# Boundary Value Problems for First-Order Impulsive Functionalq-Integrodifference Equations

Authors: Jessada Tariboon; Sotiris K. Ntouyas
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We discuss the existence and uniqueness of solutions for a first-order boundary value problem for impulsive functionalq
k-integrodifference equations. The main results are obtained with the aid of some classical fixed point theorems. Illustrative examples are also presented.


---

## Body

## 1.  Introduction

In this paper, we study the boundary value problem for impulsive functionalq
k-integro-difference equation of the following the form:
(1)
D
q
k
x
(
t
)
=
f
(
t
,
x
(
t
)
,
x
(
θ
(
t
)
)
,
(
S
q
k
x
)
(
t
)
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
∈
J
∶
=
[
0
,
T
]
,
t
≠
t
k
,
Δ
x
(
t
k
)
=
I
k
(
x
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
α
x
(
0
)
=
β
x
(
T
)
+
∑
i
=
0
m
γ
i
∫
t
i
t
i
+
1
x
(
s
)
d
q
i
s
,
where 0
=
t
0
<
t
1
<
t
2
<
⋯
<
t
k
<
⋯
<
t
m
<
t
m
+
1
=
T, f
:
J
×
ℝ
3
→
ℝ,  θ
:
J
→
J,
(2)
(
S
q
k
x
)
(
t
)
=
∫
t
k
t
ϕ
(
t
,
s
,
x
(
s
)
)
d
q
k
s
,
k
=
0,1
,
2
,
…
,
m
,
ϕ
:
J
2
×
ℝ
→
[
0
,
∞
) is a continuous function, I
k
∈
C
(
ℝ
,
ℝ
), Δ
x
(
t
k
)
=
x
(
t
k
+
)
-
x
(
t
k
) for k
=
1,2
,
…
,
m,  x
(
t
k
+
)
=
lim
⁡
h
→
0
x
(
t
k
+
h
),  α
,
β
,
γ
i
,
i
=
0,1
,
…
,
m are real constants, and 0
<
q
k
<
1 for k
=
0,1
,
2
,
…
,
m.The notions ofq
k-derivative and q
k-integral on finite intervals were introduced recently by the authors in [1]. Their basic properties were studied and applications existence and uniqueness results were proved for initial value problems for first- and second-order impulsive q
k-difference equations. In this paper, we continue the study on this new subject by considering the boundary value problem (1).The book by Kac and Cheung [2] covers many of the fundamental aspects of the quantum calculus. In recent years, the topic of q-calculus has attracted the attention of several researchers and a variety of new results can be found in the papers [3–15] and the references cited therein. On the other hand, for some monographs on the impulsive differential equations we refer to [16–18].The rest of this paper is organized as follows. In Section2, we recall the notions of q
k-derivative and q
k-integral on finite intervals and present a preliminary result which will be used in this paper. In Section 3, we will consider the existence results for problem (1) while in Section 4, we will give examples to illustrate our main results.


## 2. Preliminaries

In this section, we recall the notions ofq
k-derivative and q
k-integral on finite intervals. For a fixed k
∈
ℕ
∪
{
0
} let J
k
:
=
[
t
k
,
t
k
+
1
]
⊂
ℝ be an interval and let 0
<
q
k
<
1 be a constant. We define q
k-derivative of a function f
:
J
k
→
ℝ at a point t
∈
J
k as follows.Definition 1.
Assumef
:
J
k
→
ℝ is a continuous function and let t
∈
J
k. Then the expression
(3)
D
q
k
f
(
t
)
=
f
(
t
)
-
f
(
q
k
t
+
(
1
-
q
k
)
t
k
)
(
1
-
q
k
)
(
t
-
t
k
)
,
t
≠
t
k
,
D
q
k
f
(
t
k
)
=
lim
⁡
t
→
t
k
D
q
k
f
(
t
)
is called the q
k-derivative of function f at t.We say thatf is q
k-differentiable on J
k provided that D
q
k
f
(
t
) exists for all t
∈
J
k. Note that if t
k
=
0 and q
k
=
q in (3), then D
q
k
f
=
D
q
f, where D
q is the well-known q-derivative of the function f
(
t
) defined by
(4)
D
q
f
(
t
)
=
f
(
t
)
-
f
(
q
t
)
(
1
-
q
)
t
.In addition, we should define the higherq
k-derivative of functions.Definition 2.
Letf
:
J
k
→
ℝ be a continuous function; we call the second-order q
k-derivative D
q
k
2
f provided that D
q
k
f is q
k-differentiable on J
k with D
q
k
2
f
=
D
q
k
(
D
q
k
f
)
:
J
k
→
ℝ. Similarly, we define higher order q
k-derivative D
q
k
n
:
J
k
→
ℝ.Theq
k-integral is defined as follows.Definition 3.
Assumef
:
J
k
→
ℝ is a continuous function. Then the q
k-integral is defined by
(5)
∫
t
k
t
‍
f
(
s
)
d
q
k
s
=
(
1
-
q
k
)
(
t
-
t
k
)
∑
n
=
0
∞
q
k
n
f
(
q
k
n
t
+
(
1
-
q
k
n
)
t
k
)
for t
∈
J
k. Moreover, if a
∈
(
t
k
,
t
), then the definite q
k-integral is defined by
(6)
∫
a
t
‍
f
(
s
)
d
q
k
s
=
∫
t
k
t
‍
f
(
s
)
d
q
k
s
-
∫
t
k
a
‍
f
(
s
)
d
q
k
s
=
(
1
-
q
k
)
(
t
-
t
k
)
∑
n
=
0
∞
‍
q
k
n
f
(
q
k
n
t
+
(
1
-
q
k
n
)
t
k
)
-
(
1
-
q
k
)
(
a
-
t
k
)
∑
n
=
0
∞
‍
q
k
n
f
(
q
k
n
a
+
(
1
-
q
k
n
)
t
k
)
.Note that ift
k
=
0 and q
k
=
q, then (5) reduces to q-integral of a function f
(
t
), defined by
(7)
∫
0
t
‍
f
(
s
)
d
q
s
=
(
1
-
q
)
t
∑
n
=
0
∞
‍
q
n
f
(
q
n
t
)
for
t
∈
[
0
,
∞
)
.For the basic properties ofq
k-derivative and q
k-integral we refer to [1].LetJ
=
[
0
,
T
], J
0
=
[
t
0
,
t
1
], and J
k
=
(
t
k
,
t
k
+
1
] for k
=
1,2
,
…
,
m. Let PC
(
J
,
ℝ
)
=
{
x
:
J
→
ℝ
:
x
(
t
) be continuous everywhere except for some t
k at which x
(
t
k
+
) and x
(
t
k
-
) exist and x
(
t
k
-
)
=
x
(
t
k
), k
=
1,2
,
…
,
m
}.  PC
(
J
,
ℝ
) is a Banach space with the norm ∥
x
∥
PC
=
sup
⁡
{
|
x
(
t
)
|
;
t
∈
J
}.We now consider the following linear case:(8)
D
q
k
x
(
t
)
=
h
(
t
)
,
t
∈
[
0
,
T
]
,
t
≠
t
k
,
Δ
x
(
t
k
)
=
I
k
(
x
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
α
x
(
0
)
=
β
x
(
T
)
+
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
x
(
s
)
d
q
i
s
,
where h
:
J
→
ℝ.Lemma 4.
Letα
≠
β
+
∑
i
=
0
m
‍
γ
i
(
t
i
+
1
-
t
i
). The unique solution of problem (8) is given by
(9)
x
(
t
)
=
β
Ω
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
β
Ω
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
1
Ω
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
h
(
s
)
d
q
i
s
d
q
i
u
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∑
0
<
t
k
<
t
‍
(
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
+
∫
t
k
t
‍
h
(
s
)
d
q
k
s
,
with ∑
i
=
a
b
‍
(
·
)
=
0 for a
>
b, where
(10)
Ω
=
1
α
-
β
-
∑
i
=
0
m
γ
i
(
t
i
+
1
-
t
i
)
.Proof.
Fort
∈
J
0,  q
0-integrating (8), it follows
(11)
x
(
t
)
=
x
0
+
∫
0
t
‍
h
(
s
)
d
q
0
s
,
which leads to
(12)
x
(
t
1
)
=
x
0
+
∫
0
t
1
‍
h
(
s
)
d
q
0
s
.
For t
∈
J
1, taking q
1-integral to (8), we have
(13)
x
(
t
)
=
x
(
t
1
+
)
+
∫
t
1
t
‍
h
(
s
)
d
q
1
s
.
Since x
(
t
1
+
)
=
x
(
t
1
)
+
I
1
(
x
(
t
1
)
), then we have
(14)
x
(
t
)
=
x
0
+
∫
0
t
1
‍
h
(
s
)
d
q
0
s
+
∫
t
1
t
‍
h
(
s
)
d
q
1
s
+
I
1
(
x
(
t
1
)
)
.
Again q
2-integrating (8) from t
2 to t, where t
∈
J
2, then
(15)
x
(
t
)
=
x
(
t
2
+
)
+
∫
t
2
t
‍
h
(
s
)
d
q
2
s
=
x
0
+
∫
0
t
1
‍
h
(
s
)
d
q
0
s
+
∫
t
1
t
2
‍
h
(
s
)
d
q
1
s
+
∫
t
2
t
‍
h
(
s
)
d
q
2
s
+
I
1
(
x
(
t
1
)
)
+
I
2
(
x
(
t
2
)
)
.
Repeating the above process, for t
∈
J, we obtain
(16)
x
(
t
)
=
x
0
+
∑
0
<
t
k
<
t
‍
(
∫
t
k
-
1
t
k
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
+
∫
t
k
t
‍
h
(
s
)
d
q
k
s
.
In particular, for t
=
T, we have
(17)
x
(
T
)
=
x
0
+
∑
k
=
1
m
‍
(
∫
t
k
-
1
t
k
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
+
∫
t
m
T
‍
h
(
s
)
d
q
m
s
.
Further, q
i-integrating (16) from t
i to t
i
+
1, it follows
(18)
∫
t
i
t
i
+
1
x
(
u
)
d
q
i
u
=
x
0
(
t
i
+
1
-
t
i
)
+
∑
k
=
1
i
‍
(
∫
t
k
-
1
t
k
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
(
t
i
+
1
-
t
i
)
+
∫
t
i
t
i
+
1
‍
∫
t
i
u
h
(
s
)
d
q
i
s
d
q
i
u
.
Applying the boundary condition of (8), one has
(19)
α
x
0
=
β
x
0
+
β
∑
k
=
1
m
‍
(
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
+
β
∫
t
m
T
‍
h
(
s
)
d
q
m
s
+
x
0
∑
i
=
0
m
‍
γ
i
(
t
i
+
1
-
t
i
)
+
∑
i
=
0
m
‍
γ
i
∑
k
=
1
i
‍
(
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
)
(
t
i
+
1
-
t
i
)
+
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
h
(
s
)
d
q
i
s
d
q
i
u
=
β
x
0
+
β
∑
k
=
1
m
‍
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
β
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
β
∫
t
m
T
‍
h
(
s
)
d
q
m
s
+
x
0
∑
i
=
0
m
‍
γ
i
(
t
i
+
1
-
t
i
)
+
∑
i
=
0
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
∑
i
=
0
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
h
(
s
)
d
q
i
s
d
q
i
u
.
Since T
=
t
m
+
1 and ∑
i
=
a
b
‍
(
·
)
=
0 for a
>
b, we have
(20)
x
0
(
α
-
β
-
∑
i
=
0
m
‍
γ
i
(
t
i
+
1
-
t
i
)
)
=
β
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
β
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
h
(
s
)
d
q
i
s
d
q
i
u
.
Therefore,
(21)
x
0
=
β
Ω
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
β
Ω
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
1
Ω
∑
i
=
0
m
‍
γ
i
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
h
(
s
)
d
q
i
s
d
q
i
u
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
∫
t
k
-
1
t
k
‍
h
(
s
)
d
q
k
-
1
s
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
.
Substituting the constant x
0 into (16), we obtain (9) as requested.


## 3. Main Results

In view of Lemma4, we define an operator 𝒦
:
PC
(
J
,
ℝ
)
→
PC
(
J
,
ℝ
) by
(22)
(
𝒦
x
)
(
t
)
=
β
Ω
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
β
Ω
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
1
Ω
∑
i
=
0
m
‍
γ
i
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
d
q
i
s
d
q
i
u
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∫
t
k
t
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
x
)
(
s
)
)
d
q
k
s
+
∑
0
<
t
k
<
t
(
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
∫
t
k
-
1
t
k
‍
)
.It should be noticed that problem (1) has solutions if and only if the operator 𝒦 has fixed points.Our first result is an existence and uniqueness result for the impulsive boundary value problem (1) by using Banach's contraction mapping principle.Further, for convenience we set(23)
Λ
1
=
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
[
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
(
L
1
+
L
2
)
(
t
k
-
t
k
-
1
)
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
+
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
[
ϕ
0
L
3
(
t
i
+
1
-
t
i
)
3
1
+
q
i
+
q
i
2
(
L
1
+
L
2
)
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
+
ϕ
0
L
3
(
t
i
+
1
-
t
i
)
3
1
+
q
i
+
q
i
2
]
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
[
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
(
L
1
+
L
2
)
(
t
k
-
t
k
-
1
)
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
L
4
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
(
|
β
|
+
|
Ω
|
)
L
4
|
Ω
|
,
(24)
Λ
2
=
|
β
|
+
|
Ω
|
|
Ω
|
M
1
∑
k
=
1
m
+
1
‍
(
t
k
-
t
k
-
1
)
+
M
1
|
Ω
|
∑
i
=
0
m
|
γ
i
|
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
M
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
t
k
-
t
k
-
1
)
+
M
2
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
(
|
β
|
+
|
Ω
|
)
M
2
|
Ω
|
.Theorem 5.
Assume the following.(
H
1)
The functionϕ
:
J
2
×
ℝ
→
ℝ is continuous and there exists a constant ϕ
0
>
0 such that (25)
|
ϕ
(
t
,
s
,
y
)
-
ϕ
(
t
,
s
,
z
)
|
≤
ϕ
0
|
y
-
z
|
,
for each t
,
s
∈
J and y
,
z
∈
ℝ.
(H
2)
The functionf
:
J
×
ℝ
3
→
ℝ is continuous and there exist constants L
1
,
L
2
,
L
3
>
0 such that (26)
|
f
(
t
,
y
1
,
y
2
,
y
3
)
-
f
(
t
,
z
1
,
z
2
,
z
3
)
|
≤
L
1
|
y
1
-
z
1
|
+
L
2
|
y
2
-
z
2
|
+
L
3
|
y
3
-
z
3
|
,
for each t
∈
J and y
i
,
z
i
,
∈
ℝ,  i
=
1,2
,
3.
(H
3)
The functionsI
k
:
ℝ
→
ℝ are continuous and there exists a constant L
4
>
0 such that (27)
|
I
k
(
y
)
-
I
k
(
z
)
|
≤
L
4
|
y
-
z
|
,
for each y
,
z
∈
ℝ, k
=
1,2
,
…
,
m.
If(28)
Λ
1
≤
δ
<
1
,
where Λ
1 is defined by (23), then the boundary value problem (1) has a unique solution on J.Proof.
Firstly, we transform the boundary value problem (1) into a fixed point problem, x
=
𝒦
x, where the operator 𝒦 is defined by (22). Using the Banach contraction principle, we will show that 𝒦 has a fixed point which is the unique solution of the boundary value problem (1).
LetM
1 and M
2 be nonnegative constants such that sup
⁡
t
∈
J
|
f
(
t
,
0,0
,
0
)
|
=
M
1 and sup
⁡
{
|
I
k
(
0
)
|
:
k
=
1,2
,
…
,
m
}  =  M
2. By choosing a positive constant r as
(29)
r
≥
Λ
2
1
-
ɛ
,
where δ
≤
ɛ
<
1 and Λ
2 defined by (24), we will show that 𝒦
B
r
⊂
B
r, where a suitable ball B
r is defined by B
r
=
{
x
∈
PC
(
J
,
ℝ
)
:
∥
x
∥
≤
r
}. For x
∈
B
r, we have
(30)
|
𝒦
x
(
t
)
|
≤
|
β
|
|
Ω
|
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
|
β
|
|
Ω
|
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
|
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
|
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
|
I
k
(
x
(
t
k
)
)
|
+
∫
t
k
t
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
x
)
(
s
)
)
|
d
q
k
s
+

∑
0
<
t
k
<
t
‍
(

∫
t
k
-
1
t
k
‍

|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s

+
|
I
k
(
x
(
t
k
)
)
|
∫
t
k
-
1
t
k
‍
)
≤
|
β
|
+
|
Ω
|
|
Ω
|
×
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
|
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
|
I
k
(
x
(
t
k
)
)
|
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
|
≤
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k

(
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
-
f
(
s
,
0,0
,
0
)
|

+
|
f
(
s
,
0,0
,
0
)
|

(
S
q
k
-
1
x
)
(
S
q
k
-
1
x
)
)
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍

(
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
-
f
(
s
,
0,0
,
0
)
|
+
|
f
(
s
,
0,0
,
0
)
|

(
S
q
i
x
)
)
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍

(
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
-
f
(
s
,
0,0
,
0
)
(
S
q
k
-
1
x
)
|
+
|
f
(
s
,
0,0
,
0
)
|
(
S
q
k
-
1
x
)
)
d
q
k
-
1
s
+

1
|
Ω
|
∑
i
=
1
m
‍

∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
|
I
k
(
x
(
t
k
)
)
-
I
k
(
0
)
|

+
|
I
k
(
0
)
|
)
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
(
|
I
k
(
x
(
t
k
)
)
-
I
k
(
0
)
|
+
|
I
k
(
0
)
|
)
≤
|
β
|
+
|
Ω
|
|
Ω
|
×
∑
k
=
1
m
+
1

‍
∫
t
k
-
1
t
k
‍

[
r
(

L
1
+
L
2
+
ϕ
0
L
3
∫
t
k
-
1
s
‍

d
q
k
-
1
u
)

+
M
1
]
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|

×
∫
t
i
t
i
+
1
‍

∫
t
i
u
‍
[
r
(
L
1
+
L
2
+
ϕ
0
L
3
∫
t
i
s
‍
d
q
i
v
)
+
M
1
]
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
[
r
(
L
1
+
L
2
+
ϕ
0
L
3
∫
t
k
-
1
s
‍
d
q
k
-
1
u
)
+
M
1
]
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
r
L
4
+
M
2
)
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
(
r
L
4
+
M
2
)
=
r
Λ
1
+
Λ
2
≤
r
,
which yields 𝒦
B
r
⊂
B
r.
For anyx
,
y
∈
PC
(
J
,
ℝ
) and for each t
∈
J, we have
(31)
|
𝒦
x
(
t
)
-
𝒦
y
(
t
)
|
≤
|
β
|
+
|
Ω
|
|
Ω
|
×
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍

|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
-
f
(
s
,
y
(
s
)
,
y
(
θ
(
s
)
)
,
(
S
q
k
-
1
y
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
+
∑
i
=
0
m
‍
|
γ
i
|
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
-
f
(
s
,
y
(
s
)
,
y
(
θ
(
s
)
)
,
(
S
q
i
y
)
(
s
)
)
|
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
×
∫
t
k
-
1
t
k
‍
-
f
(
s
,
y
(
s
)
,
y
(
θ
(
s
)
)
,
(
S
q
k
-
1
y
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
|
I
k
(
x
(
t
k
)
)
-
I
k
(
y
(
t
k
)
)
|
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
-
I
k
(
y
(
t
k
)
)
|
≤
|
β
|
+
|
Ω
|
|
Ω
|
∥
x
-
y
∥
×
∑
k
=
1
m
+
1
‍
[
(
L
1
+
L
2
)
(
t
k
-
t
k
-
1
)
+
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
∥
x
-
y
∥
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
[
(
L
1
+
L
2
)
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
∥
x
-
y
∥
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
+
ϕ
0
L
3
(
t
i
+
1
-
t
i
)
3
1
+
q
i
+
q
i
2
]
+
∥
x
-
y
∥
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
[
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
(
L
1
+
L
2
)
(
t
k
-
t
k
-
1
)
+
∥
x
-
y
∥
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
ϕ
0
L
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
L
4
∥
x
-
y
∥
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
(
|
β
|
+
|
Ω
|
)
L
4
|
Ω
|
∥
x
-
y
∥
=
Λ
1
∥
x
-
y
∥
,
which implies that ∥
𝒦
x
-
𝒦
y
∥
≤
Λ
1
∥
x
-
y
∥. Since Λ
1
<
1, 𝒦 is a contraction. Therefore, by Banach's contraction mapping principle, we conclude that 𝒦 has a fixed point which is the unique solution of problem (1).The second existence result is based on Krasnoselskii's fixed point theorem.Lemma 6 ((Krasnoselskii's fixed point theorem) [19]).
LetM be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A, B be the operators such that (a) A
x
+
B
y
∈
M whenever x
,
y
∈
M; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists z
∈
M such that z
=
A
z
+
B
z.We use the following notations:(32)
Λ
3
=
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
(
t
k
-
t
k
-
1
)
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
t
k
-
t
k
-
1
)
,
(33)
Λ
4
=
m
|
β
|
N
|
Ω
|
+
N
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
N
.Theorem 7.
Letf
:
J
×
ℝ
→
ℝ be a continuous function. Suppose that (
H
3
) holds. In addition, we suppose the following: (
H
4)
|
f
(
t
,
z
1
,
z
2
,
z
3
)
|
≤
μ
(
t
), ∀
(
t
,
z
1
,
z
2
,
z
3
)
∈
J
×
ℝ
3, and μ
∈
C
(
J
,
ℝ
+
),
(H
5)
there exists a constantN
>
0 such that |
I
k
(
x
)
|
≤
N for all x
∈
ℝ,  for k
=
1,2
,
…
,
m.Then the impulsive functional q
k-integrodifference boundary value problem (1) has at least one solution on J provided that
(34)
(
|
β
|
+
|
Ω
|
|
Ω
|
)
m
L
4
+
L
4
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
<
1
.Proof.
Letsup
⁡
t
∈
J
|
μ
(
t
)
|
=
∥
μ
∥. By choosing a suitable ball B
R
=
{
x
∈
PC
(
J
,
ℝ
)
:
∥
x
∥
≤
R
}, where
(35)
R
≥
∥
μ
∥
Λ
3
+
Λ
4
,
and Λ
3, Λ
4 are defined by (32) and (33), respectively; we define the operators 𝒜
1 and 𝒜
2 on B
R by
(36)
(
𝒜
1
x
)
(
t
)
=
β
Ω
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
1
Ω
∑
i
=
0
m
‍
γ
i
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
d
q
i
s
d
q
i
u
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
∑
0
<
t
k
<
t
‍
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
∫
t
k
t
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
x
)
(
s
)
)
d
q
k
s
,
(
𝒜
2
x
)
(
t
)
=
β
Ω
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∑
0
<
t
k
<
t
‍
I
k
(
x
(
t
k
)
)
.
For any x
,
y
∈
B
R, we have
(37)
∥
𝒜
1
x
+
𝒜
2
y
∥
≤
∥
μ
∥
[
|
β
|
|
Ω
|
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
∫
t
k
-
1
t
k
‍
d
q
k
-
1
s
+
∑
k
=
1
m
‍
∫
t
k
-
1
t
k
‍
d
q
k
-
1
s
+
∫
t
m
T
‍
d
q
k
s
|
β
|
|
Ω
|
∑
k
=
1
m
+
1
‍
]
+
m
|
β
|
N
|
Ω
|
+
N
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
N
=
∥
μ
∥
[
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
(
t
k
-
t
k
-
1
)
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
t
k
-
t
k
-
1
)
]
+
m
|
β
|
N
|
Ω
|
+
N
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
+
m
N
≤
R
.
This implies that 𝒜
1
x
+
𝒜
2
y
∈
B
R.
To show that𝒜
2 is a contraction, for x
,
y
∈
PC
(
J
,
ℝ
), we have
(38)
∥
𝒜
2
x
-
𝒜
2
y
∥
≤
|
β
|
|
Ω
|
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
-
I
k
(
y
(
t
k
)
)
|
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
|
I
k
(
x
(
t
k
)
)
-
I
k
(
y
(
t
k
)
)
|
+
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
-
I
k
(
y
(
t
k
)
)
|
≤
[
(
|
β
|
+
|
Ω
|
|
Ω
|
)
m
L
4
+
L
4
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
]
∥
x
-
y
∥
.
From (34), it follows that 𝒜
2 is a contraction.
Next, the continuity off implies that operator 𝒜
1 is continuous. Further, 𝒜
1 is uniformly bounded on B
R by
(39)
∥
𝒜
1
x
∥
≤
∥
μ
∥
Λ
3
.
Now we will prove the compactness of 𝒜
1. Setting sup
⁡
(
t
,
z
1
,
z
2
,
z
3
)
∈
J
×
B
R
3
|
f
(
t
,
z
1
,
z
2
,
z
3
)
|
=
f
*
<
∞, then for each τ
1
,
τ
2
∈
(
t
l
,
t
l
+
1
) for some l
∈
{
0,1
,
…
,
m
} with τ
2
>
τ
1, we have
(40)
|
(
𝒜
1
x
)
(
τ
2
)
-
(
𝒜
1
x
)
(
τ
1
)
|
=
|
∫
t
l
τ
2
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
l
x
)
(
s
)
)
d
q
l
s
-
∫
t
l
τ
1
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
l
x
)
(
s
)
)
d
q
l
s
|
≤
|
τ
2
-
τ
1
|
f
*
.
As τ
1
→
τ
2, the right hand side above tends to zero independently on x. Therefore, the operator 𝒜
1 is equicontinuous. Since 𝒜
1 maps bounded subsets into relatively compact subsets, it follows that 𝒜
1 is relative compact on B
R. Hence, by the Arzelá-Ascoli theorem, 𝒜
1 is compact on B
R. Thus, all the assumptions of Lemma 6 are satisfied. Hence, by the conclusion of Lemma 6, the impulsive functional q
k-integrodifference boundary value problem (1) has at least one solution on J.Our third existence result is based on Leray-Schauder degree theory. Before proving the result, we set(41)
Λ
5
=
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
[
ξ
1
(
t
k
-
t
k
-
1
)
+
ξ
2
ξ
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
[
ξ
1
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
ξ
2
ξ
3
(
t
i
+
1
-
t
i
)
3
1
+
q
i
+
q
i
2
]
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
[
ξ
1
(
t
k
-
t
k
-
1
)
+
ξ
2
ξ
3
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
1
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
ξ
4
+
m
(
|
β
|
+
|
Ω
|
)
ξ
4
|
Ω
|
,
(42)
Λ
6
=
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
+
1
‍
[
ξ
2
Q
2
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
+
Q
1
(
t
k
-
t
k
-
1
)
]
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
[
ξ
2
Q
2
(
t
i
+
1
-
t
i
)
3
1
+
q
i
+
q
i
2
+
Q
1
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
[
ξ
2
Q
2
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
+
Q
1
(
t
k
-
t
k
-
1
)
(
t
k
-
t
k
-
1
)
2
1
+
q
k
-
1
]
+
L
4
|
Ω
|
∑
i
=
1
m
‍
i
|
γ
i
|
(
t
i
+
1
-
t
i
)
Q
3
+
m
(
|
β
|
+
|
Ω
|
)
Q
3
|
Ω
|
.Theorem 8.
Assume thatf
:
J
×
ℝ
3
→
ℝ and ϕ
:
J
2
×
ℝ
→
ℝ are continuous functions. In addition we suppose the following. (H
6)
There exist constantsξ
1
,
ξ
2
>
0 and Q
1
≥
0 such that
(43)
|
f
(
t
,
z
1
,
z
2
,
z
3
)
|
≤
ξ
1
|
z
1
|
+
ξ
2
|
z
3
|
+
Q
1
h
h
h
h
h
h
h
h
h
h
h
∀
(
t
,
z
1
,
z
2
,
z
3
)
∈
J
×
ℝ
3
.
(H
7)
There exist constantsξ
3
>
0 and Q
2
≥
0 such that
(44)
|
ϕ
(
t
,
s
,
z
)
|
≤
ξ
3
|
z
|
+
Q
2
∀
(
t
,
s
,
z
)
∈
J
2
×
ℝ
.
(H
8)
There exist constantsξ
4
>
0 and Q
3
≥
0 such that
(45)
|
I
k
(
z
)
|
≤
ξ
4
|
z
|
+
Q
3
∀
z
∈
ℝ
,
k
=
1,2
,
…
,
m
.If
(46)
Λ
5
<
1
,
where Λ
5 is given by (41), then the impulsive functional q
k-integrodifference boundary value problem (1) has at least one solution on J.Proof.
We define an operator𝒦
:
PC
(
J
,
ℝ
)
→
PC
(
J
,
ℝ
) as in (22) and consider the fixed point problem:
(47)
x
=
𝒦
x
.
We are going to prove that there exists a fixed point x
∈
PC
(
J
,
ℝ
) satisfying (47). It is sufficient to show that 𝒦
:
B
¯
ρ
→
PC
(
J
,
ℝ
) satisfies
(48)
x
≠
λ
𝒦
x
,
∀
x
∈
∂
B
ρ
,
∀
λ
∈
[
0,1
]
,
where B
ρ
=
{
x
∈
PC
(
J
,
ℝ
)
:
max
⁡
t
∈
J
|
x
(
t
)
|
<
ρ
,
ρ
>
0
}. We define
(49)
H
(
λ
,
x
)
=
λ
𝒦
x
,
x
∈
PC
(
J
,
ℝ
)
,
λ
∈
[
0,1
]
.
It is easy to see that the operator 𝒦 is continuous, uniformly bounded, and equicontinuous. Then, by the Arzelá-Ascoli Theorem, a continuous map h
λ defined by h
λ
(
x
)
=
x
-
H
(
λ
,
x
)
=
x
-
λ
𝒦
x is completely continuous. If (48) is true, then the following Leray-Schauder degrees are well defined and by the homotopy invariance of topological degree, it follows that
(50)
deg
⁡
(
h
λ
,
B
ρ
,
0
)
=
deg
⁡
(
I
-
λ
𝒦
,
B
ρ
,
0
)
=
deg
⁡
(
h
1
,
B
ρ
,
0
)
=
deg
⁡
(
h
0
,
B
ρ
,
0
)
=
deg
⁡
(
I
,
B
ρ
,
0
)
=
1
≠
0
,
0
∈
B
ρ
,
where I denotes the identity operator. By the nonzero property of Leray-Schauder degree, h
1
(
x
)
=
x
-
𝒦
x
=
0 for at least one x
∈
B
ρ. In order to prove (48), we assume that x
=
λ
𝒦
x for some λ
∈
[
0,1
]. Then
(51)
|
𝒦
x
(
t
)
|
≤
sup
⁡
t
∈
J
{
β
Ω
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
β
Ω
∑
k
=
1
m
‍
I
k
(
x
(
t
k
)
)
+
1
Ω
∑
i
=
0
m
‍
γ
i
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
d
q
i
s
d
q
i
u
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
1
Ω
∑
i
=
1
m
‍
∑
k
=
1
i
‍
γ
i
(
t
i
+
1
-
t
i
)
I
k
(
x
(
t
k
)
)
+
∫
t
k
t
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
x
)
(
s
)
)
d
q
k
s
+

∑
0
<
t
k
<
t
‍

(

∫
t
k
-
1
t
k
‍
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
d
q
k
-
1
s
+
I
k
(
x
(
t
k
)
)
∫
t
k
-
1
t
k
‍

)

∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
}
≤
|
β
|
+
|
Ω
|
|
Ω
|
×
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
i
x
)
(
s
)
)
|
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍
|
f
(
s
,
x
(
s
)
,
x
(
θ
(
s
)
)
,
(
S
q
k
-
1
x
)
(
s
)
)
|
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
|
I
k
(
x
(
t
k
)
)
|
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
|
I
k
(
x
(
t
k
)
)
|
≤
|
β
|
+
|
Ω
|
|
Ω
|
×
∑
k
=
1
m
+
1
‍
∫
t
k
-
1
t
k
‍
(
∫
t
k
-
1
s
‍
ξ
1
∥
x
∥
+
ξ
2
∫
t
k
-
1
s
‍
(
ξ
3
∥
x
∥
+
Q
2
)
d
q
k
-
1
v
+
Q
1
)
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
0
m
‍
|
γ
i
|
×
∫
t
i
t
i
+
1
‍
∫
t
i
u
‍
(
∫
t
k
-
1
s
‍
ξ
1
∥
x
∥
+
ξ
2
∫
t
k
-
1
s
‍
(
ξ
3
∥
x
∥
+
Q
2
)
d
q
k
-
1
v
+
Q
1

)
d
q
i
s
d
q
i
u
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
×
∫
t
k
-
1
t
k
‍

(

ξ
1
∥
x
∥
+
ξ
2
∫
t
k
-
1
s
‍

(
ξ
3
∥
x
∥
+
Q
2
)
d
q
k
-
1
v
+
Q
1
)
d
q
k
-
1
s
+
1
|
Ω
|
∑
i
=
1
m
‍
∑
k
=
1
i
‍
|
γ
i
|
(
t
i
+
1
-
t
i
)
(
ξ
4
∥
x
∥
+
Q
3
)
+
|
β
|
+
|
Ω
|
|
Ω
|
∑
k
=
1
m
‍
(
ξ
4
∥
x
∥
+
Q
3
)
=
Λ
5
∥
x
∥
+
Λ
6
,
which implies that
(52)
∥
x
∥
≤
Λ
6
1
-
Λ
5
.
If ρ
=
Λ
6
/
(
1
-
Λ
5
)
+
1, inequality (48) holds. This completes the proof.


## 4. Examples

In this section, we will give some examples to illustrate our main results.Example 1.
Consider the following boundary value problem for nonlinear first-order impulsive functionalq
k-integrodifference equation:
(53)
D
(
1
/
2
)
sin
⁡
(
(
(
k
+
1
)
/
6
)
π
)
x
(
t
)
=
t
2
sin
⁡
π
t
(
2
t
+
4
)
2
|
x
|
|
x
|
+
1
-
3
t
x
(
t
/
2
)
2
(
t
+
3
)
2
+
t
2
2
(
e
t
+
1
)
2
∫
t
k
t
‍
2
t
-
s
4
e
t
-
s
x
(
s
)
d
(
1
/
2
)
sin
⁡
(
(
(
k
+
1
)
/
6
)
π
)
s
,
+
t
2
2
(
e
t
+
1
)
2
∫
t
k
t
‍
2
t
-
s
4
e
t
-
s
x
(
s
)
t
∈
J
,
t
≠
t
k
,
Δ
x
(
t
k
)
=
|
x
(
t
k
)
|
2
(
k
+
3
)
+
|
x
(
t
k
)
|
,
t
k
=
k
5
,
k
=
1,2
,
…
,
4
,
x
(
0
)
=
1
3
x
(
1
)
+
∑
i
=
0
4
‍
(
1
i
+
2
)
∫
t
i
t
i
+
1
‍
x
(
s
)
d
(
1
/
2
)
sin
⁡
(
(
(
i
+
1
)
/
6
)
π
)
s
.SetJ
=
[
0,1
],  q
k
=
(
1
/
2
)
sin
(
(
k
+
1
)
π
/
6
) for k
=
0,1
,
…
,
4, γ
i
=
1
/
(
i
+
2
) for i
=
0,1
,
…
,
4, m
=
4, T
=
1, θ
(
t
)
=
t
/
2,
(54)
f
(
t
,
x
,
x
(
θ
)
,
(
S
q
k
x
)
)
=
t
2
sin
⁡
π
t
(
2
t
+
4
)
2
|
x
|
|
x
|
+
1
-
3
t
x
(
t
/
2
)
2
(
t
+
3
)
2
+
t
2
2
(
e
t
+
1
)
2
∫
t
k
t
‍
2
t
-
s
4
e
t
-
s
x
(
s
)
d
(
1
/
2
)
sin
(
(
(
k
+
1
)
/
6
)
π
)
s
,
and I
k
(
x
)
=
|
x
(
t
k
)
|
/
(
2
(
k
+
3
)
+
|
x
(
t
k
)
|
).Since(55)
|
ϕ
(
t
,
s
,
y
)
-
ϕ
(
t
,
s
,
z
)
|
≤
1
2
|
y
-
z
|
,
|
f
(
t
,
y
1
,
y
2
,
y
3
)
-
f
(
t
,
z
1
,
z
2
,
z
3
)
|
≤
1
9
|
y
1
-
z
1
|
+
1
6
|
y
2
-
z
2
|
+
1
8
|
y
3
-
z
3
|
,
|
I
k
(
y
)
-
I
k
(
z
)
|
≤
1
8
|
y
-
z
|
,
then (
H
1
)
–
(
H
3
) are satisfied with ϕ
0
=
1
/
2, L
1
=
1
/
9, L
2
=
1
/
6, L
3
=
1
/
8, and L
4
=
1
/
8. We can show that Λ
1
≈
0.9517257476
<
1. Hence, by Theorem 5, the boundary value problem (53) has a unique solution on [
0,1
].Example 2.
Consider the following boundary value problem for nonlinear first-order impulsive functionalq
k-integrodifference equation:
(56)
D
(
k
+
1
)
/
e
k
+
1
x
(
t
)
=
t
2
cos
⁡
π
t
(
4
t
+
3
)
2
x
x
2
+
2
+
(
t
+
2
)
2
sin
2
(
x
(
2
t
3
)
)
+
5
cos
⁡
⁡
π
t
3
(
e
t
+
4
)
2
∫
t
k
t
‍
sin
2
(
t
-
s
)
(
e
t
-
s
+
1
)
2
x
(
s
)
d
(
k
+
1
)
/
e
k
+
1
s
,
+
5
cos
⁡
⁡
π
t
3
(
e
t
+
4
)
2
∫
t
k
t
‍
sin
2
(
t
-
s
)
(
e
t
-
s
+
1
)
2
t
∈
J
=
[
0,1
]
,
t
≠
t
k
,
Δ
x
(
t
k
)
=
|
x
(
t
k
)
|
5
(
k
+
4
)
+
|
x
(
t
k
)
|
+
3
sin
⁡
π
x
(
t
k
)
,
|
x
(
t
k
)
|
5
(
k
+
4
)
+
|
x
(
t
k
)
|
t
k
=
k
9
,
k
=
1,2
,
…
,
8
,
2
x
(
0
)
=
1
4
x
(
1
)
+
∑
i
=
0
8
‍
(
i
+
1
i
+
4
)
∫
t
i
t
i
+
1
‍
x
(
s
)
d
(
i
+
1
)
/
e
i
+
1
s
.Setq
k
=
(
k
+
1
)
/
(
e
k
+
1
) for k
=
0,1
,
…
,
8,  γ
i
=
(
i
+
1
)
/
(
i
+
4
) for i
=
0,1
,
…
,
8,   m
=
8,  T
=
1,
(57)
f
(
t
,
x
,
x
(
θ
)
,
(
S
q
k
x
)
)
=
t
2
cos
⁡
⁡
π
t
(
4
t
+
3
)
2
x
x
2
+
2
+
(
t
+
2
)
2
sin
2
(
x
(
2
t
3
)
)
+
5
cos
⁡
⁡
π
t
3
(
e
t
+
4
)
2
∫
t
k
t
‍
sin
2
(
t
-
s
)
(
e
t
-
s
+
1
)
2
x
(
s
)
d
(
k
+
1
)
/
e
k
+
1
s
,
θ
(
t
)
=
2
t
/
3, and I
k
(
x
)
=
(
|
x
(
t
k
)
|
/
(
5
(
k
+
5
)
+
|
x
(
t
k
)
|
)
)
+
3
sin
⁡
π
x
(
t
k
).Since(58)
|
f
(
t
,
x
,
x
(
θ
)
,
(
S
q
k
x
)
)
|
≤
1
18
|
x
|
+
1
15
|
S
q
k
x
|
+
9
,
|
ϕ
(
t
,
s
,
x
)
|
≤
(
1
/
4
)
|
x
|, and |
I
k
(
x
)
|
≤
(
1
/
25
)
|
x
|
+
3, then (
H
6
)
–
(
H
8
) are satisfied with ξ
1
=
1
/
18, ξ
2
=
1
/
15, ξ
3
=
1
/
4, ξ
4
=
1
/
25, Q
1
=
9, Q
2
=
0, and Q
3
=
3. We can show that Λ
5
≈
0.9134109736
<
1. Hence, by Theorem 8, the boundary value problem (56) has at least one solution on [
0,1
].



---

*Source: 374565-2014-03-20.xml*