)
x
2
(
t
)
(
1
-
x
2
(
t
-
τ
2
)
M
2
)
,
t
>
0
,
where r
22
(
t
) is a positive function, τ
2
>
0, which is the gross of economy of the tour domain x
2
(
t
)
>
0 for t
>
τ
2, and M
2 is the upper limit of the gross of economy of the tour domain x
2
(
t
); that is, if x
2
(
t
)
>
M
2, then x
2
(
t
) will decreasing.In fact,x
1
(
t
), x
2
(
t
) influence each other. As the nuclear-electricity station begins running, the development of x
1
(
t
) and x
2
(
t
) will grow. If x
2
(
t
) increases or decreases, then x
1
(
t
) will also increase or decrease. But, if x
2
(
t
-
τ
2
)
>
M
2, then x
1
(
t
) decreases. From (1), it follows that
(3)
x
1
′
(
t
)
=
r
11
(
t
)
x
1
(
t
)
(
1
-
x
1
(
t
-
τ
1
)
M
1
)
+
r
12
(
t
)
x
2
(
t
)
(
1
-
x
2
(
t
-
τ
2
)
M
2
)
,
t
>
0
,
where r
1
i
(
t
)
>
0, i
=
1,2.Similarly,(4)
x
2
′
(
t
)
=
r
21
(
t
)
x
1
(
t
)
(
1
-
x
1
(
t
-
τ
1
)
M
1
)
+
r
22
(
t
)
x
2
(
t
)
(
1
-
x
2
(
t
-
τ
2
)
M
2
)
,
t
>
0
,
where r
2
i
(
t
)
>
0, i
=
1,2.According to (3)-(4), the the gross of area economy can be described as
(5)
x
1
′
(
t
)
=
r
11
(
t
)
x
1
(
t
)
(
1
-
x
1
(
t
-
τ
1
)
M
1
)
+
r
12
(
t
)
x
2
(
t
)
(
1
-
x
2
(
t
-
τ
2
)
M
2
)
,
t
>
0
,
x
2
′
(
t
)
=
r
21
(
t
)
x
1
(
t
)
(
1
-
x
1
(
t
-
τ
1
)
M
1
)
+
r
22
(
t
)
x
2
(
t
)
(
1
-
x
2
(
t
-
τ
2
)
M
2
)
,
t
>
0
,
where r
i
j
(
t
)
>
0, i
=
1,2, j
=
1,2, M
1
>
0, M
2
>
0, and τ
j
≥
0, j
=
1,2.


## 3. The Predigestion of Model (5)

It is obvious that the equilibrium points of (5) are M
=
(
M
1
,
M
2
)
T and (
0,0
)
T, and M
=
(
M
1
,
M
2
)
T is a positive equilibrium, so we only consider the positive equilibrium M
=
(
M
1
,
M
2
)
T. Let
(6)
1
-
x
i
(
t
)
M
i
=
y
i
(
t
)
,
i
=
1,2
.
Then
(7)
x
i
(
t
)
=
M
i
(
1
-
y
i
(
t
)
)
,
i
=
1,2
.Sincex
i
(
t
)
≥
0, i
=
1,2, then y
i
(
t
)
≤
1, i
=
1,2. It is obvious that lim
⁡
t
→
∞
x
i
(
t
)
=
M
i
(
i
=
1,2
) if and only if lim
⁡
t
→
∞
y
i
(
t
)
=
0
(
i
=
1,2
,
). It is obvious that lim
⁡
t
→
∞
x
(
t
)
=
M if and only if lim
⁡
t
→
∞
y
(
t
)
=
0.Substituting (7) into (5), we obtain
(8)
y
1
′
(
t
)
=
-
(
M
2
M
1
r
11
(
t
)
y
1
(
t
-
τ
1
)
(
1
-
y
1
(
t
)
)
+
M
2
M
1
r
12
(
t
)
y
2
(
t
-
τ
2
)
(
1
-
y
2
(
t
)
)
)
,
t
>
0
,
y
2
′
(
t
)
=
-
(
M
1
M
2
r
21
(
t
)
y
1
(
t
-
τ
1
)
(
1
-
y
1
(
t
)
)
+
r
22
(
t
)
y
2
(
t
-
τ
2
)
(
1
-
y
2
(
t
)
)
M
1
M
2
)
,
t
>
0
.
It is easy to show that (
0,0
)
T is an equilibrium point of (8). Let
(9)
y
(
t
)
=
(
y
1
(
t
)
,
y
2
(
t
)
)
T
,
(10)
R
(
t
)
=
(
r
11
(
t
)
M
2
M
1
r
12
(
t
)
M
1
M
2
r
21
(
t
)
r
22
(
t
)
)
,
(11)
F
(
y
(
t
)
,
y
(
t
-
τ
1
)
,
y
(
t
-
τ
2
)
)
=
(
y
1
(
t
-
τ
1
)
(
1
-
y
1
(
t
)
)
,
y
2
(
t
-
τ
2
)
(
1
-
y
2
(
t
)
)
)
T
.
Then (8) can be rewritten as
(12)
y
′
(
t
)
+
R
(
t
)
F
(
y
(
t
)
,
y
(
t
-
τ
1
)
,
y
(
t
-
τ
2
)
)
=
0
,
t
>
0
.
Let τ
=
max
⁡
{
τ
1
,
τ
2
}. We consider the following initial condition:
(13)
y
(
t
)
=
Φ
(
t
)
,
t
0
-
τ
≤
t
≤
t
0
.
Therefore, system (12) is equivalent to system (5); by the time, it is equivalent to (8) also. Therefore, we only need to investigate system (12)-(13).It is obvious that vector functionF satisfies the following:(i)
ify
(
t
)
>
0, then F
>
0,
(ii)
ify
(
t
)
<
0, then F
<
0.


## 4. The Continuous Solutions of Model (12)

Integrating both sides of (12) from t
0 to t, it follows that
(14)
y
(
t
)
=
Φ
(
t
0
)
-
∫
t
0
t
‍
R
(
s
)
F
(
y
(
s
)
,
y
(
s
-
τ
1
)
,
y
(
s
-
τ
2
)
)
d
s
,
t
>
t
0
,
where τ
=
max
⁡
{
τ
1
,
τ
2
}. Then y
(
t
) is a continuous solution of (12)-(13) if and only if it is a continuous solution (14).Theorem 1.
Assume thatr
i
j
(
t
)
(
i
,
j
=
1,2
,
3
) are continuous and positive function. Suppose that there exists a positive constant L such that
(15)
∥
Φ
∥
2
-
2
∥
Φ
∥
+
1
≤
L
2
≤
∥
Φ
∥
2
-
∥
Φ
∥
,
(16)
∫
t
0
t
‍
∥
R
(
s
)
∥
d
s
≤
L
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
)
,
t
0
≤
t
≤
T
,
where
(17)
∥
R
(
t
)
∥
=
r
11
(
t
)
+
M
2
M
1
r
12
(
t
)
+
M
1
M
2
r
21
(
t
)
+
r
22
(
t
)
,
t
0
≤
t
≤
T
.
Then (12) has a continuous solution in [
t
0
,
T
].Proof.
Sincey
(
t
) is a continuous solution of (12)-(13) if and only if it is a continuous solution (14), we only need to show that (14) has a continuous solution.
Let(18)
K
=
{
y
:
y
∈
C
(
[
t
0
-
τ
,
T
]
,
R
2
)
,
∥
y
(
t
)
-
Φ
∥
≤
L
}
.
Define an operator T as follows:
(19)
T
y
(
t
)
=
{
Φ
,
t
0
-
τ
≤
t
≤
t
0
,
Φ
-
∫
t
0
t
R
(
s
)
F
(
y
(
s
)
,
y
(
s
-
τ
1
)
,
y
(
s
-
τ
2
)
)
d
s
,
t
0
≤
t
≤
T
.
It is easy to show that K is a nonempty closed convex set. Now, we prove that T maps K into itself; that is, for any y
∈
K, T
y
∈
K.
From (18), it is easy to show that
(20)
∥
y
i
(
t
)
-
Φ
i
∥
≤
L
,
∥
y
i
(
t
-
τ
i
)
-
Φ
i
∥
≤
L
,
i
=
1,2
,
which implies
(21)
∥
Φ
i
∥
-
L
≤
y
i
(
t
)
≤
∥
Φ
i
∥
+
L
,
∥
Φ
i
∥
-
L
≤
y
i
(
t
-
τ
i
)
≤
∥
Φ
i
∥
+
L
,
1
-
∥
Φ
i
∥
-
L
≤
1
-
y
i
(
t
)
≤
1
-
∥
Φ
i
∥
+
L
,
i
=
1,2
.
Combining (11) and (12) with (21), we have
(22)
∥
F
(
y
(
s
)
,
y
(
s
-
τ
1
)
,
y
(
s
-
τ
2
)
)
∥
≤
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
)
.
From (14), (16), (22), and it follows that
(23)
∥
(
T
y
)
(
t
)
-
Φ
∥
=
∥
(
T
y
)
(
t
)
-
(
T
Φ
)
∥
=
∥
∫
t
0
t
‍
R
(
s
)
F
(
y
(
s
)
,
y
(
s
-
τ
1
)
,
y
(
s
-
τ
2
)
)
d
s
∥
≤
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
)
∫
t
0
t
‍
∥
R
(
s
)
∥
d
s
≤
L
,
which means that T maps K into itself.
Defining a sequence{
y
k
(
t
)
} in K,
(24)
y
0
(
t
)
≡
Φ
,
t
0
-
τ
≤
t
≤
T
,
y
k
+
1
(
t
)
=
(
T
y
k
)
(
t
)
,
t
0
-
τ
≤
t
≤
T
,
k
=
0,1
,
…
.
It is easy to show that{
y
k
(
t
)
} is a sequence of measurable functions, and y
k
(
t
)
∈
K. Thus there exists a subsequence {
y
k
i
(
t
)
} of {
y
k
(
t
)
} and a function y
~
(
t
)
∈
K such that
(25)
lim
⁡
t
→
∞
y
k
i
(
t
)
=
y
~
(
t
)
,
t
0
-
τ
≤
t
≤
T
.
By the continuity of{
y
k
(
t
)
}, then the function y
~
(
t
) is continuous for t
0
-
τ
≤
t
≤
T. It follows from Lebesgue's dominated convergence theorem that y
~
(
t
) satisfies
(26)
y
~
(
t
)
=
{
Φ
,
t
0
-
τ
≤
t
≤
t
0
,
Φ
-
∫
t
0
t
‍
R
(
s
)
F
(
y
~
(
s
)
,
y
~
(
s
-
τ
1
)
,
y
~
(
s
-
τ
2
)
)
d
s
,
t
0
≤
t
≤
T
,
which means thaty
~
(
t
) is a continuous solution of (14) for t
∈
[
t
0
-
τ
,
T
]. Using the extension of solution repeatedly, y
~
(
t
) satisfies
(27)
y
~
(
t
)
=
{
Φ
,
t
0
-
τ
≤
t
≤
t
0
,
Φ
-
∫
t
0
t
‍
R
(
s
)
F
(
y
~
(
s
)
,
y
~
(
s
-
τ
1
)
,
y
~
(
s
-
τ
2
)
)
d
s
,
t
>
t
0
,
which is a continuous solution of (14) for t
≥
t
0. The proof is complete.Remark 2.
From conditions (15) and (16), we may choose the number 1
-
∥
Φ
∥
+
L small enough such that L
/
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
) large enough. For example, we take L
=
90.001, ∥
Φ
∥
=
100; then L
/
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
)
≈
470. In fact, if 1
-
∥
Φ
∥
+
L
→
0, then L
/
(
∥
Φ
∥
+
L
)
(
1
-
∥
Φ
∥
+
L
)
→
∞, which implies that it is possible that ∫
t
0
t
‍
∥
R
(
s
)
∥
d
s
=
∞. Therefore, we only need to control the number 1
-
∥
Φ
∥
+
L and not ∫
t
0
t
‍
∥
R
(
s
)
∥
d
s.


## 5. Approximation of Model (12)

Definition 3.
A solution of system (12) is oscillatory if and only if there is a nontrivial component of y
i
(
t
)
(
i
=
1,2
) being oscillatory; one calls system (12) oscillatory if every solution of (12) is oscillatory.Remark 4.
If a vector solutiony
(
t
) is nonoscillatory, then every nontrivial component of (12) is eventually positive or eventually negative.Definition 5.
A solution of system (12) is oscillatory strongly if and only if every nontrivial component y
i
(
t
)
(
i
=
1,2
) is oscillatory; one calls system (12) oscillatory strongly if every solution of (12) is oscillatory strongly.For convenience, we give some hypothesis as follows:(H):r
i
j
(
t
)
(
i
,
j
=
1,2
) are positive continuous functions, do not vanish to zero, and satisfy
(28)
lim
⁡
t
→
∞
r
12
(
t
)
r
11
(
t
)
<
1
,
lim
⁡
t
→
∞
r
21
(
t
)
r
22
(
t
)
<
1
,
lim
⁡
t
→
∞
r
i
j
(
t
)
>
0
(
i
,
j
=
1,2
)
.Usually, we consider the influences of nuclear-electricity industry on area economy after the nuclear-electricity station has been established, and the nuclear-electricity station can be running many years. So, we may ignore the time delaysτ
1 and τ
2. Thus, we take τ
1
=
τ
2
=
0 in this section, and (12) can be rewritten as follows:
(29)
y
1
′
(
t
)
=
-
(
M
2
M
1
r
11
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
+
M
2
M
1
r
12
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
)
,
t
>
0
,
(30)
y
2
′
(
t
)
=
-
(
M
1
M
2
r
21
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
+
r
22
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
M
2
M
1
)
,
t
>
0
.We have the following main theorem.Theorem 6.
Assume thatτ
1
=
τ
2
=
0. Further, suppose that (H) holds. Then, every solution y
(
t
) of (12) satisfies lim
⁡
t
→
∞
y
(
t
)
=
0.Proof.
We divide the proof into the following two steps.
Step  1. Prove that all solutions of (29)-(30) are nonoscillatory.
Assume thaty
(
t
)
=
(
y
1
(
t
)
,
y
2
(
t
)
)
T is an oscillatory solution of (29)-(30); for convenience, we suppose that y
1
(
t
) is oscillatory.
Suppose thaty
2
(
t
) is also oscillatory. By defining oscillatory, there exist two sequences {
t
1
k
} and {
t
2
k
}
(
k
=
1,2
,
…
,
) such that y
1
(
t
1
k
)
=
0, y
2
(
t
2
k
)
=
0. Then, there exist two points t
1
k
i
<
t
2
k
j or t
2
k
j
<
t
1
k
i such that
(31)
y
1
(
t
1
k
i
)
=
0
,
y
2
(
t
2
k
j
)
=
0
,
y
1
(
t
)
≠
0
,
y
2
(
t
)
≠
0
,
for
t
1
k
i
<
t
<
t
2
k
j
or
t
2
k
j
<
t
<
t
1
k
i
.
For convenience, we suppose that t
1
k
i
<
t
2
k
j(the proof of the case t
2
k
j
<
t
1
k
i is similar). Assume that y
2
(
t
1
k
i
)
≠
0; for convenience, we assume that y
2
(
t
1
k
i
)
>
0 (the case of y
2
(
t
1
k
i
)
<
0 is similar); then we have from (29)-(30) that y
1
′
(
t
1
k
i
)
<
0, y
2
′
(
t
1
k
i
)
<
0, which follows that
(32)
y
1
(
t
)
<
0
,
y
2
(
t
)
>
0
for
t
1
k
i
<
t
<
t
2
k
j
.
Ify
1
(
t
2
k
j
)
≠
0, then y
1
(
t
2
k
j
)
<
0; it implies from (30) that y
2
′
(
t
2
k
j
)
>
0, which contradicts the fact that y
2
(
t
2
k
j
)
=
0 and (32).
Thus, the pointt
2
k
j is a repeated zero point of y
1
(
t
) and y
2
(
t
). We have from (29)-(30) that y
1
′
(
t
2
k
)
=
0, y
2
′
(
t
2
k
)
=
0, which means that the point t
2
k is a maximum point or minimum point of y
1
(
t
) and y
2
(
t
). We consider the following four cases:(a)
t
2
k is a maximum point of y
1
(
t
) and a maximum point of y
2
(
t
);
(b)
t
2
k is a maximum point of y
1
(
t
) and a minimum point of y
2
(
t
);
(c)
t
2
k is a minimum point of y
1
(
t
) and a maximum point of y
2
(
t
);
(d)
t
2
k is a minimum point of y
1
(
t
) and a minimum point of y
2
(
t
).
We only prove case (a) and case (b); since the proof of (b) is similar to (a), thus, we omit it.
First, we prove case (a). In the right neighborhood oft
2
k, y
1
′
(
t
)
<
0, y
2
′
(
t
)
<
0 and y
1
(
t
)
<
0, y
2
(
t
)
<
0, which contradicts (29)-(30); thus, case (a) is impossible.
Now, we prove case (b). In the right neighborhood oft
2
k, y
1
′
(
t
)
<
0, y
2
′
(
t
)
>
0 and y
1
(
t
)
<
0, y
2
(
t
)
>
0, which follows from (H) and (29) that
(33)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
>
-
r
11
(
t
)
M
1
r
12
(
t
)
M
2
y
1
(
t
)
(
1
-
y
1
(
t
)
)
,
for t near the right neighborhood of t
1
k. Substituting the above into the right-hand side of (30), one gets
(34)
M
1
M
2
r
21
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
+
r
22
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
>
M
1
M
2
r
21
(
t
)
(
1
-
r
11
(
t
)
r
22
(
t
)
r
12
(
t
)
r
21
(
t
)
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
>
0
,
for t near the right neighborhood of t
1
k. This means from (30) that y
2
′
(
t
)
<
0 for t near the right neighborhood of t
1
k, which contradicts t
1
k a minimum point of y
2
(
t
).
Similarly, it is easy to show that cases (c) and (d) are impossible.
By the available, the functiony
2
(
t
) is eventually positive or eventually negative. Suppose that y
2
(
t
) is eventually negative; by the definition of oscillatory again, there exists a sequence {
t
k
⋆
} such that t
k
⋆ is a local minimum point of y
1
(
t
); that is, y
1
(
t
k
⋆
)
≤
0, y
1
′
(
t
k
⋆
)
=
0, y
2
(
t
k
⋆
)
<
0. From (29), one gets
(35)
0
=
y
1
′
(
t
k
⋆
)
=
-
(
M
2
M
1
r
11
(
t
k
⋆
)
y
1
(
t
k
⋆
)
(
1
-
y
1
(
t
k
⋆
)
)
+
M
2
M
1
r
12
(
t
k
⋆
)
y
2
(
t
k
⋆
)
(
1
-
y
2
(
t
k
⋆
)
)
)
>
0
,
which derive a contradiction. Similarly, we can prove the case of that y
2
(
t
) is eventually positive. Thus, the function y
1
(
t
) is nonoscillatory.
Similarly, we can prove thaty
2
(
t
) is nonoscillatory. Therefore, y
1
(
t
)
,
y
2
(
t
) are all nonoscillatory.
Step  2 (approximation of model (29)-(30)). Since y
(
t
) is a nonoscillatory solution of (12), then every component y
i
(
t
)
(
i
=
1,2
) is nonoscillatory, which means that y
i
(
t
)
(
i
=
1,2
) are eventually positive or eventually negative. Then there exists T
>
0 large enough, such that, when t
>
T, y
i
(
t
)
(
i
=
1,2
) satisfy the following four possible cases:(A)
y
1
(
t
)
>
0
,
y
2
(
t
)
>
0,
(B)
y
1
(
t
)
>
0
,
y
2
(
t
)
<
0,
(C)
y
1
(
t
)
<
0
,
y
2
(
t
)
>
0,
(D)
y
1
(
t
)
<
0
,
y
2
(
t
)
<
0.
For the convenience, we only prove case (A) and case (B), the proof of the rest is similar.
Case (A).   Consider y
1
(
t
)
>
0
,
y
2
(
t
)
>
0.
Fromy
1
(
t
)
>
0
,
y
2
(
t
)
>
0 and (29)-(30), it follows that y
1
′
(
t
)
<
0, y
2
′
(
t
)
<
0, which implies that y
1
(
t
), y
2
(
t
) are decreasing; that is, for any t
>
t
0, y
1
(
t
)
<
y
1
(
t
0
), y
2
(
t
)
<
y
2
(
t
0
). Using (29)-(30) again, one gets
(36)
y
1
′
(
t
)
=
-
(
r
11
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
)
+
M
2
M
1
r
12
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
≤
-
r
11
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
0
)
)
,
y
2
′
(
t
)
=
-
(
M
1
M
2
r
21
(
t
)
y
1
(
t
)
(
1
-
y
1
(
t
)
)
)
+
r
22
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
≤
-
r
22
(
t
)
y
2
(
t
)
(
1
-
y
2
(
t
0
)
)
,
t
≥
t
0
,
that is,
(37)
y
1
′
(
t
)
y
1
(
t
)
≤
-
(
1
-
y
1
(
t
0
)
)
r
11
(
t
)
,
y
2
′
(
t
)
y
2
(
t
)
≤
-
(
1
-
y
2
(
t
0
)
)
r
22
(
t
)
,
t
≥
t
0
.
Integrating both sides of the above from t
0 to t, it follows that
(38)
y
1
(
t
)
≤
y
1
(
t
0
)
e
-
(
1
-
y
1
(
t
0
)
)
∫
t
0
t
‍
r
11
(
s
)
d
s
,
y
2
(
t
)
≤
y
2
(
t
0
)
e
-
(
1
-
y
2
(
t
0
)
)
∫
t
0
t
‍
r
22
(
s
)
d
s
,
t
≥
t
0
,
which means
(39)
lim
⁡
t
→
∞
y
1
(
t
)
=
0
,
lim
⁡
t
→
∞
y
2
(
t
)
=
0
.
Case (B).   Consider y
1
(
t
)
>
0
,
y
2
(
t
)
<
0.
From (29) and (30), it is impossible that y
1
′
(
t
) and y
2
′
(
t
) are all positive, or y
1
′
(
t
) and y
2
′
(
t
) are all negative. So we consider the following three subcases:(B1)y
1
′
(
t
) is oscillatory, or y
2
′
(
t
) is oscillatory;
(B2)y
1
′
(
t
) is eventually negative and y
2
′
(
t
) is eventually positive;
(B3)y
1
′
(
t
) is eventually positive and y
2
′
(
t
) is eventually negative.
Subcase (B1).  y
1
′
(
t
) is oscillatory, or y
2
′
(
t
) is oscillatory.
Suppose thaty
2
′
(
t
) is oscillatory (the case that y
1
′
(
t
) is oscillatory is similar), then for t
¯ large enough such that y
2
′
(
t
¯
)
≤
0. By (H), we have
(40)
r
12
(
t
¯
)
r
11
(
t
¯
)
≤
c
<
1
,
r
21
(
t
¯
)
r
22
(
t
¯
)
≤
c
<
1
.
Combining (30) to (40) and noticing that 0
<
y
1
≤
1, one gets
(41)
y
2
(
t
¯
)
(
1
-
y
2
(
t
¯
)
)
≥
-
M
1
r
21
(
t
¯
)
M
2
r
22
(
t
¯
)
y
1
(
t
¯
)
(
1
-
y
1
(
t
¯
)
)
≥
-
M
1
M
2
c
y
1
(
t
¯
)
(
1
-
y
1
(
t
¯
)
)
.
From (40) and y
2
(
t
¯
)
≤
0, then y
1
(
t
¯
)
(
1
-
y
1
(
t
¯
)
)
≥
(
M
2
/
M
1
c
)
|
y
2
(
t
¯
)
(
1
-
y
2
(
t
¯
)
)
|, which follows from (29) that
(42)
y
1
′
(
t
¯
)
<
0
.
Sincey
2
′
(
t
) is oscillatory, then, there exists a sequence {
t
¯
2
k
} such that they are the minimum points of y
2
(
t
), and y
2
′
(
t
¯
2
k
)
=
0.
Ify
1
′
(
t
)
≤
0 eventually, then lim
⁡
t
→
∞
y
1
′
(
t
)
=
0 and lim
⁡
t
→
∞
y
1
(
t
) exists, which implies from (30) that lim
⁡
t
→
∞
y
2
′
(
t
¯
2
k
) exists. Let
(43)
lim
⁡
t
→
∞
y
1
(
t
)
=
a
,
lim
⁡
t
→
∞
y
2
(
t
¯
2
k
)
=
b
,
lim
⁡
t
→
∞
r
i
j
(
t
)
=
r
i
j
,
i
,
j
=
1,2
.
Then, we have from (29) and (30) that
(44)
0
=
r
11
a
(
1
-
a
)
+
M
2
M
1
r
12
b
(
1
-
b
)
,
0
=
M
1
M
2
r
21
a
(
1
-
a
)
+
r
22
b
(
1
-
b
)
,
which implies that a
=
0 and b
=
0. Therefore,
(45)
lim
⁡
t
→
∞
y
1
(
t
)
=
0
,
lim
⁡
t
→
∞
y
2
(
t
)
=
0
.
Ify
1
′
(
t
) is also oscillatory, then there exists a sequence {
t
¯
1
k
} which are maximum points of y
1
(
t
), and y
1
′
(
t
¯
1
k
)
=
0. From (29) and (30), y
2
′
(
t
¯
1
k
)
>
0, and for any t
¯
2
i
∈
(
t
¯
1
k
,
t
¯
1
k
+
1
), t
¯
1
j
∈
(
t
¯
2
k
,
t
¯
2
k
+
1
). For convenience, we suppose that
(46)
t
¯
11
<
t
¯
21
<
t
¯
12
<
⋯
<
t
¯
1
k
<
t
¯
2
k
<
t
¯
1
k
+
1
<
⋯
.
Thus, for any t
¯
2
k large enough,
(47)
y
2
(
t
¯
2
k
)
(
1
-
y
2
(
t
¯
2
k
)
)
=
-
M
1
r
21
(
t
¯
2
k
)
M
2
r
22
(
t
¯
2
k
)
y
1
(
t
¯
2
k
)
(
1
-
y
1
(
t
¯
2
k
)
)
≥
-
M
1
M
2
c
y
1
(
t
¯
1
k
)
(
1
-
y
1
(
t
¯
1
k
)
)
.
Also
(48)
y
1
(
t
¯
1
k
)
(
1
-
y
1
(
t
¯
1
k
)
)
=
-
M
2
r
12
(
t
¯
1
k
)
M
1
r
11
(
t
¯
1
k
)
y
2
(
t
¯
1
k
)
(
1
-
y
2
(
t
¯
2
k
)
)
≤
-
M
2
M
1
c
y
2
(
t
¯
2
k
-
1
)
(
1
-
y
2
(
t
¯
2
k
-
1
)
)
.
Substituting (48) into (47), one gets
(49)
y
2
(
t
¯
2
k
)
(
1
-
y
2
(
t
¯
2
k
)
)
≥
c
2
y
2
(
t
¯
2
k
-
1
)
(
1
-
y
2
(
t
¯
2
k
-
1
)
)
.
Repeating the above computing, we have
(50)
y
2
(
t
¯
2
k
)
(
1
-
y
2
(
t
¯
2
k
)
)
≥
c
k
y
2
(
t
¯
21
)
(
1
-
y
2
(
t
¯
21
)
)
.
Noticing that 0
<
c
<
1, we have
(51)
lim
⁡
k
→
∞
y
2
(
t
¯
2
k
)
(
1
-
y
2
(
t
¯
2
k
)
)
=
0
,
which follows that lim
⁡
t
→
∞
y
2
(
t
)
=
0, or lim
⁡
t
→
∞
y
2
(
t
)
=
1. If lim
⁡
t
→
∞
y
1
(
t
)
=
1, then lim
⁡
t
→
∞
x
1
(
t
)
=
0; it is impossible. Thus, lim
⁡
t
→
∞
y
2
(
t
)
=
0. Similarly, lim
⁡
t
→
∞
y
1
(
t
)
=
0. Therefore, (39) holds.
Subcase (B2).  y
1
′
(
t
) is eventually negative, and y
2
′
(
t
) is eventually positive.
Noticing thaty
1
(
t
)
>
0 and y
2
(
t
)
<
0, then lim
⁡
t
→
∞
y
1
′
(
t
)
=
0
,
lim
⁡
t
→
∞
y
2
′
(
t
)
=
0, which follows that lim
⁡
t
→
∞
y
1
(
t
), lim
⁡
t
→
∞
y
2
(
t
) exist. Let
(52)
lim
⁡
t
→
∞
y
1
(
t
)
=
a
¯
,
lim
⁡
t
→
∞
y
2
(
t
¯
2
k
)
=
b
¯
.
According to (29), (30) and using a similar method of (44), we obtain
(53)
lim
⁡
t
→
∞
y
1
(
t
)
=
0
,
lim
⁡
t
→
∞
y
2
(
t
)
=
0
.
Subcase (B3).  y
1
′
(
t
) is eventually positive and y
2
′
(
t
) is eventually negative.
Fromy
1
′
(
t
) being eventually positive and y
1
(
t
)
≤
1, then lim
⁡
t
→
∞
y
1
(
t
) exists; let lim
⁡
t
→
∞
y
1
(
t
)
=
a
~. Since y
2
′
(
t
) is eventually negative, we have from (41) that
(54)
y
2
(
t
)
(
1
-
y
2
(
t
)
)
≥
-
M
1
M
2
c
a
~
(
1
-
a
~
)
,
for t large enough, which implies that
(55)
y
2
(
t
)
≥
-
M
1
M
2
c
a
~
(
1
-
a
~
)
.
Thus, lim
⁡
t
→
∞
y
2
(
t
) exists; let lim
⁡
t
→
∞
y
1
(
t
)
=
b
~; then we have from (29) and (30) that
(56)
0
=
r
11
a
~
(
1
-
a
~
)
+
M
2
M
1
r
12
b
~
(
1
-
b
~
)
,
0
=
M
1
M
2
r
21
a
~
(
1
-
a
~
)
+
r
22
b
~
(
1
-
b
~
)
,
thus, b
~
=
0 which contradicts the fact that y
2
′
(
t
)
≤
0.
By the similar method of Case (A) and Case (B), we can prove Case (C) and Case (D). Concluding the above,lim
⁡
t
→
∞
y
1
(
t
)
=
0, lim
⁡
t
→
∞
y
2
(
t
)
=
0. The proof is complete.


## 6. Conclusion

From Theorem6, when t
→
∞, the solution y
(
t
) of (12) satisfies lim
⁡
t
→
∞
y
(
t
)
=
0, which follows that the solution x
(
t
)
=
(
x
1
(
t
)
,
x
2
(
t
)
)
T of (5) satisfies lim
⁡
t
→
∞
x
1
(
t
)
=
M
1, lim
⁡
t
→
∞
x
2
(
t
)
=
M
2. This means that the gross of economy of the service domain x
1
(
t
) and the gross of economy of the tour domain x
2
(
t
) will tend to their equilibrium point M
1 and M
2, respectively, when time t is large enough.



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*Source: 792307-2014-02-18.xml*