# Solvability for a Fractional Order Three-Point Boundary Value System at Resonance

Authors: Zigen Ouyang; Hongliang Liu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

A class of fractional order three-point boundary value system with resonance is investigated in this paper. Using some techniques of inequalities, a completely new method is incorporated. We transform the problem into an integral equation with a pair of undetermined parameters. The topological degree theory is applied to determine the particular value of the parameters so that the system has a solution.


---

## Body

## 1. Introduction

In this paper, we consider the following fractional differential system:(1)
D
0
+
α
X
(
t
)
+
f
(
t
,
X
(
t
)
)
=
0
,
0
<
t
<
1
,
X
=
(
x
1
(
t
)
,
x
2
(
t
)
)
,
X
(
0
)
=
Y
(
0
)
=
0
,
X
(
1
)
=
1
η
α
-
1
X
(
η
)
,
0
<
η
<
1
,
where D
0
+
α is standard Riemann-Liouville fractional derivative of order 1
<
α
≤
2, 0
<
η
<
1 and f
=
(
f
1
,
f
2
) is a nonlinear two-dimension continuous vector function.In the last few decades, many authors have focused on the dynamics of differential equations [1–7]; most of them have investigated fractional differential equations which have been applied in many fields such as physics, mechanics, chemistry, and engineering; see [8–13]. In particular, the positive solutions of the boundary value problem have attracted many authors’ attention [14–25].Recently, the existence of solutions of three-point boundary value problem(2)
D
0
+
α
u
(
t
)
+
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
β
u
(
η
)
,
0
<
η
<
1
,
where D
0
+
α is standard Riemann-Liouville fractional derivative of order 1
<
α
≤
2 has been studied by many authors under the case that β
η
<
1. They obtained some nice results by using some fixed point theorems; see [26–28].In [29], Ahmad and Nieto considered the existence results for following three-point boundary value problem for a coupled system of nonlinear fractional differential equations given by
(3)
D
0
+
α
u
(
t
)
=
f
(
t
,
v
(
t
)
,
D
0
+
p
v
(
t
)
)
=
0
,
0
<
t
<
1
,
D
0
+
β
v
(
t
)
=
g
(
t
,
u
(
t
)
,
D
0
+
q
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
=
γ
u
(
η
)
,
v
(
0
)
=
0
,
v
(
1
)
=
γ
v
(
η
)
,
where 1
<
α
,
β
<
2,  p
,
q
,
γ
>
0,  0
<
η
<
1,  α
-
q
≥
1,  β
-
p
≥
1,  γ
η
α
-
1
<
1,  γ
η
β
-
1
<
1,  D
0
+
α is standard Riemann-Liouville fractional derivative and f
,
g
:
[
0,1
]
×
R
×
R
→
R are given continuous functions. An existence result was proved in their paper by applying the Schauder fixed point theorem.However, few authors have investigated fractional differential boundary value problems with resonance [1, 2, 30–32].In this paper, we establish some sufficient conditions for the existence of the boundary value system (1) by using intermediate value theorems. To present the main results, we assume that f
(
t
,
X
)
=
(
f
1
(
t
,
X
)
,
f
2
(
t
,
X
)
) satisfies the following.(
H
)
f
(
t
,
X
)
∈
C
(
[
0,1
]
×
R
×
R
,
R
×
R
), X
=
(
x
1
,
x
2
)
∈
R
×
R. Suppose that there exist nonnegative functions a
i
(
t
), b
i
j
(
t
)    (
i
,
j
=
1,2
), with b
11
(
t
)
>
0, b
22
(
t
)
>
0, b
i
j
(
t
)
(
i
,
j
=
1,2
,
i
≠
j
)
≤
b
11
(
t
)
,
b
22
(
t
) for any t
∈
[
0,1
] such that
(4)
|
f
i
(
t
,
t
α
-
1
X
)
|
≤
a
i
(
t
)
+
b
i
1
(
t
)
|
x
1
|
p
i
+
b
i
2
(
t
)
|
x
2
|
q
i
,
i
=
1,2
,
where0
≤
p
i
,
q
i
≤
1  (
i
=
1,2
), q
1
<
p
1, and p
2
<
q
2. For any real numbers a and b, the functions f
i
(
t
,
t
α
-
1
(
u
,
v
)
)
(
i
=
1,2
) satisfy
(5)
lim
⁡
v
→
+
∞
f
1
(
t
,
t
α
-
1
(
u
,
v
)
)
>
-
∞
,
lim
⁡
v
→
-
∞
f
1
(
t
,
t
α
-
1
(
u
,
v
)
)
<
+
∞
,
vv
for
any
u
∈
R
,
t
∈
(
0,1
]
,
(6)
lim
⁡
v
→
+
∞
f
2
(
t
,
t
α
-
1
(
u
,
v
)
)
>
-
∞
,
lim
⁡
v
→
-
∞
f
2
(
t
,
t
α
-
1
(
u
,
v
)
)
<
+
∞
,
vv
for
any
v
∈
R
,
t
∈
(
0,1
]
.
Furthermore, assume that(7)
lim
⁡
v
→
+
∞
f
1
(
t
,
t
α
-
1
(
v
,
u
(
v
)
)
)
=
+
∞
,
for
any
u
(
v
)
≥
-
|
v
|
,
t
∈
(
0,1
]
,
lim
⁡
v
→
-
∞
f
1
(
t
,
t
α
-
1
(
v
,
u
(
v
)
)
)
=
-
∞
,
w
for
any
u
(
v
)
≤
|
v
|
,
t
∈
(
0,1
]
,
(8)
lim
⁡
v
→
+
∞
f
2
(
t
,
t
α
-
1
(
u
(
v
)
,
v
)
)
=
+
∞
,
for
any
u
(
v
)
≥
-
|
v
|
,
t
∈
(
0,1
]
,
lim
⁡
v
→
-
∞
f
2
(
t
,
t
α
-
1
(
u
(
v
)
,
v
)
)
=
-
∞
,
w
for
any
u
(
v
)
≤
|
v
|
,
t
∈
(
0,1
]
.
We have the following results.Theorem 1.
Assume that(
H
) holds. If
(9)
max
⁡
1
≤
i
≤
2
{
∫
0
1
‍
G
*
(
s
,
s
)
(
b
i
1
(
s
)
+
b
i
2
(
s
)
)
d
s
}
<
1
,
where
(10)
G
*
(
s
,
s
)
=
1
Γ
(
α
)
(
1
-
η
α
-
1
)
×
{
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
,
0
≤
s
≤
η
,
(
1
-
s
)
α
-
1
,
η
≤
s
≤
1
,
then (1) has at least one solution in [
0,1
].Also, we consider the following special case of(
H
) as follows.(H
~)
f
(
t
,
X
)
∈
C
(
[
0,1
]
×
R
×
R
,
R
×
R
), X
=
(
x
1
,
x
2
)
∈
R
×
R. Suppose that there exist nonnegative functions a
i
(
t
), b
i
j
(
t
)    (
i
,
j
=
1,2
), with b
11
(
t
)
>
0, b
22
(
t
)
>
0, b
i
j
(
t
)
(
i
,
j
=
1,2
,
i
≠
j
)
≤
b
11
(
t
), b
22
(
t
) for any t
∈
[
0,1
] such that
(11)
|
f
i
(
t
,
t
α
-
1
X
)
|
≤
a
i
(
t
)
+
b
i
1
(
t
)
|
x
1
|
p
i
+
b
i
2
(
t
)
|
x
2
|
q
i
,
i
=
1,2
,
where0
≤
p
i
,
q
i
≤
1  (
i
=
1,2
), q
1
<
p
1, and p
2
<
q
2. The functions f
i
(
t
,
t
α
-
1
(
u
,
v
)
)
(
i
=
1,2
) satisfy
(12)
lim
⁡
v
→
±
∞
|
f
1
(
t
,
t
α
-
1
(
u
,
v
)
)
|
<
∞
for
any
u
∈
R
,
t
∈
(
0,1
]
,
(13)
lim
⁡
u
→
±
∞
|
f
2
(
t
,
t
α
-
1
(
u
,
v
)
)
|
<
∞
for
any
v
∈
R
,
t
∈
(
0,1
]
.
Furthermore, assume that (7) and (8) hold.From Theorem1, we have the following corollary.Corollary 2.
Assume that(
H
~
) and (9) hold; then (1) has at least one solution in [
0,1
].


## 2. Some Lemmas

In this section, we first introduce some definitions and preliminary facts and some lemmas which will be used in this paper.Definition 3 (see [21]).
The fractional integral of orderα
>
0 of a function y
:
(
0
,
∞
)
→
R is given by
(14)
I
0
+
α
y
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
y
(
s
)
d
s
provided that the right integral converges.Definition 4 (see [21]).
The standard Riemann-Liouville fractional derivative of orderα
>
0 of a continuous function y
:
(
0
,
∞
)
→
R is given by
(15)
D
0
+
α
y
(
t
)
=
1
Γ
(
n
-
α
)
d
n
d
t
n
∫
0
t
‍
(
t
-
s
)
n
-
α
-
1
y
(
s
)
d
s
,
where n
=
[
α
]
+
1, provided that the right integral converges.Lemma 5 (see [21]).
Assume thatu
∈
C
(
0,1
)
∩
L
(
0,1
) with a fractional derivative of order α
>
0 that belongs to C
(
0,1
)
∩
L
(
0,1
).Then
(16)
I
0
+
α
D
0
+
α
y
(
t
)
=
y
(
t
)
+
C
1
t
α
-
1
+
C
2
t
α
-
2
+
⋯
+
C
n
t
α
-
n
,
for some C
i
∈
R, i
=
1,2
,
…
,
n, where n is the smallest integer greater than or equal to α.The following lemma is a fixed point theorem in a particular Banach space:(17)
Ω
=
{
(
x
(
t
)
,
y
(
t
)
)
∣
x
(
t
)
,
y
(
t
)
∈
C
(
[
0,1
]
,
R
)
}
,
equipped with the norm
(18)
∥
(
x
(
t
)
,
y
(
t
)
)
∥
=
max
⁡
{
max
⁡
t
∈
[
0,1
]
|
x
(
t
)
|
,
max
⁡
t
∈
[
0,1
]
|
y
(
t
)
|
}
.It is easy to show that ifX
(
t
)
∈
Ω, then t
α
-
1
X
(
t
)
∈
Ω.Lemma 6 (see [33]).
LetX be a Banach space with C
⊂
X closed and convex. Assume that U is a relatively open subset of C with 0
∈
U and T
:
U
¯
→
C is completely continuous. Then either(i)
T has a fixed point in U
¯, or
(ii)
there exist anu
∈
∂
U and γ
∈
(
0,1
) with u
=
γ
T
u.To use this lemma to prove our main result, we need transfer (1) into an integral operator.Lemma 7 (see [34]).
Problem (1) is equivalent to the following integral equation:
(19)
X
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
X
(
s
)
)
d
s
+
X
(
1
)
t
α
-
1
,
where
(20)
G
(
t
,
s
)
=
{
(
t
α
-
1
(
1
-
s
)
α
-
1
-
t
α
-
1
(
η
-
s
)
α
-
1
-
(
1
-
η
α
-
1
)
(
t
-
s
)
α
-
1
)
×
(
Γ
(
α
)
(
1
-
η
α
-
1
)
)
-
1
,
t
α
-
1
(
1
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
s
≤
min
⁡
{
t
,
η
}
≤
1
;
t
α
-
1
(
1
-
s
)
α
-
1
-
t
α
-
1
(
η
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
t
≤
s
≤
η
≤
1
;
t
α
-
1
(
1
-
s
)
α
-
1
-
(
1
-
η
α
-
1
)
(
t
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
t
α
-
1
(
1
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
η
≤
s
≤
t
≤
1
;
t
α
-
1
(
1
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
max
⁡
{
t
,
η
}
≤
s
≤
1
.Lemma 8 (see [34]).
For any(
t
,
s
)
∈
[
0,1
]
×
[
0,1
], G
(
t
,
s
) is continuous, and G
(
t
,
s
)
>
0 for any (
t
,
s
)
∈
(
0,1
)
×
(
0,1
).Let(21)
G
(
t
,
s
)
=
t
α
-
1
G
*
(
t
,
s
)
,
where
(22)
G
*
(
t
,
s
)
=
{
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
-
(
1
-
η
α
-
1
)
(
1
-
s
/
t
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
s
≤
min
⁡
{
t
,
η
}
≤
1
;
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
t
≤
s
≤
η
≤
1
;
(
1
-
s
)
α
-
1
-
(
1
-
η
α
-
1
)
(
1
-
s
/
t
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
η
≤
s
≤
t
≤
1
;
(
1
-
s
)
α
-
1
Γ
(
α
)
(
1
-
η
α
-
1
)
,
0
≤
max
⁡
{
t
,
η
}
≤
s
≤
1
.Then (1) is equivalent to the following integral equation:
(23)
X
(
t
)
=
∫
0
1
‍
t
α
-
1
G
*
(
t
,
s
)
f
(
s
,
X
(
s
)
)
d
s
+
X
(
1
)
t
α
-
1
.
The new Green’s function G
*
(
t
,
s
) has the following properties.Lemma 9 (see [34]).
G
*
(
t
,
s
) is continuous for (
t
,
s
)
∈
(
0,1
)
×
(
0,1
) and
(24)
lim
⁡
t
→
0
+
G
*
(
t
,
s
)
:
=
G
*
(
0
,
s
)
=
{
1
Γ
(
α
)
(
1
-
η
α
-
1
)
×
{
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
}
,
0
≤
s
≤
η
;
1
Γ
(
α
)
(
1
-
η
α
-
1
)
(
1
-
s
)
α
-
1
,
η
≤
s
≤
1
.
Furthermore, G
*
(
t
,
s
)
>
0 for (
t
,
s
)
∈
(
0,1
)
×
(
0,1
).Lemma 10 (see [34]).
G
*
(
t
,
s
) is nonincreasing with respect to t
∈
[
0,1
] for any s
∈
(
0,1
). In particular, for any s
∈
[
0,1
], ∂
G
*
(
t
,
s
)
/
∂
t
≤
0, and ∂
G
*
(
t
,
s
)
/
∂
t
=
0 for t
∈
[
0
,
s
]. That is, G
*
(
1
,
s
)
≤
G
*
(
t
,
s
)
≤
G
*
(
s
,
s
), where
(25)
G
*
(
1
,
s
)
=
1
Γ
(
α
)
(
1
-
η
α
-
1
)
×
{
η
α
-
1
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
,
0
≤
s
≤
η
;
η
α
-
1
(
1
-
s
)
α
-
1
,
η
≤
s
≤
1
,
G
*
(
s
,
s
)
=
1
Γ
(
α
)
(
1
-
η
α
-
1
)
×
{
(
1
-
s
)
α
-
1
-
(
η
-
s
)
α
-
1
,
0
≤
s
≤
η
;
(
1
-
s
)
α
-
1
,
η
≤
s
≤
1
.
Let
(26)
X
(
t
)
=
t
α
-
1
Y
(
t
)
.
Then X
(
1
)
=
Y
(
1
), and (23) gives
(27)
Y
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
(
s
,
s
α
-
1
Y
(
s
)
)
d
s
+
Y
(
1
)
.
Let
(28)
W
(
t
)
=
Y
(
t
)
-
Y
(
1
)
.
Then Y
(
t
)
=
W
(
t
)
+
Y
(
1
), and W
(
1
)
=
Y
(
1
)
-
Y
(
1
)
=
0. From (27), (28) can be rewritten as
(29)
W
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
(
s
,
s
α
-
1
(
W
(
s
)
+
Y
(
1
)
)
)
d
s
with W
(
1
)
=
0. Now the integral equation (27) is equivalent to (29). It can be seen from (29) that the solution W
(
t
) of (29) is dependent on the value Y
(
1
). Now, instead of (29), we replace Y
(
1
) with a real vector κ
=
(
μ
,
ν
) and consider
(30)
W
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
)
)
d
s
.
For anyκ
=
(
μ
,
ν
), let
(31)
K
=
{
W
(
t
)
=
(
w
1
(
t
)
,
w
2
(
t
)
)
∈
Ω
}
,
equipped with the norm ∥
W
(
t
)
∥
=
max
⁡
{
max
⁡
t
∈
[
0,1
]
w
1
(
t
)
,
max
⁡
t
∈
[
0,1
]
w
2
(
t
)
}. Define an operator T in K as follows:
(32)
T
W
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
)
)
d
s
.
Using a similar method of Lemmas 3.5 and  3.6 in [34], we obtain that T is completely continuous in K, and (30) has at least a solution W
(
t
) for any given real constant vector κ; the solution W
(
t
) is dependent on the given vector κ. We note the solution W
(
t
)
:
=
W
κ
(
t
).


## 3. The Proof of Theorem1

From Lemma10, for any real vector κ, the integral equation (30) has at least a solution W
(
t
). Therefore, to show that problem (1) has a solution, it remains to show that there exists a κ
=
(
μ
,
ν
), such that W
(
1
)
=
0, or Y
(
1
)
=
κ
=
(
μ
,
ν
).In what follows, we will use the method of topological degree to prove our main result.LetD be an open subset of the plane R
2 with the boundary ∂
D being a simple closed curve; T
~ is a continuous mapping from D
¯
=
D
∪
∂
D to R
2. Let (
c
,
d
)
∈
R
2. Denote by A a variable point on the boundary ∂
D. As A traverses the boundary, assume that its image T
~
(
A
) traces out a closed curve that does not pass through the point (
c
,
d
). As in complex analysis, we can define the winding number of this curve with respect to (
c
,
d
), by measuring the total change of the argument of the vector joining (
c
,
d
) and the variable point T
~
(
A
). For two-dimensional space, this number is equivalent to the topological degree of the mapping T
~ at (
c
,
d
).We introduce a proposition from [8] as follows.Proposition 11.
If the degree of a continuous mappingT
~ with respect to a point (
c
,
d
) is nonzero, then the equation T
~
(
μ
,
ν
)
=
(
c
,
d
) has a solution (
μ
,
ν
)
∈
D.From Section2, for any parameters μ, ν, κ
=
(
μ
,
ν
), there exists a solution W
κ
(
t
) of (30). At the point t
=
1, we denote w
1
(
1
)
:
=
θ, w
2
(
1
)
:
=
ϑ. It is obvious that the parameters θ, ϑ depend on the parameters μ, ν, so we define a map T
~ as follows:(33)
T
~
(
μ
,
ν
)
=
(
θ
,
ϑ
)
.Therefore, if we can find a domainD with its boundary as a closed curve L, so that its image T
~
(
L
) contains the point (
0,0
) in it, then it is implied by Proposition 11 that there exists a point κ
0
=
(
μ
0
,
ν
0
) in D such that T
~
(
μ
0
,
ν
0
)
=
(
0,0
). Thus, the function Y
(
t
)
=
W
κ
0
(
t
)
+
κ
0
=
W
(
t
)
+
Y
(
1
) is a solution of (29), where
(34)
W
κ
0
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
(
s
,
s
α
-
1
(
W
κ
0
(
s
)
+
κ
0
)
)
d
s
.We now proceed to find suchL. For convenience, we take a curve L
=
P
Q
R
S, where P
=
(
-
μ
*
,
-
ν
*
), Q
=
(
-
μ
*
,
ν
*
), R
=
(
μ
~
*
,
ν
*
), S
=
(
μ
~
*
,
-
ν
*
), and P
Q, Q
R, R
S, and S
P are a part of line. The image T
~
(
P
Q
R
S
)
=
P
′
Q
′
R
′
S
′. We want to show that the point (
θ
,
ϑ
)
=
(
0,0
) is inside the closed curve P
′
Q
′
R
′
S
′ as the parameters μ
*, μ
~
*, and ν
* are large enough. In fact, we will prove that the line P
′
Q
′(R
′
S
′) lies in the left (right) side of the ϑ-axis, and the line Q
′
R
′(S
′
P
′) lies above (under) the θ-axis as μ
*, μ
~
*, and ν
* are large enough.Let(35)
a
1
=
∫
0
1
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
,
b
11
=
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
,
b
12
=
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
,
a
2
=
∫
0
1
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
,
b
21
=
∫
0
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
,
b
22
=
∫
0
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
.From (9) and (H), a
1
,
a
2
≥
0 and 0
<
b
11
,
b
12
,
b
21
,
b
22
<
1, and b
12
<
b
11, b
21
<
b
22, we may take μ
*, μ
~
*, and ν
* large enough satisfying
(36)
μ
*
=
b
12
b
11
ν
*
,
(37)
μ
~
*
=
b
22
b
21
ν
*
.
Then, the points P, Q, R, and S can be expressed as follows:
(38)
P
=
(
-
b
12
b
11
ν
*
,
-
ν
*
)
,
Q
=
(
-
b
12
b
11
ν
*
,
ν
*
)
,
R
=
(
b
22
b
21
ν
*
,
ν
*
)
,
S
=
(
b
22
b
21
ν
*
,
-
ν
*
)
.Now the proof of Theorem1 is reduced as the following lemmas.Lemma 12.
Suppose that(
H
) and (9) hold. Then, for ν
* large enough, P
′ lies in the third quadrant.Proof.
From (30), we have
(39)
W
κ
(
1
)
:
=
W
(
1
)
=
∫
0
1
‍
G
*
(
1
,
s
)
f
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
)
)
d
s
.
By the definition of θ, ϑ in (33), we may rewrite (39) as follows:
(40)
θ
κ
=
w
1
(
1
)
=
∫
0
1
‍
G
*
(
1
,
s
)
f
1
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
)
)
d
s
,
ϑ
κ
=
w
2
(
1
)
=
∫
0
1
‍
G
*
(
1
,
s
)
f
2
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
)
)
d
s
.
Letκ
*
=
(
-
μ
*
,
-
ν
*
)
=
(
-
(
b
12
/
b
11
)
ν
*
,
-
ν
*
). From (40), we have
(41)
θ
κ
*
=
w
1
(
1
)
=
∫
0
1
‍
G
*
(
1
,
s
)
f
1
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
*
)
)
d
s
,
ϑ
κ
*
=
w
2
(
1
)
=
∫
0
1
‍
G
*
(
1
,
s
)
f
2
(
s
,
s
α
-
1
(
W
(
s
)
+
κ
*
)
)
d
s
.
Now we will show that θ
κ
*
,
ϑ
κ
*
→
-
∞ as ν
*
→
∞. We only show that lim
⁡
ν
*
→
∞
θ
κ
*
=
-
∞ and the proof of lim
⁡
ν
*
→
∞
ϑ
κ
*
=
-
∞ is similar. Assume on the contrary that lim
⁡
¯
ν
*
→
∞
θ
κ
*
=
l
>
-
∞. Thus, there exists a sequence {
κ
n
}
=
{
(
μ
n
,
ν
n
)
}, μ
n
=
(
b
12
/
b
11
)
ν
n
<
0 such that lim
⁡
ν
n
→
-
∞
θ
n
=
l
>
-
∞.
Recall that(42)
w
1
κ
n
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
,
w
2
κ
n
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
2
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
.
Now we claim that it is impossible to have(43)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
0
,
∀
t
∈
[
0,1
]
as -
ν
n is sufficiently large. Indeed, assume that (43) is true. Then, by the first equation of (42), we have
(44)
w
1
κ
n
(
t
)
≤
0
for all t
∈
[
0,1
]. Therefore, we obtain
(45)
lim
⁡
μ
n
→
-
∞
(
w
1
κ
n
(
t
)
+
μ
n
)
=
lim
⁡
ν
n
→
-
∞
(
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
)
=
-
∞
for t
∈
[
0,1
]. We define some sets as follows:
(46)
A
n
=
{
t
∈
[
0,1
]
:
|
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
|
≥
w
2
κ
n
(
t
)
+
ν
n
}
,
B
n
=
{
t
∈
[
0,1
]
:
|
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
|
<
w
2
κ
n
(
t
)
+
ν
n
}
,
C
n
=
{
t
∈
[
0,1
]
:
w
2
κ
n
(
t
)
+
ν
n
≥
0
}
.
We have assumed in (7) that
(47)
lim
⁡
v
→
-
∞
f
1
(
t
,
t
α
-
1
(
v
,
u
(
v
)
)
)
=
-
∞
for
any
u
(
v
)
≤
|
v
|
,
t
∈
(
0,1
]
.
It is easy to show from (42), (
H
), and our assumption that the set B
n is not empty, and B
n
⊂
C
n. We have the following:
(48)
lim
⁡
ν
n
→
-
∞
∥
w
2
κ
n
(
t
)
∥
B
n
=
lim
⁡
ν
n
→
-
∞
∥
w
2
κ
n
(
t
)
∥
C
n
=
lim
⁡
ν
n
→
-
∞
max
⁡
t
∈
B
n
w
2
κ
n
(
t
)
=
+
∞
.
Using conditions (6) and (8), we have from (45) that there exists a constant l such that
(49)
f
2
(
t
,
t
α
-
1
(
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
,
w
2
κ
n
(
t
)
+
ν
n
)
)
<
l
for t
∈
[
0,1
]
∖
C
n and any n large enough. From the second formula of (42), (45)–(49), one gets
(50)
w
2
κ
n
(
t
)
≤
(
∫
[
0,1
]
∖
C
n
‍
+
∫
C
n
∩
A
n
‍
+
∫
C
n
∩
B
n
‍
)
G
*
(
s
,
s
)
×
f
2
(
s
,
s
α
-
1
(
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
,
w
2
κ
n
(
s
)
+
ν
n
)
)
d
s
≤
l
∫
[
0,1
]
∖
C
n
‍
G
*
(
s
,
s
)
d
s
+
∫
C
n
∩
B
n
‍
G
*
(
s
,
s
)
v
v
v
v
v
v
v
×
f
2
(
s
,
s
α
-
1
(
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
,
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
w
2
κ
n
(
s
)
(
+
ν
n
)
b
12
b
11
)
)
d
s
=
l
∫
[
0,1
]
∖
C
n
‍
G
*
(
s
,
s
)
d
s
+
∫
B
n
‍
G
*
(
s
,
s
)
v
v
v
v
v
×
f
2
(
s
,
s
α
-
1
(
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
,
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
w
2
κ
n
(
s
)
+
ν
n
b
12
b
11
)
)
d
s
≤
l
∫
[
0,1
]
∖
C
n
‍
G
*
(
s
,
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
+
∥
w
1
κ
n
(
t
)
+
b
21
b
11
ν
n
∥
B
n
p
2
∫
0
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
+
∥
w
2
κ
n
(
t
)
+
ν
n
∥
B
n
q
2
∫
0
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
≤
l
∫
[
0,1
]
∖
C
n
‍
G
*
(
s
,
s
)
d
s
+
a
2
+
(
b
21
+
b
22
)
∥
w
2
κ
n
(
t
)
∥
C
n
,
&
G
*
(
s
,
s
)
d
s
+
a
2
+
a
2
a
2
+
+
(
b
21
+
b
22
)
t
∈
C
n
,
which implies that
(51)
∥
w
2
κ
n
(
t
)
∥
C
n
≤
l
∫
[
0,1
]
∖
C
n
‍
G
*
(
s
,
s
)
d
s
+
a
2
1
-
b
21
-
b
22
<
∞
.
It contradicts (48).
Thus, for any-
ν
n large enough, there exists some t
∈
(
0,1
], such that
(52)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
.
Now we define(53)
I
n
=
{
t
∈
[
0,1
]
:
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
}
,
I
n
′
=
{
t
∈
[
0,1
]
:
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
}
.
Then, I
n is not empty.
We can further divide the setI
n into two sets I
~
n and I
n
^, and divide the set I
n
′ into two sets I
n
′
~ and I
n
′
^ as follows:(54)
I
~
n
=
{
t
∈
I
n
∣
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
≤
0
}
,
I
n
^
=
{
t
∈
I
n
∣
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
>
0
}
,
I
n
′
~
=
{
t
∈
I
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
≤
0
}
,
I
n
′
^
=
{
t
∈
I
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
>
0
}
.
It is easy to know that I
~
n
∩
I
n
^
=
ϕ, I
n
′
~
∩
I
n
′
^
=
ϕ and I
n
=
I
~
n
∪
I
n
^, I
n
′
=
I
n
′
~
∪
I
n
′
^.
We claim that the setI
n
^ is not empty for -
ν
n large enough. Otherwise, the function f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
) is bounded from above. In fact, assume that f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
) is unbounded from above for -
ν
n large enough; then we have from (
H
) that there exist a sequence {
t
i
} and a subsequence {
ν
n
i
} of {
ν
n
} such that
(55)
lim
⁡
ν
n
i
→
-
∞
w
2
κ
n
i
(
t
i
)
=
∞
,
lim
⁡
ν
n
i
→
-
∞
|
w
1
κ
n
i
(
t
i
)
+
b
12
b
11
ν
n
i
|
≤
lim
⁡
ν
n
i
→
-
∞
(
w
2
κ
n
i
(
t
i
)
+
ν
n
i
)
=
+
∞
.
Using a similar method of (51), we can derive a contradiction. Therefore, f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
) is bounded from above. From (42), w
1
κ
n
(
t
) is bounded from above, which implies that w
1
κ
n
(
t
)
+
(
b
12
/
b
11
)
ν
n
→
-
∞ as ν
n
→
-
∞. If B
n
=
ϕ (where B
n is defined in (46)), then lim
⁡
ν
n
→
-
∞
θ
n
=
-
∞, which contradicts our assumption. Thus, B
n
≠
ϕ. Using a similar method of getting (51) also gives a contradiction. Therefore, I
^
n is not empty.
Similarly as getting (51) again, we conclude that the function f
i
(
t
,
t
α
-
1
X
) is bounded above by a constant for t
∈
[
0,1
] and x
i
∈
(
-
∞
,
0
]
(
i
=
1,2
). From the condition (
H
), if w
1
κ
n
(
t
)
+
ν
n
>
0  (or w
2
κ
n
(
t
)
+
μ
n
>
0) and f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0 (or f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0), then w
2
κ
n
(
t
)
+
ν
n (or w
1
κ
n
(
t
)
+
μ
n
) is also bounded from above by a constant for t
∈
[
0,1
]. Therefore, from the definition of I
~
n, I
n
′
~, there exists a constant M
>
1, independent of t and ν
n such that
(56)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
M
,
for
t
∈
I
~
n
,
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
M
,
for
t
∈
I
n
′
~
,
w
2
κ
n
(
t
)
+
ν
n
≤
M
for
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
,
t
∈
I
n
^
,
w
1
κ
n
(
t
)
+
b
12
b
11
ν
n
≤
M
,
for
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
,
t
∈
I
n
′
^
.
Let
(57)
M
¯
1
(
κ
n
)
=
max
⁡
t
∈
I
n
w
1
κ
n
(
t
)
.
From the definitions of I
~
n and I
n
^, we have
(58)
M
¯
1
(
κ
n
)
=
max
⁡
t
∈
I
n
^
w
1
κ
n
(
t
)
=
∥
w
1
κ
n
(
t
)
∥
I
n
.
Since I
n
^ is not empty, it follows that M
¯
1
(
κ
n
)
→
∞ as ν
n
→
-
∞. Recall from (9) and (35) that b
i
j
<
1
(
i
,
j
=
1,2
). Therefore, we can choose ν
n
1
>
0 large enough so that
(59)
M
¯
1
(
κ
n
)
>
max
⁡
{
1
,
P
1
,
P
2
}
for ν
n
<
-
ν
n
1, where
(60)
P
1
=
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
1
-
b
11
,
(61)
P
2
=
(
M
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
-
b
22
+
b
12
)
+
b
12
(
1
-
b
22
+
b
21
)
+
(
a
2
+
b
22
)
b
12
+
(
a
1
+
b
11
)
(
1
-
b
22
)
∫
0
1
‍
)
×
(
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
)
-
1
.
Now, for later use, for any integral in a domainA
(62)
∫
A
‍
G
*
(
s
,
s
)
b
i
j
(
s
)
g
(
s
)
d
s
,
for
g
(
s
)
>
0
,
i
,
j
=
1,2
,
we define a subset (
A
)
1 as
(63)
(
A
)
1
=
{
t
∈
A
∣
g
(
t
)
≥
1
}
.
Thus, the integral in (62) can be rewritten as
(64)
∫
A
‍
G
*
(
s
,
s
)
b
i
j
(
s
)
g
(
s
)
d
s
=
∫
(
A
)
1
‍
G
*
(
s
,
s
)
b
i
j
(
s
)
g
(
s
)
d
s
+
∫
A
∖
(
A
)
1
G
*
(
s
,
s
)
b
i
j
(
s
)
g
(
s
)
d
s
From (
H
), (42), and the definitions of I
~
n, I
n
^ and I
n
′
~, I
n
′
^, for ν
n
<
-
ν
n
1, we have
(65)
w
1
κ
n
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
≤
∫
I
n
‍
G
*
(
t
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
≤
∫
I
~
n
‍
G
*
(
s
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
+
∫
I
n
^
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
I
n
^
‍
G
*
(
s
,
s
)
(
b
11
(
s
)
|
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
|
p
1
v
v
v
v
v
v
v
v
v
v
v
w
v
+
b
12
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
1
|
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
|
p
1
)
d
s
≤
∫
I
~
n
‍
G
*
(
s
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
+
∫
I
n
^
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
I
n
^
∖
(
I
n
^
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
|
p
1
d
s
+
∫
(
I
n
^
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
|
p
1
d
s
+
∫
(
I
n
^
∩
I
n
′
)
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
1
d
s
+
∫
(
I
n
^
∩
(
[
0,1
]
∖
I
n
′
)
)
∪
(
(
I
n
^
∩
I
n
′
)
∖
(
I
n
^
∩
I
n
′
)
1
)
‍
G
*
(
s
,
s
)
b
12
(
s
)
&
2222222222222222222222
×
|
w
2
κ
n
(
s
)
+
ν
n
|
q
1
d
s
,
which yields from (56) and the definition in (63) that
(66)
w
1
κ
n
(
t
)
≤
∫
I
~
n
‍
G
*
(
s
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
+
∫
I
n
^
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
I
n
^
∖
(
I
n
^
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
+
∫
(
I
n
^
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
+
b
12
b
11
ν
n
|
d
s
+
∫
(
I
n
^
∩
I
n
′
)
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
d
s
+
M
∫
(
I
n
^
∩
(
[
0,1
]
∖
I
n
′
)
)
∪
(
(
I
n
^
∩
I
n
′
)
∖
(
I
n
^
∩
I
n
′
)
1
)
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
.
Further, one gets from (56) that
(67)
w
1
κ
n
(
t
)
≤
∫
0
1
‍
G
*
(
s
,
s
)
M
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
+
M
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
(
M
¯
1
(
κ
n
)
-
b
12
b
11
∥
ν
n
∥
)
+
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
(
∥
w
2
κ
n
(
t
)
∥
I
n
^
∩
I
n
′
+
∥
ν
n
∥
)
=
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
11
M
¯
1
(
κ
n
)
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
n
^
∩
I
n
′
,
which gives
(68)
M
¯
1
(
κ
n
)
<
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
11
M
¯
1
(
κ
n
)
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
n
^
∩
I
n
′
.
That is,
(69)
M
¯
1
(
κ
n
)
<
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
n
^
∩
I
n
′
1
-
b
11
.
IfI
n
^
∩
I
n
′
=
ϕ, then we have from (69) that
(70)
M
¯
1
(
κ
n
)
<
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
1
-
b
11
,
which contradicts (59).
IfI
n
^
∩
I
n
′
≠
ϕ, using a similar method of (69), we can estimate w
2
κ
n
(
t
) as
(71)
∥
w
2
κ
n
(
t
)
∥
I
n
^
∩
I
n
′
≤
∥
w
2
κ
n
(
t
)
∥
I
n
′
<
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
21
)
+
a
2
+
b
22
+
b
21
M
¯
(
κ
n
)
1
-
b
22
.
Substituting this into (69), we obtain
(72)
M
¯
(
κ
n
)
<
(
M
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
-
b
22
+
b
12
)
+
b
12
(
1
-
b
22
+
b
21
)
)
ww
.
+
(
a
2
+
b
22
)
b
12
+
(
a
1
+
b
11
)
(
1
-
b
22
)

(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
-
b
22
+
b
12
)
+
b
12
(
1
-
b
22
+
b
21
)
)
)
×
(
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
)
-
1
,
which finally contradicts (59). Therefore, our result is proved.
Similarly, we can show thatlim
⁡
ν
*
→
∞
ϑ
κ
*
=
-
∞. Thus, the point T
~
(
-
μ
*
,
-
ν
*
) lies in the third quadrant. The proof is completed.Lemma 13.
Suppose that(
H
) and (9) hold. Then, for ν
*
>
0 large enough, Q
′ lies in the second quadrant.Proof.
It suffices to show thatlim
⁡
ν
*
→
∞
ϑ
*
=
∞ and lim
⁡
ν
*
→
∞
θ
*
=
-
∞.
First, we claim thatlim
⁡
ν
*
→
∞
ϑ
*
=
∞. On the contrary, we assume that there exists a sequence {
κ
n
}
=
{
(
μ
n
,
ν
n
)
}
=
{
(
-
(
b
12
/
b
11
)
ν
n
,
ν
n
)
} such that lim
⁡
ν
n
→
∞
ϑ
n
=
l
<
∞. By a similar method in Lemma 12, we know it is impossible to have
(73)
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≥
0
,
∀
t
∈
[
0,1
]
as ν
n is sufficiently large.
Now, for largeν
n, we define
(74)
J
n
=
{
t
∈
[
0,1
]
:
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
}
,
J
n
′
=
{
t
∈
[
0,1
]
:
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
}
.
Then, J
n
′ is not empty.
As in Lemma12, we can further divide the set J
n into two sets J
n
~ and J
n
^ and divide the set J
n
′ into two sets J
n
′
~ and J
n
′
^ as follows:
(75)
J
n
~
=
{
t
∈
J
n
∣
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
≥
0
}
,
J
n
^
=
{
t
∈
J
n
∣
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
<
0
}
,
J
n
′
~
=
{
t
∈
J
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
≥
0
}
,
J
n
′
^
=
{
t
∈
J
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
<
0
}
.
Then J
n
~
∩
J
n
^
=
ϕ,  J
n
′
~
∩
J
n
′
^
=
ϕ and J
n
=
J
n
~
∪
J
n
^,  J
n
′
=
J
n
′
~
∪
J
n
′
^.
Using a similar method as in the proof of Lemma12, we can show that the set J
n
′
^ is not empty. Furthermore, the function f
i
(
t
,
t
α
-
1
X
) is bounded below by a constant for t
∈
[
0,1
] and x
i
∈
[
0
,
∞
)
(
i
=
1,2
). If w
2
κ
n
(
t
)
+
μ
n
<
0 (or w
1
κ
n
(
t
)
+
ν
n
<
0
) and f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0 (or f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
), then w
1
κ
n
(
t
)
+
ν
n (or w
2
κ
n
(
t
)
+
μ
n
) is also bounded below by a constant for t
∈
[
0,1
]. From the definition of J
n
~, J
n
′
~ and the condition (
H
), there exists a constant M
~
<
-
1, independent of t and ν
n such that
(76)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≥
M
~
,
for
t
∈
J
n
~
,
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≥
M
~
,
for
t
∈
J
n
′
~
,
w
2
κ
n
(
t
)
+
ν
n
≥
M
~
for
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
,
t
∈
J
n
^
,
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
≥
M
~
,
for
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
,
t
∈
J
n
′
^
.
Let
(77)
m
¯
2
(
κ
n
)
=
min
⁡
t
∈
J
n
′
w
2
κ
n
(
t
)
.
From the definitions of J
n
′
~ and J
n
′
^, we have
(78)
m
¯
2
(
κ
n
)
=
min
⁡
t
∈
J
n
′
^
w
2
κ
n
(
t
)
=
-
∥
w
2
κ
n
(
t
)
∥
J
n
′
,
and it follows that m
¯
2
(
κ
n
)
→
-
∞ as ν
n
→
∞. Therefore, we can choose ν
n
1 large enough so that
(79)
m
¯
2
(
κ
n
)
<
min
⁡
{
-
1
,
Q
1
,
Q
2
}
for ν
n
>
ν
n
1, where
(80)
Q
1
=
M
~
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
+
b
21
)
-
a
2
-
b
22
1
-
b
22
,
Q
2
=
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
-
b
11
-
b
21
)
+
b
21
(
1
-
b
11
-
b
21
)
(
1
-
b
22
)
(
1
-
b
11
)
-
b
21
b
12
M
~
-
(
a
2
+
b
22
)
(
1
-
b
11
)
+
(
a
1
+
b
11
)
b
21
(
1
-
b
22
)
(
1
-
b
11
)
-
b
21
b
12
.
Notice thatb
12
<
b
11, b
21
<
b
22. From (
H
), (42), and the definitions of J
n
~, J
n
^ and J
n
′
~, J
n
′
^, for ν
n
>
ν
n
1, we have
(81)
w
2
κ
n
(
t
)
≥
∫
J
n
′
‍
G
*
(
s
,
s
)
f
2
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
≥
∫
J
n
′
~
‍
G
*
(
s
,
s
)
f
2
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
(
b
21
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
vvvvvvvvvvvv
+
b
22
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
2
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
)
d
s
≥
∫
J
n
′
~
‍
G
*
(
s
,
s
)
f
2
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
-
∫
J
n
′
^
∩
J
n
‍
G
*
(
s
,
s
)
b
21
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
d
s
-
∫
J
n
′
^
∩
(
[
0,1
]
∖
J
n
)
‍
G
*
(
s
,
s
)
b
21
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
b
22
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
2
d
s
.
Thus,
(82)
w
2
κ
n
(
t
)
≥
∫
J
n
′
~
‍
G
*
(
s
,
s
)
f
2
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
-
∫
(
J
n
′
^
∩
J
n
)
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
d
s
-
∫
(
J
n
′
^
∩
(
[
0,1
]
∖
J
n
)
)
∪
(
(
J
n
′
^
∩
J
n
)
∖
(
J
n
′
^
∩
J
n
)
1
)
‍
G
*
(
s
,
s
)
b
21
(
s
)
∫
(
J
n
′
^
∩
(
[
0,1
]
∖
J
n
)
)
∪
(
(
J
n
′
^
∩
J
n
)
∖
(
J
n
′
^
∩
J
n
)
1
)
‍
×
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
2
d
s
-
∫
(
J
n
′
^
)
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
2
d
s
-
∫
J
n
′
^
∖
(
J
n
′
^
)
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
2
d
s
,
which follows from (76) and the definition of (63) that
(83)
w
2
κ
n
(
t
)
≥
M
~
∫
J
n
′
~
‍
G
*
(
s
,
s
)
d
s
-
∫
J
n
′
^
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
+
M
~
∫
J
n
′
^
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
-
∫
(
J
n
′
^
∩
J
n
)
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
×
∥
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
∥
(
J
n
′
^
∩
J
n
)
1
-
∫
(
J
n
′
^
)
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
∥
w
2
κ
n
(
t
)
+
ν
n
∥
J
n
′
^
-
∫
J
n
′
^
∖
(
J
n
′
^
)
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
≥
M
~
∫
0
1
‍
G
*
(
s
,
s
)
d
s
-
∫
0
1
‍
G
*
(
s
,
s
)
a
2
(
s
)
d
s
+
M
~
∫
0
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
-
∫
0
1
‍
G
*
(
s
,
s
)
b
21
(
s
)
d
s
×
(
∥
w
1
κ
n
(
t
)
∥
J
n
′
^
∩
J
n
+
∥
b
12
b
11
ν
n
∥
)
-
∫
0
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
-
∫
0
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
(
-
m
¯
2
(
κ
n
)
-
∥
ν
n
∥
)
-
∫
0
1
‍
G
*
(
s
,
s
)
b
22
(
s
)
d
s
=
M
~
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
21
)
-
a
2
-
b
22
+
∥
ν
n
∥
(
b
11
b
22
-
b
12
b
21
)
b
11
-
b
21
∥
w
1
κ
n
(
t
)
∥
J
n
′
^
∩
J
n
+
b
22
m
¯
2
(
κ
n
)
.
Thus,
(84)
m
¯
2
(
κ
n
)
>
M
~
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
21
)
-
a
2
-
b
22
-
b
21
∥
w
1
κ
n
(
t
)
∥
J
n
′
^
∩
J
n
+
b
22
m
¯
2
(
κ
n
)
,
which implies that
(85)
m
¯
2
(
κ
n
)
>
M
~
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
21
)
-
a
2
-
b
22
-
b
21
∥
w
1
κ
n
(
t
)
∥
J
n
′
^
∩
J
n
1
-
b
22
.
IfJ
n
′
^
∩
J
n
=
ϕ, from (85), we have
(86)
m
¯
2
(
κ
n
)
>
M
~
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
21
)
-
a
2
-
b
22
1
-
b
22
,
which contradicts (79).
IfJ
n
′
^
∩
J
n
≠
ϕ. Using a similar method of (85), we have
(87)
-
∥
w
1
κ
n
(
t
)
∥
J
n
′
^
∩
J
n
≥
-
∥
w
1
κ
n
(
t
)
∥
J
n
>
M
~
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
-
a
1
-
b
11
-
b
12
m
¯
2
(
κ
n
)
1
-
b
11
.
Substituting it into (85), we obtain
(88)
m
¯
2
(
κ
n
)
>
∫
0
1
‍
G
*
(
s
,
s
)
d
s
(
1
-
b
11
-
b
21
)
+
b
21
(
1
-
b
11
-
b
21
)
(
1
-
b
22
)
(
1
-
b
11
)
-
b
21
b
12
M
~
-
(
a
2
+
b
22
)
(
1
-
b
11
)
+
(
a
1
+
b
11
)
b
21
(
1
-
b
22
)
(
1
-
b
11
)
-
b
21
b
12
,
which also contradicts (79). Therefore, lim
⁡
ν
*
→
∞
ϑ
=
∞.
Now, we show thatlim
⁡
ν
*
→
∞
θ
=
-
∞.
On the contrary, assume that there exists a vector sequence{
κ
n
}
=
{
(
μ
n
,
ν
n
)
} such that μ
n
=
-
(
b
12
/
b
11
)
ν
n and
(89)
lim
⁡
ν
n
→
∞
θ
n
=
l
>
-
∞
.
Similarly as before, it is impossible to have(90)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
0
,
∀
t
∈
[
0,1
]
as ν
n is sufficiently large.
Now for largeν
n, we define
(91)
I
¯
n
=
{
t
∈
[
0,1
]
:
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
}
,
I
¯
n
′
=
{
t
∈
[
0,1
]
:
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
>
0
}
.
Then, I
¯
n is not empty.
We can further divide the setI
¯
n into two sets I
¯
~
n and I
¯
^
n and divide the set I
¯
n
′ into two sets I
¯
~
n
′ and I
¯
^
n
′ as follows:
(92)
I
¯
~
n
=
{
t
∈
I
¯
n
∣
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
≤
0
}
,
I
¯
^
n
=
{
t
∈
I
n
∣
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
>
0
}
,
I
¯
~
n
′
=
{
t
∈
I
¯
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
≤
0
}
,
I
¯
^
n
′
=
{
t
∈
I
¯
n
′
∣
w
2
κ
n
(
t
)
+
ν
n
>
0
}
.
It is easy to know that I
¯
~
n
∩
I
¯
^
n
=
ϕ, I
¯
~
n
′
∩
I
¯
^
n
′
=
ϕ and I
¯
n
=
I
¯
~
n
∪
I
¯
^
n, I
¯
n
′
=
I
¯
~
n
′
∪
I
¯
^
n
′.
Using a similar method of the proof of Lemma12, we obtain that the set I
¯
^
n is not empty. Furthermore, there exists a constant M
^
>
1, independent of t and ν
n such that
(93)
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
M
^
,
for
t
∈
I
¯
~
n
,
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
≤
M
^
,
for
t
∈
I
¯
~
n
′
,
w
2
κ
n
(
t
)
+
ν
n
≤
M
^
,
for
f
2
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
,
t
∈
I
¯
^
n
,
w
1
κ
n
(
t
)
-
b
12
b
11
ν
n
≤
M
^
,
for
f
1
(
t
,
t
α
-
1
(
W
κ
n
(
t
)
+
κ
n
)
)
<
0
,
t
∈
I
¯
^
n
′
.
Let
(94)
M
^
1
(
κ
n
)
=
max
⁡
t
∈
I
¯
n
w
1
κ
n
(
t
)
.
From the definitions of I
¯
~
n and I
¯
^
n, we have
(95)
M
^
1
(
κ
n
)
=
max
⁡
t
∈
I
¯
^
n
w
1
κ
n
(
t
)
=
∥
w
1
κ
n
(
t
)
∥
I
¯
n
.
Since I
¯
^
n is not empty, it follows that M
^
1
(
κ
n
)
→
∞ as ν
n
→
∞. Therefore, we can choose ν
n
1 large enough so that
(96)
M
^
1
(
κ
n
)
>
max
⁡
{
1
,
P
^
1
,
P
^
2
}
,
for ν
n
>
ν
n
1, where
(97)
P
^
1
=
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
1
-
b
11
,
P
^
2
=
∫
0
1
‍
G
*
(
s
,
s
)
d
s
[
(
1
-
b
22
)
+
b
12
]
+
b
12
(
1
-
b
22
+
b
21
)
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
M
^
+
(
a
2
+
b
22
)
b
12
+
(
a
1
+
b
11
)
(
1
-
b
22
)
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
.
From(
H
) and (42), we have
(98)
w
1
κ
n
(
t
)
=
∫
0
1
‍
G
*
(
t
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
≤
∫
I
¯
~
n
‍
G
*
(
s
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
+
∫
I
¯
^
n
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
(
I
¯
^
n
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
1
d
s
+
∫
I
¯
^
n
∖
(
I
¯
^
n
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
1
d
s
+
∫
(
I
¯
^
n
∩
I
¯
n
′
)
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
|
w
2
κ
n
(
s
)
+
ν
n
|
q
1
d
s
+
∫
(
I
¯
^
n
∩
(
[
0,1
]
∖
I
¯
n
′
)
)
∪
(
(
I
¯
^
n
∩
I
¯
n
′
)
∖
(
I
¯
^
n
∩
I
¯
n
′
)
1
)
‍
G
*
(
s
,
s
)
b
12
(
s
)
vvvvvvvvvvvvvvvvvvvvvvivvvv
×
|
w
2
κ
n
(
s
)
+
ν
n
|
q
1
d
s
,
which follows from (93) and the definition in (63) that
(99)
w
1
κ
n
(
t
)
≤
∫
0
1
‍
G
*
(
s
,
s
)
M
^
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
M
^
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
+
∫
I
¯
^
n
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
(
m
^
1
(
κ
n
)
-
∥
b
12
b
11
ν
n
∥
)
+
∫
I
¯
^
n
∖
(
I
¯
^
n
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
+
∫
(
I
¯
^
n
∩
I
¯
n
′
)
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
×
(
∥
w
2
κ
n
(
t
)
∥
(
I
¯
^
n
∩
I
¯
n
′
)
1
+
∥
ν
n
∥
)
.
Thus,
(100)
w
1
κ
n
(
t
)
≤
∫
0
1
‍
G
*
(
s
,
s
)
M
^
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
M
^
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
(
m
^
1
(
κ
n
)
-
∥
b
12
b
11
ν
n
∥
)
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
×
(
∥
w
2
κ
n
(
t
)
∥
(
I
¯
^
n
∩
I
¯
n
′
)
1
+
∥
ν
n
∥
)
=
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
11
M
^
1
(
κ
n
)
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
,
which implies that
(101)
M
^
1
(
κ
n
)
<
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
11
M
^
1
(
κ
n
)
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
.
Therefore, we have
(102)
M
^
1
(
κ
n
)
<
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
1
-
b
11
.
IfI
¯
^
n
∩
I
¯
n
′
=
ϕ, then we have from (76) that
(103)
M
^
1
(
κ
n
)
<
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
1
-
b
11
,
which contradicts (102).
IfI
¯
^
n
∩
I
¯
n
′
≠
ϕ. Using a similar method to that in Lemma 12, we have
(104)
M
^
1
(
κ
n
)
<
∫
0
1
‍
G
*
(
s
,
s
)
d
s
[
(
1
-
b
22
)
+
b
12
]
+
b
12
(
1
-
b
22
+
b
21
)
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
M
^
+
(
a
2
+
b
22
)
b
12
+
(
a
1
+
b
11
)
(
1
-
b
22
)
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
,
which also contradicts (96). Thus, the point T
~
(
-
μ
*
,
ν
*
)
=
(
-
(
b
12
/
b
11
)
ν
*
,
ν
*
) lies in the second quadrant. The proof is completed.Lemma 14.
Suppose that(
H
) and (9) hold. Then, for ν
* large enough, the line P
′
Q
′ lies in the left of ϑ-axis.Proof.
For any pointA
(
θ
,
ϑ
) in P
′
Q
′, it suffices to show that θ
→
-
∞ as ν
*
→
∞ for any ν
∈
[
-
ν
*
,
ν
*
].
On the contrary, we assume that there exists a vector sequence{
κ
n
}
=
{
(
μ
n
,
ν
n
)
} satisfying μ
n
=
-
(
b
12
/
b
11
)
ν
n and a point ν
(
ν
n
)
∈
[
-
ν
n
,
ν
n
] such that θ
(
-
(
b
12
/
b
11
)
ν
n
,
ν
(
ν
n
)
)
→
l
>
-
∞ as ν
n
→
∞. We define some sets I
¯
n, I
¯
~
n, I
¯
^
n and I
¯
n
′, I
¯
~
n
′, I
¯
^
n
′, and some numbers M
^, M
^
1
(
κ
n
) as in Lemma 13. Using a similar method of the proof of Lemma 13, we have
(105)
w
1
κ
n
(
t
)
≤
∫
I
¯
~
n
‍
G
*
(
s
,
s
)
f
1
(
s
,
s
α
-
1
(
W
κ
n
(
s
)
+
κ
n
)
)
d
s
+
∫
I
¯
^
n
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
∫
(
I
¯
^
n
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
1
d
s
+
∫
I
¯
^
n
∖
(
I
¯
^
n
)
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
|
w
1
κ
n
(
s
)
-
b
12
b
11
ν
n
|
p
1
d
s
+
∫
(
I
¯
^
n
∩
I
¯
n
′
)
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
|
w
2
κ
n
(
s
)
+
ν
(
ν
n
)
|
q
1
d
s
+
∫
(
I
¯
^
n
∩
(
[
0,1
]
∖
I
¯
n
′
)
)
∪
(
(
I
¯
^
n
∩
I
¯
n
′
)
∖
(
I
¯
^
n
∩
I
¯
n
′
)
1
)
‍
G
*
(
s
,
s
)
b
12
(
s
)
vvvvvvvvvvvvvvvvvvvvvvvvvvv
×
|
w
2
κ
n
(
s
)
+
ν
(
ν
n
)
|
q
1
d
s
.
It follows from (93)–(96) that(106)
w
1
κ
n
(
t
)
≤
∫
0
1
‍
G
*
(
s
,
s
)
M
^
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
a
1
(
s
)
d
s
+
M
^
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
(
M
^
1
(
κ
n
)
-
∥
b
12
b
11
ν
n
∥
)
+
∫
0
1
‍
G
*
(
s
,
s
)
b
11
(
s
)
d
s
+
∫
0
1
‍
G
*
(
s
,
s
)
b
12
(
s
)
d
s
×
(

∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′

+
∥
ν
(
ν
n
)
∥
)
.
Notice that -
ν
n
≤
ν
(
ν
n
)
≤
ν
n; from (35), one gets
(107)
w
1
κ
n
(
t
)
≤
∫
0
1
‍
G
*
(
s
,
s
)
M
^
d
s
+
a
1
+
M
^
b
12
+
b
11
+
b
11
(
M
^
1
(
κ
n
)
-
∥
b
12
b
11
ν
n
∥
)
+
b
12
(
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
+
∥
ν
n
∥
)
=
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
11
M
^
1
(
κ
n
)
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
,
which implies that
(108)
M
^
1
(
κ
n
)
<
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
+
b
12
∥
w
2
κ
n
(
t
)
∥
I
¯
^
n
∩
I
¯
n
′
1
-
b
11
.
It is easy to show that(109)
M
^
1
(
κ
n
)
<
M
^
(
∫
0
1
‍
G
*
(
s
,
s
)
d
s
+
b
12
)
+
a
1
+
b
11
1
-
b
11
for I
¯
^
n
∩
I
¯
n
′
=
ϕ, which contradicts (102), and
(110)
M
^
1
(
κ
n
)
<
∫
0
1
‍
G
*
(
s
,
s
)
d
s
[
(
1
-
b
22
)
+
b
12
]
+
b
12
(
1
-
b
22
)
+
b
12
b
21
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
M
^
+
(
a
2
+
b
22
)
b
12
+
(
a
1
+
b
11
)
(
1
-
b
22
)
(
1
-
b
11
)
(
1
-
b
22
)
-
b
12
b
21
for I
¯
^
n
∩
I
¯
n
′
≠
ϕ, which contradicts (96) also. Thus, the line P
′
Q
′ lies in the left of ϑ-axis. The proof is completed.Similar to the proof of Lemma12, we can show that the image point R
′ of the point R lies in the first quadrant. From (37), we have μ
~
*
=
(
b
22
/
b
21
)
ν
*. Using a similar method of Lemma 13, we can show that the image point S
′ of the point S lies in the fourth quadrant.Using the conditions (36) and (37), similar to Lemma 14, we can show that the image line Q
′
R
′ of the line Q
R lies above the θ-axis, R
′
S
′ lies in the right of the ϑ-axis, and S
′
P
′ lies under the θ-axis. Therefore, we have the following lemmas.Lemma 15.
Suppose that(
H
) and (9) hold. For ν
* large enough, Q
′
R
′ lies above the θ-axis, Q
′ lies in the second quadrant, and R
′ lies in the first quadrant.Lemma 16.
Suppose that(
H
) and (9) hold. For ν
* large enough, R
′
S
′ lies in the right of the ϑ-axis and S
′ lies in the fourth quadrant.Lemma 17.
Suppose that(
H
) and (9) hold. For ν
* large enough, S
′
P
′ lies below the θ-axis.Proof of Theorem1.
From Lemmas12–17, when μ
*, ν
*, and μ
~
* are large enough and satisfy (36) and (37), then the image P
′
Q
′
R
′
S
′ of the curve P
Q
R
S will contain the zero in it. From Proposition 11, it follows that there exists a vector κ
0
=
(
μ
0
,
ν
0
) such that the solution W
κ
0
(
t
) of (30) satisfies W
κ
0
(
1
)
=
0, which implies that the integral equation (27) has a solution Y
(
t
). From (26), it follows that (19) has a solution. Therefore, the problem (1) has at least one solution. The proof is completed.


## 4. Examples

Example 1.
Consider the following boundary value system:(111)
D
0
+
3
/
2
x
(
t
)
+
t
1
/
3
2
x
1
/
3
(
t
)
+
t
1
/
3
4
y
1
/
3
(
t
)
1
+
|
y
1
/
3
(
t
)
|
+
t
=
0
,
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
t
∈
(
0,1
)
,
D
0
+
3
/
2
y
(
t
)
+
t
1
/
3
3
x
1
/
3
(
t
)
1
+
|
x
1
/
3
(
t
)
|
+
t
1
/
3
2
y
1
/
3
(
t
)
+
t
4
=
0
,
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
i
t
∈
(
0,1
)
,
x
(
0
)
=
y
(
0
)
=
0
,
x
(
1
)
=
2
1
/
2
x
(
1
2
)
,
y
(
1
)
=
2
1
/
2
y
(
1
2
)
,
where
(112)
f
1
(
t
,
t
α
-
1
(
x
,
y
)
)
=
f
1
(
t
,
t
1
/
2
(
x
,
y
)
)
=
t
1
/
2
2
x
1
/
3
(
t
)
+
t
1
/
2
4
y
1
/
3
(
t
)
1
+
t
1
/
6
|
y
1
/
3
(
t
)
|
+
t
,
f
2
(
t
,
t
α
-
1
(
x
,
y
)
)
=
f
2
(
t
,
t
1
/
2
(
x
,
y
)
)
=
t
1
/
2
3
x
1
/
3
(
t
)
1
+
t
1
/
6
|
x
1
/
3
(
t
)
|
+
t
1
/
2
2
y
1
/
3
(
t
)
+
t
4
.
It is obvious that
(113)
|
f
1
(
t
,
t
1
/
2
(
x
,
y
)
)
|
≤
t
1
/
2
2
|
x
(
t
)
|
1
/
3
+
t
1
/
2
4
|
y
(
t
)
|
1
/
3
+
t
,
|
f
2
(
t
,
t
1
/
2
(
x
,
y
)
)
|
≤
t
1
/
2
3
|
x
(
t
)
|
1
/
3
+
t
1
/
2
2
|
y
(
t
)
|
1
/
3
+
t
4
,
where
(114)
b
11
(
t
)
=
t
1
/
2
2
,
b
12
(
t
)
=
t
1
/
2
4
4
,
b
21
(
t
)
=
t
1
/
2
3
,
b
22
(
t
)
=
t
1
/
2
2
,
a
1
(
t
)
=
t
,
a
2
(
t
)
=
t
4
,
η
=
1
2
,
p
1
=
p
2
=
q
1
=
q
2
=
1
3
.
It is easy to check that the function f satisfies (8)–(12). Notice that
(115)
G
*
(
s
,
s
)
=
1
Γ
(
1
/
2
)
(
1
-
(
1
/
2
)
1
/
2
)
×
{
(
1
-
s
)
1
/
2
-
(
1
2
-
s
)
1
/
2
,
0
≤
s
≤
1
2
,
(
1
-
s
)
1
/
2
,
1
2
≤
s
≤
1
,
b
21
(
t
)
,
b
12
(
t
)
<
b
11
(
t
)
,
b
22
(
t
)
=
max
⁡
i
,
j
=
1,2
{
b
i
j
(
t
)
}
=
t
1
/
2
2
.Consider(116)
max
⁡
1
≤
i
≤
2
∫
0
1
‍
G
*
(
s
,
s
)
(
b
i
1
(
s
)
+
b
i
2
(
s
)
)
d
s
≤
∫
0
1
‍
G
*
(
s
,
s
)
(
s
1
/
2
2
+
s
1
/
2
2
)
d
s
=
∫
0
1
‍
G
*
(
s
,
s
)
s
1
/
2
d
s
=
1
Γ
(
1
/
2
)
(
1
-
(
1
/
2
)
1
/
2
)
×
(
∫
0
1
‍
[
(
1
-
s
)
s
]
1
/
2
d
s
-
∫
0
1
/
2
‍
[
(
1
2
-
s
)
s
]
1
/
2
d
s
)
=
1
π
(
1
-
2
/
2
)
(
π
8
-
π
32
)
≈
0.2945
1.2533
<
1
which satisfies (9). Therefore, all conditions of Theorem 1 hold and thus the problem (111) has at least a solution.



---

*Source: 419514-2014-06-11.xml*