# Solvability of a Class of Higher-Order Fractional Two-Point Boundary Value Problem

Authors: Xiangshan Kong; Haitao Li
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

This paper investigates the solvability of a class of higher-order fractional two-point
boundary value problem (BVP), and presents several new results. First, Green’s function
of the considered BVP is obtained by using the property of Caputo derivative. Second, based
on Schaefer’s fixed point theorem, the solvability of the considered BVP is studied, and
a sufficient condition is presented for the existence of at least one solution. Finally, an
illustrative example is given to support the obtained new results.


---

## Body

## 1. Introduction

Fractional differential equations (FDEs) have been of great interest recently. This is due to the development of the theory of fractional calculus itself as well as its applications [1–3]. Apart from diverse areas of mathematics, fractional differential equations arise in engineering, technology, biology, chemical process, and many other branches of science. The first issue for the theory of fractional differential equations is the existence of solutions to kinds of boundary value problems (BVPs), which has been studied recently by many scholars [4–18], and lots of excellent results have been obtained by means of fixed-point theorems, Leray-Schauder theory, upper and lower solutions technique, and so forth.It should be pointed out that as an important branch of fractional differential equations, higher-order FDEs have been studied in a series of recent works [19–23]. In [19], higher-order fractional heat-type equations were investigated and some interesting properties on the solution to this type of equations were presented. Goodrich [21] considered another kind of higher-order fractional differential equations and presented some results on the existence of one positive solution.In this paper, we study the following higher-order fractional two-point boundary value problem:(1)
c
D
0
+
α
x
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
[
0,1
]
,
x
(
0
)
=
x
(
1
)
=
x
′
′
(
0
)
=
x
′
′
(
1
)
=
0
,
where 3
<
α
<
4, f
:
[
0,1
]
×
R
→
R, and c
D
0
+
α is the Caputo derivative. To our best knowledge, there are fewer results on the existence of solutions to BVP (1). Firstly, we establish Green’s function for BVP (1) by using the property of Caputo derivative. Secondly, based on Schaefer’s fixed point theorem, we present a sufficient condition for the existence of at least one solution. Throughout this paper, we assume that the nonlinearity f
:
[
0,1
]
×
R
→
R is continuous. Moreover, let E
=
C
[
0,1
] with the norm ∥
x
∥
∞
=
max
t
∈
[
0,1
]
|
x
(
t
)
|. Then, E is a Banach space.The rest of this paper is structured as follows. Section2 contains some preliminaries on the Caputo derivative. Section 3 investigates the existence of solutions to BVP (1) and presents the main results of this paper.


## 2. Preliminaries

In this section, we give some necessary preliminaries on the Caputo derivative, which will be used in the sequel. For details, please refer to [1–3] and the references therein.Definition 1 (see [3]).
The Riemann-Liouville fractional integral of orderα
>
0 of a function y
:
(
0
,
+
∞
)
→
ℝ is given by
(2)
I
0
+
α
y
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
y
(
s
)
d
s
,
provided that the right side is pointwise defined on (
0
,
+
∞
).Definition 2 (see [3]).
The Caputo fractional derivative of orderα
>
0 of a continuous function y
:
(
0
,
+
∞
)
→
ℝ is given by
(3)
c
D
0
+
α
y
(
t
)
=
1
Γ
(
n
-
α
)
∫
0
t
‍
y
(
n
)
(
s
)
(
t
-
s
)
α
-
n
+
1
d
s
,
where n
=
[
α
]
+
1, provided that the right side is pointwise defined on (
0
,
+
∞
).One can easily obtain the following property from the definition of Caputo derivative.Proposition 3 (see [3]).
Letα
>
0. Assume that y, c
D
0
+
α
y
∈
L
(
0,1
). Then the following equality holds:
(4)
I
0
+
α
c
D
0
+
α
y
(
t
)
=
y
(
t
)
+
C
0
+
C
1
t
+
⋯
+
C
n
-
1
t
n
-
1
,
for some C
i
∈
R, i
=
0,1
,
…
,
n
-
1, where n
=
[
α
]
+
1.Finally, we give Schaefer’s fixed point theorem.Lemma 4 (see [24]).
LetE be a Banach space, and H
:
E
→
E is a completely continuous operator. If the set S
:
=
{
x
∈
E
∣
x
=
λ
H
x
,
λ
∈
(
0,1
)
}, is bounded, then H has at least a fixed point.


## 3. Main Results

In this section, we first convert BVP (1) into an equivalent operator equation and then present some new results on the existence of solutions to BVP (1).Firstly, we convert BVP (1) into an equivalent operator equation.Lemma 5.
Givenh
∈
C
[
0,1
], the unique solution of
(5)
c
D
0
+
α
x
(
t
)
=
h
(
t
)
,
t
∈
[
0,1
]
,
x
(
0
)
=
x
(
1
)
=
x
′
′
(
0
)
=
x
′
′
(
1
)
=
0
is
(6)
x
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
h
(
s
)
d
s
,
where
(7)
G
(
t
,
s
)
=
1
Γ
(
α
)
{
(
t
-
s
)
α
-
1
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
(
α
-
1
)
×
(
t
-
t
3
)
(
1
-
s
)
α
-
3
,
0
≤
s
≤
t
≤
1
,
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
×
(
α
-
1
)
(
t
-
t
3
)
(
1
-
s
)
α
-
3
,
0
≤
t
<
s
≤
1
.Proof.
Assume thatx
(
t
) satisfies (5). Then, from Proposition 3, we have
(8)
x
(
t
)
=
I
0
+
α
h
(
t
)
+
c
0
+
c
1
t
+
c
2
t
2
+
c
3
t
3
,
c
0
,
c
1
,
c
2
,
c
3
∈
R
.
From the boundary value conditionx
(
0
)
=
x
(
1
)
=
0, one can see that
(9)
c
0
=
0
,
c
1
+
c
2
+
c
3
=
-
I
0
+
α
h
(
t
)
|
t
=
1
=
-
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
h
(
s
)
d
s
.
Thus, we have
(10)
x
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
h
(
s
)
d
s
+
c
1
t
+
c
2
t
2
+
c
3
t
3
,
which together with the boundary value condition x
′′
(
0
)
=
x
′′
(
1
)
=
0 yields that
(11)
x
′
(
t
)
=
1
Γ
(
α
-
1
)
×
∫
0
t
‍
(
t
-
s
)
α
-
2
h
(
s
)
d
s
+
c
1
+
2
c
2
t
+
3
c
3
t
2
,
x
′′
(
t
)
=
1
Γ
(
α
-
2
)
∫
0
t
‍
(
t
-
s
)
α
-
3
h
(
s
)
d
s
+
2
c
2
+
6
c
3
t
,
c
2
=
0
,
c
3
=
-
1
6
1
Γ
(
α
-
2
)
∫
0
1
‍
(
1
-
s
)
α
-
3
h
(
s
)
d
s
,
c
1
=
1
6
1
Γ
(
α
-
2
)
∫
0
1
‍
(
1
-
s
)
α
-
3
h
(
s
)
d
s
-
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
h
(
s
)
d
s
.
Therefore, the unique solution of (5) is
(12)
x
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
h
(
s
)
d
s
+
1
6
1
Γ
(
α
-
2
)
∫
0
1
‍
t
(
1
-
s
)
α
-
3
h
(
s
)
d
s
-
1
Γ
(
α
)
∫
0
1
‍
t
(
1
-
s
)
α
-
1
h
(
s
)
d
s
-
1
6
1
Γ
(
α
-
2
)
∫
0
1
‍
t
3
(
1
-
s
)
α
-
3
h
(
s
)
d
s
=
1
Γ
(
α
)
∫
0
t
‍
[
1
6
(
t
-
s
)
α
-
1
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
×
1
6
(
α
-
1
)
(
t
-
t
3
)
(
1
-
s
)
α
-
3
]
h
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
1
‍
[
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
×
1
6
(
α
-
1
)
(
t
-
t
3
)
(
1
-
s
)
α
-
3
]
h
(
s
)
d
s
.
The proof is completed.Lemma 6.
x
(
t
)
∈
E is a solution to BVP (1), if and only if x
(
t
)
=
F
x
(
t
), where
(13)
F
x
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
x
(
s
)
)
d
s
.Our main result, based on Schaefer’s fixed point theorem, is stated as follows.Theorem 7.
Letf
:
[
0,1
]
×
R
→
R be continuous. Assume that(H) there exist nonnegative functionsa
,
b
∈
C
[
0,1
] such that
(14)
|
f
(
t
,
x
)
|
≤
a
(
t
)
+
b
(
t
)
|
x
|
,
∀
t
∈
[
0,1
]
,
x
∈
R
.Then, BVP (1) has at least one solution, provided that
(15)
1
Γ
(
α
+
1
)
[
2
+
3
27
α
(
α
-
1
)
]
∥
b
∥
∞
<
1
.Proof.
We divide the proof into the following two steps.
Step1. F
:
C
[
0,1
]
→
C
[
0,1
] is completely continuous.
LetΩ
⊂
C
[
0,1
] be an open bounded subset. By the continuity of f, we can get that F is continuous and F
(
Ω
¯
) is bounded. Moreover, there exists a constant T
>
0 such that |
f
(
t
,
x
)
|
≤
T, ∀
x
∈
Ω
¯, t
∈
[
0,1
]. Thus, in view of the Arzelá-Ascoli theorem, we need only to prove that F
(
Ω
¯
)
⊂
C
[
0,1
] is equicontinuous.
For0
≤
t
1
≤
t
2
≤
1, x
∈
Ω
¯, we have
(16)
|
F
x
(
t
2
)
-
F
x
(
t
1
)
|
=
|
∫
0
1
‍
G
(
t
2
,
s
)
f
(
s
,
x
(
s
)
)
d
s
-
∫
0
1
‍
G
(
t
1
,
s
)
f
(
s
,
x
(
s
)
)
d
s
|
≤
T
∫
0
1
‍
|
G
(
t
2
,
s
)
-
G
(
t
1
,
s
)
|
d
s
=
T
∫
0
t
1
‍
|
G
(
t
2
,
s
)
-
G
(
t
1
,
s
)
|
d
s
+
T
∫
t
2
1
‍
|
G
(
t
2
,
s
)
-
G
(
t
1
,
s
)
|
d
s
+
T
∫
t
1
t
2
‍
|
G
(
t
2
,
s
)
-
G
(
t
1
,
s
)
|
d
s
=
T
Γ
(
α
)
∫
0
t
1
‍
|
(
t
2
-
s
)
α
-
1
-
(
t
1
-
s
)
α
-
1
+
t
1
(
1
-
s
)
α
-
1
-
t
2
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
(
α
-
1
)
×
(
t
2
-
t
2
3
-
t
1
+
t
1
3
)
(
1
-
s
)
α
-
3
|
d
s
+
T
Γ
(
α
)
∫
1
t
2
‍
|
t
1
(
1
-
s
)
α
-
1
-
t
2
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
×
(
α
-
1
)
(
t
2
-
t
2
3
-
t
1
+
t
1
3
)
(
1
-
s
)
α
-
3
1
6
|
d
s
+
T
Γ
(
α
)
∫
t
1
t
2
‍
|
1
6
(
t
2
-
s
)
α
-
1
+
t
1
(
1
-
s
)
α
-
1
-
t
2
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
(
α
-
1
)
×
(
t
2
-
t
2
3
-
t
1
+
t
1
3
)
(
1
-
s
)
α
-
3
1
6
|
d
s
≤
T
Γ
(
α
)
∫
0
t
1
‍
|
1
6
(
t
2
-
s
)
α
-
1
-
(
t
1
-
s
)
α
-
1
+
(
1
-
s
)
α
-
1
(
t
1
-
t
2
)
+
1
6
(
α
-
2
)
(
α
-
1
)
×
[
(
t
2
-
t
1
)
+
(
t
1
3
-
t
2
3
)
]
(
1
-
s
)
α
-
3
1
6
|
d
s
+
T
Γ
(
α
)
∫
1
t
2
‍
|
(
1
-
s
)
α
-
1
(
t
1
-
t
2
)
+
1
6
(
α
-
2
)
(
α
-
1
)
×
[
(
t
2
-
t
1
)
+
(
t
1
3
-
t
2
3
)
]
(
1
-
s
)
α
-
3
1
6
|
d
s
+
T
Γ
(
α
)
[
3
+
2
3
(
α
-
2
)
(
α
-
1
)
]
×
(
t
2
-
t
1
)
⟶
0
,
t
2
-
t
1
⟶
0
,
which implies that F
(
Ω
¯
)
⊂
C
[
0,1
] is equicontinuous.
Step2. A priori bounds.
Set(17)
Ω
=
{
x
∈
C
[
0,1
]
∣
x
=
λ
F
x
,
λ
∈
(
0,1
)
}
.
Now it remains to show that the setΩ is bounded.
Forx
∈
Ω, we get x
(
t
)
=
λ
F
x
(
t
). Thus, form (
H
), we obtain that
(18)
|
x
(
t
)
|
=
λ
|
F
x
(
t
)
|
≤
|
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
x
(
s
)
)
d
s
|
≤
|
∫
0
1
‍
G
(
t
,
s
)
d
s
|
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
=
1
Γ
(
α
)
|
∫
0
t
‍
[
1
6
(
t
-
s
)
α
-
1
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
(
α
-
1
)
(
t
-
t
3
)
(
1
-
s
)
α
-
3
]
d
s
+
∫
t
1
‍
[
-
t
(
1
-
s
)
α
-
1
+
1
6
(
α
-
2
)
(
α
-
1
)
∫
0
t
‍
×
(
t
-
t
3
)
(
1
-
s
)
α
-
3
1
6
]
d
s
|
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
=
1
Γ
(
α
)
|
1
α
t
α
+
1
α
t
(
1
-
t
)
α
-
1
α
t
+
1
6
(
α
-
2
)
(
α
-
1
)
(
t
-
t
3
)
1
α
-
2
-
1
6
(
α
-
2
)
(
α
-
1
)
(
t
-
t
3
)
1
α
-
2
(
1
-
t
)
α
-
2
-
1
α
t
(
1
-
t
)
α
+
1
6
(
α
-
2
)
(
α
-
1
)
×
(
t
-
t
3
)
1
α
-
2
(
1
-
t
)
α
-
2
|
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
=
1
Γ
(
α
)
|
1
α
t
α
-
1
α
t
+
1
6
(
α
-
2
)
(
α
-
1
)
(
t
-
t
3
)
1
α
-
2
|
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
=
1
Γ
(
α
+
1
)
|
t
α
-
t
+
1
6
α
(
α
-
1
)
(
t
-
t
3
)
|
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
≤
1
Γ
(
α
+
1
)
[
|
t
α
|
+
|
t
|
+
1
6
α
(
α
-
1
)
|
t
-
t
3
|
]
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
≤
1
Γ
(
α
+
1
)
[
2
+
3
27
α
(
α
-
1
)
]
×
(
∥
a
∥
∞
+
∥
b
∥
∞
∥
x
∥
∞
)
,
∀
t
∈
[
0,1
]
.
In view of (15), we can see that there exists a constant M
>
0 such that
(19)
∥
x
∥
∞
≤
M
.
As a consequence of Schaefer’s fixed point theorem, we deduce thatF has a fixed point which is the solution to BVP (1). The proof is completed.Finally, we give an illustrative example to support our new results.Example 8.
Consider the following BVP:(20)
c
D
0
+
7
/
2
x
(
t
)
=
2
t
+
4
t
2
sin
x
(
t
)
,
t
∈
[
0,1
]
,
x
(
0
)
=
x
(
1
)
=
x
′
′
(
0
)
=
x
′
′
(
1
)
=
0
.
A simple calculation shows that(
Γ
(
α
+
1
)
)
/
(
2
+
(
3
/
27
)
α
(
α
-
1
)
)
≈
4.5413.
Then one can see that(21)
1
Γ
(
(
7
/
2
)
+
1
)
[
2
+
3
27
7
2
(
7
2
-
1
)
]
∥
4
t
2
∥
∞
<
1
,
which implies that (15) holds. By Theorem 7, BVP (20) has at least one solution.



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*Source: 871908-2014-01-02.xml*