# Solutions of Second-Orderm-Point Boundary Value Problems for Impulsive Dynamic Equations on Time Scales

Authors: Xue Xu; Yong Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We study a general second-orderm-point boundary value problems for nonlinear singular impulsive dynamic equations on time scales u
Δ
∇
(
t
)
+
a
(
t
)
u
Δ
(
t
)
+
b
(
t
)
u
(
t
)
+
q
(
t
)
f
(
t
,
u
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
t
≠
t
k
,
u
Δ
(
t
k
+
)
=
u
Δ
(
t
k
)
-
I
k
(
u
(
t
k
)
), and k
=
1,2
,
…
,
n
,
u
(
ρ
(
0
)
)
=
0
,
u
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
α
i
u
(
η
i
)
‍
. The existence and uniqueness of positive solutions are established by using the mixed monotone fixed point theorem on cone and Krasnosel’skii fixed point theorem. In this paper, the function items may be singular in its dependent variable. We present examples to illustrate our results.


---

## Body

## 1. Introduction

The theory of dynamic equations on time scales unifies the well-known analogies in the concept of difference equations and differential equations. In the past few years, the boundary value problems of dynamic equations on time scales have been studied by many authors (see [1–16] and references cited therein). Some classical tools have been used in the literature to study dynamic equations. These classical tools include the coincidence degree theory [11, 12], the method of upper and lower solutions [7, 10], and some fixed point theorems in cones for completely continuous operators [1–5, 9, 13–16]. Recently, multiple-point boundary value problems on time scale have been studied for instance [4, 5, 12].In 2008, Lin and Du [5] studied the m-point boundary value problem for second-order dynamic equations on time scales:
(1)
u
Δ
∇
(
t
)
+
f
(
t
,
u
)
=
0
,
t
∈
(
0
,
T
)
∈
T
,
u
(
0
)
=
0
,
u
(
T
)
=
∑
i
=
1
m
-
2
‍
k
i
u
(
ξ
i
)
,
where T is a time scale. By using Green’s function and the Leggett-Williams fixed point theorem in an appropriate cone, the existence of at least three positive solutions of the problem is obtained.In 2009, Topal and Yantir [4] studied the general second-order nonlinear m-point boundary value problems:
(2)
u
Δ
∇
(
t
)
+
a
(
t
)
u
Δ
(
t
)
+
b
(
t
)
u
(
t
)
+
λ
h
(
t
)
f
(
t
,
u
(
t
)
)
=
0
,
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
n
i
t
∈
(
0,1
)
,
u
(
ρ
(
0
)
)
=
0
,
u
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
‍
α
i
u
(
η
i
)
with no singularity. The authors deal with the determining the value of λ, the existence of multiple positive solutions of (2) is obtained by using the Krasnosel’skii and Legget-William fixed point theorems.Impulsive differential equations are now recognized as an excellent source of models for simulating processes and phenomena observed in control theory, physics, chemistry, population dynamics, biotechnology, industrial robotics, optimal control, and so forth. In recent years, impulsive differential equations have become a very active area of research. In this paper, we consider the following impulsive singular dynamic equations on time scales:(3)
u
Δ
∇
(
t
)
+
a
(
t
)
u
Δ
(
t
)
+
b
(
t
)
u
(
t
)
+
q
(
t
)
f
(
t
,
u
(
t
)
)
=
0
,
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
k
k
k
l
n
t
∈
(
0,1
)
,
t
≠
t
k
,
u
Δ
(
t
k
+
)
=
u
Δ
(
t
k
)
-
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
n
,
u
(
ρ
(
0
)
)
=
0
,
u
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
‍
α
i
u
(
η
i
)
,
where α
≥
0
,
0
<
η
i
<
η
i
+
1
<
1, for all i
=
1,2
,
…
,
m
-
2, I
k
,
f
,
q
,
a, and b satisfy the following:(C1)
f
∈
C
(
[
ρ
(
0
)
,
σ
(
1
)
]
×
(
0
,
+
∞
)
,
[
0
,
+
∞
)
) and f
(
t
,
u
) may be singular at u
=
0, I
k
∈
C
(
[
0
,
+
∞
)
,
[
0
,
+
∞
)
);
(C2)
q
∈
C
(
(
0,1
)
,
[
0
,
+
∞
)
) and there exists t
0
∈
(
0,1
) such that q
(
t
0
)
>
0, q
(
t
) may be singular at t
=
0,1;
(C3)
a
∈
C
(
[
0,1
]
,
[
0
,
+
∞
)
), b
∈
C
(
[
0,1
]
,
(
-
∞
,
0
]
).The main theorems of this paper complement the very little existence results devoted to impulsive dynamic equations on a time scale. We will prove our two existence results for the problem (3) by using mixed monotone fixed point theorem on cone [17] and Krasnosel’skii fixed point theorem [18]. This paper is organized as follows. In Section 2, starting with some preliminary lemmas, we state a mixed monotone fixed point theorem on cone and Krasnosel’skii fixed point theorem. In Section 3, we give the main result which states the sufficient conditions for the m-point boundary value problem (3) to have existence of positive solutions. We also present examples to illustrate our results work.


## 2. Preliminaries

In this section we state the preliminary information that we need to prove the main results. From Lemmas 2.1 and 2.2 in [4], we have the following lemma.Lemma 1.
Assume that (C3) holds. Then the equations(4)
ϕ
1
Δ
∇
(
t
)
+
a
(
t
)
ϕ
1
Δ
(
t
)
+
b
(
t
)
ϕ
1
(
t
)
=
0
,
t
∈
(
0,1
)
,
ϕ
1
(
ρ
(
0
)
)
=
0
,
ϕ
1
(
σ
(
1
)
)
=
1
,
ϕ
2
Δ
∇
(
t
)
+
a
(
t
)
ϕ
2
Δ
(
t
)
+
b
(
t
)
ϕ
2
(
t
)
=
0
,
t
∈
(
0,1
)
,
ϕ
2
(
ρ
(
0
)
)
=
1
,
ϕ
2
(
σ
(
1
)
)
=
0
have unique solutions ϕ
1 and ϕ
2, respectively, and
(a)ϕ
1 is strictly increasing on [
ρ
(
0
)
,
σ
(
1
)
]; (b) ϕ
2 is strictly decreasing on [
ρ
(
0
)
,
σ
(
1
)
].For the rest of the paper we need the following assumption:(C4)
0
<
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
<
1.Lemma 2.
Assume that (C3) and (C4) hold. Letx
∈
C
[
ρ
(
0
)
,
σ
(
1
)
]. Then boundary value problem
(5)
x
Δ
∇
(
t
)
+
a
(
t
)
x
Δ
(
t
)
+
b
(
t
)
x
(
t
)
+
x
(
t
)
=
0
,
t
∈
(
0,1
)
,
x
(
ρ
(
0
)
)
=
0
,
x
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
‍
α
i
x
(
η
i
)
is equivalent to the integral equation
(6)
x
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
H
(
t
,
s
)
p
(
s
)
x
(
s
)
∇
s
+
A
ϕ
1
(
t
)
,
where
(7)
p
(
t
)
=
e
a
(
ρ
(
t
)
,
ρ
(
0
)
)
,
A
=
1
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
∑
i
=
1
m
-
2
‍
α
i
∫
ρ
(
0
)
σ
(
1
)
H
(
η
i
,
s
)
p
(
s
)
x
(
s
)
∇
s
,
H
(
t
,
s
)
=
1
ϕ
1
Δ
(
ρ
(
0
)
)
{
ϕ
1
(
s
)
ϕ
2
(
t
)
,
s
≤
t
,
ϕ
1
(
t
)
ϕ
2
(
s
)
,
t
≤
s
.Lemma 3.
Green’s functionH
(
t
,
s
) has the properties
(8)
H
(
t
,
s
)
≤
H
(
t
,
t
)
,
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
H
(
t
,
t
)
H
(
s
,
s
)
≤
H
(
t
,
s
)
≤
H
(
s
,
s
)
,
H
(
t
,
t
)
≤
ϕ
1
(
t
)
∥
ϕ
2
∥
ϕ
1
Δ
(
ρ
(
0
)
)
.Lemma 4.
Assume that (C3) and (C4) hold. Letu
∈
C
[
ρ
(
0
)
,
σ
(
1
)
] be a solution of the boundary value problem (3) if and only if u is a solution of the following impulsive integral equation:
(9)
u
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
G
(
t
,
s
)
p
(
s
)
q
(
s
)
f
(
s
,
u
(
s
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
,
where
(10)
G
(
t
,
s
)
=
H
(
t
,
s
)
+
1
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
∑
i
=
1
m
-
2
‍
α
i
H
(
η
i
,
s
)
ϕ
1
(
t
)
.Lemma 5.
Green’s functionG
(
t
,
s
) defined by (10) has the properties
(11)
G
0
(
t
)
G
*
(
s
)
≤
G
(
t
,
s
)
≤
G
*
(
s
)
,
G
(
t
,
s
)
≤
ϕ
1
(
t
)
C
(
s
)
,
where
(12)
C
(
s
)
=
∥
ϕ
2
∥
ϕ
1
Δ
(
ρ
(
0
)
)
+
1
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
∑
i
=
1
m
-
2
‍
α
i
H
(
η
i
,
s
)
,
G
*
(
s
)
=
H
(
s
,
s
)
+
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
∑
i
=
1
m
-
2
‍
α
i
H
(
η
i
,
s
)
,
G
0
(
t
)
=
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
H
(
t
,
t
)
.Lemma 6.
Assume that (C1)–(C4) hold. Then the solution of (3) satisfies u
(
t
)
≥
G
0
(
t
)
∥
u
∥.The proofs of the Lemmas3–6 can be obtained easily by Lemmas 1 and 2.For our constructions, we will consider the Banach spaceE
=
C
[
ρ
(
0
)
,
σ
(
1
)
] equipped with standard norm ∥
u
∥
=
max
⁡
⁡
ρ
(
0
)
≤
t
≤
σ
(
1
)
|
u
(
t
)
|
,
u
∈
E. We define a cone K by
(13)
K
=
{
u
∈
X
∣
u
(
t
)
≥
G
0
(
t
)
∥
u
∥
,
t
∈
[
ρ
(
0
)
,
σ
(
1
)
]
}
.
From Lemmas 4 and 5, we define the integral operator T
:
K
→
E by
(14)
T
u
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
G
(
t
,
s
)
p
(
s
)
q
(
s
)
f
(
s
,
u
(
s
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
.
Then, it is clear that the solutions of (3) are the fixed points of the operator T.Thus, from Lemma4, standard arguments show that T
(
K
)
⊂
K and T is completely continuous.The following content will play major role in our next analysis.LetP be a normal cone of a Banach space E, and let e
∈
P with ∥
e
∥
≤
1
,
e
≠
θ. Define
(15)
Q
e
=
{
x
∈
P
∣
x
≠
θ
,
there
exist
constants
m
,
M
>
0
Q
e
=
5
such  that
m
e
≤
x
≤
M
e
}
.Definition 7 (see [17]).
AssumeS
:
Q
e
×
Q
e
→
Q
e. S is said to be mixed monotone if S
(
x
,
y
) is nondecreasing in x and nonincreasing in y, that is, if x
1
≤
x
2
(
x
1
,
x
2
∈
Q
e
) implies S
(
x
1
,
y
)
≤
S
(
x
2
,
y
) for any y
∈
Q
e and y
1
≤
y
2
(
y
1
,
y
2
∈
Q
e
) implies S
(
x
,
y
1
)
≥
S
(
x
,
y
2
) for any x
∈
Q
e. x
*
∈
Q
e is said to be a fixed point of S if S
(
x
*
,
x
*
)
=
x
*.Theorem 8 (see [17, 19]).
Suppose thatS
:
Q
e
×
Q
e
→
Q
e is a mixed monotone operator and ∃ a constant α
,
0
≤
α
<
1, such that
(16)
S
(
t
x
,
1
t
y
)
≥
t
α
S
(
x
,
y
)
,
∀
x
,
y
∈
Q
e
,
0
<
t
<
1
.
Then S has a unique fixed point x
*
∈
Q
e. Moreover, for any (
x
0
,
y
0
)
∈
Q
e
×
Q
e,
(17)
x
n
=
S
(
x
n
-
1
,
y
n
-
1
)
,
y
n
=
S
(
y
n
-
1
,
x
n
-
1
)
,
n
=
1,2
,
…
satisfy
(18)
x
n
⟶
x
*
,
y
n
⟶
x
*
,
where
(19)
∥
x
n
-
x
*
∥
=
o
(
1
-
r
α
n
)
,
∥
y
n
-
x
*
∥
=
o
(
1
-
r
α
n
)
,
0
<
r
<
1, and r is a constant from (
x
0
,
y
0
).Theorem 9 (see [18]).
LetX be a Banach space, and let P
⊂
X be a cone in X. Assume that Ω
1
,
Ω
2 are open subsets of X with 0
∈
Ω
1
⊂
Ω
¯
1
⊂
Ω
2, and let S
:
P
→
P be a completely continuous operator such that either(1)
∥
S
w
∥
≤
∥
w
∥, w
∈
P
∩
∂
Ω
1, ∥
S
w
∥
≥
∥
w
∥, w
∈
P
∩
∂
Ω
2, or
(2)
∥
S
w
∥
≥
∥
w
∥, w
∈
P
∩
∂
Ω
1, ∥
S
w
∥
≤
∥
w
∥
w
∈
P
∩
∂
Ω
2.Then S has a fixed point in P
∩
Ω
¯
2
∖
Ω
1.


## 3. Main Results

First, by using Theorem8 we establish the following main result.Theorem 10.
Suppose that conditions(
C
1
)
–
(
C
4
) hold and(C5)
f
(
t
,
u
)
=
h
0
(
u
)
+
g
0
(
u
), I
k
(
u
)
=
h
k
(
u
)
+
g
k
(
u
) (k
=
0,1
,
2
,
…
,
n) and
(20)
g
k
:
(
0
,
+
∞
)
⟶
(
0
,
+
∞
)
i
s
c
o
n
t
i
n
u
o
u
s
a
n
d
n
o
n
i
n
c
r
e
a
s
i
n
g
;
h
k
:
[
0
,
+
∞
)
⟶
[
0
,
+
∞
)
i
s
c
o
n
t
i
n
u
o
u
s
a
n
d
n
o
n
d
e
c
r
e
a
s
i
n
g
fork
=
0,1
,
2
,
…
,
m.
(C6)
There existsα
∈
(
0,1
) such that
(21)
g
k
(
t
-
1
x
)
≥
t
α
g
k
(
x
)
,
(22)
h
k
(
t
x
)
≥
t
α
h
k
(
x
)
,
for anyt
∈
(
0,1
) and, x
>
0, k
=
0,1
,
2
,
…
,
n.
(C7)
Consider thatq
∈
C
(
(
0,1
)
,
(
0
,
∞
)
) satisfies
(23)
∫
0
1
C
(
s
)
ϕ
1
-
α
(
s
)
p
(
s
)
q
(
s
)
d
s
<
+
∞
.Then (3) has an unique positive solution u
*
(
t
).Proof.
Since (21) holds, let t
-
1
x
=
y; one has
(24)
g
k
(
y
)
≥
t
α
g
k
(
t
y
)
.
Then
(25)
g
k
(
t
y
)
≤
1
t
α
g
k
(
y
)
,
for
t
∈
(
0,1
)
,
y
>
0
.
Lety
=
1. The above inequality is
(26)
g
k
(
t
)
≤
1
t
α
g
k
(
1
)
,
for
t
∈
(
0,1
)
.
From (21), (25), and (26), one has
(27)
g
k
(
t
-
1
x
)
≥
t
α
g
k
(
x
)
,
g
k
(
1
t
)
≥
t
α
g
k
(
1
)
,
g
k
(
t
x
)
≤
1
t
α
g
k
(
x
)
,
g
k
(
t
)
≤
1
t
α
g
k
(
1
)
,
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
for
t
∈
(
0,1
)
,
x
>
0
.
Similarly, from (22), one has
(28)
h
k
(
t
x
)
≥
t
α
h
k
(
x
)
,
h
k
(
t
)
≥
t
α
h
k
(
1
)
,
l
l
l
l
l
for
t
∈
(
0,1
)
,
x
>
0
,
k
=
0,1
,
…
,
m
.
Let t
=
1
/
x
,
x
>
1; one has
(29)
h
k
(
x
)
≤
x
α
h
k
(
1
)
,
for
x
≥
1
,
k
=
0,1
,
…
,
m
.
Let e
(
t
)
=
ϕ
1
(
t
)
/
∥
ϕ
1
∥; it is clear that ∥
e
∥
≤
1. We define
(30)
Q
e
=
{
x
∈
C
[
0,1
]
∣
1
M
e
(
t
)
≤
x
(
t
)
≤
M
e
(
t
)
,
t
∈
[
0,1
]
}
,
where M
>
1 is chosen such that
(31)
M
>
max
⁡
{
[
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
1
)
+
e
-
α
(
s
)
g
0
(
1
)
)
∇
s
+
∑
k
=
1
n
‍
∥
ϕ
1
∥
C
(
t
k
)
×
(
h
k
(
1
)
+
e
-
α
(
t
k
)
g
k
(
1
)
)
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
(
h
0
(
1
)
]
1
/
(
1
-
α
)
;
[
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
×
∑
i
=
1
m
-
2
‍
α
i
∫
ρ
(
0
)
σ
(
1
)
G
(
η
i
,
s
)
p
(
s
)
q
(
s
)
×
(
e
α
(
s
)
h
0
(
1
)
+
g
0
(
1
)
)
∇
s
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
]
-
(
1
/
(
1
-
α
)
)
}
.
For any x
,
y
∈
Q
e, we define
(32)
T
(
x
,
y
)
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
G
(
t
,
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
x
(
s
)
)
+
g
0
(
y
(
s
)
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
(
h
k
(
x
(
t
k
)
)
+
g
k
(
x
(
t
k
)
)
)
.
Let x
,
y
∈
Q
e. On one hand, from (27), (28), and (29), for k
=
0,1
,
2
,
…
,
m, we have
(33)
g
k
(
y
(
t
)
)
≤
g
k
(
1
M
e
(
t
)
)
≤
M
α
e
-
α
(
t
)
g
k
(
1
)
,
h
k
(
x
(
t
)
)
≤
h
k
(
M
e
(
t
)
)
≤
h
k
(
M
)
≤
M
α
h
k
(
1
)
,
g
k
(
y
(
t
)
)
≥
g
k
(
M
e
(
t
)
)
≥
g
k
(
M
)
≥
g
k
(
1
1
/
M
)
≥
1
M
α
g
k
(
1
)
,
h
k
(
x
(
t
)
)
≥
h
k
(
1
M
e
(
t
)
)
≥
e
α
(
t
)
h
k
(
1
M
)
≥
e
α
(
t
)
1
M
α
h
k
(
1
)
;
then
(34)
1
M
α
(
e
α
(
t
)
h
k
(
1
)
+
g
k
(
1
)
)
≤
h
k
(
x
(
t
)
)
+
g
k
(
y
(
t
)
)
≤
M
α
(
h
k
(
1
)
+
e
-
α
(
t
)
g
k
(
1
)
)
,
t
∈
(
0,1
)
.
Thus, for anyx
,
y
∈
Q
e, we have
(35)
T
(
x
,
y
)
(
t
)
≥
e
(
t
)
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
×
∑
i
=
1
m
-
2
‍
α
i
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
η
i
,
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
x
(
s
)
)
+
g
0
(
y
(
s
)
)
)
∇
s
≥
e
(
t
)
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
×
∑
i
=
1
m
-
2
‍
α
i
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
η
i
,
s
)
p
(
s
)
q
(
s
)
1
M
α
×
(
e
α
(
s
)
h
0
(
1
)
+
g
0
(
1
)
)
∇
s
≥
1
M
α
e
(
t
)
∥
ϕ
1
∥
1
-
∑
i
=
1
m
-
2
α
i
ϕ
1
(
η
i
)
×
∑
i
=
1
m
-
2
‍
α
i
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
η
i
,
s
)
p
(
s
)
q
(
s
)
×
(
e
α
(
s
)
h
0
(
1
)
+
g
0
(
1
)
)
∇
s
≥
1
M
e
(
t
)
,
T
(
x
,
y
)
(
t
)
≤
e
(
t
)
[
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
(
h
0
(
x
(
s
)
)
∑
k
=
1
n
‍
+
g
0
(
x
(
s
)
)
∇
s
+
∑
k
=
1
n
‍
∥
ϕ
1
∥
C
(
t
k
)
)
×
(
h
k
(
x
(
t
k
)
)
+
g
k
(
x
(
t
k
)
)
)
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
(
h
0
(
x
(
s
)
)
]
≤
e
(
t
)
[
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
M
α
×
(
h
0
(
1
)
+
e
-
α
(
s
)
g
0
(
1
)
)
∇
s
+
∑
k
=
1
n
‍
∥
ϕ
1
∥
C
(
t
k
)
M
α
×
(
h
k
(
1
)
+
e
-
α
(
t
k
)
g
k
(
1
)
)
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
M
α
(
h
0
(
1
)
]
≤
M
α
e
(
t
)
[
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
1
)
+
e
-
α
(
s
)
g
0
(
1
)
)
∇
s
+
∑
k
=
1
n
‍
∥
ϕ
1
∥
C
(
t
k
)
×
(
h
k
(
1
)
+
e
-
α
(
t
k
)
g
k
(
1
)
)
∫
ρ
(
0
)
σ
(
1
)
∥
ϕ
1
∥
C
(
s
)
p
(
s
)
q
(
s
)
]
≤
M
e
(
t
)
,
for
t
∈
[
0,1
]
.
So, T is well defined and T
(
Q
e
×
Q
e
)
⊂
Q
e.
Next, for anyl
∈
(
0,1
), one has
(36)
T
(
l
x
,
l
-
1
y
)
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
t
,
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
l
x
(
s
)
)
+
g
0
(
l
-
1
y
(
s
)
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
(
h
k
(
l
x
(
t
k
)
)
+
g
k
(
l
-
1
y
(
t
k
)
)
)
≥
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
t
,
s
)
p
(
s
)
q
(
s
)
×
(
l
α
h
0
(
x
(
s
)
)
+
l
α
g
0
(
y
(
s
)
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
(
l
α
h
k
(
x
(
t
k
)
)
+
l
α
g
k
(
y
(
t
k
)
)
)
≥
l
α
{
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
t
,
s
)
p
(
s
)
q
(
s
)
×
(
h
0
(
x
(
s
)
)
+
g
0
(
y
(
s
)
)
)
∇
s
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
(
h
k
(
x
(
t
k
)
)
+
g
k
(
y
(
t
k
)
)
)
}
=
l
α
T
(
x
,
y
)
(
t
)
,
for
t
∈
[
0,1
]
.
So the conditions of Theorems8 hold. Therefore there exists a unique u
*
∈
Q
e such that T
(
u
*
,
u
*
)
=
u
*. This completes the proof of Theorem 10.Example 11.
Consider the following singular boundary value problem:(37)
u
Δ
∇
(
t
)
+
a
(
t
)
u
Δ
(
t
)
+
b
(
t
)
u
(
t
)
+
(
μ
x
α
+
x
-
β
)
=
0
,
k
k
k
k
k
k
k
k
k
k
k
k
l
k
k
k
k
k
k
k
k
k
k
k
j
n
i
t
∈
(
0,1
)
,
t
≠
t
k
,
u
Δ
(
t
k
+
)
=
u
Δ
(
t
k
)
-
a
k
(
u
a
(
t
k
)
)
,
k
=
1,2
,
…
,
n
;
u
(
ρ
(
0
)
)
=
0
,
u
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
‍
α
i
u
(
η
i
)
,
where α
,
β
>
0
,
max
⁡
{
α
,
β
}
<
1
,
μ
≥
0,0
<
η
i
<
η
i
+
1
<
1, for
all
i
=
1,2
,
…
,
m
-
2, and a and b satisfy a
∈
C
(
[
0,1
]
,
[
0
,
+
∞
)
), b
∈
C
(
[
0,1
]
,
(
-
∞
,
0
]
).
Letα
0
=
max
⁡
⁡
{
α
,
β
}, h
(
x
)
=
μ
x
a, g
(
x
)
=
x
-
b, and q
(
t
)
=
1. Thus
(38)
h
(
l
x
)
=
μ
l
α
x
α
≥
l
α
0
h
(
x
)
,
g
(
l
-
1
x
)
=
l
β
x
-
β
≥
l
α
0
g
(
x
)
,
∫
0
1
G
-
α
0
(
s
,
s
)
d
s
<
+
∞
.
Applying Theorem 10, we can find that the above equation has a unique solution u
*
(
t
).In the next, using Theorem9 we establish the following main result.Theorem 12.
Suppose that conditions(
C
1
)
–
(
C
4
) hold and the following conditions are satisfied:
(39)
f
(
t
,
u
)
≤
g
0
(
u
)
+
h
0
(
u
)
,
I
k
(
u
)
≤
g
k
(
u
)
+
h
k
(
u
)
(
k
=
1
,
2
,
…
,
n
)
o
n
[
0,1
]
×
(
0
,
∞
)
w
i
t
h
g
i
>
0
(
i
=
0
,
1,2
,
…
,
n
)
c
o
n
t
i
n
u
o
u
s
a
n
d
n
o
n
i
n
c
r
e
a
s
i
n
g
o
n
(
0
,
∞
)
,
h
i
≥
0
(
i
=
0
,
1
,
2
,
…
,
n
)
c
o
n
t
i
n
u
o
u
s
o
n
[
0
,
∞
)
,
h
i
g
i
(
i
=
0
,
1
,
2
,
…
,
n
)
n
o
n
d
e
c
r
e
a
s
i
n
g
o
n
(
0
,
∞
)
,
(40)
∃
K
0
w
i
t
h
g
i
(
x
y
)
≤
K
0
g
i
(
x
)
g
i
(
y
)
,
∀
x
>
0
,
y
>
0
,
i
=
0,1
,
2
,
…
,
n
,
(41)
a
0
=
∫
ρ
(
0
)
σ
(
1
)
G
*
(
s
)
p
(
s
)
q
(
s
)
g
0
(
G
0
(
s
)
)
∇
s
+
∑
k
=
1
n
‍
G
*
(
t
k
)
g
k
(
G
0
(
t
k
)
)
<
∞
,
(42)
∃
r
>
0
w
i
t
h
r
max
⁡
0
≤
i
≤
n
⁡
{
g
i
(
r
)
+
h
i
(
r
)
}
>
K
0
a
0
;
(43)
t
h
e
r
e
e
x
i
s
t
s
ρ
(
0
)
<
θ
<
σ
(
1
)
-
ρ
(
0
)
2
(
c
h
o
o
s
e
a
n
d
f
i
x
i
t
)
a
n
d
a
c
o
n
t
i
n
u
o
u
s
,
n
o
n
i
n
c
r
e
a
s
i
n
g
f
u
n
c
t
i
o
n
g
i
:
(
0
,
∞
)
⟶
(
0
,
∞
)
,
a
c
o
n
t
i
n
u
o
u
s
f
u
n
c
t
i
o
n
h
¯
i
:
[
0
,
∞
)
⟶
(
0
,
∞
)
w
i
t
h
h
¯
i
g
¯
i
n
o
n
d
e
c
r
e
a
s
i
n
g
o
n
(
0
,
∞
)
(
i
=
0,1
,
2
,
…
,
n
)
,
a
n
d
w
i
t
h
f
(
t
,
u
)
≥
g
¯
0
(
u
)
+
h
¯
0
(
u
)
,
I
k
(
u
)
≥
g
¯
k
(
u
)
+
h
¯
k
(
u
)
(
k
=
1
,
…
,
n
)
f
o
r
(
t
,
u
)
∈
[
θ
,
σ
(
1
)
-
θ
]
×
(
0
,
∞
)
,
(44)
∃
0
<
R
1
<
r
<
R
2
w
i
t
h
(
j
=
1,2
)
,
R
j
(
min
⁡
0
≤
i
≤
n
⁡
{
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
R
j
)
×
[
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
min
⁡
0
≤
i
≤
n
⁡
{
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
R
j
)
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥

R
j
ϕ
1
×
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
min
⁡
0
≤
i
≤
n
⁡
{
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
R
j
)
)
-
1
)
]
}
)
-
1
<
μ
b
¯
0
,
b
¯
0
=
∫
θ
σ
(
1
)
-
θ
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
∇
s
+
∑
k
=
1
n
‍
G
(
σ
,
t
k
)
.
Here G
(
t
,
s
) is Green’s function and
(45)
∫
θ
σ
(
1
)
-
θ
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
d
s
=
sup
⁡
t
∈
[
0,1
]
∫
θ
σ
(
1
)
-
θ
‍
G
(
t
,
s
)
p
(
s
)
q
(
s
)
d
s
.
Then (3) has two nonnegative solutions u
i with R
1
<
∥
u
1
∥
<
r
<
∥
u
2
∥
<
R
2 and u
i
(
t
)
>
0 for t
∈
(
0,1
)
,
i
=
1,2.Proof.
First we will show that there exists a solutionu
2 to (3) with u
2
(
t
)
>
0 for t
∈
(
0,1
) and r
<
∥
u
2
∥
<
R
2. Let
(46)
Ω
1
=
{
u
∈
E
:
∥
u
∥
<
r
}
,
Ω
2
=
{
u
∈
E
:
∥
u
∥
<
R
2
}
.
We now show that
(47)
∥
T
u
∥
<
∥
u
∥
for
K
∩
∂
Ω
1
.
To see this, letu
∈
K
∩
∂
Ω
1. Then ∥
u
∥
=
∥
u
∥
[
0,1
]
=
r and u
(
t
)
≥
G
0
(
t
)
r for t
∈
[
0,1
]. So for t
∈
[
0,1
] we have
(48)
g
i
(
u
)
+
h
i
(
u
)
=
g
i
(
u
)
(
1
+
h
i
(
u
)
g
i
(
u
)
)
≤
g
i
(
G
0
(
t
)
r
)
(
1
+
h
i
(
r
)
g
i
(
r
)
)
≤
K
0
g
i
(
r
)
(
1
+
h
i
(
r
)
g
i
(
r
)
)
g
i
(
G
0
(
t
)
)
≤
K
0
(
g
i
(
r
)
+
h
i
(
r
)
)
g
i
(
G
0
(
t
)
)
≤
K
0
max
⁡
0
≤
i
≤
m
⁡
{
g
i
(
r
)
+
h
i
(
r
)
}
g
i
(
G
0
(
t
)
)
,
for
i
=
0,1
,
2
,
…
,
n
.
Then
(49)
(
T
u
)
(
t
)
=
∫
ρ
(
0
)
σ
(
1
)
G
(
t
,
s
)
p
(
s
)
q
(
s
)
f
(
(
s
,
u
(
s
)
)
)
∇
s
llllllllllll
+
∑
k
=
1
n
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
llllllllllll
≤
∫
ρ
(
0
)
σ
(
1
)
G
*
(
s
)
p
(
s
)
q
(
s
)
(
g
0
(
u
(
s
)
)
+
h
0
(
u
(
s
)
)
)
∇
s
llllllllllllll
l
+
∑
k
=
1
n
‍
G
*
(
t
k
)
(
g
k
(
u
(
s
)
)
+
h
k
(
u
(
s
)
)
)
lllllllllll
≤
K
0
[
∫
ρ
(
0
)
σ
(
1
)
‍
G
*
(
s
)
p
(
s
)
q
(
s
)
g
0
(
G
0
(
s
)
)
∇
s
llllllllllllkkklll
+
∑
k
=
1
n
‍
G
*
(
t
k
)
g
k
(
G
0
(
t
k
)
)
]
k
k
k
k
k
k
k
k
k
k
×
max
⁡
0
≤
i
≤
m
⁡
{
g
i
(
r
)
+
h
i
(
r
)
}
.
This together with (42) yields
(50)
∥
T
u
∥
=
∥
T
u
∥
[
0,1
]
<
r
=
∥
u
∥
,
so (47) is satisfied.
Next we show(51)
∥
T
u
∥
>
∥
u
∥
for
K
∩
∂
Ω
2
.
To see this let u
∈
K
∩
∂
Ω
2 so |
|
u
|
|
=
|
|
u
|
|
[
0,1
]
=
R
2 and u
(
t
)
≥
G
0
(
t
)
R
2 for t
∈
[
0,1
].
We have(52)
g
¯
i
(
u
(
s
)
)
+
h
¯
i
(
u
(
s
)
)
=
g
¯
i
(
u
(
s
)
)
[
1
+
h
¯
i
(
u
(
s
)
)
g
¯
i
(
u
(
s
)
)
]
≥
g
¯
i
(
R
2
)
[
1
+
h
¯
i
(
G
0
(
s
)
R
2
)
g
¯
i
(
G
0
(
s
)
R
2
)
]
≥
g
¯
i
(
R
2
)
×
[
R
2
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
)
-
1
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
)
-
1
)
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
2
ϕ
1
×
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
×
(
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
2
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
)
-
1
)
]
for
s
∈
[
θ
,
σ
(
1
)
-
θ
]
.
Then
(53)
(
T
u
)
(
σ
)
≥
∫
ρ
(
0
)
σ
(
1
)
‍
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
[
g
¯
0
(
u
(
s
)
)
+
h
¯
0
(
u
(
s
)
)
]
∇
s
+
∑
k
=
1
n
‍
G
(
σ
,
t
k
)
[
g
¯
k
(
u
(
t
k
)
)
+
h
¯
k
(
u
(
t
k
)
)
]
≥
{
∫
θ
σ
(
1
)
-
θ
‍
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
∇
s
+
∑
k
=
1
n
‍
G
(
σ
,
t
k
)
}
×
min
⁡
0
≤
i
≤
m
⁡
{
ϕ
1
Δ
∥
ϕ
1
∥
)
)
-
1
)
]
}
R
2
g
¯
i
(
R
2
)
[
R
2
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
)
-
1

ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
-
1
)
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
2
ϕ
1
(
θ
)
×
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
ϕ
2
(
σ
(
1
)
-
θ
)
)
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
2
ϕ
1
(
θ
)
ϕ
2
×
ϕ
1
Δ
∥
ϕ
1
∥
(
σ
(
1
)
-
θ
)
)
)
-
1
)
]
}
.
This together with (44) yields
(54)
(
T
u
)
(
σ
)
>
R
2
=
∥
u
∥
.
Thus ∥
T
u
∥
>
∥
u
∥, so (51) holds.
Now Theorem9 implies that T has a fixed point u
2
∈
K
∩
(
Ω
-
2
∖
Ω
1
); that is, r
≤
∥
u
2
∥
=
∥
u
2
∥
[
0,1
]
≤
R and u
2
(
t
)
≥
q
(
t
)
r for t
∈
[
0,1
]. It follows from (47) and (51) that ∥
u
2
∥
≠
r
,
∥
u
2
∥
≠
R
2, so we have r
<
∥
u
2
∥
<
R
2.
Similarly, if we put(55)
Ω
1
=
{
u
∈
E
:
∥
u
∥
<
R
1
}
,
Ω
2
=
{
u
∈
E
:
∥
u
∥
<
r
}
,
we can show that there exists a solution u
1 to (3) with u
1
(
t
)
>
0 for t
∈
(
0,1
) and R
1
<
∥
u
1
∥
<
r. This completes the proof of Theorem 12.Similar to the proof of Theorem12, we have the following result.Theorem 13.
Suppose that conditions(
C
1
)
–
(
C
4
) and (39)–(43) hold. In addition suppose that
(56)
∃
0
<
R
1
<
r
w
i
t
h
R
1
(
min
⁡
0
≤
i
≤
n
⁡
{
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
R
1
)
×
[
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
1
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
1
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
-
1
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
]
}
)
-
1
<
μ
b
¯
0
,
b
¯
0
=
∫
θ
σ
(
1
)
-
θ
‍
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
∇
s
+
∑
k
=
1
n
‍
G
(
σ
,
t
k
)
.
Then (3) has a nonnegative solution u
1 with R
1
<
∥
u
1
∥
<
r and u
1
(
t
)
>
0 for t
∈
(
0,1
).Remark 14.
If in (56) we have R
1
>
r, then (3) has a nonnegative solution u
2 with r
<
∥
u
2
∥
<
R
2 and u
2
(
t
)
>
0 for t
∈
(
0,1
).It is easy to use Theorem12 and Remark 14 to write theorems which guarantee the existence of more than two solutions to (3). We state one such result.Theorem 15.
Suppose that conditions (C1)–(C4); (39)–(41); and (43) hold. Assume that ∃
m
∈
{
1,2
,
…
} and constants R
j
,
r
j
(
j
=
1
,
…
,
m
), with r
1
>
b
0, and
(57)
0
<
R
1
<
r
1
<
R
2
<
r
2
<
⋯
<
R
m
<
r
m
.
In addition suppose for each j
=
1
,
…
,
m that
(58)
r
j
max
⁡
0
≤
i
≤
n
⁡
{
g
j
(
r
j
)
+
h
j
(
r
j
)
}
>
K
0
2
a
0
,
R
j
(
min
⁡
0
≤
i
≤
n
⁡
{
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
)
-
1
)
g
¯
i
(
R
j
)
×
[
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
)
-
1
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
j
ϕ
1
(
θ
)
×
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
j
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
)
-
1
)
]
}
)
-
1
<
μ
b
¯
0
,
b
¯
0
=
∫
θ
σ
(
1
)
-
θ
‍
G
(
σ
,
s
)
p
(
s
)
q
(
s
)
∇
s
+
∑
k
=
1
n
‍
G
(
σ
,
t
k
)
hold. Then (3) has nonnegative solutions y
1
,
…
,
y
m with y
j
(
t
)
>
0 for t
∈
(
0,1
).Example 16.
Consider the boundary value problem(59)
u
Δ
∇
(
t
)
+
a
(
t
)
u
Δ
(
t
)
+
b
(
t
)
u
(
t
)
+
μ
0
(
u
-
α
0
(
t
)
(
1
+
|
sin
t
|
)
+
u
β
0
(
t
)
(
1
+
|
cos
⁡
t
|
)
)
=
0
,
t
∈
(
0,1
)
,
t
≠
t
k
,
u
Δ
(
t
k
+
)
=
u
Δ
(
t
k
)
-
μ
k
(
u
-
β
k
(
t
k
)
+
u
β
k
(
t
k
)
)
,
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
l
k
k
m
i
k
k
=
1,2
,
…
,
n
;
u
(
ρ
(
0
)
)
=
0
,
u
(
σ
(
1
)
)
=
∑
i
=
1
m
-
2
‍
α
i
u
(
η
i
)
,
where 0
<
α
i
<
1
<
β
i  (
i
=
0,1
,
2
,
…
,
n
), and max
⁡
0
≤
i
≤
n
⁡
{
μ
i
}
∈
(
0
,
μ
) is such that
(60)
μ
≤
1
4
a
1
,
and here
(61)
a
1
=
∫
ρ
(
0
)
σ
(
1
)
‍
G
*
(
s
)
p
(
s
)
g
0
(
G
0
(
s
)
)
∇
s
+
∑
k
=
1
n
‍
G
*
(
t
k
)
g
k
(
G
0
(
t
k
)
)
<
∞
.Then (59) has two solutions u
1
,
u
2 with u
1
(
t
)
>
0
,
u
2
(
t
)
>
0 for t
∈
(
0,1
).To see this we will apply Theorem12 with (here 0
<
R
1
<
1
<
R
2 will be chosen below)
(62)
g
i
(
u
)
=
1
2
g
¯
i
(
u
)
=
2
u
-
α
i
,
h
i
(
u
)
=
1
2
h
¯
i
(
u
)
=
2
u
β
i
,
k
l
l
l
l
k
l
l
i
=
0,1
,
2
,
…
,
n
,
θ
=
3
ρ
(
0
)
+
σ
(
1
)
4
,
q
(
t
)
=
μ
0
,
K
0
=
1
.
Clearly (39)–(41) and (43) hold, and
(63)
a
0
=
μ
0
∫
ρ
(
0
)
σ
(
1
)
‍
G
*
(
s
)
p
(
s
)
g
0
(
G
(
s
,
s
)
)
∇
s
+
∑
k
=
1
n
‍
μ
k
G
*
(
t
k
)
g
k
(
G
0
(
t
k
)
)
<
∞
.
Now (42) holds with r
=
1 since
(64)
r
max
⁡
0
≤
i
≤
n
⁡
{
g
i
(
r
)
+
h
i
(
r
)
}
=
1
4
≥
μ
a
1
>
K
0
2
a
0
.
Finally, let
(65)
g
¯
i
*
(
R
j
)
[
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
)
)
-
1
1
+
(
h
¯
i
*
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
g
¯
i
*
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
-
1
)
]
=
min
⁡
0
≤
i
≤
n
⁡
{
g
¯
i
(
R
j
)
[
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
×
R
j
ϕ
1
(
θ
)
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
-
1
)
]
}
and notice that (44) is satisfied for R
1 small and R
2 large since
(66)
R
j
(
min
⁡
0
≤
i
≤
n
⁡
{
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
R
j
)
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
∥
ϕ
1
∥
)
)
-
1
)
]
×
[
1
+
(
h
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
∥
ϕ
1
∥
)
)
-
1
)
×
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
×
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
g
¯
i
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
R
j
ϕ
1
(
θ
)
ϕ
2
×
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
∥
ϕ
1
∥
)
)
-
1
)
]
}
)
-
1
≤
R
j
1
+
α
i
*
×
(
1
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
+
(
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
ϕ
1
(
θ
)
×
ϕ
2
(
σ
(
1
)
-
θ
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
(
α
i
*
+
β
i
*
)
×

R
j
(
α
i
*
+
β
i
*
)
ϕ
1
Δ
(
ρ
(
0
)
)
∥
ϕ
1
∥
∥
ϕ
2
∥
)
-
1
⟶
0
,
as R
1
→
0
,
R
2
→
∞, since b
>
1. Thus all the conditions of Theorem 12 are satisfied so existence is guaranteed.



---

*Source: 867018-2014-04-10.xml*