# A New Impulsive Multi-Orders Fractional Differential Equation Involving Multipoint Fractional Integral Boundary Conditions

Authors: Guotao Wang; Sanyang Liu; Dumitru Baleanu; Lihong Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

A new impulsive multi-orders fractional differential equation is studied. The existence and uniqueness results are obtained for a nonlinear problem with fractional integral boundary conditions by applying standard fixed point theorems. An example for the illustration of the main result is presented.


---

## Body

## 1. Introduction

Nowadays, fractional differential equations have attracted a lot of attention due to its wide range of applications in many practical problems such as in physics, engineering, economics, and so on; see [1–5].Impulsive differential equations have extensively been studied in the past two decades. Indeed impulsive differential equations are used to describe the dynamics of processes in which sudden, discontinuous jumps occur. Such processes are naturally seen in harvesting, earthquakes, diseases, and so forth. Recently, fractional impulsive differential equations have attracted the attention of many researchers. For the general theory and applications of such equations we refer the interested reader to see [6–24] and the references therein.In this paper, we investigate a new impulsive nonlinear differential equation involvingmulti-orders fractional derivatives and deviating argument. Precisely, we consider the following multipoint fractional integral boundary value problem:
(1)
C
D
t
k
+
α
k
u
(
t
)
=
f
(
t
,
u
(
t
)
,
u
(
θ
(
t
)
)
)
,
1
<
α
k
≤
2
,
t
,
,
u
(
t
)
k
=
0,1
,
2
,
…
,
p
,
t
∈
J
′
,
Δ
u
(
t
k
)
=
I
k
(
u
(
t
k
)
)
,
Δ
u
′
(
t
k
)
=
I
k
*
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
p
,
u
(
0
)
=
∑
k
=
0
p
‍
λ
k
J
t
k
+
β
k
u
(
η
k
)
,
u
′
(
0
)
=
0
,
t
k
<
η
k
<
t
k
+
1
,
where C
D
t
k
+
α
k is the Caputo fractional derivative of order α
k and J
t
k
+
β
k is fractional Riemann-Liouville integral of order β
k
>
0, f
∈
C
(
J
×
R
×
R
,
R
), I
k, I
k
*
∈
C
(
R
,
R
), deviating argument θ
∈
C
(
J
,
J
), J
=
[
0
,
T
]
(
T
>
0
), 0
=
t
0
<
t
1
<
⋯
<
t
k
<
⋯
<
t
p
<
t
p
+
1
=
T, J
′
=
J
/
{
t
1
,
t
2
,
…
,
t
p
}, and Δ
u
(
t
k
)
=
u
(
t
k
+
)
-
u
(
t
k
-
), where u
(
t
k
+
) and u
(
t
k
-
) denote the right and the left limits of u
(
t
) at t
=
t
k
(
k
=
1,2
,
…
,
p
), respectively. Δ
u
′
(
t
k
) have a similar meaning for u
′
(
t
).The paper is organized as follows. Section2 gives some definitions and necessary lemmas, while the main results are presented in Section 3.


## 2. Preliminaries

Let us fixJ
0
=
[
0
,
t
1
], J
k
-
1
=
(
t
k
-
1
,
t
k
], and k
=
1,2
,
…
,
p
+
1 with t
p
+
1
=
T and introduce a Banach space:
(2)
P
C
(
J
,
R
)
=
{
u
:
J
⟶
R
∣
u
∈
C
(
J
k
)
,
k
=
0,1
,
…
,
p
,
u
(
t
k
+
)
exist
,
k
=
1,2
,
…
,
p
.
}
,
with the norm ∥
u
∥
=
sup
⁡
t
∈
J
|
u
(
t
)
|.For the reader’s convenience, we present some necessary definitions from fractional calculus theory and lemmas.Definition 1.
The Riemann-Liouville fractional integral of orderα for a function f
:
[
d
,
∞
)
→
R is defined as
(3)
J
d
+
α
f
(
t
)
=
1
Γ
(
α
)
∫
d
t
‍
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
α
>
0
,
provided the integral exists.Definition 2.
The Caputo fractional derivative of orderα for a function f
:
[
d
,
∞
)
→
R is defined by
(4)
C
D
d
+
α
f
(
t
)
=
1
Γ
(
n
-
α
)
∫
d
t
‍
(
t
-
s
)
n
-
α
-
1
f
(
n
)
(
s
)
d
s
,
&
(
t
-
)
n
-
α
-
1
d
n
=
[
α
]
+
1
,
where [
α
] denotes the integer part of real number α.Lemma 3.
For a giveny
∈
C
[
0
,
T
], a function u is a solution of the following impulsive boundary value problem:
(5)
C
D
t
k
+
α
k
u
(
t
)
=
y
(
t
)
,
1
<
α
k
≤
2
,
k
=
0,1
,
2
,
…
,
p
,
t
∈
J
′
,
Δ
u
(
t
k
)
=
I
k
(
u
(
t
k
)
)
,
Δ
u
′
(
t
k
)
=
I
k
*
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
p
,
u
(
0
)
=
∑
k
=
0
p
‍
λ
k
J
t
k
+
β
k
u
(
η
k
)
,
u
′
(
0
)
=
0
,
if and only if u is a solution of the impulsive fractional integral equation
(6)
u
(
t
)
=
{
1
Γ
(
α
0
)
∫
0
t
‍
(
t
-
s
)
α
0
-
1
y
(
s
)
d
s
+
A
,
t
∈
J
0
;
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
y
(
s
)
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
k
k
k
k
k
k
k
k
k
k
k
k
k
k
+
I
i
*
(
u
(
t
i
)
)
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
A
,
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
t
∈
J
k
,
k
=
1,2
,
…
,
p
,
where
(7)
A
=
(
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
)
-
1
×
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
y
(
s
)
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
}
.Proof.
Letu be a solution of (5). For any t
∈
J
0, we have
(8)
u
(
t
)
=
J
0
α
0
y
(
t
)
-
c
1
-
c
2
t
=
1
Γ
(
α
0
)
∫
0
t
‍
(
t
-
s
)
α
0
-
1
y
(
s
)
d
s
-
c
1
-
c
2
t
,
t
∈
J
0
,
for some c
1
,
c
2
∈
R. Differentiating (8), we get
(9)
u
′
(
t
)
=
1
Γ
(
α
0
-
1
)
∫
0
t
‍
(
t
-
s
)
α
0
-
2
y
(
s
)
d
s
-
c
2
,
t
∈
J
0
.
Ift
∈
J
1, then
(10)
u
(
t
)
=
1
Γ
(
α
1
)
∫
t
1
t
‍
(
t
-
s
)
α
1
-
1
y
(
s
)
d
s
-
d
1
-
d
2
(
t
-
t
1
)
,
u
′
(
t
)
=
1
Γ
(
α
1
-
1
)
∫
t
1
t
‍
(
t
-
s
)
α
1
-
2
y
(
s
)
d
s
-
d
2
,
for some d
1
,
d
2
∈
R. Thus,
(11)
u
(
t
1
-
)
=
1
Γ
(
α
0
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
1
y
(
s
)
d
s
-
c
1
-
c
2
t
1
,
u
(
t
1
+
)
=
-
d
1
,
u
′
(
t
1
-
)
=
1
Γ
(
α
0
-
1
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
2
y
(
s
)
d
s
-
c
2
,
u
′
(
t
1
+
)
=
-
d
2
.
Using the impulse conditions
(12)
Δ
u
(
t
1
)
=
u
(
t
1
+
)
-
u
(
t
1
-
)
=
I
1
(
u
(
t
1
)
)
,
Δ
u
′
(
t
1
)
=
u
′
(
t
1
+
)
-
u
′
(
t
1
-
)
=
I
1
*
(
u
(
t
1
)
)
,
we find that
(13)
-
d
1
=
1
Γ
(
α
0
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
1
y
(
s
)
d
s
-
c
1
-
c
2
t
1
+
I
1
(
u
(
t
1
)
)
,
-
d
2
=
1
Γ
(
α
0
-
1
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
2
y
(
s
)
d
s
-
c
2
+
I
1
*
(
u
(
t
1
)
)
.
Consequently,
(14)
u
(
t
)
=
1
Γ
(
α
1
)
∫
t
1
t
‍
(
t
-
s
)
α
1
-
1
y
(
s
)
d
s
+
1
Γ
(
α
0
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
1
y
(
s
)
d
s
+
t
-
t
1
Γ
(
α
0
-
1
)
∫
0
t
1
‍
(
t
1
-
s
)
α
0
-
2
y
(
s
)
d
s
+
I
1
(
u
(
t
1
)
)
+
(
t
-
t
1
)
I
1
*
(
u
(
t
1
)
)
-
c
1
-
c
2
t
,
+
I
1
(
u
(
t
1
)
)
+
(
t
-
t
1
)
I
1
*
(
u
(
t
1
)
)
ll
c
1
-
t
∈
J
1
.
By a similar process, we can get
(15)
u
(
t
)
=
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
y
(
s
)
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
-
c
1
-
c
2
t
,
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
t
t
t
t
t
t
t
(
s
)
d
s
I
i
*
t
∈
J
k
,
k
=
1,2
,
…
,
p
.
The boundary conditionu
′
(
0
)
=
0 implies c
2
=
0. For t
∈
J
k, we have
(16)
J
t
k
+
β
k
u
(
t
)
=
∫
t
k
t
‍
(
t
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
y
(
s
)
d
s
+
∑
i
=
1
k
‍
(
t
-
t
k
)
β
k
Γ
(
β
k
+
1
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
-
1
‍
(
t
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
-
c
1
(
t
-
t
k
)
β
k
Γ
(
β
k
+
1
)
,
∑
k
=
0
p
‍
λ
k
J
t
k
+
β
k
u
(
η
k
)
=
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
y
(
s
)
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
-
∑
k
=
0
p
‍
c
1
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
.
Applying the boundary conditionu
(
0
)
=
∑
k
=
0
p
λ
k
J
t
k
+
β
k
u
(
η
k
), then
(17)
-
c
1
=
(
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
)
-
1
×
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
y
(
s
)
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
y
(
s
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
y
(
s
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
}
.
Substituting the value of c
i
(
i
=
1,2
) in (8) and (15), we obtain (6). Conversely, assume that u is a solution of the impulsive fractional integral equation (6); then by a direct computation, it follows that the solution given by (6) satisfies (5). This completes the proof.


## 3. Main Results

Define an operatorG
:
P
C
(
J
,
R
)
→
P
C
(
J
,
R
) by
(18)
G
u
(
t
)
=
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
+
∑
i
=
1
k
‍
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
+
I
i
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
+
I
i
*
(
u
(
t
i
)
)
]
+
(
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
)
-
1
×
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
×
11111
1
+
I
i
(
u
(
t
i
)
)
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
×
1111
1
+
I
i
*
(
u
(
t
i
)
)
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
d
s
×
1111
1
+
I
i
*
(
u
(
t
i
)
)
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
∑
k
=
0
p
‍
}
.Notice that problem (1) has a solution if and only if the operator G has a fixed point.For convenience, we will give some notations:(19)
T
*
=
max
⁡
0
≤
i
≤
p
{
T
α
i
}
,
Γ
*
=
min
⁡
0
≤
i
≤
p
{
Γ
(
α
i
)
}
,
Λ
1
=
∑
k
=
0
p
‍
λ
k
T
α
k
+
β
k
Γ
(
α
k
+
β
k
+
1
)
,
Λ
2
=
∑
k
=
1
p
‍
λ
k
T
β
k
Γ
(
β
k
+
1
)
,
Λ
3
=
∑
k
=
1
p
‍
λ
k
T
β
k
Γ
(
β
k
+
2
)
,
Δ
=
|
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
|
-
1
,
Υ
=
3
p
T
*
Γ
*
+
Δ
[
Λ
1
+
(
2
p
-
1
)
T
*
Γ
*
Λ
2
+
p
T
*
Γ
*
Λ
3
]
,
μ
(
x
)
=
Υ
∥
x
∥
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
.Theorem 4.
Assume the following.(H1)
There exists a nonnegative functiona
(
t
)
∈
L
(
0
,
T
) such that
(20)
|
f
(
t
,
u
,
v
)
|
≤
a
(
t
)
+
ξ
1
|
u
|
ρ
+
ξ
2
|
v
|
ϱ
,
0
<
ρ
,
ϱ
<
1
,
where ξ
1
,
ξ
2 are nonnegative constants.
(H2)
There exist positive constantsL
2 and L
3 such that
(21)
|
I
k
(
u
)
|
≤
L
2
,
|
I
k
*
(
u
)
|
≤
L
3
,
f
o
r
t
∈
J
,
u
∈
R
,
k
=
1,2
,
…
,
p
.Then problem (1) has at least one solution.Proof.
Firstly, we will prove thatG
:
P
C
(
J
,
R
)
→
P
C
(
J
,
R
) is a completely continuous operator. Obviously, the continuity of functions f, I
k, and I
k
* ensures the continuity of operator G.
LetΩ
⊂
P
C
(
J
,
R
) be bounded. Then, there exist positive constants L
i
>
0
(
i
=
1,2
,
3
) such that |
f
(
t
,
u
)
|
≤
L
1, |
I
k
(
u
)
|
≤
L
2 and |
I
k
*
(
u
)
|
≤
L
3 for all u
∈
Ω. Thus, for any u
∈
Ω, we have
(22)
|
G
u
(
t
)
|
≤
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
∑
i
=
1
k
‍
+
|
I
i
(
u
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
|
I
i
*
(
u
(
t
i
)
)
|
∑
i
=
1
k
-
1
‍
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∑
i
=
1
k
‍
]
+
|
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
|
-
1
×
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
+
|
I
i
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∑
i
=
1
k
‍
]
}
≤
L
1
(
t
-
t
k
)
α
k
Γ
(
α
k
+
1
)
+
∑
i
=
1
k
‍
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
L
2
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
Δ
{
L
1
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
α
k
+
β
k
Γ
(
α
k
+
β
k
+
1
)
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
L
2
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
[
L
1
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
}
≤
L
1
∑
i
=
1
p
+
1
‍
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
p
L
2
+
T
L
1
∑
i
=
1
p
-
1
‍
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
(
p
-
1
)
T
L
3
+
T
L
1
∑
i
=
1
p
‍
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
p
T
L
3
+
Δ
{
L
1
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
α
k
+
β
k
Γ
(
α
k
+
β
k
+
1
)
+
20
Δ
+
L
1
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
β
k
+
1
)
Γ
(
α
i
-
1
+
1
)
+
20
Δ
+
L
2
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
+
20
Δ
+
L
1
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
β
k
+
1
)
Γ
(
α
i
-
1
)
+
20
Δ
+
L
3
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
+
20
Δ
+
L
1
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
β
k
+
2
)
Γ
(
α
i
-
1
)
+
20
Δ
+
L
3
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
}
≤
L
1
∑
i
=
1
p
+
1
‍
T
*
Γ
*
+
p
L
2
+
T
L
1
∑
i
=
1
p
-
1
‍
T
*
Γ
*
+
(
p
-
1
)
T
L
3
+
T
L
1
∑
i
=
1
p
‍
T
*
Γ
*
+
p
T
L
3
+
Δ
{
L
1
∑
k
=
0
p
‍
λ
k
T
α
k
+
β
k
Γ
(
α
k
+
β
k
+
1
)
+
p
L
1
1
+
Δ
×
∑
k
=
1
p
‍
λ
k
T
β
k
T
*
Γ
(
β
k
+
1
)
Γ
*
+
p
L
2
∑
k
=
1
p
‍
λ
k
T
β
k
Γ
(
β
k
+
1
)
1
+
Δ
+
(
p
-
1
)
L
1
∑
k
=
1
p
‍
λ
k
T
β
k
T
*
Γ
(
β
k
+
1
)
Γ
*
1
+
Δ
+
(
p
-
1
)
L
3
∑
k
=
1
p
‍
λ
k
T
β
k
+
1
Γ
(
β
k
+
1
)
1
+
Δ
+
p
L
1
∑
k
=
1
p
‍
λ
k
T
β
k
T
*
Γ
(
β
k
+
2
)
Γ
*
+
p
L
3
∑
k
=
1
p
‍
λ
k
T
β
k
+
1
Γ
(
β
k
+
2
)
}
≤
{
3
p
T
*
Γ
*
+
Δ
[
Λ
1
+
(
2
p
-
1
)
T
*
Γ
*
Λ
2
+
p
T
*
Γ
*
Λ
3
]
}
L
1
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
∶
=
Υ
L
1
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
,
which implies
(23)
∥
G
u
∥
≤
Υ
L
1
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
∶
=
L
(
constant
)
.
On the other hand, for any t
∈
J
k, 0
≤
k
≤
p, we have
(24)
|
(
G
u
)
′
(
t
)
|
≤
∫
t
k
t
‍
(
t
-
s
)
α
k
-
2
Γ
(
α
k
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
|
d
s
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∑
i
=
1
k
‍
]
≤
L
1
∫
t
k
t
‍
(
t
-
s
)
α
k
-
2
Γ
(
α
k
-
1
)
d
s
+
∑
i
=
1
p
‍
[
L
1
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
d
s
+
L
3
]
≤
T
α
k
-
1
L
1
Γ
(
α
k
)
+
p
[
L
1
max
⁡
0
≤
i
≤
p
T
α
i
-
1
min
⁡
0
≤
i
≤
p
⁡
Γ
(
α
i
)
+
L
3
]
≤
(
p
+
1
)
L
1
T
*
T
Γ
*
+
p
L
3
:
=
L
(
constant
)
.
Hence, for τ
1
,
τ
2
∈
J
k with τ
1
≤
τ
2 and 0
≤
k
≤
p, we have
(25)
|
(
G
u
)
(
τ
2
)
-
(
G
u
)
(
τ
1
)
|
≤
∫
τ
1
τ
2
‍
|
(
G
u
)
′
(
s
)
|
d
s
≤
L
(
t
2
-
t
1
)
.
This implies that G
u is equicontinuous on all J
k
,
k
=
0,1
,
2
,
…
,
p. Consequently, Arzela-Ascoli theorem ensures the operator G
:
P
C
(
J
,
R
)
→
P
C
(
J
,
R
) is a completely continuous operator.
Next, we will show that the operatorG maps B into B. For that, let us choose R
≥
max
⁡
{
3
μ
,
(
3
Υ
ξ
1
)
(
1
/
(
1
-
ρ
)
)
,
(
3
Υ
ξ
2
)
(
1
/
(
1
-
ϱ
)
)
} and define a ball B
=
{
u
∈
P
C
(
J
,
R
)
:
∥
u
∥
≤
R
}. For any u
∈
B, by the conditions (H
1) and (H
2), we have
(26)
|
G
u
(
t
)
|
≤
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
+
+
1
∑
i
=
1
k
‍
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
+
11
∑
i
=
1
k
‍
+
|
I
i
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
∑
i
=
1
k
-
1
‍
(
t
k
-
hh
t
i
)
∫
t
i
-
1
t
i
‍
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
+
|
I
i
*
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
∑
i
=
1
k
-
1
‍
(
t
k
h
-
t
i
)
∫
t
i
-
1
t
i
‍
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
∑
i
=
1
k
‍
(
t
-
t
k
)
+
|
I
i
*
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
|
1
-
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
|
-
1
×
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
×
∑
k
=
0
p
‍
h
λ
k
∫
t
k
η
k
‍
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
+
+
+
+
+
h
+
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
+
+
+
+
+
|
I
i
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
+
+
+
+
h
h
h
+
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
+
+
+
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
+
+
+
×
[
a
(
s
)
+
ξ
1
|
u
(
s
)
|
ρ
+
ξ
2
|
u
(
θ
(
s
)
)
|
ϱ
]
d
s
+
+
+
+
+
+
|
I
i
*
(
u
(
t
i
)
)
|
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
∑
k
=
0
p
‍
}
≤
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
-
t
k
)
α
k
Γ
(
α
k
+
1
)
+
∑
i
=
1
k
‍
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
L
2
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
×
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
×
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
Δ
{
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
∑
k
=
0
p
‍
λ
k
(
η
k
-
t
k
)
α
k
+
β
k
Γ
(
α
k
+
β
k
+
1
)
+
Δ
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
+
Δ
×
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
Δ
×
×
(
t
i
-
t
i
-
1
)
α
i
-
1
Γ
(
α
i
-
1
+
1
)
+
L
2
]
+
Δ
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
+
Δ
×
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
Δ
×
×
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
+
Δ
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
+
Δ
×
[
[
∥
a
∥
+
ξ
1
∥
u
∥
ρ
+
ξ
2
∥
u
∥
ϱ
]
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
Δ
×
×
(
t
i
-
t
i
-
1
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
L
3
]
}
≤
Υ
∥
a
∥
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
+
Υ
ξ
1
∥
u
∥
ρ
+
Υ
ξ
2
∥
u
∥
ϱ
≤
μ
(
a
)
+
Υ
ξ
1
∥
u
∥
ρ
+
Υ
ξ
2
∥
u
∥
ϱ
.
Thus,
(27)
∥
G
u
∥
≤
μ
(
a
)
+
Υ
ξ
1
∥
u
∥
ρ
+
Υ
ξ
2
∥
u
∥
ϱ
≤
R
3
+
R
3
+
R
3
=
R
.
This implies G
:
B
→
B. Hence, we conclude that G
:
B
→
B is completely continuous. It follows from the Schauder fixed point theorem that the operator G has at least one fixed point. That is, problem (1) has at least one solution in B.Theorem 5.
Suppose that there exist a nonnegative functionM
∈
C
(
J
,
R
+
) and nonnegative constants N
,
K such that
(28)
|
f
(
t
,
u
)
-
f
(
t
,
v
)
|
≤
M
(
t
)
|
u
-
v
|
,
|
I
k
(
u
)
-
I
k
(
v
)
|
≤
N
|
u
-
v
|
,
|
I
k
*
(
u
)
-
I
k
*
(
v
)
|
≤
K
|
u
-
v
|
,
for t
∈
J, u
,
v
∈
R and k
=
1,2
,
…
,
p. Furthermore, the assumption μ
(
M
)
<
1 holds. Then problem (1) has a unique solution.Proof.
Foru
,
v
∈
P
C
(
J
,
R
), we have
(29)
|
(
G
u
)
(
t
)
-
(
G
v
)
(
t
)
|
≤
∫
t
k
t
‍
(
t
-
s
)
α
k
-
1
Γ
(
α
k
)
≤
∫
t
k
t
‍
×
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
∑
i
=
1
k
‍
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
+
∑
i
=
1
k
‍
∫
t
i
-
1
t
i
‍
×
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s

+
∑
i
=
1
k
‍
∫
t
i
-
1
t
i
‍
+
|
I
i
(
u
(
t
i
)
)
-
I
i
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
]
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
×
×
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
∑
i
=
1
k
-
1
‍
g
(
t
k
-
t
i
)
×
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
∑
i
=
1
k
-
1
‍
(
t
k
-
t
i
)
+
|
I
i
*
(
u
(
t
i
)
)
-
I
i
*
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
i
=
1
k
‍
(
t
-
t
k
)
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
+
1
+
∑
i
=
1
k
-
1
‍
(
t
-
t
)
×
×
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
+
∑
i
=
1
k
-
1
‍
h
h
h
(
t
-
t
i
)
×
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
∑
i
=
1
k
-
1
‍
(
t
-
t
i
)
+
|
I
i
*
(
u
(
t
i
)
)
-
I
i
*
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
Δ
{
∑
k
=
0
p
‍
λ
k
∫
t
k
η
k
‍
(
η
k
-
s
)
α
k
+
β
k
-
1
Γ
(
α
k
+
β
k
)
×
+
∑
k
=
0
p
‍
λ
k
×
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
×
+
1
∑
k
=
0
p
‍
λ
k
-
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
+
+
×
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
+
+
+
|
I
i
(
u
(
t
i
)
)
-
I
i
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
-
1
‍
λ
k
(
η
k
-
t
k
)
β
k
(
t
k
-
t
i
)
Γ
(
β
k
+
1
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
+
hh
1
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
+
+
+
|
I
i
*
(
u
(
t
i
)
)
-
I
i
*
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
+
∑
k
=
1
p
‍
∑
i
=
1
k
‍
λ
k
(
η
k
-
t
k
)
β
k
+
1
Γ
(
β
k
+
2
)
×
[
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
|
f
(
s
,
u
(
s
)
,
u
(
θ
(
s
)
)
)
+
h
1
h
+
×
∫
t
i
-
1
t
i
‍
(
t
i
-
s
)
α
i
-
1
-
1
Γ
(
α
i
-
1
)
-
f
(
s
,
v
(
s
)
,
v
(
θ
(
s
)
)
)
|
d
s
+
+
+
|
I
i
*
(
u
(
t
i
)
)
-
I
i
*
(
v
(
t
i
)
)
|
(
t
i
-
s
)
α
i
-
1
-
2
Γ
(
α
i
-
1
-
1
)
]
}
≤
{
Υ
∥
M
∥
+
(
1
+
Λ
2
)
p
L
2
+
[
(
p
-
1
)
T
Λ
2
+
p
T
Λ
3
]
L
3
}
∥
u
-
v
∥
=
μ
(
M
)
∥
u
-
v
∥
.
As μ
(
M
)
<
1, we have ∥
G
u
-
G
v
∥
<
∥
u
-
v
∥. Therefore, G is a contraction. It follows from the Banach contraction mapping principle that problem (1) has a unique solution.Example 6.
Forα
0
=
5
/
4, α
1
=
8
/
5, β
0
=
1
/
2, β
1
=
5
/
3, λ
0
=
2
/
5, λ
1
=
3
/
7, η
0
=
1
/
2, η
1
=
4
/
5, 0
<
ρ, ϱ
<
1, and t
1
=
3
/
4, we consider the following impulsive multi-orders fractional differential equation:
(30)
C
D
t
k
+
α
k
u
(
t
)
=
e
t
sin
2
[
3
u
(
t
)
+
e
(
1
/
2
)
u
(
t
)
]
2
+
u
4
(
t
)
+
cos
⁡
(
2
t
+
5
)
3
+
u
2
(
t
)
|
u
(
t
)
|
ρ
+
arctan
2
u
(
t
)
3
|
u
(
t
3
)
|
ϱ
,
+
ii
iiii
cos
⁡
⁡
(
2
t
+
5
)
3
+
u
2
(
t
)
0
<
t
<
1
,
t
≠
3
4
,
k
=
0,1
,
Δ
u
(
3
4
)
=
11
sin
2
u
(
3
4
)
,
Δ
u
′
(
3
4
)
=
|
u
(
3
/
4
)
|
2
(
1
+
|
u
(
3
/
4
)
|
)
,
u
(
0
)
=
∑
k
=
0
1
‍
λ
k
J
t
k
+
β
k
u
(
η
k
)
+
1
2
,
u
′
(
0
)
=
0
.
Observe that
(31)
|
f
(
t
,
u
,
v
)
|
=
|
e
t
sin
2
[
3
u
+
e
(
1
/
2
)
u
]
2
+
u
4
+
cos
⁡
(
2
t
+
5
)
3
+
u
2
|
u
|
ρ
+
arctan
2
u
3
|
v
|
ϱ
e
t
sin
2
[
3
u
+
e
(
1
/
2
)
u
]
2
+
u
4
|
≤
e
t
2
+
1
3
|
u
|
ρ
+
π
12
|
v
|
ϱ
.
Clearly, a
(
t
)
=
e
t
/
2, ξ
1
=
1
/
3, ξ
2
=
π
/
12, L
2
=
11, and L
3
=
1
/
2 and the conditions of Theorem 4 hold. Thus, by Theorem 4, the impulsive multi-orders fractional boundary value problem (30) has at least one solution.



---

*Source: 932747-2014-06-16.xml*