# Nonlocal Boundary Value Problem for Nonlinear Impulsiveq

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

---

## Abstract

A nonlinear impulsive integrodifference equation within the frame ofq
k-quantum calculus is investigated by applying using fixed point theorems. The conditions for existence and uniqueness of solutions are obtained.


---

## Body

## 1. Introduction

Recently, by introducing and applying the fractional difference operators to real world problems (see, e.g., [1–7] and the references therein) we revitalized the importance of the quantum calculus [8]. However the real world phenomena are usually described by complex model based involving different types of operators. In this way we hope to understand deeper the dynamics of complex or hypercomplex systems and to reveal their hidden aspects.On this line of thought in this paper, we study the existence and uniqueness of solutions for nonlinearq
k-integrodifference equation with nonlocal boundary condition and impulses:
(1)
D
q
k
u
(
t
)
=
f
(
t
,
u
(
t
)
)
+
I
t
k
q
k
g
(
t
,
u
(
t
)
)
,
000000000
0
<
q
k
<
1
,
t
∈
J
′
,
Δ
u
(
t
k
)
=
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
u
(
0
)
=
h
(
u
)
+
u
0
,
u
0
∈
ℝ
,
where D
q
k, I
t
k
q
k are q
k-derivatives and q
k-integrals (
k
=
0,1
,
2
,
…
,
m
), respectively. f, g
∈
C
(
J
×
ℝ
,
ℝ
), I
k, h
∈
C
(
ℝ
,
ℝ
), J
=
[
0
,
T
]
(
T
>
0
), 0
=
t
0
<
t
1
<
⋯
<
t
k
<
⋯
<
t
m
<
t
m
+
1
=
T, J
′
=
J
∖
{
t
1
,
t
2
,
…
,
t
m
}, 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
,
…
,
m
), respectively.


## 2. Preliminaries

Let us setJ
0
=
[
0
,
t
1
], J
1
=
(
t
1
,
t
2
]
,
…
,
J
m
-
1
=
(
t
m
-
1
,
t
m
], J
m
=
(
t
m
,
T
] and introduce the space:
(2)
P
C
(
J
,
ℝ
)
=
{
u
:
J
⟶
ℝ
∣
u
∈
C
(
J
k
)
,
k
=
0,1
,
…
,
m
,
0000
and
u
(
t
k
+
)
exist
,
k
=
1,2
,
…
,
m
}
,
with the norm ∥
u
∥
=
sup
⁡
t
∈
J
⁡
|
u
(
t
)
|. Then, P
C
(
J
,
ℝ
) is a Banach space.For convenience, let us recall some basic concepts ofq
k-calculus [9].For0
<
q
k
<
1 and t
∈
J
k, we define the q
k-derivatives of a real valued continuous function f as
(3)
D
q
k
f

(
t
)
=
f

(
t
)
-
f

(
q
k
t
+
(
1
-
q
k
)
t
k
)
(
1
-
q
k
)
(
t
-
t
k
)
,
D
q
k
f

(
t
k
)
=
lim
⁡
t
→
t
k
D
q
k
f

(
t
)
.
Higher order q
k-derivatives are given by
(4)
D
q
k
0
f

(
t
)
=
f

(
t
)
,
D
q
k
n
f

(
t
)
=
D
q
k
D
q
k
n
-
1
f

(
t
)
,
000000000000000000000000000
n
∈
ℕ
,
t
∈
J
k
.
The q
k-integral of a function f is defined by
(5)
t
k
I
q
k
f
(
t
)
≔
∫
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
)
,
00000000000000000000
t
∈
J
k
,
provided the series converges. If a
∈
(
t
k
,
t
) and f is defined on the interval (
t
k
,
t
), then
(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
.
Observe that
(7)
D
q
k
(
I
t
k
q
k
f

(
t
)
)
=
D
q
k
∫
t
k
t
f

(
s
)
d
q
k
s
=
f

(
t
)
,
I
t
k
q
k
(
D
q
k
f

(
t
)
)
=
∫
t
k
t
D
q
k
f

(
s
)
d
q
k
s
=
f

(
t
)
,
I
a
q
k
(
D
q
k
f

(
t
)
)
=
∫
a
t
D
q
k
f

(
s
)
d
q
k
s
=
f

(
t
)
-
f

(
a
)
,
0000000000000000000000000000000
a
∈
(
t
k
,
t
)
.
For t
∈
J
k, the following reversing order of q
k-integration holds
(8)
∫
t
k
t
‍
∫
t
k
s
‍
f

(
r
)
d
q
k
r
d
q
k
s
=
∫
t
k
t
‍
∫
q
k
r
+
(
1
-
q
k
)
t
k
t
‍
f

(
r
)
d
q
k
s
d
q
k
r
.Note that ift
k
=
0 and q
k
=
q in (3) and (5), then D
q
k
f
=
D
q
f, I
t
k
q
k
f
=
I
0
q
f, where D
q and I
0
q are the well-known q-derivative and q-integral of the function f
(
t
) defined by
(9)
D
q
f

(
t
)
=
f

(
t
)
-
f

(
q
t
)
(
1
-
q
)
t
,
I
0
q
f

(
t
)
=
∫
0
t
‍
f

(
s
)
d
q
s
=
∑
n
=
0
∞
‍
t
(
1
-
q
)
q
n
f
(
t
q
n
)
.Lemma 1.
For giveny
q
k
∈
C
(
J
,
ℝ
), the function u
∈
P
C
(
J
,
ℝ
) is a solution of the impulsive q
k-integrodifference equation
(10)
D
q
k
u
(
t
)
=
y
q
k
(
t
)
,
0
<
q
k
<
1
,
t
∈
J
′
,
Δ
u
(
t
k
)
=
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
u
(
0
)
=
h
(
u
)
+
u
0
,
u
0
∈
ℝ
,
if and only if u satisfies the q
k-integral equation
(11)
u
(
t
)
=
{
∫
0
t
y
q
0
(
s
)
d
q
0
s
+
h
(
u
)
+
u
0
,
t
∈
J
0
;
∫
t
k
t
y
q
k
(
s
)
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
y
q
i
(
s
)
d
q
i
s
+
∑
i
=
1
k
‍
I
i
(
u
(
t
i
)
)
+
h
(
u
)
+
u
0
,
t
∈
J
k
.Proof.
Letu be a solution of q
k-difference equation (10). For t
∈
J
0, applying the operator I
0
q
0 on both sides of D
q
0
u
(
t
)
=
y
q
0
(
t
), we have
(12)
u
(
t
)
=
u
(
0
)
+
I
0
q
0
y
q
0
(
t
)
=
u
(
0
)
+
∫
0
t
‍
y
q
0
(
s
)
d
q
0
s
.
Thus,
(13)
u
(
t
1
-
)
=
u
(
0
)
+
∫
0
t
1
‍
y
q
0
(
s
)
d
q
0
s
.
Similarly, fort
∈
J
1, applying the operator I
t
1
+
q
1 on both sides of D
q
1
u
(
t
)
=
y
q
1
(
t
), then
(14)
u
(
t
)
=
u
(
t
1
+
)
+
∫
t
1
t
‍
y
q
1
(
s
)
d
q
1
s
.
In view of Δ
u
(
t
1
)
=
u
(
t
1
+
)
-
u
(
t
1
-
)
=
I
1
(
u
(
t
1
)
), it holds
(15)
u
(
t
)
=
u
(
0
)
+
∫
t
1
t
‍
y
q
1
(
s
)
d
q
1
s
+
∫
0
t
1
‍
y
q
0
(
s
)
d
q
0
s
+
I
1
(
u
(
t
1
)
)
,
∀
t
∈
J
1
.
Repeating the above process, we can get(16)
u
(
t
)
=
u
(
0
)
+
∫
t
k
t
‍
y
q
k
(
s
)
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
y
q
i
(
s
)
d
q
i
s
+
∑
i
=
1
k
‍
I
i
(
u
(
t
i
)
)
,
t
∈
J
k
.
Using the boundary value condition given in (10), it follows
(17)
u
(
t
)
=
∫
t
k
t
‍
y
q
k
(
s
)
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
y
q
i
(
s
)
d
q
i
s
+
∑
i
=
1
k
‍
I
i
(
u
(
t
i
)
)
+
h
(
u
)
+
u
0
,
t
∈
J
k
.
Conversely, assume thatu satisfies the impulsive q
k-integral equation (11); applying D
q
k on both sides of (11) and substituting t
=
0 in (11), then (10) holds. This completes the proof.


## 3. Main Results

Lettingy
q
k
(
t
)
=
f
(
t
,
u
(
t
)
)
+
I
t
k
q
k
g
(
t
,
u
(
t
)
), in view of Lemma 1, we introduce an operator Q
:
P
C
(
J
,
ℝ
)
→
P
C
(
J
,
ℝ
) as
(18)
(
Q
u
)
(
t
)
=
∫
t
k
t
[
f
(
s
,
u
(
s
)
)
+
∫
t
k
s
‍
g
(
r
,
u
(
r
)
)
d
q
k
r
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
f
(
s
,
u
(
s
)
)
+
∫
t
i
s
‍
g
(
r
,
u
(
r
)
)
d
q
i
r
]
0000000000
×
(
s
)
d
q
i
s
+
∑
i
=
1
k
‍
I
i
(
u
(
t
i
)
)
+
h
(
u
)
+
u
0
.
By reversing the order of integration, we obtain
(19)
(
Q
u
)
(
t
)
=
∫
t
k
t
[
f
(
s
,
u
(
s
)
)
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
000000
×
g
(
s
,
u
(
s
)
)
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
f
(
s
,
u
(
s
)
)
000000000000
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
000000000000
×
g
(
s
,
u
(
s
)
)
]
d
q
i
s
+
∑
i
=
1
k
‍
I
i
(
u
(
t
i
)
)
+
h
(
u
)
+
u
0
.
Then, the impulsive q
k-integrodifference equation (1) has a solution if and only if the operator equation u
=
Q
u has a fixed point.In order to prove the existence of solutions for (1), we need the following known result [10].Theorem 2.
LetE be a Banach space. Assume that T
:
E
→
E is a completely continuous operator and the set V
=
{
x
∈
E
∣
x
=
μ
T
x
,
0
<
μ
<
1
} is bounded. Then T has a fixed point in E.Theorem 3.
Assume the following.(
H
1
)
There exist nonnegative bounded functionsM
i
(
t
)
(
i
=
1,2
,
3,4
) such that
(20)
|
f
(
t
,
u
)
|
≤
M
1
(
t
)
+
M
2
(
t
)
|
u
|
,
|
g
(
t
,
u
)
|
≤
M
3
(
t
)
+
M
4
(
t
)
|
u
|
,
for any t
∈
J, u
∈
ℝ.
(
H
2
)
There exist positive constantsL
¯
,
L
~ such that
(21)
|
I
k
(
u
)
|
≤
L
¯
,
|
h
(
u
)
|
≤
L
~
,
for any u
∈
ℝ, k
=
1,2
,
…
,
m.Then problem (1) has at least one solution provided
(22)
sup
⁡
t
∈
J
[
T
M
2
(
t
)
+
M
4
(
t
)
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
<
1
.Proof.
Firstly, we prove the operatorQ
:
P
C
(
J
,
ℝ
)
→
P
C
(
J
,
ℝ
) is completely continuous. Clearly, continuity of the operator Q follows from the continuity of f, g, I
k, and h. Let Ω
⊂
P
C
(
J
,
ℝ
) be bounded. Then ∀
t
∈
J,  u
∈
Ω; there exist positive constants L
i
(
i
=
1,2
,
3,4
) such that |
f
(
t
,
u
)
|
≤
L
1, |
g
(
t
,
u
)
|
≤
L
2, |
I
k
(
u
)
|
≤
L
3, |
h
(
u
)
|
≤
L
4. Thus
(23)
|
(
Q
u
)
(
t
)
|
≤
∫
t
k
t
‍
[
|
f
(
s
,
u
(
s
)
)
|
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
00000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
|
f
(
s
,
u
(
s
)
)
|
00000000000
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
00000000000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
i
s
+
∑
i
=
1
k
‍
|
I
i
(
u
(
t
i
)
)
|
+
|
h
(
u
)
|
+
|
u
0
|
≤
∫
t
k
t
[
L
1
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
L
2
]
d
q
k
s
‍
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
L
1
+
[
(
t
i
+
1
-
t
i
)
000000000000000
-
q
i
(
s
-
t
i
)
]
L
2
]
d
q
i
s
+
∑
i
=
1
k
‍
L
3
+
L
4
+
|
u
0
|
≤
L
1
(
t
-
t
k
)
+
L
2
(
t
-
t
k
)
2
1
+
q
k
+
∑
i
=
0
k
-
1
‍
[
L
1
(
t
i
+
1
-
t
i
)
+
L
2
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
+
m
L
3
+
L
4
+
|
u
0
|
≤
T
L
1
+
L
2
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
m
L
3
+
L
4
+
|
u
0
|
≔
ℒ
(
constant
)
.
This implies ∥
Q
u
∥
≤
ℒ.
Furthermore, for anyt
′,  t
′′
∈
J
k
(
k
=
0,1
,
2
,
…
,
m
) satisfying t
′
<
t
′′, we have
(24)
|
(
Q
u
)
(
t
′′
)
-
(
Q
u
)
(
t
′
)
|
≤
|
∫
t
k
t
′′
‍
[
|
f
(
s
,
u
(
s
)
)
|
0000000
+
[
(
t
′′
-
t
k
)
-
q
k
(
s
-
t
k
)
]
0000000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
k
s
000
-
∫
t
k
t
′
‍
[
|
f
(
s
,
u
(
s
)
)
|
000000
+
[
(
t
′
-
t
k
)
-
q
k
(
s
-
t
k
)
]
000000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
k
s
∫
t
k
t
′′
‍
|
≤
∫
t
′
t
′′
‍
|
f
(
s
,
u
(
s
)
)
00000
+
[
(
t
′′
-
t
k
)
-
q
k
(
s
-
t
k
)
]
0000000
×
g
(
s
,
u
(
s
)
)
|
d
q
k
s
0000
+
∫
t
k
t
′
‍
(
t
′′
-
t
′
)
00000000000
×
|
g
(
s
,
u
(
s
)
)
|
d
q
k
s
≤
∫
t
′
t
′′
[
|
f
(
s
,
u
(
s
)
)
|
+
(
t
′′
-
t
k
)
000000
×
|
g
(
s
,
u
(
s
)
)
|

|
f
(
s
,
u
(
s
)
)
|
+
(
t
′′
-
t
k
)
]
d
q
k
s
+
L
2
(
t
′
-
t
k
)
(
t
′′
-
t
′
)
≤
[
L
1
+
L
2
(
t
′′
-
t
k
)
]
(
t
′′
-
t
′
)
+
L
2
(
t
′
-
t
k
)
(
t
′′
-
t
′
)
.
As t
′
→
t
′′, the right hand side of the above inequality tends to zero. Thus, Q
(
Ω
) is relatively compact. As a consequence of Arzela Ascoli's theorem, Q is a compact operator. Therefore, Q is a completely continuous operator.
Define the set𝒲
=
{
u
∈
P
C
(
J
,
ℝ
)
∣
u
=
λ
Q
u
,
0
<
λ
<
1
}.
Next, we show𝒲 is bounded. Let u
∈
𝒲; then u
=
λ
Q
u, 0
<
λ
<
1. For any t
∈
J, by conditions (
H
1
) and (
H
2
), we have
(25)
|
u
(
t
)
|
=
λ
|
(
Q
u
)
(
t
)
|
≤
∫
t
k
t
‍
[
|
f
(
s
,
u
(
s
)
)
|
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
000000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
|
f
(
s
,
u
(
s
)
)
|
000000000000
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
000000000000
×
|
g
(
s
,
u
(
s
)
)
|
]
d
q
i
s
+
∑
i
=
1
k
‍
|
I
i
(
u
(
t
i
)
)
|
+
|
h
(
u
)
|
+
|
u
0
|
≤
∫
t
k
t
‍
[
[
M
3
(
s
)
+
M
4
(
s
)
|
u
(
s
)
|
]
M
1
(
s
)
+
M
2
(
s
)
|
u
(
s
)
|
000000
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
000000
×
[
M
3
(
s
)
+
M
4
(
s
)
|
u
(
s
)
|
]
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
M
1
(
s
)
+
M
2
(
s
)
|
u
(
s
)
|
000000000000
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
000000000000
×
[
M
3
(
s
)
+
M
4
(
s
)
|
u
(
s
)
|
]
]
d
q
i
s
+
∑
i
=
1
k
‍
L
¯
+
L
~
+
|
u
0
|
≤
(
M
1
+
M
2
∥
u
∥
)
(
t
-
t
k
)
+
(
M
3
+
M
4
∥
u
∥
)
×
(
t
-
t
k
)
2
1
+
q
k
+
∑
i
=
0
k
-
1
‍
[
(
t
i
+
1
-
t
i
)
2
1
+
q
i
(
M
1
+
M
2
∥
u
∥
)
(
t
i
+
1
-
t
i
)
×
(
t
-
t
k
)
2
1
+
q
k
+
∑
i
=
0
k
-
1
‍
00
+
(
M
3
+
M
4
∥
u
∥
)
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
+
m
L
¯
+
L
~
+
|
u
0
|
≤
(
M
1
+
M
2
∥
u
∥
)
T
+
(
M
3
+
M
4
∥
u
∥
)
×
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
m
L
¯
+
L
~
+
|
u
0
|
≤
M
1
T
+
M
3
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
+
m
L
¯
+
L
~
+
|
u
0
|
+
[
M
2
T
+
M
4
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
∥
u
∥
,
which implies
(26)
∥
u
∥
≤
M
1
T
+
M
3
∑
i
=
0
m
(
(
t
i
+
1
-
t
i
)
2
/
(
1
+
q
i
)
)
+
m
L
¯
+
L
~
+
|
u
0
|
1
-
[
M
2
T
+
M
4
∑
i
=
0
m
(
(
t
i
+
1
-
t
i
)
2
/
(
1
+
q
i
)
)
]
≔
constant
.
So, the set 𝒲 is bounded. Thus, Theorem 2 ensures the impulsive q
k-integrodifference equation (1) has at least one solution.Corollary 4.
Assume the following.(
H
3
)
There exist nonnegative constantsL
i
(
i
=
1,2
,
3,4
) such that
(27)
|
f
(
t
,
u
)
|
≤
L
1
,
|
g
(
t
,
u
)
|
≤
L
2
,
|
I
k
(
u
)
|
≤
L
3
,
|
h
(
u
)
|
≤
L
4
,
for anyt
∈
J, u
∈
ℝ, k
=
1,2
,
…
,
m.
Then problem (1) has at least one solution.Theorem 5.
Assume the following.(
H
4
)
There exist nonnegative bounded functionsM
(
t
) and N
(
t
) such that
(28)
|
f
(
t
,
u
)
-
f
(
t
,
v
)
|
≤
M
(
t
)
|
u
-
v
|
,
|
g
(
t
,
u
)
-
g
(
t
,
v
)
|
≤
N
(
t
)
|
u
-
v
|
,
fort
∈
J
,
u
,
v
∈
ℝ.
(
H
5
)
There exist positive constantsK
,
G such that
(29)
|
I
k
(
u
)
-
I
k
(
v
)
|
≤
K
|
u
-
v
|
,
|
h
(
u
)
-
h
(
v
)
|
≤
G
|
u
-
v
|
,
foru
,
v
∈
ℝ and k
=
1,2
,
…
,
m.
(
H
6
)
𝒦
=
sup
⁡
t
∈
J
⁡
[
M
(
t
)
t
+
m
K
+
G
+
N
(
t
)
∑
i
=
0
m
(
(
t
i
+
1
-
t
i
)
2
/
(
1
+
q
i
)
)
]
<
1.
Then problem (1) has a unique solution.Proof.
Denotesup
⁡
t
∈
J
⁡
|
M
(
t
)
|
=
M, sup
⁡
t
∈
J
⁡
|
N
(
t
)
|
=
N. For ∀
u
,
v
∈
P
C
(
J
,
ℝ
), by (
H
4
) and (
H
5
), we have
(30)
|
(
Q
u
)
(
t
)
-
(
Q
v
)
(
t
)
|
≤
∫
t
k
t
‍
[
|
f
(
s
,
u
(
s
)
)
-
f
(
s
,
v
(
s
)
)
|
00000000
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
00000000
×
|
g
(
s
,
u
(
s
)
)
-
g
(
s
,
v
(
s
)
)
|
]
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
|
f
(
s
,
u
(
s
)
)
-
f
(
s
,
v
(
s
)
)
|
0000000000000
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
0000000000000
×
|
g
(
s
,
u
(
s
)
)
-
g
(
s
,
v
(
s
)
)
|
]
d
q
i
s
+
∑
i
=
1
k
‍
|
I
i
(
u
(
t
i
)
)
-
I
i
(
v
(
t
i
)
)
|
+
|
h
(
u
)
-
h
(
v
)
|
≤
∫
t
k
t
‍
[
M
(
s
)
+
[
(
t
-
t
k
)
-
q
k
(
s
-
t
k
)
]
N
(
s
)
]
000000
×
|
(
u
-
v
)
(
s
)
|
d
q
k
s
+
∑
i
=
0
k
-
1
‍
∫
t
i
t
i
+
1
‍
[
M
(
s
)
+
[
(
t
i
+
1
-
t
i
)
-
q
i
(
s
-
t
i
)
]
000000000000000
×
N
(
s
)
]
|
(
u
-
v
)
(
s
)
|
d
q
i
s
+
∑
i
=
1
k
‍
K
|
(
u
-
v
)
(
t
i
)
|
+
G
|
u
-
v
|
≤
[
M
(
t
-
t
k
)
+
N
(
t
-
t
k
)
2
1
+
q
k
00000
+
∑
i
=
0
k
-
1
‍
[
M
(
t
i
+
1
-
t
i
)
+
N
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
00000
+
m
K
+
G
(
t
-
t
k
)
2
1
+
q
k
]
∥
u
-
v
∥
≤
[
M
t
+
m
K
+
G
+
N
∑
i
=
0
m
‍
(
t
i
+
1
-
t
i
)
2
1
+
q
i
]
∥
u
-
v
∥
≤
𝒦
∥
u
-
v
∥
.
As𝒦
<
1 by (
H
6
), then ∥
Q
u
-
Q
v
∥
<
∥
u
-
v
∥. Therefore, Q is a contractive map. Thus, the conclusion of the Theorem 5 follows by Banach contraction mapping principle.


## 4. Example

Consider the following nonlinearq
k-integrodifference equation with impulses
(31)
D
1
/
(
2
+
k
)
u
(
t
)
=
8
+
3
t
+
ln
⁡
⁡
(

1
+
5
t
3
+
t
2
5
|
u
(
t
)
|
)
+
∫
1
/
(
1
+
2
k
)
t
‍
[
10
s
+
s
3
3
sin
⁡
u
(
s
)
]
d
1
/
(
2
+
k
)
s
,
00000000000000000
t
∈
[
0,1
]
,
t
≠
1
1
+
2
k
,
Δ
u
(
1
1
+
2
k
)
=
cos
⁡
⁡
(
u
(
1
1
+
2
k
)
)
,
k
=
1,2
,
…
,
6
,
u
(
0
)
=
5
+
e
-
u
2
(
1
/
2
)
.
Obviously, q
k
=
1
/
(
2
+
k
)
(
k
=
0,1
,
2
,
…
,
6
), t
k
=
1
/
(
1
+
2
k
)
(
k
=
1,2
,
…
,
6
), f
(
t
,
u
)
=
8
+
3
t + ln
⁡
⁡
(
1
+
5
t
3
+
(
t
2
/
5
)
|
u
|
), g
(
t
,
u
)
=
10
t
+
(
t
3
/
3
)
sin
⁡
u, I
k
(
u
)
=
cos
⁡
⁡
u, and h
(
u
)
=
e
-
u
2.By a simple calculation, we can get(32)
|
f
(
t
,
u
)
|
≤
8
+
3
t
+
5
t
3
+
t
2
5
|
u
|
,
|
g
(
t
,
u
)
|
≤
10
t
+
t
3
3
|
u
|
,
|
I
k
(
u
)
|
≤
1
,
|
h
(
u
)
|
≤
1
.TakeM
1
(
t
)
=
8
+
3
t
+
5
t
3, M
2
(
t
)
=
t
2
/
5,  M
3
(
t
)
=
10
t, M
4
(
t
)
=
t
3
/
3, and L
¯
=
L
~
=
1. Then all conditions of Theorem 3 hold. By Theorem 3, nonlinear impulsive q
k-integrodifference (31) has at least one solution.



---

*Source: 478185-2014-03-10.xml*