# Existence Results for a Coupled System of Nonlinear Fractional Differential Equations in Banach Spaces

Authors: Yuping Cao; Chuanzhi Bai
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We investigate boundary value problems for a coupled system of nonlinear fractional differential equations involving Caputo derivative in Banach spaces. A generalized singular type coupled Gronwall inequality system is given to obtain an important a priori bound. Existence results are obtained by using fixed point theorems and an example is given to illustrate the results.


---

## Body

## 1. Introduction

Fractional differential equations involving the Riemann-Liouville fractional derivative or the Caputo fractional derivative have been shown to be very useful in the study of models of many phenomena in various fields of science and engineering, such as physics, chemistry, biology, signal and image processing, biophysics, blood flow phenomena, control theory, economics, aerodynamics and fitting of experimental data. For more details, see, for example, [1–6].In recent years, many researchers paid much attention to the coupled system of fractional differential equations due to its applications in differential fields. The reader is referred to the papers [7–11] and the references cited therein.Up to now, there are fewer results of fractional differential equations with boundary conditions in infinite dimensional spaces than in finite dimensional spaces. Recently, Wang et al. [12] investigated the existence and uniqueness of solutions for a fractional boundary value problem involving the Caputo derivative in Banach space as follows:
(1)
D
t
α
0
C
y
t
=
f
t
,
y
t
,
0
<
α
<
1
,
t
∈
J
=
0
,
T
,
a
x
0
+
b
x
T
=
c
,
which extended the earlier work [13], where D
t
α
0
C is the Caputo fractional derivative of order α, f
:
J
×
X
→
X, where X is a Banach spaces and a
,
b, and c are real constants with a
+
b
≠
0.To the best of our knowledge, there is no effort being made in the literature to study the existence of solutions for a coupled system of fractional boundary value problems involving the Caputo derivative in Banach space. Motivated by the above-mentioned works, in this paper, we study a coupled system of fractional differential equations with boundary conditions of the type(2)
D
t
α
0
C
x
t
=
f
t
,
x
t
,
y
t
,
0
<
α
<
1
,
t
∈
J
=
0
,
T
,
D
t
β
0
C
y
t
=
g
t
,
x
t
,
y
t
,
0
<
β
<
1
,
t
∈
J
=
0
,
T
,
a
x
0
+
b
x
T
=
p
,
c
y
0
+
d
y
T
=
q
,
where 0
C
D
t
α and 0
C
D
t
β are the Caputo fractional derivatives of order α and β, respectively, f
,
g
:
J
×
X
×
X
→
X, where X is a Banach spaces and a, b, c, d, p, and q are real constants with a
+
b
≠
0 and c
+
d
≠
0. We will apply Schaefer fixed point theorem, nonlinear alternative of Leray-Schauder type, and a new singular coupled Gronwall inequality system given by us to establish the existence of solutions for BVP (2).This paper is organized as follows. In Section2, we recall some preliminary facts that we need in the sequel, and we give a generalized singular type coupled Gronwall inequality system which can be used to obtain an important a priori bound. In Section 3, we give two existence results of the problem (2) which is based on two fixed point theorems, respectively. Finally, an example is given to illustrate the results in Section 4.


## 2. Preliminaries

For the convenience of the reader, we first briefly recall some definitions of fractional calculus; for more details, see [1, 2, 5], for example.Definition 1.
The Riemann-Liouville fractional integral of orderα
>
0 of a function u
:
(
0
,
∞
)
→
R is given by
(3)
I
0
+
α
u
t
=
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
u
s
d
s
provided that the right side is pointwise defined on (
0
,
∞
), where Γ
(
·
) is the Gamma function.Definition 2.
The Caputo fractional derivative of orderγ
>
0 of a function u
:
(
0
,
∞
)
→
R can be written as
(4)
D
t
γ
0
C
u
t
=
1
Γ
n
-
γ
∫
0
t
‍
u
n
s
t
-
s
γ
+
1
-
n
d
s
,
n
-
1
<
γ
<
n
.Definition 3.
The Mittag-Leffler function in two parameters is defined as(5)
E
α
,
β
z
=
∑
k
=
0
∞
‍
z
k
Γ
k
α
+
β
,
where α
>
0,  β
>
0, and z
∈
C; C denotes the complex plane. In particular, for β
=
1, one has
(6)
E
α
,
1
z
=
E
α
z
=
∑
k
=
0
∞
‍
z
k
Γ
k
α
+
1
.
The Laplace transform of Mittag-Leffler function is(7)
L
t
β
-
1
E
α
,
β
-
λ
t
α
=
s
α
-
β
s
α
+
λ
,
R
s
>
λ
1
/
α
,
where t and s are, respectively, the variables in the time domain and Laplace domain; L
{
·
} stands for the Laplace transform.Throughout this paper, letC
(
J
,
X
) be the Banach space of all continuous functions from J into X with the norm x
:
=
sup
⁡
{
x
(
t
)
:
t
∈
J
}. Let E
=
C
(
J
,
X
)
×
C
(
J
,
X
) be the Banach space endowed with the norm as follows:
(8)
x
,
y
=
max
⁡
x
,
y
,
∀
x
,
y
∈
E
.Now, we give the definition of the solution for problem (2).Definition 4.
A(
u
,
v
)
∈
E is said to be a solution of a coupled system of fractional BVP (2) if (
u
,
v
) satisfies the system 0
C
D
t
α
x
(
t
)
=
f
(
t
,
x
(
t
)
,
y
(
t
)
), 0
C
D
t
β
y
(
t
)
=
g
(
t
,
x
(
t
)
,
y
(
t
)
) on J and the conditions a
x
(
0
)
+
b
x
(
T
)
=
p, c
y
(
0
)
+
d
y
(
T
)
=
q.By Lemma 3.2 in [14], we have the following.Lemma 5.
(
x
,
y
)
∈
E is a solution of the fractional integral system
(9)
x
(
t
)
=
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
1
a
+
b
×
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
p
,
y
t
=
1
Γ
β
∫
0
t
‍
t
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
1
c
+
d
×
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
q
.Wang et al. in [15] gave a generalized Gronwall inequality as follows.Lemma 6.
Lety
∈
C
(
J
,
X
) satisfy the following inequality:
(10)
y
t
≤
a
+
b
∫
0
t
‍
y
θ
λ
1
d
θ
+
c
∫
0
T
‍
y
θ
λ
2
d
θ
+
d
∫
0
t
‍
y
θ
λ
3
d
θ
+
e
∫
0
T
‍
y
θ
λ
4
d
θ
,
t
∈
J
,
where λ
1
,
λ
3
∈
[
0,1
], λ
2
,
λ
4
∈
(
0,1
), a
,
b
,
c
,
d
,
e
≥
0 are constants, and y
θ
B
=
sup
⁡
0
≤
s
≤
θ
y
(
s
). Then there exists a constant M
*
>
0 such that
(11)
y
t
≤
M
*
.By using the above generalized Gronwall inequality, we now give the following generalized singular type coupled Gronwall inequality system.Lemma 7.
Letx
,
y
∈
C
(
J
,
X
) satisfy the following inequality system:
(12)
x
t
≤
a
1
+
b
1
∫
0
t
‍
t
-
s
α
-
1
x
s
λ
1
d
s
+
c
1
∫
0
t
‍
t
-
s
α
-
1
y
s
λ
2
d
s
+
d
1
∫
0
T
‍
T
-
s
α
-
1
x
s
λ
1
d
s
+
e
1
∫
0
T
‍
T
-
s
α
-
1
y
s
λ
2
d
s
,
y
t
≤
a
2
+
b
2
∫
0
t
‍
t
-
s
β
-
1
x
s
μ
1
d
s
+
b
2
∫
0
t
‍
t
-
s
β
-
1
y
s
μ
2
d
s
+
d
2
∫
0
T
‍
T
-
s
β
-
1
x
s
μ
1
d
s
+
e
2
∫
0
T
‍
T
-
s
β
-
1
y
s
μ
2
d
s
,
where α, β
∈
(
0,1
), λ
i, μ
i
∈
[
0,1
/
2
), and a
i
,
b
i
,
c
i
,
d
i
,
e
i
≥
0 (i
=
1,2) are constants. Then there exists a constant M
*
>
0 such that
(13)
x
t
≤
M
*
,
y
t
≤
M
*
.Proof.
Let(14)
u
t
=
1
,
x
t
≤
1
,
x
t
,
x
t
>
1
,
v
t
=
1
,
y
t
≤
1
,
y
t
,
y
t
>
1
.
By (12), we have
(15)
u
t
≤
a
1
+
1
+
b
1
∫
0
t
‍
t
-
s
α
-
1
u
s
λ
1
d
s
+
c
1
∫
0
t
‍
t
-
s
α
-
1
v
s
λ
2
d
s
+
d
1
∫
0
T
‍
T
-
s
α
-
1
u
s
λ
1
d
s
+
e
1
∫
0
T
‍
T
-
s
α
-
1
v
s
λ
2
d
s
≤
A
1
+
B
1
∫
0
t
‍
t
-
s
α
-
1
u
s
λ
+
v
s
λ
d
s
+
C
1
∫
0
T
‍
T
-
s
α
-
1
u
s
λ
+
v
s
λ
d
s
,
(16)
v
t
≤
a
2
+
1
+
b
2
∫
0
t
‍
t
-
s
β
-
1
u
s
μ
1
d
s
+
c
2
∫
0
t
‍
t
-
s
β
-
1
v
s
μ
2
d
s
+
d
2
∫
0
T
‍
T
-
s
β
-
1
u
s
μ
1
d
s
+
e
2
∫
0
T
‍
T
-
s
β
-
1
v
s
μ
2
d
s
≤
A
2
+
B
2
∫
0
t
‍
t
-
s
β
-
1
u
s
μ
+
v
s
μ
d
s
+
C
2
∫
0
T
‍
T
-
s
β
-
1
u
s
μ
+
v
s
μ
d
s
,
where A
i
=
a
i
+
1, B
i
=
max
⁡
{
b
i
,
c
i
}, C
i
=
max
⁡
{
d
i
,
e
i
}
(
i
=
1,2
), λ
=
max
⁡
{
λ
1
,
λ
2
}, and μ
=
max
⁡
{
μ
1
,
μ
2
}. It is easy to know that
(17)
a
θ
+
b
θ
≤
2
a
+
b
θ
,
θ
∈
0,1
,
a
,
b
≥
0
.
Adding (15) to (16), we get by Cauchy inequality and (17) that (θ
=
max
⁡
{
λ
,
μ
})
(18)
u
t
+
v
t
≤
A
1
+
A
2
+
∫
0
t
‍
B
1
t
-
s
α
-
1
+
B
2
t
-
s
β
-
1
×
u
s
θ
+
v
s
θ
d
s
+
∫
0
T
‍
C
1
T
-
s
α
-
1
+
C
2
T
-
s
β
-
1
×
u
s
θ
+
v
s
θ
d
s
≤
A
1
+
A
2
+
∫
0
t
‍
B
1
t
-
s
α
-
1
+
B
2
t
-
s
β
-
1
2
d
s
1
/
2
·
∫
0
t
‍
u
s
θ
+
v
s
θ
2
d
s
1
/
2
+
∫
0
T
‍
C
1
T
-
s
α
-
1
+
C
2
T
-
s
β
-
1
2
d
s
1
/
2
·
∫
0
T
‍
u
s
θ
+
v
s
θ
2
d
s
1
/
2
≤
A
1
+
A
2
+
∫
0
t
B
1
2
t
-
s
2
α
-
1
+
2
B
1
B
2
t
-
s
α
+
β
-
2
+
B
2
2
t
-
s
2
β
-
1
d
s
∫
T
1
/
2
·
2
∫
0
t
‍
u
s
+
v
s
2
θ
d
s
+
∫
0
T
C
1
2
T
-
s
2
α
-
1
+
2
C
1
C
2
T
-
s
α
+
β
-
2
+
C
2
2
T
-
s
2
β
-
1
T
-
s
2
α
-
1
d
s
∫
T
1
/
2
·
2
∫
0
T
‍
u
s
+
v
s
2
θ
d
s
≤
A
1
+
A
2
+
2
B
1
2
2
α
-
1
T
2
α
-
1
+
2
B
1
B
2
α
+
β
-
1
T
α
+
β
-
1
+
B
2
2
2
β
-
1
T
2
β
-
1
1
/
2
×
∫
0
t
‍
u
s
+
v
s
2
θ
d
s
+
2
C
1
2
2
α
-
1
T
2
α
-
1
+
2
C
1
C
2
α
+
β
-
1
T
α
+
β
-
1
+
C
2
2
2
β
-
1
T
2
β
-
1
1
/
2
×
∫
0
T
‍
u
s
+
v
s
2
θ
d
s
.
From Lemma 6, we obtain that there exists M
*
>
0 such that u
(
t
)
+
v
(
t
)
≤
M
*. Thus,
(19)
x
t
≤
u
t
+
v
t
≤
M
*
,
y
t
≤
u
t
+
v
t
≤
M
*
.Theorem 8 (Schaefer’s fixed point theorem [16]).
LetF
:
X
→
X completely continuous operator. If the set
(20)
E
F
=
x
∈
X
:
x
=
λ
F
x
f
o
r
s
o
m
e
λ
∈
0,1
is bounded, then F has fixed points.Theorem 9 (Nonlinear alternative of Leray-Schauder type [17]).
LetX be a Banach space, D a closed, convex subset of X, U an open subset of D, and 0
∈
D. Assume that A
:
U
¯
→
D is a continuous and compact map. Then either(i)
Ahas fixed points or
(ii)
there existsu
∈
∂
U and λ
∈
(
0,1
) with u
=
λ
A
(
u
).


## 3. Main Results

In order to obtain main result, we make the following assumptions.(H1)
The functionsf
,
g
:
J
×
X
×
X
→
X are continuous.
(H2)
There exist constantsλ
1
,
λ
2
,
μ
1
,
μ
2
∈
[
0,1
/
2
) such that
(21)
f
t
,
x
,
y
≤
M
1
1
+
x
λ
1
+
y
λ
2
,
g
t
,
x
,
y
≤
M
2
1
+
x
μ
1
+
y
μ
2
,
∀
t
∈
J
,
x
,
y
∈
X
.
(H3)
For eacht
∈
J, the sets
(22)
K
1
=
t
-
s
α
-
1
∫
f
s
,
x
s
,
y
s
:
∫
x
,
y
∈
C
J
,
X
,
s
∈
0
,
t
,
K
2
=
t
-
s
β
-
1
g
s
,
x
s
,
y
s
:
t
-
s
β
-
1
x
,
y
∈
C
J
,
X
,
s
∈
0
,
t
are relatively compact.Define the operatorF
:
E
→
E as follows:
(23)
F
x
,
y
t
:
=
F
1
x
,
y
t
,
F
2
x
,
y
t
,
where
(24)
F
1
x
,
y
t
=
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
1
a
+
b
×
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
p
,
F
2
x
,
y
t
=
1
Γ
β
∫
0
t
‍
t
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
1
c
+
d
×
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
q
.It is easy to know that the existence of solution of the coupled fractional BVP (2) is equivalent to the operator F having a fixed point on E.Theorem 10.
Suppose that (H1)–(H3) hold. Then the coupled fractional BVP (2) has at least one solution on J.Proof.
We will use Schaefer’s fixed point theorem to prove thatF has a fixed point. The proof is divided into several steps.
Firstly,F is continuous. Let {
(
x
n
,
y
n
)
} be a sequence such that (
x
n
,
y
n
)
→
(
x
,
y
) in E. For each t
∈
J, we have
(25)
F
1
x
n
,
y
n
t
-
F
1
x
,
y
t
≤
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
×
f
s
,
x
n
s
,
y
n
s
-
f
s
,
x
s
,
y
s
d
s
+
b
a
+
b
Γ
α
×
∫
0
T
‍
T
-
s
α
-
1
×
f
s
,
x
n
s
,
y
n
s
-
f
s
,
x
s
,
y
s
d
s
≤
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
d
s
+
b
a
+
b
∫
0
T
‍
T
-
s
α
-
1
d
s
×
f
·
,
x
n
,
y
n
-
f
·
,
x
,
y
≤
T
α
Γ
α
+
1
1
+
b
a
+
b
f
·
,
x
n
,
y
n
-
f
·
,
x
,
y
.
Similarly, we obtain
(26)
F
2
x
n
,
y
n
t
-
F
2
x
,
y
t
≤
T
β
Γ
β
+
1
1
+
d
c
+
d
g
·
,
x
n
,
y
n
-
g
·
,
x
,
y
.
Since f, g are continuous ((H1)), we have by (25) and (26) that
(27)
F
1
x
n
,
y
n
-
F
1
x
,
y
⟶
0
,
F
2
x
n
,
y
n
-
F
2
x
,
y
⟶
0
as
n
⟶
∞
.
Thus, we get
(28)
F
x
n
,
y
n
-
F
x
,
y
⟶
0
,
as
n
⟶
∞
.
Secondly, we will prove thatF maps bounded sets into bounded sets in E.
By (H2), for anym
>
0, we have for each t
∈
J and (
x
,
y
)
∈
B
m
=
{
(
x
,
y
)
∈
E
:
(
x
,
y
)
≤
m
} that
(29)
F
1
x
,
y
t
≤
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
+
b
a
+
b
Γ
α
×
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
+
p
a
+
b
≤
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
1
+
x
s
λ
1
+
y
s
λ
2
d
s
+
b
M
1
a
+
b
Γ
α
×
∫
0
T
‍
T
-
s
α
-
1
1
+
x
s
λ
1
+
y
s
λ
2
d
s
+
p
a
+
b
=
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
d
s
+
b
M
1
a
+
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
d
s
+
p
a
+
b
+
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
x
s
λ
1
d
s
+
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
y
s
λ
2
d
s
+
b
M
1
a
+
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
x
s
λ
1
d
s
+
b
M
1
a
+
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
y
s
λ
2
d
s
≤
M
1
T
α
Γ
α
+
1
+
b
M
1
T
α
a
+
b
Γ
α
+
1
+
p
a
+
b
+
M
1
T
α
m
λ
1
Γ
α
+
1
+
M
1
T
α
m
λ
2
Γ
α
+
1
+
b
M
1
T
α
m
λ
1
a
+
b
Γ
α
+
1
+
|
b
|
M
1
T
α
m
λ
2
a
+
b
Γ
α
+
1
:
=
N
1
,
which implies that F
1
(
x
,
y
)
≤
N
1. Similarly, we can obtain that F
2
(
x
,
y
)
≤
N
2, where
(30)
N
2
=
M
2
T
β
Γ
β
+
1
+
d
M
2
T
β
c
+
d
Γ
β
+
1
+
q
c
+
d
+
M
2
T
β
m
μ
1
Γ
β
+
1
+
M
2
T
β
m
μ
2
Γ
β
+
1
+
d
M
2
T
β
m
μ
1
c
+
d
Γ
β
+
1
+
d
M
2
T
β
m
μ
2
c
+
d
Γ
β
+
1
.
Thus, we get
(31)
F
x
,
y
≤
max
⁡
N
1
,
N
2
.
Thirdly,F maps bounded sets into equicontinuous sets of E. Let 0
≤
t
1
<
t
2
≤
T, (
x
,
y
)
∈
B
m. By (H2), we have
(32)
F
1
x
,
y
t
2
-
F
1
x
,
y
t
1
≤
1
Γ
α
∫
0
t
1
‍
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
×
f
s
,
x
s
,
y
s
d
s
+
1
Γ
α
∫
t
1
t
2
‍
t
2
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
≤
M
1
Γ
α
∫
0
t
1
‍
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
×
1
+
x
s
λ
1
+
y
s
λ
2
d
s
+
M
1
Γ
α
∫
t
1
t
2
‍
t
2
-
s
α
-
1
1
+
x
s
λ
1
+
y
s
λ
2
d
s
≤
M
1
Γ
α
1
+
m
λ
1
+
m
λ
2
∫
0
t
1
‍
t
1
-
s
α
-
1
-
t
2
-
s
α
-
1
d
s
+
M
1
Γ
α
1
+
m
λ
1
+
m
λ
2
∫
t
1
t
2
‍
t
2
-
s
α
-
1
d
s
≤
M
1
Γ
α
+
1
1
+
m
λ
1
+
m
λ
2
×
t
1
α
-
t
2
α
+
2
t
2
-
t
2
α
⟶
0
,
as
t
2
⟶
t
1
.
Similarly, we obtain
(33)
F
2
x
,
y
t
2
-
F
2
x
,
y
t
1
≤
M
2
Γ
β
+
1
1
+
m
μ
1
+
m
μ
2
×
t
1
β
-
t
2
β
+
2
t
2
-
t
2
β
⟶
0
,
as
t
2
⟶
t
1
.
Hence, F is equicontinuous.
Let{
(
x
n
,
y
n
)
}, n
=
1,2
,
…, be a sequence on B
m, and
(34)
F
1
x
n
,
y
n
t
=
G
1
x
n
,
y
n
t
+
G
2
x
n
,
y
n
T
,
t
∈
J
,
F
2
x
n
,
y
n
t
=
H
1
x
n
,
y
n
t
+
H
2
x
n
,
y
n
T
,
t
∈
J
,
where
(35)
G
1
x
n
,
y
n
t
=
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
n
s
,
y
n
s
d
s
,
t
∈
J
,
(36)
G
2
x
n
,
y
n
T
=
-
1
a
+
b
×
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
n
s
,
y
n
s
d
s
-
p
,
H
1
x
n
,
y
n
t
=
1
Γ
β
∫
0
t
‍
t
-
s
β
-
1
g
s
,
x
n
s
,
y
n
s
d
s
,
t
∈
J
,
H
2
x
n
,
y
n
T
=
-
1
c
+
d
×
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
g
s
,
x
n
s
,
y
n
s
d
s
-
q
.
According to the condition (H3) and Mazur Lemma [18], we know that conv
¯
K
1 is compact. For any t
*
∈
J,
(37)
G
1
x
n
,
y
n
t
*
=
1
Γ
α
∫
0
t
*
‍
t
*
-
s
α
-
1
f
s
,
x
n
s
,
y
n
s
d
s
=
t
*
Γ
α
ξ
n
,
where
(38)
ξ
n
=
lim
⁡
k
→
∞
∑
i
=
1
k
‍
1
k
t
*
-
i
t
*
k
α
-
1
f
i
t
*
k
,
x
n
i
t
*
k
,
y
n
i
t
*
k
.
Since conv
¯
K
1 is convex and compact, we have that ξ
n
∈
conv
¯
K
1. Thus, for any t
*
∈
J, the set {
G
1
(
x
n
,
y
n
)
(
t
*
)
} is relatively compact. From Ascoli-Arzela theorem [17], every {
G
1
(
x
n
,
y
n
)
(
t
)
} contains a uniformly convergent subsequence {
G
1
(
x
n
k
,
y
n
k
)
(
t
)
}, k
=
1,2
,
…, on J. Hence, the set {
G
1
(
x
,
y
)
:
(
x
,
y
)
∈
B
m
} is relatively compact. Similarly, one can obtain that {
G
2
(
x
n
,
y
n
)
(
T
)
} contains a uniformly convergent subsequence {
G
2
(
x
n
k
,
y
n
k
)
(
T
)
}, k
=
1,2
,
…. Thus, the set {
F
1
(
x
,
y
)
:
(
x
,
y
)
∈
B
m
} is relatively compact. Similar to the above process, we can get that the set {
F
2
(
x
,
y
)
:
(
x
,
y
)
∈
B
m
} is relatively compact. Thus, the set {
F
(
x
,
y
)
,
(
x
,
y
)
∈
B
m
} is relatively compact.
From the above three steps, we can conclude thatF is continuous and completely compact.
Finally, we will show that the set(39)
E
F
=
x
,
y
∈
E
:
x
,
y
=
λ
F
x
,
y
,
for
some
λ
∈
0,1
is bounded.
Let(
x
,
y
)
∈
E
(
F
); then (
x
,
y
)
=
λ
F
(
x
,
y
) for some λ
∈
(
0,1
). Hence, for any t
∈
J, we obtain
(40)
x
t
=
λ
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
1
a
+
b
a
a
a
a
a
a
×
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
-
p
,
y
t
=
λ
1
Γ
β
∫
0
t
‍
t
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
1
c
+
d
×
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
g
s
,
x
s
,
y
s
d
s
-
q
.
For each t
∈
J, we obtain
(41)
x
t
≤
M
1
T
α
Γ
α
+
1
+
b
M
1
T
α
a
+
b
Γ
α
+
1
+
p
a
+
b
+
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
x
s
λ
1
d
s
+
M
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
y
s
λ
2
d
s
+
M
1
b
a
+
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
x
s
λ
1
d
s
+
M
1
b
a
+
b
Γ
α
∫
0
T
‍
T
-
s
α
-
1
y
s
λ
2
d
s
,
y
t
≤
M
2
T
β
Γ
β
+
1
+
d
M
2
T
β
c
+
d
Γ
β
+
1
+
q
c
+
d
+
M
2
Γ
β
∫
0
t
‍
t
-
s
β
-
1
x
s
μ
1
d
s
+
M
2
Γ
β
∫
0
t
‍
t
-
s
β
-
1
y
s
μ
2
d
s
+
M
2
d
c
+
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
x
s
μ
1
d
s
+
M
2
d
c
+
d
Γ
β
∫
0
T
‍
T
-
s
β
-
1
y
s
μ
2
d
s
.
By Lemma 7, there exists a M
*
>
0 such that
(42)
x
t
≤
M
*
,
y
t
≤
M
*
,
t
∈
J
.
Hence for any t
∈
J, we obtain
(43)
x
,
y
=
max
⁡
x
,
y
≤
M
*
,
which implies that the set E
(
F
) is bounded. From Theorem 8 (Schaefer’s fixed point theorem), we have that F has a fixed point which is a solution of the fractional BVP (2).Next, we give the second result of this paper, which applies Theorem9. We firstly introduce the following assumption.(H4)
There exist functionsh
i
,
k
i
∈
L
∞
(
J
,
X
) and nondecreasing functions φ
i
,
ψ
i
:
X
→
X (i
=
1,2) such that
(44)
f
t
,
x
,
y
≤
h
1
t
φ
1
x
+
h
2
t
φ
2
y
,
for
t
∈
J
,
x
,
y
∈
E
,
g
t
,
x
,
y
≤
k
1
t
ψ
1
x
+
k
2
t
ψ
2
y
,
for
t
∈
J
,
x
,
y
∈
E
.For convenience, let(45)
A
1
=
T
α
Γ
α
max
⁡
1
,
b
a
+
b
h
1
L
∞
+
h
2
L
∞
,
(46)
A
2
=
T
β
Γ
β
max
⁡
1
,
d
c
+
d
k
1
L
∞
+
k
2
L
∞
,
(47)
Φ
t
=
max
⁡
max
⁡
φ
1
t
,
φ
2
t
,
max
⁡
ψ
1
t
,
ψ
2
t
.Theorem 11.
Let (H1), (H3), and (H4) hold. Assume that there existsr
>
0, with
(48)
r
A
Φ
r
+
B
>
1
,
where
(49)
A
=
max
⁡
A
1
,
A
2
,
B
=
max
⁡
p
a
+
b
,
q
c
+
d
.
Then problem (2) has at least one solution.Proof.
Firstly, we prove thatF maps sets into bounded sets in E. Let D be a bounded subset of E. For each t
∈
J, and (
x
,
y
)
∈
D with (
x
,
y
)
≤
r (r
>
0), we have
(50)
F
1
x
,
y
t
≤
1
Γ
α
∫
0
t
‍
t
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
+
|
b
|
a
+
b
Γ
α
×
∫
0
T
‍
T
-
s
α
-
1
f
s
,
x
s
,
y
s
d
s
+
p
a
+
b
≤
1
Γ
α
h
1
L
∞
φ
1
x
+
h
2
L
∞
φ
2
y
×
∫
0
t
‍
t
-
s
α
-
1
d
s
+
b
a
+
b
Γ
α
×
h
1
L
∞
φ
1
x
+
h
2
L
∞
φ
2
y
×
∫
0
T
‍
T
-
s
α
-
1
d
s
+
p
a
+
b
≤
T
α
Γ
α
max
⁡
1
,
b
a
+
b
h
1
L
∞
+
h
2
L
∞
×
max
⁡
φ
1
x
,
y
,
φ
2
x
,
y
+
p
a
+
b
,
which implies that
(51)
F
1
x
,
y
≤
A
1
max
⁡
φ
1
x
,
y
,
φ
2
x
,
y
+
p
a
+
b
,
where A
1 is as in (45). Similarly, we have
(52)
F
2
x
,
y
≤
A
2
max
⁡
ψ
1
x
,
y
,
ψ
2
x
,
y
+
q
c
+
d
,
where A
2 is as in (46). Combining (47), (51), and (52), we obtain
(53)
F
x
,
y
≤
A
Φ
x
,
y
+
B
≤
A
Φ
r
+
B
,
where A and B are as in (49). This implies that F
(
D
) is bounded in E.
Secondly, we claim thatF is continuous and completely continuous. The proof of this claim is the same as the corresponding part in the proof of Theorem 10 by the conditions (H1), (H3), and (H4).
Finally, let(
x
,
y
)
=
ν
F
(
x
,
y
) for some ν
∈
(
0,1
). Then for any t
∈
J, we have by (53) that
(54)
x
,
y
A
Φ
x
,
y
+
B
≤
1
.
By (48), we know that there exists r
>
0 such that (
x
,
y
)
≠
r. Let
(55)
U
=
x
,
y
∈
E
:
x
,
y
<
r
.
From the choice of U, there is no (
x
,
y
)
∈
∂
U such that (
x
,
y
)
=
ν
F
(
x
,
y
) for some ν
∈
(
0,1
). Therefore, Theorem 9 guarantees that F has a fixed point (
x
,
y
)
∈
U
¯ which is a solution of (2). This completes the proof.


## 4. An Example

In this section, we give an example to illustrate the main results.Example 1.
Consider the following fractional boundary value problem:(56)
D
t
α
0
C
x
t
=
1
+
t
x
t
λ
1
+
t
2
y
t
λ
2
1
+
e
t
1
+
x
t
1
+
y
t
,
0
<
α
<
1
,
t
∈
J
:
=
0,1
,
D
t
β
0
C
y
t
=
t
+
sin
t
x
t
μ
1
+
cos
⁡
t
y
t
μ
2
e
t
2
+
x
t
1
+
y
t
,
0
<
β
<
1
,
t
∈
J
:
=
0,1
,
x
0
+
2
x
1
=
0
,
y
0
-
y
1
=
0
.
Set(57)
f
t
,
x
,
y
=
1
+
t
x
λ
1
+
t
2
y
λ
2
1
+
e
t
1
+
x
1
+
y
,
t
,
x
,
y
∈
J
×
1,2
×
1,2
,
g
t
,
x
,
y
=
t
+
sin
t
x
μ
1
+
cos
⁡
t
y
μ
2
e
t
2
+
x
1
+
y
,
t
,
x
,
y
∈
J
×
1,2
×
1,2
.
For each t
∈
J and (
x
,
y
)
∈
[
1,2
]
×
[
1,2
], we have
(58)
f
t
,
x
,
y
≤
1
+
t
x
λ
1
+
t
2
y
λ
2
1
+
e
t
1
+
x
1
+
y
≤
1
8
1
+
x
λ
1
+
y
λ
2
,
g
t
,
x
,
y
≤
1
+
sin
t
x
μ
1
+
cos
⁡
t
y
μ
2
e
t
2
+
x
1
+
y
≤
1
6
1
+
x
μ
1
+
y
μ
2
.
On the other hand, we easily see that
(59)
∫
0
t
‍
t
-
s
α
-
1
1
+
s
x
s
λ
1
+
s
2
y
s
λ
2
1
+
e
s
1
+
x
s
1
+
y
s
d
s
≤
∫
0
t
‍
t
-
s
α
-
1
1
8
+
1
4
s
+
1
4
s
2
d
s
=
1
8
α
t
α
+
Γ
α
Γ
3
/
2
4
Γ
α
+
3
/
2
t
α
+
1
/
2
+
Γ
α
Γ
3
4
Γ
α
+
3
t
α
+
2
≤
1
8
α
+
Γ
α
Γ
3
/
2
4
Γ
α
+
3
/
2
+
Γ
α
2
Γ
α
+
3
,
∫
0
t
‍
t
-
s
β
-
1
s
+
sin
s
x
s
μ
1
+
cos
⁡
s
y
s
μ
2
e
s
2
+
x
s
1
+
y
s
≤
∫
0
t
‍
t
-
s
β
-
1
s
6
+
1
2
+
1
3
d
s
=
Γ
β
Γ
2
6
Γ
β
+
2
t
β
+
1
+
5
6
β
t
β
≤
Γ
(
β
)
6
Γ
(
β
+
2
)
+
5
6
β
.
Thus the sets
(60)
K
1
=
t
-
s
α
-
1
1
+
s
x
s
λ
1
+
s
2
y
s
λ
2
1
+
e
s
1
+
x
s
1
+
y
s
:
x
,
y
∈
C
J
,
1,2
,
s
∈
0
,
t
1
+
s
x
s
λ
1
+
s
2
y
s
λ
2
1
+
e
s
1
+
x
s
1
+
y
s
,
K
2
=
t
-
s
β
-
1
s
+
sin
s
x
s
μ
1
+
cos
⁡
s
y
s
μ
2
e
s
2
+
x
s
1
+
y
s
:
x
,
y
∈
C
J
,
1,2
,
s
∈
0
,
t
s
+
sin
s
x
s
μ
1
+
cos
⁡
s
y
s
μ
2
e
s
2
+
x
s
1
+
y
s
are bounded and closed which implies that K
1 and K
2 are compact. Hence, all the assumptions in Theorem 10 are satisfied. By Theorem 10, the fractional boundary value problem (56) has at least one solution.



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*Source: 463148-2014-11-18.xml*