# Two Positive Solutions of Third-Order BVP with Integral Boundary Condition and Sign-Changing Green's Function

Authors: Bing-Wei Niu; Jian-Ping Sun; Qiu-Yan Ren
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We are concerned with the following third-order boundary value problem with integral boundary condition:u
′
′
′
(
t
)
=
f
(
t
,
u
(
t
)
)
,
t
∈
[
0,1
]
,  u
′
(
0
)
=
u
(
1
)
=
0
,
u
′
′
(
η
)
+
∫
α
β
‍
u
(
t
)
d
t
=
0
, where 1
/
2
<
α
≤
β
≤
1
,
α
+
β
≤
4
/
3, and η
∈
(
1
/
2
,
α
]. Although the corresponding Green's function is sign-changing, we still obtain the existence of at least two positive and decreasing solutions under some suitable conditions on f by using the two-fixed-point theorem due to Avery and Henderson. An example is also included to illustrate the main results obtained.


---

## Body

## 1. Introduction

Third-order differential equations arise from a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity driven flows, and so on [1].Recently, the existence of single or multiple positive solutions for some third-order three-point boundary value problems (BVPs for short) and for some third-order BVPs with integral boundary conditions has received much attention from many authors; see [2–15] and the references therein. It is necessary to point out that all the abovementioned papers are achieved when the corresponding Green’s functions are nonnegative, which is a very important condition.In 2008, Palamides and Smyrlis [16] studied the existence of at least one positive solution to the singular third-order three-point BVP with an indefinitely signed Green’s function:(1)
u
′
′
′
t
=
a
t
f
t
,
u
t
,
t
∈
0,1
,
u
0
=
u
1
=
u
′
′
η
=
0
,where η
∈
17
/
24
,
1. Their technique was a combination of the Guo-Krasnoselskii fixed point theorem and properties of the corresponding vector field.Inspired greatly by [16], Sun and Zhao discussed the third-order three-point BVP with sign-changing Green’s function:(2)
u
′
′
′
t
=
f
t
,
u
t
,
t
∈
0,1
,
u
′
0
=
u
1
=
u
′
′
η
=
0
.They proved the existence of at least one or three positive solutions when η
∈
(
1
/
2
,
1
) by using the Guo-Krasnoselskii and Leggett-Williams fixed point theorems in [17, 18] and obtained the existence of a positive solution when η
∈
[
2
-
2
,
1
) by using iterative technique in [19]. For similar results, one can refer to [20, 21].It is worth mentioning that there are other types of works on sign-changing Green’s functions which prove the existence of sign-changing solutions, positive in some cases; see Infante and Webb’s papers [22–24].In this paper, we consider the following third-order BVP involving integral boundary condition and sign-changing Green’s function:(3)
u
′
′
′
t
=
f
t
,
u
t
,
t
∈
0,1
,
u
′
0
=
u
1
=
0
,
u
′
′
η
+
∫
α
β
‍
u
t
d
t
=
0
.Throughout this paper, we always assume that 1
/
2
<
α
≤
β
≤
1
,
α
+
β
≤
4
/
3, and η
∈
(
1
/
2
,
α
]. By imposing suitable conditions on f, we obtain the existence of at least two positive and decreasing solutions for the BVP (3). The main tool used is the well-known Avery-Henderson two-fixed-point theorem.To end this section, we state some fundamental definitions and the two-fixed-point theorem due to Avery and Henderson [25].LetP be a cone in a real Banach space E.Definition 1.
A functionalψ
:
P
→
R is said to be increasing on P provided ψ
(
x
)
≤
ψ
(
y
) for all x
,
y
∈
P with x
≤
y, where x
≤
y if and only if y
-
x
∈
P.Definition 2.
Letγ
:
P
→
[
0
,
+
∞
) be continuous. For each d
>
0, one defines the set(4)
P
γ
,
d
=
u
∈
P
:
γ
u
<
d
.Theorem 3.
Letψ and γ be increasing, nonnegative, and continuous functionals on P, and let ω be a nonnegative continuous functional on P with ω
(
0
)
=
0 such that, for some c
>
0 and M
>
0,(5)
γ
u
≤
ω
u
≤
ψ
u
,
u
≤
M
γ
ufor all u
∈
P
(
γ
,
c
)
¯. Suppose there exist a completely continuous operator A
:
P
(
γ
,
c
)
¯
→
P and 0
<
a
<
b
<
c such that(6)
ω
λ
u
≤
λ
ω
u
f
o
r
0
≤
λ
≤
1
,
u
∈
∂
P
ω
,
b
,and(1)
γ
(
A
u
)
>
c for all u
∈
∂
P
(
γ
,
c
);
(2)
ω
(
A
u
)
<
b for all u
∈
∂
P
(
ω
,
b
);
(3)
P
(
ψ
,
a
)
≠
∅, and ψ
(
A
u
)
>
a for all u
∈
∂
P
(
ψ
,
a
).Then A has at least two fixed points u
1 and u
2 in P
(
γ
,
c
)
¯ such that(7)
a
<
ψ
u
1
w
i
t
h
ω
u
1
<
b
,
b
<
ω
u
2
w
i
t
h
γ
u
2
<
c
.


## 2. Main Results

For anyy
∈
C
[
0,1
], we consider the BVP(8)
u
′
′
′
t
=
y
t
,
t
∈
0,1
,
u
′
0
=
u
1
=
0
,
u
′
′
η
+
∫
α
β
‍
u
t
d
t
=
0
.After a direct computation, one may obtain the expression of Green’s functionG
(
t
,
s
) of the BVP (8) as follows:(9)
G
t
,
s
=
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
K
t
,
s
+
0
,
s
≥
η
,
3
1
-
t
2
ρ
,
s
<
η
,where(10)
ρ
=
6
-
3
β
-
α
+
β
3
-
α
3
,
K
t
,
s
=
1
2
1
-
t
1
+
t
-
2
s
,
0
≤
s
≤
t
≤
1
,
1
-
s
2
,
0
≤
t
≤
s
≤
1
.Remark 4.
Consider the following:ρ
>
0.Remark 5.
Consider the following:0
≤
K
(
t
,
s
)
≤
(
1
/
2
)
(
1
-
s
)
2 for (
t
,
s
)
∈
[
0,1
]
×
[
0,1
].Remark 6.
G
(
t
,
s
) has the following properties:(11)
G
t
,
s
≥
0
f
o
r
0
≤
s
<
η
,
G
t
,
s
≤
0
f
o
r
η
≤
s
≤
1
.Moreover, for s
≥
η,(12)
m
a
x
G
t
,
s
:
t
∈
0,1
=
G
1
,
s
=
0
,
min
⁡
G
t
,
s
:
t
∈
0,1
=
G
0
,
s
=
-
3
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
s
2and, for s
<
η,(13)
max
⁡
G
t
,
s
:
t
∈
0,1
=
G
0
,
s
=
3
ρ
-
3
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
s
2
,
m
i
n
G
t
,
s
:
t
∈
0,1
=
G
1
,
s
=
0
.In the remainder of this paper, we always assume thatf
:
[
0,1
]
×
[
0
,
+
∞
)
→
[
0
,
+
∞
) is continuous and satisfies the following two conditions:(
C
1
) for each u
∈
[
0
,
+
∞
), the mapping t
↦
f
(
t
,
u
) is decreasing;
(
C
2
) for each t
∈
[
0,1
], the mapping u
↦
f
(
t
,
u
) is increasing.Let real Banach spaceC
[
0,1
] be equipped with the norm u
=
max
t
∈
[
0,1
]
⁡
u
(
t
) and(14)
P
=
a
n
d
m
i
n
t
∈
1
-
θ
,
θ
u
t
≥
θ
∗
u
u
∈
C
0,1
:
u
t
i
s
n
o
n
n
e
g
a
t
i
v
e
a
n
d
d
e
c
r
e
a
s
i
n
g
o
n
0,1
,
a
n
d
m
i
n
t
∈
1
-
θ
,
θ
u
t
≥
θ
∗
u
,where θ
∈
(
1
/
2
,
η
)
a
n
d
θ
∗
=
(
η
-
θ
)
/
η. Then P is a cone in C
[
0,1
]. Now, we define an operator A as follows:(15)
A
u
t
=
∫
0
1
‍
G
t
,
s
f
s
,
u
s
d
s
,
u
∈
P
,
t
∈
0,1
.Obviously, if u is a fixed point of A in P, then u is a nonnegative and decreasing solution of the BVP (3).Lemma 7.
A
:
P
→
P is completely continuous.Proof.
Letu
∈
P. Then, for t
∈
[
0
,
η
], we have (16)
A
u
t
=
∫
0
t
‍
3
1
-
t
2
ρ
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
t
1
+
t
-
2
s
+
3
1
-
t
2
ρ
3
1
-
t
2
ρ
∫
α
β
‍
×
f
s
,
u
s
d
s
+
∫
t
η
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
s
2
+
3
1
-
t
2
ρ
×
f
s
,
u
s
d
s
+
∫
η
1
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
s
2
×
f
s
,
u
s
d
s
.In view of u
∈
P, 1
/
2
<
η
≤
α
,
β
+
α
≤
4
/
3, Remark 5, and (
C
1
) and (
C
2
), we get (17)
A
u
′
t
=
∫
0
t
‍
6
t
ρ
∫
α
β
‍
K
τ
,
s
d
τ
+
t
-
s
-
6
t
ρ
f
s
,
u
s
d
s
+
6
t
ρ
∫
t
η
‍
∫
α
β
‍
K
τ
,
s
d
τ
-
1
f
s
,
u
s
d
s
+
6
t
ρ
∫
η
1
‍
∫
α
β
‍
K
τ
,
s
d
τ
f
s
,
u
s
d
s
≤
∫
0
t
‍
3
t
ρ
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
+
t
-
s
-
6
t
ρ
×
f
s
,
u
s
d
s
+
3
t
ρ
∫
t
η
‍
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
-
2
f
s
,
u
s
d
s
+
3
t
β
-
α
ρ
∫
η
1
‍
1
-
s
2
f
s
,
u
s
d
s
=
1
ρ
∫
η
1
‍
-
3
t
β
-
α
2
-
β
-
α
+
ρ
×
∫
0
t
‍
s
f
s
,
u
s
d
s
-
t
∫
t
η
‍
ρ
+
3
s
β
-
α
2
-
β
-
α
f
s
,
u
s
d
s
+
3
t
β
-
α
∫
η
1
‍
1
-
s
2
f
s
,
u
s
d
s
≤
f
η
,
u
η
ρ
×
∫
η
1
‍
-
3
t
β
-
α
2
-
β
-
α
+
ρ
×
∫
0
t
‍
s
d
s
-
t
∫
t
η
‍
ρ
+
3
s
β
-
α
2
-
β
-
α
d
s
+
3
t
β
-
α
∫
η
1
‍
1
-
s
2
d
s
=
t
f
η
,
u
η
2
ρ
×
-
ρ
2
η
-
t
-
3
β
-
α
2
-
β
-
α
η
2
+
2
β
-
α
1
-
η
3
≤
t
f
η
,
u
η
2
ρ
×
-
3
β
-
α
2
-
β
-
α
η
2
+
2
β
-
α
1
-
η
2
=
β
-
α
t
f
η
,
u
η
2
ρ
×
3
β
+
α
-
4
η
2
+
2
1
-
2
η
≤
0
,
t
∈
0
,
η
,
A
u
′
′
t
=
∫
0
t
‍
6
ρ
∫
α
β
‍
K
τ
,
s
d
τ
+
1
-
6
ρ
f
s
,
u
s
d
s
+
6
ρ
∫
t
η
‍
∫
α
β
‍
K
τ
,
s
d
τ
-
1
f
s
,
u
s
d
s
+
6
ρ
∫
η
1
‍
∫
α
β
‍
K
τ
,
s
d
τ
f
s
,
u
s
d
s
≤
∫
0
t
‍
3
ρ
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
+
1
-
6
ρ
f
s
,
u
s
d
s
+
3
ρ
∫
t
η
‍
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
-
2
f
s
,
u
s
d
s
+
3
β
-
α
ρ
∫
η
1
‍
1
-
s
2
f
s
,
u
s
d
s
=
1
ρ
-
3
β
-
α
2
-
β
-
α
∫
0
t
‍
s
f
s
,
u
s
d
s
-
∫
t
η
‍
ρ
+
3
s
β
-
α
2
-
β
-
α
f
s
,
u
s
d
s
+
3
β
-
α
∫
η
1
‍
1
-
s
2
f
s
,
u
s
d
s
≤
f
η
,
u
η
ρ
×
-
3
β
-
α
2
-
β
-
α
∫
0
t
‍
s
d
s
-
∫
t
η
‍
ρ
+
3
s
β
-
α
2
-
β
-
α
d
s
+
3
β
-
α
∫
η
1
‍
1
-
s
2
d
s
=
f
η
,
u
η
2
ρ
×
1
-
η
3
-
2
ρ
η
-
t
-
3
β
-
α
2
-
β
-
α
η
2
+
2
β
-
α
1
-
η
3
η
2
≤
β
-
α
f
η
,
u
η
2
ρ
×
-
3
2
-
β
-
α
η
2
+
2
1
-
η
2
=
β
-
α
f
η
,
u
η
2
ρ
×
3
β
+
α
-
4
η
2
+
2
1
-
2
η
≤
0
,
t
∈
0
,
η
.For t
∈
(
η
,
1
], we have (18)
A
u
t
=
∫
0
η
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
t
1
+
t
-
2
s
+
3
1
-
t
2
ρ
f
s
,
u
s
d
s
+
∫
η
t
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
t
1
+
t
-
2
s
3
1
-
t
2
ρ
∫
α
β
‍
f
s
,
u
s
d
s
+
∫
t
1
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
s
2
3
1
-
t
2
ρ
∫
α
β
‍
f
s
,
u
s
d
s
.In view of u
∈
P, 1
/
2
<
η
≤
α
,
β
+
α
≤
4
/
3, Remark 5, and (
C
1
) and (
C
2
), we get (19)
A
u
′
t
=
∫
0
η
‍
6
t
ρ
∫
α
β
‍
K
τ
,
s
d
τ
+
t
-
s
-
6
t
ρ
f
s
,
u
s
d
s
+
∫
η
t
‍
6
t
ρ
∫
α
β
‍
K
τ
,
s
d
τ
+
t
-
s
f
s
,
u
s
d
s
+
6
t
ρ
∫
t
1
‍
∫
α
β
‍
K
τ
,
s
d
τ
f
s
,
u
s
d
s
≤
∫
0
η
‍
3
t
ρ
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
+
t
-
s
-
6
t
ρ
3
t
ρ
∫
α
β
‍
f
s
,
u
s
d
s
+
∫
η
t
‍
3
t
β
-
α
ρ
1
-
s
2
+
t
-
s
f
s
,
u
s
d
s
+
3
t
β
-
α
ρ
∫
t
1
‍
1
-
s
2
f
s
,
u
s
d
s
=
1
ρ
-
3
t
β
-
α
2
-
β
-
α
+
ρ
∫
0
η
‍
s
f
s
,
u
s
d
s
+
3
t
β
-
α
∫
η
1
‍
1
-
s
2
f
s
,
u
s
d
s
+
ρ
∫
η
t
‍
t
-
s
f
s
,
u
s
d
s
≤
f
η
,
u
η
ρ
×
-
3
t
β
-
α
2
-
β
-
α
+
ρ
∫
0
η
‍
s
d
s
+
3
t
β
-
α
∫
η
1
‍
1
-
s
2
d
s
+
ρ
∫
η
t
‍
t
-
s
d
s
=
t
f
η
,
u
η
2
ρ
×
-
3
β
-
α
2
-
β
-
α
η
2
+
2
β
-
α
1
-
η
3
+
ρ
t
-
2
η
η
2
≤
t
f
η
,
u
η
2
ρ
×
-
3
β
-
α
2
-
β
-
α
η
2
+
2
β
-
α
1
-
η
2
+
ρ
1
-
2
η
η
2
=
t
f
η
,
u
η
2
ρ
×
β
-
α
3
β
+
α
-
4
η
2
+
1
-
2
η
2
β
-
α
+
ρ
≤
0
,
t
∈
η
,
1
.So, it follows from A
u
′
(
t
)
≤
0 for t
∈
[
0,1
] that (
A
u
)
(
t
) is decreasing on [
0,1
], which together with (
A
u
)
(
1
)
=
0 implies that (
A
u
)
(
t
)
≥
0, t
∈
[
0,1
]. At the same time, since A
u
′
′
(
t
)
≤
0 for t
∈
[
0
,
η
], we know that (
A
u
)
(
t
) is concave on [
0
,
η
]. Therefore, (20)
A
u
t
≥
1
-
t
η
A
u
0
+
t
η
A
u
η
≥
η
-
t
η
A
u
,
t
∈
0
,
η
.Consequently,(21)
m
i
n
t
∈
1
-
θ
,
θ
A
u
t
=
A
u
θ
≥
η
-
θ
η
A
u
=
θ
∗
A
u
.This shows that A
:
P
→
P. Furthermore, although G
(
t
,
s
) is not continuous, it follows from known textbook results, for example, [26], that A
:
P
→
P is completely continuous.Lemma 8.
For anyu
∈
P and t
∈
[
1
-
θ
,
θ
],(22)
∫
0
1
-
θ
‍
G
t
,
s
f
s
,
u
s
d
s
+
∫
θ
η
‍
G
t
,
s
f
s
,
u
s
d
s
+
∫
η
1
‍
G
t
,
s
f
s
,
u
s
d
s
≥
0
.Proof.
Letu
∈
P. Then it follows from θ
<
η
≤
α, β
+
α
≤
4
/
3, Remarks 5 and 6, and (
C
1
) and (
C
2
) that, for any t
∈
[
1
-
θ
,
θ
], (23)
∫
0
1
-
θ
‍
G
t
,
s
f
s
,
u
s
d
s
+
∫
θ
η
‍
G
t
,
s
f
s
,
u
s
d
s
+
∫
η
1
‍
G
t
,
s
f
s
,
u
s
d
s
≥
∫
0
1
-
θ
‍
G
t
,
s
f
s
,
u
s
d
s
+
∫
η
1
‍
G
t
,
s
f
s
,
u
s
d
s
≥
f
η
,
u
η
∫
0
1
-
θ
‍
G
t
,
s
d
s
+
∫
η
1
‍
G
t
,
s
d
s
=
f
η
,
u
η
×
∫
0
1
-
θ
‍
-
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
-
1
2
1
-
t
1
+
t
-
2
s
+
3
1
-
t
2
ρ
d
s
-
∫
η
1
‍
3
1
-
t
2
ρ
∫
α
β
‍
K
τ
,
s
d
τ
+
1
2
1
-
s
2
d
s
≥
f
η
,
u
η
×
∫
0
1
-
θ
‍
-
3
1
-
t
2
2
ρ
∫
α
β
‍
1
-
τ
1
+
τ
-
2
s
d
τ
-
1
2
1
-
t
1
+
t
-
2
s
+
3
1
-
t
2
ρ
d
s
-
3
1
-
t
2
β
-
α
+
ρ
2
ρ
∫
η
1
‍
1
-
s
2
d
s
=
f
η
,
u
η
2
ρ
×
1
-
η
3
3
3
2
1
-
t
2
β
-
α
2
-
β
-
α
1
-
θ
2
+
ρ
1
-
t
1
-
θ
2
-
3
1
-
t
2
β
-
α
+
ρ
1
-
η
3
3
≥
f
η
,
u
η
2
ρ
×
1
-
θ
3
3
3
2
1
-
t
2
β
-
α
2
-
β
-
α
1
-
θ
2
+
ρ
1
-
t
1
-
θ
2
-
3
1
-
t
2
β
-
α
+
ρ
1
-
θ
3
3
=
1
-
θ
2
f
η
,
u
η
2
ρ
×
1
-
t
2
β
-
α
3
-
3
2
β
+
α
-
1
-
θ
+
ρ
1
-
t
-
ρ
3
1
-
θ
≥
1
-
θ
2
f
η
,
u
η
2
ρ
×
1
-
t
2
β
-
α
2
-
3
2
β
+
α
+
2
ρ
3
1
-
θ
≥
0
.In what follows, we impose conditions onf such that the BVP (3) has at least two positive and decreasing solutions. For convenience, we denote(24)
H
1
=
∫
1
-
θ
θ
‍
G
θ
,
s
d
s
,
H
2
=
∫
0
η
‍
G
θ
,
s
d
s
,
H
3
=
∫
1
-
θ
θ
‍
G
1
-
θ
,
s
d
s
.Theorem 9.
Suppose that there exist numbersa
,
b
,
a
n
d
c with 0
<
a
<
b
<
θ
∗
c such that(25)
f
θ
,
c
>
c
H
1
,
(26)
f
0
,
b
θ
∗
<
b
H
2
,
(27)
f
θ
,
θ
∗
a
>
a
H
3
.Then the BVP (3) has at least two positive and decreasing solutions.Proof.
First, we define the increasing, nonnegative, and continuous functionalsγ
,
ω, and ψ on P as follows:(28)
γ
u
=
m
i
n
t
∈
1
-
θ
,
θ
u
t
=
u
θ
,
ω
u
=
m
a
x
t
∈
θ
,
1
u
t
=
u
θ
,
ψ
u
=
m
a
x
t
∈
1
-
θ
,
1
u
t
=
u
1
-
θ
.Obviously, for any u
∈
P, γ
(
u
)
=
ω
(
u
)
≤
ψ
(
u
). At the same time, for each u
∈
P, in view of γ
(
u
)
=
min
t
∈
[
1
-
θ
,
θ
]
⁡
u
(
t
)
≥
θ
∗
u, we have(29)
u
≤
1
θ
∗
γ
u
f
o
r
u
∈
P
.In addition, we also note that(30)
ω
λ
u
=
λ
ω
u
f
o
r
0
≤
λ
≤
1
,
u
∈
P
.
Next, we assert thatγ
(
A
u
)
>
c for all u
∈
∂
P
(
γ
,
c
).
To prove this, letu
∈
∂
P
(
γ
,
c
); that is, u
∈
P
a
n
d
γ
(
u
)
=
u
(
θ
)
=
c. Then(31)
u
t
≥
u
θ
=
c
,
t
∈
1
-
θ
,
θ
.
In view of Lemma8, (25), (31), and(
C
1
) and (
C
2
), we have (32)
γ
A
u
=
A
u
θ
=
∫
0
1
‍
G
θ
,
s
f
s
,
u
s
d
s
≥
∫
1
-
θ
θ
‍
G
θ
,
s
f
s
,
u
s
d
s
≥
∫
1
-
θ
θ
‍
G
θ
,
s
f
θ
,
c
d
s
>
c
H
1
∫
1
-
θ
θ
‍
G
θ
,
s
d
s
=
c
.
Then, we assert thatω
(
A
u
)
<
b for all u
∈
∂
P
(
ω
,
b
).
To see this, supposeu
∈
∂
P
(
ω
,
b
); that is, u
∈
P
a
n
d
ω
(
u
)
=
b. Since u
≤
(
1
/
θ
∗
)
γ
(
u
)
=
(
1
/
θ
∗
)
ω
(
u
), we have(33)
0
≤
u
t
≤
u
≤
b
θ
∗
,
t
∈
0
,
η
.In view of Remark 6, (26), (33), and (
C
1
) and (
C
2
), we get (34)
ω
A
u
=
A
u
θ
=
∫
0
1
‍
G
θ
,
s
f
s
,
u
s
d
s
≤
∫
0
η
‍
G
θ
,
s
f
s
,
u
s
d
s
≤
∫
0
η
‍
G
θ
,
s
f
0
,
b
θ
∗
d
s
<
b
H
2
∫
0
η
‍
G
θ
,
s
d
s
=
b
.
Finally, we assert thatP
(
ψ
,
a
)
≠
∅, and ψ
(
A
u
)
>
a for all u
∈
∂
P
(
ψ
,
a
).
In fact, the constant functiona
/
2
∈
P
(
ψ
,
a
). Moreover, for u
∈
∂
P
(
ψ
,
a
), that is, u
∈
P
a
n
d
ψ
(
u
)
=
u
(
1
-
θ
)
=
a. Then(35)
u
t
≥
θ
∗
u
≥
θ
∗
u
1
-
θ
=
θ
∗
a
,
t
∈
1
-
θ
,
θ
.In view of Lemma 8, (27), (35), and (
C
1
) and (
C
2
), we get (36)
ψ
A
u
=
A
u
1
-
θ
=
∫
0
1
‍
G
1
-
θ
,
s
f
s
,
u
s
d
s
≥
∫
1
-
θ
θ
‍
G
1
-
θ
,
s
f
s
,
u
s
d
s
≥
∫
1
-
θ
θ
‍
G
1
-
θ
,
s
f
θ
,
θ
∗
a
d
s
>
a
H
3
∫
1
-
θ
θ
‍
G
1
-
θ
,
s
d
s
=
a
.
To sum up, all the hypotheses of Theorem3 are satisfied. Hence A has at least two fixed points; that is, the BVP (3) has at least two positive and decreasing solutions u
1 and u
2 such that(37)
a
<
m
a
x
t
∈
1
-
θ
,
1
u
1
t
w
i
t
h
m
a
x
t
∈
θ
,
1
u
1
t
<
b
,
b
<
m
a
x
t
∈
θ
,
1
u
2
t
w
i
t
h
m
i
n
t
∈
1
-
θ
,
θ
u
2
t
<
c
.


## 3. An Example

Example 1.
Consider the BVP(38)
u
′
′
′
t
=
f
t
,
u
t
,
t
∈
0,1
,
u
′
0
=
u
1
=
0
,
u
′
′
3
5
+
∫
3
/
5
11
/
15
‍
u
t
d
t
=
0
,where(39)
f
t
,
u
=
u
+
1
-
t
2
2
+
85
,
t
,
u
∈
0,1
×
0,144
,
4748
u
+
1
-
t
2
2
-
683615
,
t
,
u
∈
0,1
×
144,145
,
969
u
2
4205
+
1
-
t
2
2
,
t
,
u
∈
0,1
×
145
,
+
∞
.
Sinceη
=
α
=
3
/
5 and β
=
11
/
15, if we choose θ
=
17
/
30, then a simple calculation shows that(40)
ρ
=
19502
3375
,
θ
∗
=
1
18
,
H
1
=
262691
8775900
,
H
2
=
5106323
63186480
,
H
3
=
15256439
394915500
.Thus, if we let a
=
3
,
b
=
8, and c
=
145, then it is easy to verify that all the conditions of Theorem 9 are satisfied. So, it follows from Theorem 9 that the BVP (38) has at least two positive and decreasing solutions.



---

*Source: 491423-2015-05-19.xml*