# Existence and Uniqueness Theorems for Impulsive Fractional Differential Equations with the Two-Point and Integral Boundary Conditions

Authors: M. J. Mardanov; N. I. Mahmudov; Y. A. Sharifov
Journal: The Scientific World Journal
 (2014)

---

## Abstract

We study a boundary value problem for the system of nonlinear impulsive fractional differential equations of orderα
0
<
α
≤
1 involving the two-point and integral boundary conditions. Some new results on existence and uniqueness of a solution are established by using fixed point theorems. Some illustrative examples are also presented. We extend previous results even in the integer case α
=
1.


---

## Body

## 1. Introduction

For the last decades, fractional calculus has received a great attention because fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various processes of science and engineering. Indeed, we can find numerous applications in viscoelasticity [1–3], dynamical processes in self-similar structures [4], biosciences [5], signal processing [6], system control theory [7], electrochemistry [8], and diffusion processes [9].On the other hand, the study of dynamical systems with impulsive effects has been an object of intensive investigations in physics, biology, engineering, and so forth. The interest in the study of them is that the impulsive differential systems can be used to model processes which are subject to abrupt changes and which cannot be described by the classical differential problems (e.g., see [10–13] and references therein). Cauchy problems, boundary value problems, and nonlocal problems for impulsive fractional differential equations have been attractive to many researchers; one can see [10–22] and references therein.Fečkan et al. [22] investigated the existence and uniqueness of solutions for
(1)
D
0
+
α
c
x
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
J
′
≔
J
∖
{
t
1
,
…
,
t
p
}
,
55555555555555555555555555555
J
:
=
[
0
,
T
]
,
x
(
t
i
+
)
-
x
(
t
i
-
)
=
a
i
,
i
=
1,2
,
…
,
p
,
x
(
0
)
=
x
0
,
a
k
∈
R
,
where c
D
0
+
α denotes the Caputo fractional derivative of order α
∈
(
0,1
) and f
:
J
×
R
→
R is a given continuous function.In [21], Guo and Jiang discussed the existence of solutions for the following nonlinear fractional differential equations with boundary value conditions:
(2)
D
0
+
α
c
x
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
J
′
≔
J
∖
{
t
1
,
…
,
t
p
}
,
55555555555555555555555555555
J
:
=
[
0
,
T
]
,
x
(
t
i
+
)
-
x
(
t
i
-
)
=
I
i
(
x
(
t
i
-
)
)
,
i
=
1,2
,
…
,
p
,
a
x
(
0
)
+
b
x
(
T
)
=
c
,
where c
D
0
+
α is the Caputo fractional derivative of order α
∈
(
0,1
) with the lower limit zero, f
:
J
×
R
→
R is jointly continuous, t
k satisfy 0
=
t
<
t
1
<
⋯
<
t
p
<
t
p
+
1
=
T, x
(
t
k
+
)
=
lim
⁡
ε
→
0
+
x
(
t
k
+
ε
) and x
(
t
k
-
)
=
lim
⁡
ε
→
0
+
x
(
t
k
-
ε
) represent the right and left limits of x
(
t
) at t
=
t
k, I
k
∈
C
(
R
,
R
), and a, b, c are real constants with a
+
b
≠
0.Ashyralyev and Sharifov [20] considered nonfractional n-dimensional analogues of the problem (2) with two-point and integral boundary conditions.Motivated by the papers above, in this paper, we study impulsive fractional differential equations with the two-point and integral boundary conditions in the following form:(3)
c
D
0
+
α
x
(
t
)
=
f
(
t
,
x
(
t
)
)
,
t
∈
J
′
,
x
(
t
j
+
)
-
x
(
t
j
)
=
I
j
(
x
(
t
j
)
)
,
j
=
1,2
,
…
,
p
,
A
x
(
0
)
+
B
x
(
T
)
=
∫
0
T
‍
g
(
s
,
x
(
s
)
)
d
s
,
where A, B
∈
R
n
×
n are given matrices and det(A
+
B)≠
0. Here f
,
g
:
[
0
,
T
]
×
R
n
→
R
n and I
i
:
R
n
→
R
n are given functions.The rest of the paper is organized as follows. In Section2, we give some notations, recall some concepts, and introduce a concept of a piecewise continuous solution for our problem. In Section 3, we give two main results: the first result based on the Banach contraction principle and the second result based on the Schaefer fixed point theorem. Some examples are given in Section 4 to demonstrate the application of our main results.


## 2. Preliminaries

In this section, we introduce notations, definitions, and preliminary facts that will be used in the remainder of this paper. ByC
(
J
,
R
n
) we denote the Banach space of all continuous functions from J to R
n with the norm
(4)
∥
x
∥
C
=
max
⁡
{
|
x
(
t
)
|
:
t
∈
J
}
,
where |
·
| is the norm in space R
n. We also introduce the Banach space
(5)
P
C
(
J
,
R
n
)
=
{
x
:
J
⟶
R
n
:
x
(
t
)
∈
C
(
(
t
i
,
t
i
+
1
]
,
R
n
)
,
i
=
0,1
,
2
…
,
p
,
x
(
t
i
-
)
and
x
(
t
i
+
)
exist
i
=
1
,
…
,
p
,
and
x
(
t
i
-
)
=
x
(
t
i
)
}
,
with the norm
(6)
∥
x
∥
P
C
:
=
sup
⁡
{
|
x
(
t
)
|
:
t
∈
J
}
.
If A
∈
R
n
×
n, then ∥
A
∥ is the norm of A.Let us recall the following known definitions and results. For more details see [15, 16].Definition 1.
Ifg
∈
C
[
a
,
b
] and α
>
0, then the Riemann-Liouville fractional integral is defined by
(7)
I
a
+
α
g
(
t
)
=
1
Γ
(
α
)
∫
a
t
‍
g
(
s
)
(
t
-
s
)
1
-
α
d
s
,
where Γ
(
·
) is the Gamma function defined for any complex number z as
(8)
Γ
(
z
)
=
∫
0
∞
‍
t
z
-
1
e
-
t
d
t
.Definition 2.
The Caputo fractional derivative of orderα
>
0 of a continuous function g
:
[
a
,
b
]
→
R is defined by
(9)
c
D
a
+
α
g
(
t
)
=
1
Γ
(
n
-
α
)
∫
a
t
‍
g
(
n
)
(
s
)
(
t
-
s
)
α
-
n
+
1
d
s
,
where n
=
[
α
]
+
1 (the notation [
α
] stands for the largest integer not greater than α).Remark 3.
Under natural conditions ong
(
t
), the Caputo fractional derivative becomes the conventional integer order derivative of the function g
(
t
) as α
→
n.Remark 4.
Letα, β
>
0 and n
=
[
α
]
+
1; then the following relations hold:
(10)
D
0
+
α
c
t
β
=
Γ
(
β
)
Γ
(
β
-
α
)
t
β
-
1
,
β
>
n
,
D
0
+
α
c
t
k
=
0
,
k
=
0,1
,
2
,
…
,
n
-
1
.Lemma 5.
Forα
>
0, g
(
t
)
∈
C
[
0
,
T
]
⋂
‍
L
1
[
0
,
T
], the homogeneous fractional differential equation,
(11)
c
D
0
+
α
g
(
t
)
=
0
,
has a solution
(12)
g
(
t
)
=
c
0
+
c
1
t
+
c
2
t
2
+
·
·
·
+
c
n
-
1
t
n
-
1
,
where c
i
∈
R, i
=
0,1
,
…
,
n
-
1, and n
=
[
α
]
+
1.Lemma 6.
Assume thatg
(
t
)
∈
C
[
0
,
T
]
⋂
‍
L
1
[
0
,
T
], with derivative of order n that belongs to C
[
0
,
T
]
⋂
‍
L
1
[
0
,
T
]; then
(13)
I
0
+
α
D
0
+
α
c
g
(
t
)
=
g
(
t
)
+
c
0
+
c
1
t
+
c
2
t
2
+
·
·
·
+
c
n
-
1
t
n
-
1
,
where c
i
∈
R, i
=
0,1
,
…
,
n
-
1, and n
=
[
α
]
+
1.Lemma 7.
Letp, q
≥
0, f
∈
L
1
[
0
,
T
]. Then
(14)
I
0
+
p
I
0
+
q
f
(
t
)
=
I
0
+
p
+
q
f
(
t
)
=
I
0
+
q
I
0
+
p
f
(
t
)
is satisfied almost everywhere on [
0
,
T
]. Moreover, if f
∈
C
[
0
,
T
], then (14) is true for all t
∈
[
0
,
T
].Lemma 8.
Ifα
>
0, f
∈
C
(
[
0
,
T
]
), then c
D
0
+
α
I
0
+
α
f
(
t
)
=
f
(
t
) for all t
∈
[
0
,
T
].We define a solution problem (3) as follows.Definition 9.
A functionx
∈
P
C
(
J
,
R
n
) is said to be a solution of problem (3) if c
D
0
+
α
x
(
t
)
=
f
(
t
,
x
(
t
)
), for t
∈
[
0
,
T
], t
≠
t
i, i
=
1,2
,
…
,
p, and for each i
=
1,2
,
…
,
p, x
(
t
i
+
)
-
x
(
t
i
)
=
I
i
(
x
(
t
i
)
), 0
=
t
0
<
t
1
<
t
2
<
⋯
<
t
p
<
t
p
+
1
=
T, and the boundary conditions A
x
(
0
)
+
B
x
(
T
)
=
∫
0
T
‍
g
(
s
,
x
(
s
)
)
d
s are satisfied.We have the following result which is useful in what follows.Theorem 10.
Letf, g
∈
C
(
J
,
R
n
). Then the function x is a solution of the boundary value problem for impulsive differential equation
(15)
c
D
0
+
α
x
(
t
)
=
f
(
t
)
,
t
∈
J
′
,
x
(
t
j
+
)
-
x
(
t
j
)
=
I
j
(
x
(
t
j
)
)
,
j
=
1,2
,
…
,
p
,
A
x
(
0
)
+
B
x
(
T
)
=
∫
0
T
‍
g
(
s
)
d
s
if and only if
(16)
x
(
t
)
=
(
A
+
B
)
-
1
∫
0
T
‍
g
(
s
)
d
s
+
∑
0
<
t
j
<
T
‍
K
(
t
k
,
t
j
)
I
j
(
x
(
t
j
-
)
)
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
K
(
t
k
,
t
j
)
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
t
k
<
t
≤
t
k
+
1
,
where
(17)
K
(
t
,
τ
)
=
{
0
,
t
=
0
,
(
A
+
B
)
-
1
A
,
0
<
τ
≤
t
,
-
(
A
+
B
)
-
1
B
,
t
<
τ
≤
T
.Proof.
Assume thatx is a solution of the boundary value problem (15); then we have
(18)
x
(
t
)
=
x
(
0
)
+
1
Γ
(
α
)
∫
0
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
0
≤
t
≤
t
1
.
If t
1
<
t
≤
t
2, then
(19)
c
D
0
+
α
x
(
t
)
=
f
(
t
)
,
t
1
<
t
≤
t
2
,
x
(
t
1
+
)
-
x
(
t
1
)
=
I
1
(
x
(
t
1
)
)
.
Integrating the expression (19) from t
1 to t, one can obtain
(20)
x
(
t
)
=
x
(
t
1
+
)
+
1
Γ
(
α
)
∫
t
1
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
.
It follows that
(21)
x
(
t
)
=
x
(
t
1
)
+
I
1
(
x
(
t
1
)
)
+
1
Γ
(
α
)
∫
t
1
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
=
x
(
0
)
+
I
1
(
x
(
t
1
)
)
+
1
Γ
(
α
)
∫
0
t
1
(
t
1
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
1
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
.
Thus ift
∈
(
t
k
,
t
k
+
1
], we get
(22)
x
(
t
)
=
x
(
t
k
)
+
I
k
(
x
(
t
k
)
)
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
=
x
(
0
)
+
∑
0
<
t
k
<
t
I
k
(
x
(
t
k
)
)
+
1
Γ
(
α
)
∑
0
<
t
k
<
t
∫
t
k
-
1
t
k
(
t
k
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
where x
(
0
) is still an arbitrary constant vector. For determining x
(
0
) we use the boundary value condition A
x
(
0
)
+
B
x
(
T
)
=
∫
0
T
‍
g
(
s
)
d
s:
(23)
∫
0
T
‍
g
(
s
)
d
s
=
A
x
(
0
)
+
B
x
(
T
)
=
(
A
+
B
)
x
(
0
)
+
B
∑
0
<
t
k
<
T
I
k
(
x
(
t
k
)
)
+
1
Γ
(
α
)
∑
0
<
t
k
<
T
B
∫
t
k
-
1
t
k
(
t
k
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
B
∫
t
k
T
(
T
-
s
)
α
-
1
f
(
s
)
d
s
.
Hence, we obtain
(24)
x
(
0
)
=
(
A
+
B
)
-
1
∫
0
T
‍
g
(
s
)
d
s
-
(
A
+
B
)
-
1
B
∑
0
<
t
j
<
T
I
j
(
x
(
t
j
)
)
-
1
Γ
(
α
)
(
A
+
B
)
-
1
B
∑
0
<
t
j
≤
T
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
,
and consequently for all t
∈
(
t
k
,
t
k
+
1
]
(25)
x
(
t
)
=
(
A
+
B
)
-
1
∫
0
T
‍
g
(
s
)
d
s
-
(
A
+
B
)
-
1
B
∑
0
<
t
j
<
T
I
j
(
x
(
t
j
)
)
-
1
Γ
(
α
)
(
A
+
B
)
-
1
B
∑
0
<
t
j
≤
T
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
∑
0
<
t
j
<
t
I
j
(
x
(
t
j
)
)
+
1
Γ
(
α
)
∑
0
<
t
j
<
t
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
=
(
A
+
B
)
-
1
∫
0
T
‍
g
(
s
)
d
s
-
(
A
+
B
)
-
1
B
∑
t
k
<
t
j
<
T
I
j
(
x
(
t
j
)
)
+
(
A
+
B
)
-
1
A
∑
0
<
t
j
<
t
I
j
(
x
(
t
j
)
)
-
1
Γ
(
α
)
(
A
+
B
)
-
1
B
∑
t
k
<
t
j
≤
T
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
(
A
+
B
)
-
1
A
∑
0
<
t
j
<
t
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
=
(
A
+
B
)
-
1
∫
0
T
‍
g
(
s
)
d
s
+
∑
0
<
t
j
<
T
‍
K
(
t
k
,
t
j
)
I
j
(
x
(
t
j
-
)
)
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
K
(
t
k
,
t
j
)
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
.
Conversely, assume thatx satisfies (16). If t
∈
[
0
,
t
1
], then, using the fact that c
D
0
+
α is the left inverse of I
0
+
α, we get D
0
+
α
c
x
(
t
)
=
f
(
t
), t
0
<
t
≤
t
1. If t
∈
(
t
k
,
t
k
+
1
], k
=
1,2
,
…
,
p, then, using the fact that the Caputo derivative of a constant is equal to zero, we obtain D
0
+
α
c
x
(
t
)
=
f
(
t
), t
k
<
t
≤
t
k
+
1, and x
(
t
k
+
)
-
x
(
t
k
)
=
I
k
(
x
(
t
k
)
). The lemma is proved.Theorem 11 (see [18]).
LetX be a Banach space and W
⊂
P
C
(
J
,
X
). If the following conditions are satisfied,(1)
W is uniformly bounded subset of P
C
(
J
,
X
),
(2)
W is equicontinuous in (
t
k
,
t
k
+
1
), k
=
0,1
,
2
,
…
,
p, where t
0
=
0, t
p
+
1
=
T,
(3)
W
(
t
)
=
{
u
(
t
)
:
u
∈
W
,
t
∈
J
′
}, W
(
t
k
+
)
=
{
u
(
t
k
+
)
:
u
∈
W
}, and W
(
t
k
-
)
=
{
u
(
t
k
-
)
:
u
∈
W
} are relatively compact subsets of X,then W is a relatively compact subset of P
C
(
J
,
X
).


## 3. Main Results

Our first result is based on Banach fixed point theorem. Before stating and proving the main results, we introduce the following hypotheses.(H1)
f
,
g
:
J
×
R
n
→
R
n are continuous functions.
(H2)
There are constantsL
f
>
0 and L
g
>
0 such that
(26)
|
f
(
t
,
x
)
-
f
(
t
,
y
)
|
≤
L
f
|
x
-
y
|
,
|
g
(
t
,
x
)
-
g
(
t
,
y
)
|
≤
L
g
|
x
-
y
|
for eacht
∈
[
0
,
T
] and all x, y
∈
R
n.
(H3)
There exist constantsl
i
>
0, i
=
1,2
,
…
,
p such that
(27)
|
I
i
(
x
)
-
I
i
(
y
)
|
≤
l
i
|
x
-
y
|
for allx, y
∈
R
n.For brevity, let
(28)
L
A
B
:
=
max
⁡
(
∥
(
A
+
B
)
-
1
A
∥
,
∥
(
A
+
B
)
-
1
B
∥
)
.Theorem 12.
Assume that (H1)–(H3) hold. If(29)
L
g
∥
(
A
+
B
)
-
1
∥
T
+
L
A
B
∑
j
=
1
p
‍
l
j
+
1
Γ
(
α
+
1
)
L
f
L
A
B
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
Γ
(
α
+
1
)
L
f
<
1
,
then the boundary value problem (3) has a unique solution on J.Proof.
The proof is based on the classical Banach fixed theorem for contractions. Let us set(30)
sup
⁡
t
∈
J
⁡
|
f
(
t
,
0
)
|
=
M
f
,
sup
⁡
t
∈
J
|
g
(
t
,
0
)
|
=
M
g
,
|
I
k
(
0
)
|
=
m
k
,
δ
(
T
)
:
=
L
g
∥
(
A
+
B
)
-
1
∥
T
+
L
A
B
∑
j
=
1
p
‍
l
j
+
1
Γ
(
α
+
1
)
L
f
L
A
B
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
Γ
(
α
+
1
)
L
f
<
1
,
γ
:
=
∥
(
A
+
B
)
-
1
∥
T
M
g
+
L
A
B
∑
j
=
1
p
‍
m
j
+
1
Γ
(
α
+
1
)
M
f
L
A
B
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
Γ
(
α
+
1
)
M
f
.
It is clear that
(31)
|
f
(
t
,
x
)
|
≤
M
f
+
L
f
|
x
|
,
|
g
(
t
,
x
)
|
≤
M
g
+
L
g
|
x
|
,
|
I
k
(
x
)
|
≤
m
k
+
l
k
|
x
|
,
t
∈
J
,
x
∈
R
n
,
∥
K
(
t
k
,
t
j
)
∥
≤
L
A
B
,
k
,
j
=
0,1
,
…
,
p
+
1
.
Consider
(32)
B
r
:
=
{
x
∈
P
C
(
J
,
R
n
)
:
∥
x
∥
P
C
≤
r
}
,
where
(33)
r
≥
γ
1
-
δ
(
T
)
.
LetQ be the following operator:
(34)
(
Q
x
)
(
t
)
=
{
(
A
+
B
)
-
1
∫
0
T
g
(
s
)
d
s
+
∑
0
<
t
j
<
T
‍
K
(
t
k
,
t
j
)
I
j
(
x
(
t
j
)
)
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
K
(
t
k
,
t
j
)
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
f
(
s
)
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
t
k
<
t
≤
t
k
+
1
,
k
=
1
,
…
,
p
.
We show that Q maps B
r into B
r. It is clear that Q is well defined on P
C
(
J
,
R
n
). Moreover for x
∈
B
r and t
∈
(
t
k
,
t
k
+
1
], k
=
0
,
…
,
p, we have
(35)
|
(
Q
x
)
(
t
)
|
≤
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
g
(
s
,
x
(
s
)
)
|
d
s
+
∑
0
<
t
j
<
T
‍
∥
K
(
t
k
,
t
j
)
∥
|
I
j
(
x
(
t
j
)
)
|
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∥
K
(
t
k
,
t
j
)
∥
×
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
≤
∥
(
A
+
B
)
-
1
∥
T
(
M
g
+
L
g
r
)
+
L
A
B
∑
j
=
1
p
‍
(
m
j
+
l
j
r
)
+
1
α
Γ
(
α
)
L
A
B
(
M
f
+
L
f
r
)
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
α
Γ
(
α
)
(
M
f
+
L
f
r
)
=
γ
+
δ
(
T
)
r
≤
r
.
Consequently Q maps P
C
(
J
,
R
n
) into itself.
Next we will show thatQ is a contraction. Let x, y
∈
P
C
(
J
,
R
n
). Then, for each t
∈
(
t
k
,
t
k
+
1
], k
=
0
,
…
,
p, we have
(36)
|
(
Q
x
)
(
t
)
-
(
Q
y
)
(
t
)
|
≤
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
g
(
s
,
x
(
s
)
)
-
g
(
s
,
y
(
s
)
)
|
d
s
+
∑
0
<
t
j
<
T
‍
∥
K
(
t
k
,
t
j
)
∥
|
I
j
(
x
(
t
j
)
)
-
I
j
(
y
(
t
j
)
)
|
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∥
K
(
t
k
,
t
j
)
∥
×
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
d
s
≤
L
g
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
x
(
s
)
-
y
(
s
)
|
d
s
+
L
A
B
∑
j
=
1
p
‍
|
I
j
(
x
(
t
j
)
)
-
I
j
(
y
(
t
j
)
)
|
+
L
f
L
A
B
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
x
(
s
)
-
y
(
s
)
|
d
s
+
L
f
Γ
(
α
)
∫
t
k
t
‍
(
t
-
s
)
α
-
1
|
x
(
s
)
-
y
(
s
)
|
d
s
≤
(
L
g
∥
(
A
+
B
)
-
1
∥
T
+
L
A
B
∑
j
=
1
p
‍
l
j
+
1
α
Γ
(
α
)
L
f
L
A
B
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
α
Γ
(
α
)
L
f
)
×
∥
x
-
y
∥
P
C
.
Thus, Q is a contraction mapping on P
C
(
J
,
R
n
) due to condition (29) and the operator Q has a unique fixed point on P
C
(
J
,
R
n
) which is a unique solution to (3).The second result is based on the Schaefer fixed point theorem. We introduce the following assumptions.(H4)
There exist constantsN
f
>
0, N
g
>
0 such that |
f
(
t
,
x
)
|
≤
N
f, |
g
(
t
,
x
)
|
≤
N
g for each t
∈
J and all x
∈
R
n.
(H5)
I
k
∈
C
(
R
n
,
R
n
).Theorem 13.
Assume that (H1), (H4), and (H5) hold. Then the boundary value problem (3) has at least one solution on J.Proof.
We will use Schaefer’s fixed point theorem to prove thatQ defined by (34) has a fixed point. The proof will be given in several steps.
Step  1. Operator Q is continuous.
Let{
x
n
} be a sequence such that x
n
→
x in P
C
(
J
,
R
n
). Then, for each k
=
0,1
,
2
,
…
,
p and for all t
∈
(
t
k
,
t
k
+
1
], we have
(37)
|
(
Q
x
)
(
t
)
-
(
Q
x
n
)
(
t
)
|
≤
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
g
(
s
,
x
(
s
)
)
-
g
(
s
,
x
n
(
s
)
)
|
d
s
+
∑
0
<
t
j
<
T
‍
∥
K
(
t
k
,
t
j
)
∥
|
I
j
(
x
(
t
j
)
)
-
I
j
(
x
n
(
t
j
)
)
|
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∥
K
(
t
k
,
t
j
)
∥
×
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
x
n
(
s
)
)
|
d
s
+
1
Γ
(
α
)
∫
0
t
(
t
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
x
n
(
s
)
)
|
d
s
.
Since f, g, and I
k, k
=
0,1
,
2
,
…
,
p, are continuous functions, we have
(38)
∥
Q
x
n
-
Q
x
∥
P
C
⟶
0
as n
→
∞.
Step  2. Q maps bounded sets in bounded sets in P
C
(
J
,
R
n
).
Indeed, it is enough to show that, for anyη
>
0, there exists a positive constant l such that, for each x
∈
B
η
=
{
x
∈
P
C
(
J
,
R
n
)
:
∥
x
∥
P
C
≤
η
}, we have ∥
Q
(
x
)
∥
P
C
≤
l. By (H4), (H5) we have, for each k
=
1,2
,
…
,
p and for all t
∈
(
t
k
,
t
k
+
1
],
(39)
|
(
Q
x
)
(
t
)
|
≤
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
g
(
s
,
x
(
s
)
)
|
d
s
+
∑
0
<
t
j
<
T
‍
∥
K
(
t
k
,
t
j
)
∥
|
I
j
(
x
(
t
j
)
)
|
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∥
K
(
t
k
,
t
j
)
∥
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
≤
∥
(
A
+
B
)
-
1
∥
T
N
g
+
L
A
B
∑
j
=
1
p
‍
|
I
j
(
x
(
t
j
)
)
|
+
1
α
Γ
(
α
)
L
A
B
N
f
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
α
Γ
(
α
)
N
f
:
=
l
.
Thus
(40)
∥
Q
x
∥
P
C
≤
l
.
Step  3. Q maps bounded sets into equicontinuous sets of P
C
(
J
,
R
n
).
Letτ
1
,
τ
2
∈
(
t
k
,
t
k
+
1
], τ
1
<
τ
2
,
B
η be a bounded set of P
C
(
J
,
R
n
) as in Step  2, and let x
∈
B
η. Then
(41)
|
(
Q
x
)
(
τ
2
)
-
(
Q
x
)
(
τ
1
)
|
≤
|
1
Γ
(
α
)
∫
t
i
τ
1
‍
[
(
τ
2
-
s
)
α
-
1
-
(
τ
1
-
s
)
α
-
1
]
f
(
s
,
x
(
s
)
)
d
s
+
1
Γ
(
α
)
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
α
-
1
f
(
s
,
x
(
s
)
)
d
s
|
≤
N
f
Γ
(
α
)
∫
t
i
τ
1
‍
[
(
τ
2
-
s
)
α
-
1
-
(
τ
1
-
s
)
α
-
1
]
d
s
+
N
f
Γ
(
α
)
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
α
-
1
d
s
≤
N
f
Γ
(
α
+
1
)
[
2
(
τ
2
-
τ
1
)
α
+
τ
2
α
-
τ
1
α
]
.
As τ
1
→
τ
2, the right-hand side of the above inequality tends to zero.
As a consequence of Steps  1 to 3 together with the Arzela-Ascoli theorem (Theorem11 with X
=
R
n), we can conclude that the operator Q
:
P
C
(
J
,
R
n
)
→
P
C
(
J
,
R
n
) is completely continuous.
Step  4. One has a priori bounds.
Now it remains to show that the set(42)
Δ
=
{
x
∈
P
C
(
J
,
R
n
)
:
x
=
λ
Q
(
x
)
,
for
some
0
<
λ
<
1
}
is bounded.
Let thenx
=
λ
Q
(
x
) for some 0
<
λ
<
1. Thus, for each t
∈
(
t
k
,
t
k
+
1
], we have
(43)
|
x
(
t
)
|
=
|
λ
(
Q
x
)
(
t
)
|
≤
∥
(
A
+
B
)
-
1
∥
∫
0
T
‍
|
g
(
s
,
x
(
s
)
)
|
d
s
+
∑
0
<
t
j
<
T
‍
∥
K
(
t
k
,
t
j
)
∥
|
I
j
(
x
(
t
j
)
)
|
+
1
Γ
(
α
)
∑
0
<
t
j
≤
T
‍
∥
K
(
t
k
,
t
j
)
∥
∫
t
j
-
1
t
j
(
t
j
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
+
1
Γ
(
α
)
∫
t
k
t
(
t
-
s
)
α
-
1
|
f
(
s
,
x
(
s
)
)
|
d
s
≤
∥
(
A
+
B
)
-
1
∥
T
N
g
+
L
A
B
∑
j
=
1
p
‍
|
I
j
(
x
(
t
j
)
)
|
+
1
α
Γ
(
α
)
L
A
B
N
f
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
α
Γ
(
α
)
N
f
.
Thus
(44)
∥
x
∥
P
C
≤
∥
(
A
+
B
)
-
1
∥
T
N
g
+
L
A
B
∑
j
=
1
p
‍
|
I
j
(
x
(
t
j
)
)
|
+
1
α
Γ
(
α
)
L
A
B
N
f
∑
j
=
1
p
+
1
‍
(
t
j
-
t
j
-
1
)
α
+
T
α
α
Γ
(
α
)
N
f
.
This shows that the set Δ is bounded. As a consequence of Schaefer’s fixed point theorem, we deduce that Q has a fixed point which is a solution of the problem (3).


## 4. Examples

In this section, we give some examples to illustrate our main results.Example 1.
Consider(45)
D
0
+
α
c
x
1
(
t
)
=
cos
⁡
(
1
10
x
2
(
t
)
)
,
t
∈
(
0,2
)
∖
{
1
}
,
D
0
+
α
c
x
2
(
t
)
=
e
-
t
9
+
e
t
|
x
1
(
t
)
|
1
+
|
x
1
(
t
)
|
,
t
∈
(
0,2
)
∖
{
1
}
,
x
1
(
0
)
+
1
2
x
2
(
1
)
=
0
,
x
2
(
0
)
=
1
,
Δ
x
1
(
1
)
=
1
10
x
2
(
1
)
,
Δ
x
2
(
1
)
=
1
10
x
1
(
1
)
+
5
.
Consider boundary value problem (3) with f
1
(
t
,
x
1
,
x
2
)
=
cos
⁡
(
(
1
/
10
)
x
2
(
t
)
),  f
2
(
t
,
x
1
,
x
2
)
=
(
e
-
t
/
(
9
+
e
t
)
)
·
(
|
x
1
|
/
(
1
+
|
x
1
|
)
), and  T
=
1,  p
=
1,  I
1
(
x
1
,
x
2
)
=
[
(
1
/
10
)
x
2
(
1
/
10
)
x
1
+
5
].Evidently,(46)
A
=
I
=
(
1
0
0
1
)
,
B
=
(
0
0.5
0
0
)
,
∥
(
A
+
B
)
-
1
A
∥
=
∥
(
1
-
1
2
0
1
)
∥
=
3
2
,
∥
(
A
+
B
)
-
1
B
∥
=
∥
(
0
1
2
0
0
)
∥
=
1
2
,
L
A
B
=
max
⁡
(
3
2
,
1
2
)
=
3
2
,
and conditions (H1)–(H3) hold. We will show that condition (29) is satisfied for, say, α
=
0,2. Indeed,
(47)
L
A
B
l
1
+
2
α
+
1
α
Γ
(
α
)
L
f
L
A
B
+
2
α
α
Γ
(
α
)
L
f
=
3
2
×
1
10
+
2
α
Γ
(
α
+
1
)
×
3
10
+
2
α
Γ
(
α
+
1
)
×
1
10
<
3
2
×
1
10
+
5
4
×
2
5
=
13
20
<
1
,
where we used
(48)
Γ
(
α
+
1
)
=
Γ
(
1,2
)
=
0.92
,
2
α
Γ
(
α
+
1
)
<
1.15
0.92
=
1.25
.Then, by Theorem12, boundary value problem (45) has unique solution on [
0,2
].Example 2.
Consider(49)
D
0
+
α
c
x
1
(
t
)
=
e
-
t
9
+
e
t
|
x
1
(
t
)
|
1
+
x
2
2
(
t
)
,
t
∈
(
0,1
)
,
t
≠
0,5
D
0
+
α
c
x
2
(
t
)
=
sin
x
1
(
t
)
,
t
∈
(
0,1
)
,
t
≠
0,5
,
x
1
(
0
)
=
1
,
x
2
(
0
)
+
1
2
x
1
(
1
)
=
0
,
Δ
x
1
(
1
2
)
=
1
1
+
x
2
2
(
1
/
2
)
,
Δ
x
2
(
1
2
)
=
1
1
+
cos
⁡
2
x
1
(
1
/
2
)
.
Here0
<
α
≤
1,  f
1
(
t
,
x
1
,
x
2
)
=
e
-
t
/
(
1
+
x
2
2
(
t
)
),  f
2
(
t
,
x
1
,
x
2
)
=
sin
x
1
(
t
),  A
=
(
1
0
0
1
),  B
=
(
0
0
0.5
0
), and  T
=
1,  p
=
1,  I
1
(
x
1
,
x
2
)
=
[
1
/
(
1
+
x
2
2
)
1
/
(
1
+
cos
⁡
2
x
1
)
]. Clearly, all the conditions of Theorem 13 are satisfied (
N
f
=
1, N
g
=
0
), and consequently boundary value problem (49) has at least one solution.



---

*Source: 918730-2014-03-23.xml*