# Positive Solutions of a Third-Order Three-Point BVP with Sign-Changing Green’s Function

Authors: Li-Juan Gao; Jian-Ping Sun
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

We are concerned with the following third-order three-point boundary value problem:u
′
′
′
t
=
f
t
,
u
t
,
t
∈
0
,
1
,
u
′
0
=
u
1
=
0 and u
′
′
η
-
α
u
′
1
=
0
, where α
∈
0
,
1 and η
∈
(
14
+
α
)
/
(
24
-
3
α
)
,
1. 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.


---

## 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 to some third-order three-point boundary value problems (BVPs for short) has received much attention from many authors; see [2–11] and the references therein. However, all the above-mentioned papers are achieved when the corresponding Green’s functions are nonnegative, which is a very important condition.In 2008, Palamides and Smyrlis [12] 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.Very recently, for 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
,
Sun and Zhao 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 [13, 14] and obtained the existence of a positive solution when η
∈
[
2
-
2
,
1
) by using iterative technique in [15].In 2013, Li et al. [16] discussed the existence of a positive solution to the third-order three-point BVP with sign-changing Green’s function:
(3)
u
′′′
(
t
)
=
f
(
t
,
u
(
t
)
)
,
t
∈
[
0,1
]
,
u
′
(
0
)
=
u
(
1
)
=
0
,
u
′′
(
η
)
+
α
u
(
0
)
=
0
,
where α
∈
[
0,2
) and η
∈
[
(
121
+
24
α
-
5
)
/
3
(
4
+
α
)
,
1
). The main tool used was the Guo-Krasnoselskii fixed-point theorem.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 [17–19].Motivated greatly by the above-mentioned works, in this paper, we consider the following third-order three-point BVP:(4)
u
′′′
(
t
)
=
f
(
t
,
u
(
t
)
)
,
t
∈
[
0,1
]
,
u
′
(
0
)
=
u
(
1
)
=
0
,
u
′′
(
η
)
-
α
u
′
(
1
)
=
0
.
Throughout this paper, we always assume that α
∈
[
0,1
) and η
∈
[
(
14
+
α
)
/
(
24
-
3
α
)
,
1
). 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. Our main tool 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 [20].LetK be a cone in a real Banach space E.Definition 1.
A functionalψ
:
K
→
R is said to be increasing on K provided ψ
(
x
)
≤
ψ
(
y
) for all x
,
y
∈
K with x
≤
y, where x
≤
y if and only if y
-
x
∈
K.Definition 2.
Letγ
:
K
→
[
0
,
+
∞
) be continuous. For each d
>
0, we define the set
(5)
K
(
γ
,
d
)
=
{
u
∈
K
:
γ
(
u
)
<
d
}
.Theorem 3.
Letψ and γ be increasing, nonnegative, and continuous functionals on K, and let ω be a nonnegative continuous functional on K with ω
(
0
)
=
0 such that, for some c
>
0 and M
>
0,
(6)
γ
(
u
)
≤
ω
(
u
)
≤
ψ
(
u
)
,
∥
u
∥
≤
M
γ
(
u
)
for all u
∈
K
(
γ
,
c
)
¯. Suppose that there exist a completely continuous operator T
:
K
(
γ
,
c
)
¯
→
K and 0
<
a
<
b
<
c such that
(7)
ω
(
λ
u
)
≤
λ
ω
(
u
)
f
o
r
0
≤
λ
≤
1
,
u
∈
∂
K
(
ω
,
b
)
,
and(1)
γ
(
T
u
)
>
c for all u
∈
∂
K
(
γ
,
c
);
(2)
ω
(
T
u
)
<
b for all u
∈
∂
K
(
ω
,
b
);
(3)
K
(
ψ
,
a
)
≠
∅, and ψ
(
T
u
)
>
a for all u
∈
∂
K
(
ψ
,
a
).Then T has at least two fixed points u
1 and u
2 in K
(
γ
,
c
)
¯ such that
(8)
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. Preliminaries

For the BVP(9)
u
′′′
(
t
)
=
0
,
t
∈
[
0,1
]
,
u
′
(
0
)
=
u
(
1
)
=
0
,
u
′′
(
η
)
-
α
u
′
(
1
)
=
0
,
we have the following lemma.Lemma 4.
The BVP (9) has only trivial solution.Proof.
It is simple to check.Now, for anyy
∈
C
[
0,1
], we consider the BVP
(10)
u
′′′
(
t
)
=
y
(
t
)
,
t
∈
[
0,1
]
,
u
′
(
0
)
=
u
(
1
)
=
0
,
u
′′
(
η
)
-
α
u
′
(
1
)
=
0
.After a direct computation, one may obtain the expression of Green’s functionG
(
t
,
s
) of the BVP (10) as follows:
(11)
G
(
t
,
s
)
=
g
(
t
,
s
)
+
k
(
t
,
s
)
+
{
0
,
s
≥
η
,
1
-
t
2
2
(
1
-
α
)
,
s
<
η
,
where
(12)
g
(
t
,
s
)
=
-
α
(
1
-
t
2
)
(
1
-
s
)
2
(
1
-
α
)
-
(
1
-
s
)
2
2
,
(
t
,
s
)
∈
[
0,1
]
×
[
0,1
]
,
k
(
t
,
s
)
=
{
0
,
0
≤
t
≤
s
≤
1
,
(
t
-
s
)
2
2
,
0
≤
s
≤
t
≤
1
.It is not difficult to verify that theG
(
t
,
s
) has the following properties:
(13)
G
(
t
,
s
)
≥
0
for
0
≤
s
<
η
,
G
(
t
,
s
)
≤
0
for
η
≤
s
≤
1
.
Moreover, for s
≥
η,
(14)
max
⁡
{
G
(
t
,
s
)
:
t
∈
[
0,1
]
}
=
G
(
1
,
s
)
=
0
,
min
⁡
{
G
(
t
,
s
)
:
t
∈
[
0,1
]
}
=
G
(
0
,
s
)
=
-
α
(
1
-
s
)
2
(
1
-
α
)
-
(
1
-
s
)
2
2
and for s
<
η,
(15)
max
⁡
{
G
(
t
,
s
)
:
t
∈
[
0,1
]
}
=
G
(
0
,
s
)
=
1
-
α
(
1
-
s
)
2
(
1
-
α
)
-
(
1
-
s
)
2
2
,
min
⁡
{
G
(
t
,
s
)
:
t
∈
[
0,1
]
}
=
G
(
1
,
s
)
=
0
.Let(16)
K
0
=
{
y
∈
C
[
0,1
]
:
y
(
t
)
is
nonnegative
m
m
c
and
decreasing
on
⁡
[
0,1
]
}
.
Then K
0 is a cone in C
[
0,1
].Lemma 5.
Lety
∈
K
0 and u
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
y
(
s
)
d
s, t
∈
[
0,1
]. Then u is the unique solution of the BVP (10) and u
∈
K
0. Moreover, u
(
t
) is concave on [
0
,
η
].Proof.
Fort
∈
[
0
,
η
], we have
(17)
u
(
t
)
=
∫
0
t
‍
[
g
(
t
,
s
)
+
(
t
-
s
)
2
2
+
1
-
t
2
2
(
1
-
α
)
]
y
(
s
)
d
s
+
∫
t
η
‍
[
g
(
t
,
s
)
+
1
-
t
2
2
(
1
-
α
)
]
y
(
s
)
d
s
+
∫
η
1
‍
g
(
t
,
s
)
y
(
s
)
d
s
.
Since y
∈
K
0 and η
≥
(
14
+
α
)
/
(
24
-
3
α
) implies that η
>
α
/
(
1
+
α
), we get
(18)
u
′
(
t
)
=
-
α
t
1
-
α
∫
0
η
‍
s
y
(
s
)
d
s
-
∫
0
t
‍
s
y
(
s
)
d
s
-
t
∫
t
η
‍
y
(
s
)
d
s
+
α
t
1
-
α
∫
η
1
‍
(
1
-
s
)
y
(
s
)
d
s
≤
y
(
η
)
[
-
α
t
1
-
α
∫
0
η
‍
s
d
s
-
∫
0
t
‍
s
d
s
m
m
m
m
c
-
t
∫
t
η
‍
d
s
+
α
t
1
-
α
∫
η
1
‍
(
1
-
s
)
d
s
]
=
t
y
(
η
)
[
α
(
1
-
2
η
)
2
(
1
-
α
)
-
η
+
t
2
]
≤
t
y
(
η
)
[
α
(
1
-
2
η
)
2
(
1
-
α
)
-
η
2
]
≤
0
,
t
∈
[
0
,
η
]
.
At the same time, y
∈
K
0 and η
≥
(
14
+
α
)
/
(
24
-
3
α
)
>
1
/
2 shows that
(19)
u
′′
(
t
)
=
-
α
1
-
α
∫
0
η
‍
s
y
(
s
)
d
s
-
∫
t
η
‍
y
(
s
)
d
s
+
α
1
-
α
∫
η
1
‍
(
1
-
s
)
y
(
s
)
d
s
≤
y
(
η
)
[
-
α
1
-
α
∫
0
η
‍
s
d
s
-
∫
t
η
‍
d
s
+
α
1
-
α
∫
η
1
‍
(
1
-
s
)
d
s
]
=
y
(
η
)
[
α
(
1
-
2
η
)
2
(
1
-
α
)
-
η
+
t
]
≤
y
(
η
)
α
(
1
-
2
η
)
2
(
1
-
α
)
≤
0
,
t
∈
[
0
,
η
]
.
Fort
∈
(
η
,
1
], we have
(20)
u
(
t
)
=
∫
0
η
‍
[
g
(
t
,
s
)
+
(
t
-
s
)
2
2
+
1
-
t
2
2
(
1
-
α
)
]
y
(
s
)
d
s
+
∫
η
t
‍
[
g
(
t
,
s
)
+
(
t
-
s
)
2
2
]
y
(
s
)
d
s
+
∫
t
1
‍
g
(
t
,
s
)
y
(
s
)
d
s
.
In view of y
∈
K
0 and η
>
1
/
2, we get
(21)
u
′
(
t
)
=
-
α
t
1
-
α
∫
0
η
‍
s
y
(
s
)
d
s
+
∫
η
t
‍
(
t
-
s
)
y
(
s
)
d
s
-
∫
0
η
‍
s
y
(
s
)
d
s
+
α
t
1
-
α
∫
η
1
‍
(
1
-
s
)
y
(
s
)
d
s
≤
y
(
η
)
[
-
α
t
1
-
α
∫
0
η
‍
s
d
s
+
∫
η
t
‍
(
t
-
s
)
d
s
m
m
m
m
c
-
∫
0
η
‍
s
d
s
+
α
t
1
-
α
∫
η
1
‍
(
1
-
s
)
d
s
]
=
t
y
(
η
)
[
α
(
1
-
2
η
)
2
(
1
-
α
)
+
t
-
2
η
2
]
≤
t
y
(
η
)
[
1
-
2
η
2
(
1
-
α
)
]
≤
0
,
t
∈
(
η
,
1
]
.
Obviously,u
′′′
(
t
)
=
y
(
t
) for t
∈
[
0,1
], u
′
(
0
)
=
u
(
1
)
=
0, and u
′′
(
η
)
-
α
u
′
(
1
)
=
0. This shows that u is a solution of the BVP (10). The uniqueness follows immediately from Lemma 4. Since u
′
(
t
)
≤
0 for t
∈
[
0,1
] and u
(
1
)
=
0, we have u
(
t
)
≥
0 for t
∈
[
0,1
]. So, u
∈
K
0. In view of u
′′
(
t
)
≤
0 for t
∈
[
0
,
η
], we know that u
(
t
) is concave on [
0
,
η
].Throughout this paper, for anyu
∈
C
[
0,1
], we define ∥
u
∥
=
max
⁡
t
∈
[
0,1
]
⁡
|
u
(
t
)
|.Lemma 6.
Lety
∈
K
0. Then the unique solution u of the BVP (10) satisfies
(22)
min
⁡
t
∈
[
0
,
θ
]
u
(
t
)
≥
θ
*
∥
u
∥
,
where θ
∈
(
0
,
1
/
2
] and θ
*
=
(
η
-
θ
)
/
η.Proof.
By Lemma5, we know that u
(
t
) is concave on [
0
,
η
]; thus, for t
∈
[
0
,
η
],
(23)
u
(
t
)
≥
η
-
t
η
u
(
0
)
+
t
η
u
(
η
)
.
At the same time, it follows from u
∈
K
0 that ∥
u
∥
=
u
(
0
), which together with (23) implies that
(24)
u
(
t
)
≥
η
-
t
η
∥
u
∥
,
t
∈
[
0
,
η
]
.
Consequently,
(25)
min
⁡
t
∈
[
0
,
θ
]
u
(
t
)
=
u
(
θ
)
≥
η
-
θ
η
∥
u
∥
=
θ
*
∥
u
∥
.


## 3. Main Results

In the remainder of this paper, we always assume thatf
:
[
0,1
]
×
[
0
,
+
∞
)
→
[
0
,
+
∞
) is continuous and satisfies the following two conditions:(
H
1
)
for eachu
∈
[
0
,
+
∞
), the mapping t
↦
f
(
t
,
u
) is decreasing;
(
H
2
)
for eacht
∈
[
0,1
], the mapping u
↦
f
(
t
,
u
) is increasing.Let(26)
K
=
{
u
∈
K
0
:
min
⁡
t
∈
[
0
,
θ
]
u
(
t
)
≥
θ
*
∥
u
∥
}
.
Then it is easy to see that K is a cone in C
[
0,1
].Now, we define an operatorT as follows:
(27)
(
T
u
)
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
,
u
∈
K
,
t
∈
[
0,1
]
.
First, it is obvious that if u is a fixed point of T in K, then u is a nonnegative and decreasing solution of the BVP (4). Next, by Lemmas 5 and 6, we know that T
:
K
→
K. Furthermore, although G
(
t
,
s
) is not continuous, it follows from known textbook results, for example, see [21], that T
:
K
→
K is completely continuous.For convenience, we denote(28)
A
=
∫
0
θ
‍
G
(
η
,
s
)
d
s
,
B
=
∫
0
η
‍
G
(
θ
,
s
)
d
s
.Theorem 7.
Suppose that there exist numbersa, b, and c with 0
<
a
<
b
<
θ
*
c such that
(29)
f
(
θ
,
c
)
>
c
A
,
(30)
f
(
0
,
b
θ
*
)
<
b
B
,
(31)
f
(
θ
,
θ
*
a
)
>
a
A
.
Then the BVP (4) has at least two positive and decreasing solutions.Proof.
First, we define the increasing, nonnegative, and continuous functionalsγ, ω, and ψ on K as follows:
(32)
γ
(
u
)
=
min
⁡
t
∈
[
0
,
θ
]
u
(
t
)
=
u
(
θ
)
,
ω
(
u
)
=
max
⁡
t
∈
[
θ
,
1
]
u
(
t
)
=
u
(
θ
)
,
ψ
(
u
)
=
max
⁡
t
∈
[
0,1
]
u
(
t
)
=
u
(
0
)
.
Obviously, for any u
∈
K, γ
(
u
)
=
ω
(
u
)
≤
ψ
(
u
). At the same time, for each u
∈
K, in view of γ
(
u
)
=
min
⁡
t
∈
[
0
,
θ
]
⁡
u
(
t
)
≥
θ
*
∥
u
∥, we have
(33)
∥
u
∥
≤
1
θ
*
γ
(
u
)
for
u
∈
K
.
In addition, we also note that
(34)
ω
(
λ
u
)
=
λ
ω
(
u
)
for
0
≤
λ
≤
1
,
u
∈
K
.
Next, for anyu
∈
K, we claim that
(35)
∫
θ
1
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
0
.
In fact, it follows from (
H
1
), (
H
2
), and η
≥
(
14
+
α
)
/
(
24
-
3
α
) that
(36)
∫
θ
1
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
=
∫
θ
η
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
∫
η
1
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
f
(
η
,
u
(
η
)
)
[
∫
θ
η
‍
G
(
η
,
s
)
d
s
+
∫
η
1
‍
G
(
η
,
s
)
d
s
]
=
f
(
η
,
u
(
η
)
)
×
[
∫
η
1
‍
∫
θ
η
‍
(
g
(
η
,
s
)
+
(
η
-
s
)
2
2
+
1
-
η
2
2
(
1
-
α
)
)
d
s
m
m
m
m
m
c
+
∫
η
1
‍
g
(
η
,
s
)
d
s
(
g
(
η
,
s
)
+
(
η
-
s
)
2
2
+
1
-
η
2
2
(
1
-
α
)
)
]
=
(
1
-
η
)
f
(
η
,
u
(
η
)
)
12
(
1
-
α
)
×
[
(
4
+
2
α
)
η
2
+
(
4
-
α
)
η
-
α
-
2
+
(
3
α
-
3
α
η
-
6
)
θ
2
]
≥
(
1
-
η
)
f
(
η
,
u
(
η
)
)
12
(
1
-
α
)
×
[
(
4
+
2
α
)
η
2
+
(
4
-
7
4
α
)
η
-
1
4
α
-
7
2
]
≥
(
1
-
η
)
f
(
η
,
u
(
η
)
)
12
(
1
-
α
)
[
(
6
-
3
4
α
)
η
-
1
4
α
-
7
2
]
≥
0
.
Now, we assert thatγ
(
T
u
)
>
c for all u
∈
∂
K
(
γ
,
c
).
To prove this, letu
∈
∂
K
(
γ
,
c
); that is, u
∈
K and γ
(
u
)
=
u
(
θ
)
=
c. Then
(37)
u
(
t
)
≥
u
(
θ
)
=
c
,
t
∈
[
0
,
θ
]
.
Since (
T
u
)
(
t
) is decreasing on [
0,1
], it follows from (29), (35), (37), (
H
1
), and (
H
2
) that
(38)
γ
(
T
u
)
=
(
T
u
)
(
θ
)
≥
(
T
u
)
(
η
)
=
∫
0
1
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
∫
0
θ
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
∫
0
θ
‍
G
(
η
,
s
)
f
(
θ
,
c
)
d
s
>
c
A
∫
0
θ
‍
G
(
η
,
s
)
d
s
=
c
.
Then, we assert thatω
(
T
u
)
<
b for all u
∈
∂
K
(
ω
,
b
).
To see this, suppose thatu
∈
∂
K
(
ω
,
b
); that is, u
∈
K and ω
(
u
)
=
b. Since ∥
u
∥
≤
(
1
/
θ
*
)
γ
(
u
)
=
(
1
/
θ
*
)
ω
(
u
), we have
(39)
0
≤
u
(
t
)
≤
∥
u
∥
≤
b
θ
*
,
t
∈
[
0
,
η
]
.
In view of the properties of G
(
t
,
s
), (30), (39), (
H
1
), and (
H
2
), we get
(40)
ω
(
T
u
)
=
(
T
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
B
∫
0
η
‍
G
(
θ
,
s
)
d
s
=
b
.
Finally, we assert thatK
(
ψ
,
a
)
≠
∅ and ψ
(
T
u
)
>
a for all u
∈
∂
K
(
ψ
,
a
).
In fact, the constant functiona
/
2
∈
K
(
ψ
,
a
). Moreover, for u
∈
∂
K
(
ψ
,
a
), that is, u
∈
K and ψ
(
u
)
=
u
(
0
)
=
a, we know that
(41)
u
(
t
)
≥
θ
*
∥
u
∥
=
θ
*
u
(
0
)
=
θ
*
a
,
t
∈
[
0
,
θ
]
.
Since (
T
u
)
(
t
) is decreasing on [
0,1
], it follows from (31), (35), (41), (
H
1
), and (
H
2
) that
(42)
ψ
(
T
u
)
=
(
T
u
)
(
0
)
≥
(
T
u
)
(
η
)
=
∫
0
1
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
∫
0
θ
‍
G
(
η
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
∫
0
θ
‍
G
(
η
,
s
)
f
(
θ
,
θ
*
a
)
d
s
>
a
A
∫
0
θ
‍
G
(
η
,
s
)
d
s
=
a
.
To sum up, all the hypotheses of Theorem3 are satisfied. Hence, T has at least two fixed points u
1 and u
2; that is, the BVP (4) has at least two positive and decreasing solutions u
1 and u
2 satisfying
(43)
a
<
max
⁡
t
∈
[
0,1
]
u
1
(
t
)
with
max
⁡
t
∈
[
θ
,
1
]
u
1
(
t
)
<
b
,
b
<
max
⁡
t
∈
[
θ
,
1
]
u
2
(
t
)
with
min
⁡
t
∈
[
0
,
θ
]
u
2
(
t
)
<
c
.



---

*Source: 406815-2014-09-30.xml*