# Periodic Boundary Value Problems for a Class of Impulsive Functional Differential Equations of Hybrid Type

Authors: Guoping Chen
Journal: ISRN Applied Mathematics
 (2014)

---

## Abstract

This paper is concerned with the existence of extreme solutions of periodic
boundary value problems for a class of first-order impulsive functional differential equations
of hybrid type. We obtain the sufficient conditions for existence of extreme solutions by
using upper and lower solutions method coupled with monotone iterative technique.


---

## Body

## 1. Introduction

The theory of impulsive differential equations is now being recognized to be not only richer than the corresponding theory of differential equations without impulses but also represents a more natural framework for mathematical modeling of many real-world phenomena [1–3]. Significant progress has been made in the theory of systems of impulsive differential equations in recent years (see [4–18] and the references cited therein). It is well known that the monotone iterative technique offers an approach for obtaining approximate solutions of nonlinear differential equations; for details, see [19] and the references therein. There also exist several works devoted to the applications of this technique to periodic boundary value problems of impulsive differential equations; see [20–26]. In [27, 28], the authors introduce a new concept of upper and lower solutions for periodic boundary value problems of a class of first-order functional differential equations. In paper [23], the authors applied this new concept to study the periodic boundary value problems for first-order impulsive functional differential equations. Motivated by [23, 27, 28], we will study periodic boundary value problem for the first-order impulsive functional differential equation of hybrid type
(1)
x
′
(
t
)
=
f
(
t
,
x
(
t
)
,
x
(
θ
(
t
)
)
,
(
K
x
)
(
t
)
,
(
H
x
)
(
t
)
)
,
aaaaaaaaaa
t
≠
t
k
,
t
∈
J
=
[
0
,
T
]
,
T
>
0
,
Δ
x
(
t
k
)
=
I
k
(
x
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
x
(
0
)
=
x
(
T
)
,
where f
∈
C
(
J
×
R
4
,
R
), θ
∈
C
(
J
,
J
), I
k
∈
C
(
R
,
R
), 0
=
t
0
<
t
1
<
t
2
<
⋯
<
t
m
<
t
m
+
1
=
T, Δ
x
(
t
k
)
=
x
(
t
k
+
)
-
x
(
t
k
-
) denotes the jump of x
(
t
) at t
=
t
k, x
(
t
k
+
), and x
(
t
k
-
) represent the right and left limits of x
(
t
) at t
=
t
k, respectively. Denote J
′
=
J
∖
{
t
1
,
t
2
,
…
,
t
m
}. The integral part in (1) is defined by
(2)
(
K
x
)
(
t
)
=
∫
0
γ
(
t
)
‍
k
(
t
,
s
)
x
(
δ
(
s
)
)
d
s
,
(
H
x
)
(
t
)
=
∫
0
T
‍
h
(
t
,
s
)
x
(
σ
(
s
)
)
d
s
,
where γ
,
δ
,
σ
∈
C
(
J
,
J
), k
∈
C
(
D
,
R
+
), D
=
{
(
t
,
s
)
∈
J
×
J
:
0
≤
s
≤
γ
(
t
)
≤
T
}, h
∈
C
(
J
×
J
,
R
+
), R
+
=
[
0
,
∞
), K
0
=
max
{
k
(
t
,
s
)
:
(
t
,
s
)
∈
D
}, and H
0
=
max
{
h
(
t
,
s
)
:
(
t
,
s
)
∈
J
×
J
}.LetP
C
(
J
)
=
{
u
:
J
→
R, u is continuous for t
∈
J, t
≠
t
k, u
(
t
i
+
), and u
(
t
i
-
) exist, and u
(
t
i
-
)
=
u
(
t
i
), i
=
1,2
,
…
,
m
}. P
C
1
(
J
)
=
{
u
∈
P
C
(
J
)
:
u is continuously differentiable for t
∈
J, t
≠
t
k
}. P
C
(
J
) and P
C
1
(
J
) are Banach spaces with the norms
(3)
∥
u
∥
P
C
(
J
)
=
sup
{
|
u
(
t
)
|
:
t
∈
J
}
,
∥
u
∥
P
C
1
(
J
)
=
max
{
∥
u
∥
P
C
(
J
)
,
∥
u
′
∥
P
C
(
J
)
}
.By a solution of (1), we mean a u
∈
P
C
1
(
J
) for which problem (1) is satisfied.Note that (1) has a very general form; as special instances resulting from (1), one can have impulsive differential equations with deviating arguments and impulsive differential equations with the Volterra or Fredholm operators. For example, if f does not include K
x and H
x, then (1) reduces to
(4)
a
i
x
′
(
t
)
=
f
(
t
,
x
(
t
)
,
x
(
θ
(
t
)
)
)
,
t
≠
t
k
,
t
∈
J
=
[
0
,
T
]
,
T
>
0
,
Δ
x
(
t
k
)
=
I
k
(
x
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
x
(
0
)
=
x
(
T
)
,
periodic boundary problem for impulsive differential equations with deviating arguments, which is discussed in [22, 23]; when θ
(
t
)
=
γ
(
t
)
=
δ
(
t
)
=
σ
(
t
)
≡
t, (1) is the following periodic boundary problem for impulsive integrodifferential equations of mixed type:
(5)
x
′
(
t
)
=
f
(
t
,
x
(
t
)
,
(
K
x
)
(
t
)
,
(
H
x
)
(
t
)
)
,
t
≠
t
k
,
t
∈
J
=
[
0
,
T
]
,
T
>
0
,
Δ
x
(
t
k
)
=
I
k
(
x
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
x
(
0
)
=
x
(
T
)
;
similar problems are also discussed in [24–26].


## 2. Preliminaries

To apply the method of upper and lower solutions, we need the concept of lower solution and upper solution for (1).Definition 1.
A functionα
∈
P
C
1
(
J
) is a lower solution of (1) if there exist M
1
>
0, M
i
≥
0, i
=
2,3
,
4, and  0
≤
L
k
<
1, k
=
1,2
,
…
,
m, such that
(6)
α
′
(
t
)
≤
f
(
t
,
α
(
t
)
,
α
(
θ
(
t
)
)
,
(
K
α
)
(
t
)
,
(
H
α
)
(
t
)
)
-
a
α
(
t
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
≠
t
k
,
t
∈
J
,
Δ
α
(
t
k
)
≤
I
k
(
α
(
t
k
)
)
-
L
k
a
α
k
,
k
=
1,2
,
…
,
m
;
similarly, a function β
∈
P
C
1
(
J
) is an upper solution of (1) if
(7)
β
′
(
t
)
≥
f
(
t
,
β
(
t
)
,
β
(
θ
(
t
)
)
,
(
K
β
)
(
t
)
,
(
H
β
)
(
t
)
)
+
a
β
(
t
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
≠
t
k
,
t
∈
J
,
Δ
β
(
t
k
)
≥
I
k
(
β
(
t
k
)
)
+
L
k
a
β
k
,
k
=
1,2
,
…
,
m
,
where
(8)
a
α
(
t
)
=
{
0
,
if
α
(
0
)
≤
α
(
T
)
,
(
∫
0
γ
(
t
)
‍
M
1
t
+
M
2
θ
(
t
)
aa
+
M
3
K
0
∫
0
γ
(
t
)
‍
δ
(
s
)
d
s
a
a
+
M
4
H
0
∫
0
T
‍
σ
(
s
)
d
s
+
1
)
×
(
T
)
-
1
[
α
(
0
)
-
α
(
T
)
]
,
if
α
(
0
)
>
α
(
T
)
,
a
α
k
=
{
0
,
if
α
(
0
)
≤
α
(
T
)
,
t
k
T
[
α
(
0
)
-
α
(
T
)
]
,
if
α
(
0
)
>
α
(
T
)
,
a
β
(
t
)
=
{
0
,
if
β
(
0
)
≥
β
(
T
)
,
(
∫
0
T
‍
M
1
t
+
M
2
θ
(
t
)
aa
+
M
3
K
0
∫
0
γ
(
t
)
‍
δ
(
s
)
d
s
aa
+
M
4
H
0
∫
0
T
‍
σ
(
s
)
d
s
+
1
)
×
(
T
)
-
1
[
β
(
T
)
-
β
(
0
)
]
,
if
β
(
0
)
<
β
(
T
)
,
a
β
k
=
{
0
,
if
β
(
0
)
≥
β
(
T
)
,
t
k
T
[
β
(
T
)
-
β
(
0
)
]
,
if
β
(
0
)
<
β
(
T
)
.
In what follows, we define the set
(9)
[
u
,
v
]
=
{
w
∈
P
C
(
J
)
:
u
(
t
)
≤
w
(
t
)
≤
v
(
t
)
,
t
∈
J
}
for u
,
v
∈
P
C
(
J
) and u
(
t
)
≤
v
(
t
), t
∈
J.For the sake of convenience, we list the following conditions.(
H
1
)
Assume thatα
(
t
)
≤
β
(
t
) for t
∈
J, and there exist M
1
>
0 and M
i
≥
0, i
=
2,3
,
4, such that
(10)
f
(
t
,
x
1
,
x
2
,
x
3
,
x
4
)
-
f
(
t
,
y
1
,
y
2
,
y
3
,
y
4
)
≥
-
M
1
(
x
1
-
y
1
)
-
M
2
(
x
2
-
y
2
)
-
M
3
(
x
3
-
y
3
)
-
M
4
(
x
4
-
y
4
)
,
for α
(
t
)
≤
y
1
≤
x
1
≤
β
(
t
), α
(
θ
(
t
)
)
≤
y
2
≤
x
2
≤
β
(
θ
(
t
)
), (
K
α
)
(
t
)
≤
y
3
≤
x
3
≤
(
K
β
)
(
t
), and (
H
α
)
(
t
)
≤
y
4
≤
x
4
≤
(
H
β
)
(
t
), t
∈
J.
(
H
2
)
There existsL
k with 0
≤
L
k
<
1, k
=
1,2
,
…
,
m, such that
(11)
I
k
(
x
)
-
I
k
(
u
)
≥
-
L
k
(
x
-
u
)
,
for
α
(
t
k
)
≤
u
(
t
k
)
≤
x
(
t
k
)
≤
β
(
t
k
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
k
=
1,2
,
…
,
m
.Lemma 2 (see [5]).
Assume thats
∈
[
0
,
T
), x
∈
P
C
1
(
J
), p
,
q
∈
P
C
(
J
), a
i
≥
0, b
i, i
=
1,2
,
…
,
m, are constants such that
(12)
x
′
(
t
)
≤
p
(
t
)
x
(
t
)
+
q
(
t
)
,
t
∈
[
s
,
T
)
,
t
≠
t
k
,
x
(
t
k
+
)
≤
a
k
x
(
t
k
)
+
b
k
,
t
k
∈
[
s
,
T
)
,
k
=
1,2
,
…
,
m
;
then
(13)
x
(
t
)
≤
x
(
s
+
)
(
∏
s
<
t
k
<
t
‍
a
k
)
exp
(
∫
s
t
‍
p
(
u
)
d
u
)
+
∑
s
<
t
k
<
t
‍
(
∏
t
k
<
t
i
<
t
‍
a
i
)
exp
(
∫
t
k
t
‍
p
(
τ
)
d
τ
)
b
k
+
∫
s
t
‍
(
∏
u
<
t
k
<
t
‍
a
k
)
exp
(
∫
u
t
‍
p
(
τ
)
d
τ
)
q
(
u
)
d
u
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
t
∈
[
s
,
T
]
.Lemma 3.
Assume thatx
∈
P
C
1
(
J
), M
1
>
0, M
i
≥
0, i
=
2,3
,
4, and 0
≤
L
k
<
1, (
k
=
1,2
,
…
,
m
) such that
(14)
x
′
(
t
)
≤
-
M
1
x
(
t
)
-
M
2
x
(
θ
(
t
)
)
-
M
3
(
K
x
)
(
t
)
-
M
4
(
H
x
)
(
t
)
-
a
x
(
t
)
,
t
∈
J
,
t
≠
t
k
,
Δ
x
(
t
k
)
≤
-
L
k
x
(
t
k
)
-
L
k
a
x
k
,
k
=
1,2
,
…
,
m
,
where
(15)
a
x
(
t
)
=
{
0
,
if
x
(
0
)
≤
x
(
T
)
,
(
∫
0
T
‍
M
1
t
+
M
2
θ
(
t
)
aa
+
M
3
K
0
∫
0
γ
(
t
)
‍
δ
(
s
)
d
s
a
a
+
M
4
H
0
∫
0
T
‍
σ
(
s
)
d
s
+
1
)
×
(
T
)
-
1
[
x
(
0
)
-
x
(
T
)
]
,
if
x
(
0
)
>
x
(
T
)
,
a
x
k
=
{
0
,
if
x
(
0
)
≤
x
(
T
)
,
t
k
T
[
x
(
0
)
-
x
(
T
)
]
,
if
x
(
0
)
>
x
(
T
)
.
If
(16)
M
2
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
∫
0
T
(
∫
0
γ
(
t
)
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
aaaaaaaaaaaaaaaaa
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
∫
0
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
aaaaaaaaaaaaaaaaa
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∏
t
<
t
k
<
T
‍
∫
0
T
‍
)
d
t
≤
e
-
M
1
T
∏
j
=
1
m
‍
(
1
-
L
k
)
2
,
then x
(
t
)
≤
0 on J.Proof
Case  1  (
x
(
0
)
≤
x
(
T
)
). Let v
(
t
)
=
x
(
t
)
e
M
1
t; then v
∈
P
C
1
(
J
) and
(17)
v
′
(
t
)
≤
-
M
2
e
M
1
(
t
-
θ
(
t
)
)
v
(
θ
(
t
)
)
-
M
3
(
K
1
v
)
(
t
)
-
M
4
(
H
1
v
)
(
t
)
,
t
∈
J
′
,
Δ
v
(
t
k
)
≤
-
L
k
v
(
t
k
)
,
k
=
1,2
,
…
,
m
,
v
(
0
)
≤
v
(
T
)
e
-
M
1
T
,
where
(18)
(
K
1
v
)
(
t
)
=
∫
0
γ
(
t
)
k
1
(
t
,
s
)
v
(
δ
(
s
)
)
d
s
,
(
H
1
v
)
(
t
)
=
∫
0
T
h
1
(
t
,
s
)
v
(
σ
(
s
)
)
d
s
,
k
1
(
t
,
s
)
=
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
,
h
1
(
t
,
s
)
=
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
.
Obviously, v
(
t
)
≤
0 implies that x
(
t
)
≤
0. To show v
(
t
)
≤
0, suppose, on the contrary, that v
(
t
~
)
>
0 for some t
~
∈
J. Then there are two possible cases:(a)
v
(
t
)
≥
0 for all t
∈
J;
(b)
there existst
1
*
∈
J such that v
(
t
1
*
)
<
0.
In case (a), (17) implies that v
′
(
t
)
≤
0 for t
∈
J
′ and Δ
v
(
t
k
)
≤
0 (k
=
1,2
,
…
,
m); hence, v
(
t
) is nonincreasing in J; then v
(
0
)
≥
v
(
t
~
)
>
0 and v
(
0
)
≥
v
(
T
). On the other hand, v
(
0
)
=
x
(
0
)
≤
x
(
T
)
<
x
(
T
)
e
M
1
T
=
v
(
T
), which is a contradiction.
In case (b), lett
*
=
max
{
t
∈
J, inf
s
∈
J
v
(
s
)
=
v
(
t
)
=
-
λ, λ
>
0
}. We only consider t
*
∈
(
t
p
-
1
,
t
p
] for some p
∈
{
1,2
,
…
,
m
+
1
}, as the proof is similar for the case t
*
=
t
p
-
1
+. Next, we consider two subcases.
Subcase I  (
v
(
T
)
≥
0
). From (17), we get
(19)
v
′
(
t
)
≤
λ
M
2
e
M
1
(
t
-
θ
(
t
)
)
+
λ
M
3
∫
0
γ
(
t
)
k
1
(
t
,
s
)
d
s
+
λ
M
4
∫
0
T
h
1
(
t
,
s
)
d
s
.
From Lemma (14), we have
(20)
v
(
T
)
≤
v
(
t
*
)
∏
t
*
<
t
k
<
T
‍
(
1
-
L
k
)
+
λ
M
2
∫
t
*
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
λ
M
3
∫
t
*
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
λ
M
4
∫
t
*
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
≤
-
λ
∏
k
=
p
m
‍
(
1
-
L
k
)
+
λ
M
2
∫
t
*
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
λ
M
3
∫
t
*
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
λ
M
4
∫
t
*
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
;
hence,
(21)
v
(
T
)
<
-
λ
(
∏
k
=
1
m
‍
(
1
-
L
k
)
)
+
λ
M
2
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
λ
M
3
∫
0
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
λ
M
4
∫
0
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
This yields
(22)
∏
k
=
1
m
‍
(
1
-
L
k
)
<
M
2
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
∫
0
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
∫
0
T
(
∫
0
T
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
It contradicts (16).
Subcase II  (
v
(
T
)
<
0
). If t
*
<
t
~, similar to the subcase I, it also yields a contradiction. We assume without loss of generality that t
*
>
t
~ and t
~
∈
(
t
q
,
t
q
+
1
] for some q
∈
{
0,1
,
2
,
…
,
m
-
1
}. By Lemma 2, we have
(23)
v
(
t
~
)
≤
v
(
0
)
∏
0
<
t
k
<
t
~
‍
(
1
-
L
k
)
+
λ
M
2
∫
0
t
~
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
λ
M
3
∫
0
t
~
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
λ
M
4
∫
0
t
~
(
∫
0
T
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
Noting that v
(
0
)
≤
v
(
T
)
e
-
M
1
T, we get
(24)
0
<
v
(
t
~
)
≤
v
(
T
)
e
-
M
1
T
∏
K
=
1
q
‍
(
1
-
L
k
)
+
λ
M
2
∫
0
t
~
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
λ
M
3
∫
0
t
~
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
λ
M
4
∫
0
t
~
(
∫
0
T
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
Hence,
(25)
v
(
T
)
∏
K
=
1
q
‍
(
1
-
L
k
)
≥
-
λ
M
2
e
M
1
T
∫
0
t
~
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
-
λ
M
3
e
M
1
T
∫
0
t
~
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
-
λ
M
4
e
M
1
T
∫
0
t
~
(
∫
0
T
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
From (20) and (25), we obtain
(27)
∏
k
=
p
m
‍
(
1
-
L
k
)
(
∏
j
=
1
q
‍
(
1
-
L
k
)
)
≤
M
2
e
M
1
T
∫
0
t
~
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
e
M
1
T
∫
0
t
~
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
e
M
1
T
∫
0
t
~
‍
(
∫
0
T
‍
(
∏
t
<
t
k
<
t
~
‍
(
1
-
L
k
)
)
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
2
∏
j
=
1
q
‍
(
1
-
L
k
)
∫
t
*
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
∏
j
=
1
q
‍
(
1
-
L
k
)
∫
t
*
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
∏
j
=
1
q
‍
(
1
-
L
k
)
∫
t
*
T
‍
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
By multiplying ∏
j
=
q
+
1
m
‍
(
1
-
L
k
) on both sides of (27), we have
(28)
∏
k
=
p
m
‍
(
1
-
L
k
)
(
∏
j
=
1
m
‍
(
1
-
L
k
)
)
≤
M
2
e
M
1
T
∫
0
t
~
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
e
M
1
T
∫
0
t
~
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
e
M
1
T
∫
0
t
~
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
2
∏
j
=
1
m
‍
(
1
-
L
k
)
∫
t
*
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
∏
j
=
1
m
‍
(
1
-
L
k
)
∫
t
*
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
∏
j
=
1
m
‍
(
1
-
L
k
)
∫
t
*
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
Noting that ∏
j
=
1
m
‍
(
1
-
L
k
)
<
1, e
M
1
T
>
1, we get
(29)
∏
k
=
p
m
‍
(
1
-
L
k
)
(
∏
j
=
1
m
‍
(
1
-
L
k
)
)
<
M
2
e
M
1
T
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
e
M
1
T
∫
0
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
e
M
1
T
∫
0
T
‍
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
.
This implies
(30)
e
-
M
1
T
∏
j
=
1
m
‍
(
1
-
L
k
)
2
<
M
2
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
e
M
1
(
t
-
θ
(
t
)
)
d
t
+
M
3
∫
0
T
(
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
δ
(
s
)
)
k
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
+
M
4
∫
0
T
(
∫
0
T
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
‍
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
×
e
M
1
(
t
-
σ
(
s
)
)
h
(
t
,
s
)
d
s
∫
0
γ
(
t
)
‍
(
∏
t
<
t
k
<
T
‍
(
1
-
L
k
)
)
)
d
t
,
which contradicts (16). This completes the proof for the case x
(
0
)
≤
x
(
T
).
Case  2  (
x
(
0
)
>
x
(
T
)
). Let v
(
t
)
=
x
(
t
)
+
(
t
/
T
)
[
x
(
0
)
-
x
(
T
)
]; then v
(
0
)
=
x
(
0
)
=
v
(
T
), and
(31)
v
′
(
t
)
+
M
1
v
(
t
)
+
M
2
v
(
θ
(
t
)
)
+
M
3
(
K
v
)
(
t
)
+
M
4
(
H
v
)
(
t
)
≤
x
′
(
t
)
+
M
1
x
(
t
)
+
M
2
x
(
θ
(
t
)
)
+
M
3
(
K
x
)
(
t
)
+
M
4
(
H
x
)
(
t
)
+
(
M
1
t
+
M
2
θ
(
t
)
+
M
3
K
0
∫
0
γ
(
t
)
‍
δ
(
s
)
d
s
a
a
a
a
a
a
a
a
a
a
+
M
4
H
0
∫
0
T
‍
σ
(
s
)
d
s
+
1
)
×
(
T
)
-
1
×
[
x
(
0
)
-
x
(
T
)
]
≤
0
,
t
≠
t
k
,
t
∈
J
,
Δ
v
(
t
k
)
=
Δ
x
(
t
k
)
≤
-
L
k
x
(
t
k
)
-
L
k
t
k
T
[
x
(
0
)
-
x
(
T
)
]
=
-
L
k
v
(
t
k
)
,
k
=
1,2
,
…
,
m
.
In view of Case 1, we have v
(
t
)
≤
0, for t
∈
J. Therefore, x
(
t
)
≤
0 for t
∈
J. The proof of Lemma 3 is complete.Lemma 4.
Assume thatx
∈
P
C
1
(
J
), M
1
>
0, M
i
≥
0, i
=
2,3
,
4, η
∈
P
C
(
J
), and 0
≤
L
k
<
1, k
=
1,2
,
…
,
m. If
(32)
(
M
2
T
+
M
3
T
2
K
0
+
M
4
T
2
H
0
+
∑
k
=
1
m
‍
L
k
)
e
M
1
T
e
M
1
T
-
1
<
1
,
then the impulsive differential problem
(33)
x
′
(
t
)
=
-
M
1
x
(
t
)
-
M
2
x
(
θ
(
t
)
)
-
M
3
(
K
x
)
(
t
)
-
M
4
(
H
x
)
(
t
)
+
η
(
t
)
,
t
∈
J
,
t
≠
t
k
,
Δ
x
(
t
k
)
=
-
L
k
x
(
t
k
)
+
d
k
,
d
k
∈
R
,
k
=
1,2
,
…
,
m
,
x
(
0
)
=
x
(
T
)
,
has a unique solution.Proof.
Define a mapA
:
P
C
(
J
)
→
P
C
(
J
) by
(34)
[
A
x
]
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
(
η
(
s
)
-
M
2
x
(
θ
(
s
)
)
aaaaaaaaaaaaa
-
M
3
(
K
x
)
(
s
)
-
M
4
(
H
x
)
(
s
)
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
(
-
L
k
x
(
t
k
)
+
d
k
)
,
where
(35)
G
(
t
,
s
)
=
{
e
-
M
1
(
t
-
s
)
1
-
e
-
M
1
T
,
0
≤
s
<
t
≤
T
,
e
-
M
1
(
T
+
t
-
s
)
1
-
e
-
M
1
T
,
0
≤
t
≤
s
≤
T
.
It is easy to verify that x is a solution of (33) if and only if x is a fixed point of A. For any x
,
y
∈
P
C
(
J
), we have
(36)
|
[
A
x
]
(
t
)
-
[
A
y
]
(
t
)
|
≤
|
M
2
∫
0
T
‍
G
(
t
,
s
)
(
y
(
θ
(
s
)
)
-
x
(
θ
(
s
)
)
)
d
s
|
+
|
M
3
∫
0
T
‍
∫
0
γ
(
s
)
‍
G
(
t
,
s
)
k
(
s
,
u
)
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
(
y
(
δ
(
u
)
)
-
x
(
δ
(
u
)
)
)
d
u
d
s
∫
0
T
‍
∫
0
γ
(
s
)
‍
|
+
|
M
4
∬
0
T
G
(
t
,
s
)
h
(
s
,
u
)
a
a
a
a
a
a
a
a
a
a
a
a
a
a
×
(
y
(
σ
(
u
)
)
-
x
(
σ
(
u
)
)
)
d
u
d
s
∫
0
T
‍
∫
0
γ
(
s
)
‍
|
+
|
∑
k
=
1
m
‍
G
(
t
,
t
k
)
L
k
(
y
(
t
k
)
-
x
(
t
k
)
)
|
≤
(
M
2
T
+
M
3
T
2
K
0
+
M
4
T
2
H
0
+
∑
k
=
1
m
‍
L
k
)
×
e
M
1
T
e
M
1
T
-
1
∥
x
-
y
∥
;
this implies
(37)
∥
[
A
x
]
-
[
A
y
]
∥
≤
(
M
2
T
+
M
3
T
2
K
0
+
M
4
T
2
H
0
+
∑
k
=
1
m
‍
L
k
)
×
e
M
1
T
e
M
1
T
-
1
∥
x
-
y
∥
.
Condition (32) implies that A is a contraction mapping. Banach’s fixed point theorem implies that A has a unique fixed point, and so (33) has a unique solution. The proof is complete.


## 3. Main Results

Theorem 5.
Assume thatα and β are the lower solution and upper solution for (1) with α
≤
β on J, respectively. Further, suppose that (
H
1
), (
H
2
), (16), and (32) are satisfied. Then, (1) possesses in [
α
,
β
] the minimal and maximal solution u
,
v
∈
P
C
1
(
J
), respectively. Moreover, there exist the monotone sequences {
u
n
}
↗
u and {
v
n
}
↘
v uniformly on J, respectively.Proof.
For allψ
∈
[
α
,
β
], we consider linear impulsive differential equation
(38)
y
′
(
t
)
=
-
M
1
y
(
t
)
-
M
2
y
(
θ
(
t
)
)
-
M
3
(
K
y
)
(
t
)
-
M
4
(
H
y
)
(
t
)
+
η
(
t
)
,
t
≠
t
k
,
t
∈
J
,
Δ
y
(
t
k
)
=
I
k
(
ψ
(
t
k
)
)
-
L
k
(
y
(
t
k
)
-
ψ
(
t
k
)
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
k
=
1,2
,
…
,
m
,
y
(
0
)
=
y
(
T
)
,
where
(39)
η
(
t
)
=
f
(
t
,
ψ
(
t
)
,
ψ
(
θ
(
t
)
)
,
(
K
ψ
)
(
t
)
,
(
H
ψ
)
(
t
)
)
+
M
1
ψ
(
t
)
+
M
2
ψ
(
θ
(
t
)
)
+
M
3
(
K
ψ
)
(
t
)
+
M
4
(
H
ψ
)
(
t
)
.
By Lemma 4, one can see that (38) has a unique solution y
∈
P
C
1
(
J
). Denote that A
ψ is a unique solution of (38).
Letu
0
=
α, v
0
=
β and u
n
=
A
u
n
-
1, v
n
=
A
v
n
-
1. We will show that(a)
u
0
≤
A
u
0, A
v
0
≤
v
0,
(b)
A is nondecreasing in [
u
0
,
v
0
].
The Proof of Property (a). Let p
=
u
0
-
u
1; since u
1
(
0
)
=
u
1
(
T
), it follows that p
(
0
)
-
p
(
T
)
=
u
0
(
0
)
-
u
0
(
T
), and
(40)
p
′
(
t
)
=
u
0
′
(
t
)
-
u
1
′
(
t
)
≤
f
(
t
,
u
0
(
t
)
,
u
0
(
θ
(
t
)
)
,
(
K
u
0
)
(
t
)
,
(
H
u
0
)
(
t
)
)
-
a
u
0
(
t
)
+
M
1
(
u
1
(
t
)
)
+
M
2
(
u
1
(
θ
(
t
)
)
)
+
M
3
(
K
u
1
)
(
t
)
+
M
4
(
H
u
1
)
(
t
)
-
f
(
t
,
u
0
(
t
)
,
u
0
(
θ
(
t
)
)
,
(
K
u
0
)
(
t
)
,
(
H
u
0
)
(
t
)
)
-
M
1
u
0
(
t
)
-
M
2
u
0
(
θ
(
t
)
)
-
M
3
(
K
u
0
)
(
t
)
-
M
4
(
H
u
0
)
(
t
)
=
-
M
1
p
(
t
)
-
M
2
p
(
θ
(
t
)
)
-
M
3
(
K
p
)
(
t
)
-
M
4
(
H
p
)
(
t
)
-
a
p
(
t
)
,
t
≠
t
k
,
t
∈
J
,
Δ
p
(
t
k
)
=
Δ
u
0
(
t
k
)
-
Δ
u
1
(
t
k
)
≤
I
k
(
u
0
(
t
k
)
)
-
a
u
0
k
-
I
k
(
u
0
(
t
k
)
)
+
L
k
(
u
1
(
t
k
)
-
u
0
(
t
k
)
)
=
-
L
k
p
(
t
k
)
-
a
p
k
,
where
(41)
a
u
0
(
t
)
=
a
p
(
t
)
=
{
0
,
if
p
(
0
)
≤
p
(
T
)
,
(
∫
0
T
‍
M
1
t
+
M
2
θ
(
t
)
aa
+
M
3
K
0
∫
0
γ
(
t
)
‍
δ
(
s
)
d
s
a
+
M
4
H
0
∫
0
T
‍
σ
(
s
)
d
s
+
1
)
×
(
T
)
-
1
[
p
(
0
)
-
p
(
T
)
]
,
if
p
(
0
)
>
p
(
T
)
,
a
u
0
k
=
a
p
k
=
{
0
,
if
p
(
0
)
≤
p
(
T
)
,
t
k
T
[
p
(
0
)
-
p
(
T
)
]
,
if
p
(
0
)
>
p
(
T
)
.
By Lemma 3, we have p
(
t
)
≤
0, for t
∈
J; that is, u
0
≤
A
u
0. Analogously, it is proved that A
v
0
≤
v
0.
Next, we prove the property (b). Letψ
1, ψ
2
∈
[
u
0
,
v
0
] with ψ
1
≤
ψ
2, ω
1
=
A
ψ
1, ω
2
=
A
ψ
2, and p
=
ω
1
-
ω
2; then, from (
H
1
) and (
H
2
), we have that, for t
≠
t
k,
(42)
p
′
(
t
)
=
ω
1
′
(
t
)
-
ω
2
′
(
t
)
=
-
M
1
ω
1
(
t
)
-
M
2
ω
1
(
θ
(
t
)
)
-
M
3
K
(
ω
1
)
(
t
)
-
M
4
H
(
ω
1
)
(
t
)
+
f
(
t
,
ψ
1
(
t
)
,
ψ
1
(
θ
(
t
)
)
,
(
K
ψ
1
)
(
t
)
,
(
H
ψ
1
)
(
t
)
)
+
M
1
ψ
1
(
t
)
+
M
2
ψ
1
(
θ
(
t
)
)
+
M
3
(
K
ψ
1
)
(
t
)
+
M
4
(
H
ψ
1
)
(
t
)
+
M
1
ω
2
(
t
)
+
M
2
ω
2
(
θ
(
t
)
)
+
M
3
K
(
ω
2
)
(
t
)
+
M
4
H
(
ω
2
)
(
t
)
-
f
(
t
,
ψ
2
(
t
)
,
ψ
2
(
θ
(
t
)
)
,
(
K
ψ
2
)
(
t
)
,
(
H
ψ
2
)
(
t
)
)
-
M
1
ψ
2
(
t
)
-
M
2
ψ
2
(
θ
(
t
)
)
-
M
3
(
K
ψ
2
)
(
t
)
-
M
4
(
H
ψ
2
)
(
t
)
≤
M
1
(
ψ
2
(
t
)
-
ψ
1
(
t
)
)
+
M
2
(
ψ
2
(
θ
(
t
)
)
-
ψ
1
(
θ
(
t
)
)
)
+
M
3
(
(
K
ψ
2
)
(
t
)
-
(
K
ψ
1
)
(
t
)
)
+
M
4
(
(
H
ψ
2
)
(
t
)
-
(
H
ψ
1
)
(
t
)
)
-
M
1
ω
1
(
t
)
-
M
2
ω
1
(
θ
(
t
)
)
-
M
3
K
(
ω
1
)
(
t
)
-
M
4
H
(
ω
1
)
(
t
)
+
M
1
ψ
1
(
t
)
+
M
2
ψ
1
(
θ
(
t
)
)
+
M
3
(
K
ψ
1
)
(
t
)
+
M
4
(
H
ψ
1
)
(
t
)
+
M
1
ω
2
(
t
)
+
M
2
ω
2
(
θ
(
t
)
)
+
M
3
K
(
ω
2
)
(
t
)
+
M
4
H
(
ω
2
)
(
t
)
-
M
1
ψ
2
(
t
)
-
M
2
ψ
2
(
θ
(
t
)
)
-
M
3
(
K
ψ
2
)
(
t
)
-
M
4
(
H
ψ
2
)
(
t
)
=
-
M
1
p
(
t
)
-
M
2
p
(
θ
(
t
)
)
-
M
3
(
K
p
)
(
t
)
-
M
4
(
H
p
)
(
t
)
,
and, for t
=
t
k,
(43)
Δ
p
(
t
k
)
=
Δ
ω
1
(
t
k
)
-
Δ
ω
2
(
t
k
)
=
I
k
(
ψ
1
(
t
k
)
)
-
L
k
(
ω
1
(
t
k
)
-
ψ
1
(
t
k
)
)
-
I
k
(
ψ
2
(
t
k
)
)
+
L
k
(
ω
2
(
t
k
)
-
ψ
2
(
t
k
)
)
≤
L
k
(
ψ
2
(
t
k
)
-
ψ
1
(
t
k
)
)
-
L
k
(
ω
1
(
t
k
)
-
ψ
1
(
t
k
)
)
+
L
k
(
ω
2
(
t
k
)
-
ψ
2
(
t
k
)
)
=
-
L
k
p
(
t
k
)
;
it is clear that p
(
0
)
=
p
(
T
). It follows by Case 1 of Lemma 3 that p
(
t
)
≤
0 for t
∈
J; that is, A is monotonely nondecreasing in [
u
0
,
v
0
].
It follows, from the properties(
a
) and (
b
), that
(44)
α
≤
u
1
≤
⋯
≤
u
n
≤
v
n
⋯
≤
v
1
≤
β
,
on
J
.
By standard arguments, we conclude that there exist u and v such that
(45)
lim
n
→
∞
u
n
=
u
(
t
)
,
lim
n
→
∞
v
n
=
v
(
t
)
,
uniformly
on
J
,
u
n
(
t
)
≤
u
(
t
)
≤
v
(
t
)
≤
v
n
(
t
)
.
It is easy to show that u and v are solutions of (1) using u
n, v
n satisfy the relations
(3)
u
n
+
1
′
(
t
)
=
-
M
1
(
u
n
+
1
(
t
)
)
-
M
2
(
u
n
+
1
(
θ
(
t
)
)
)
-
M
3
(
K
u
n
+
1
)
(
t
)
-
M
4
(
H
u
n
+
1
)
(
t
)
+
f
(
t
,
u
n
(
t
)
,
u
n
(
θ
(
t
)
)
,
(
K
u
n
)
(
t
)
,
(
H
u
n
)
(
t
)
)
+
M
1
u
n
(
t
)
+
M
2
u
n
(
θ
(
t
)
)
+
M
3
(
K
u
n
)
(
t
)
+
M
4
(
H
u
n
)
(
t
)
,
t
∈
J
′
,
Δ
u
n
+
1
(
t
k
)
=
I
k
(
u
n
(
t
k
)
)
-
L
k
(
u
n
+
1
(
t
k
)
-
u
n
(
t
k
)
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
i
k
=
1,2
,
…
,
m
,
u
n
+
1
(
0
)
=
u
n
+
1
(
T
)
,
v
n
+
1
′
(
t
)
=
-
M
1
(
v
n
+
1
(
t
)
)
-
M
2
(
v
n
+
1
(
θ
(
t
)
)
)
-
M
3
(
K
v
n
+
1
)
(
t
)
-
M
4
(
H
v
n
+
1
)
(
t
)
+
f
(
t
,
v
n
(
t
)
,
v
n
(
θ
(
t
)
)
,
(
K
v
n
)
(
t
)
,
(
H
v
n
)
(
t
)
)
+
M
1
v
n
(
t
)
+
M
2
v
n
(
θ
(
t
)
)
+
M
3
(
K
v
n
)
(
t
)
+
M
4
(
H
v
n
)
(
t
)
,
t
∈
J
′
,
Δ
v
n
+
1
(
t
k
)
=
I
k
(
v
n
(
t
k
)
)
-
L
k
(
v
n
+
1
(
t
k
)
-
v
n
(
t
k
)
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
k
=
1,2
,
…
,
m
,
v
n
+
1
(
0
)
=
v
n
+
1
(
T
)
.
Taking the limit as n
→
∞ on both sides of above relations, we have
(47)
u
′
(
t
)
=
f
(
t
,
u
(
t
)
,
u
(
θ
(
t
)
)
,
(
K
u
)
(
t
)
,
(
H
u
)
(
t
)
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
t
≠
t
k
,
t
∈
J
,
Δ
u
(
t
k
)
=
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
u
(
0
)
=
u
(
T
)
,
v
′
(
t
)
=
f
(
t
,
v
(
t
)
,
v
(
θ
(
t
)
)
,
(
K
v
)
(
t
)
,
(
H
v
)
(
t
)
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
i
t
≠
t
k
,
t
∈
J
,
Δ
v
(
t
k
)
=
I
k
(
v
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
v
(
0
)
=
v
(
T
)
.
Equation (47) show that u and v are solutions of (1). Finally, we prove that, if z
∈
[
α
,
β
] is any solution of (1), then u
(
t
)
≤
z
(
t
)
≤
v
(
t
) on J. To this end, we assume, without loss of generality, that u
n
(
t
)
≤
z
(
t
)
≤
v
n
(
t
) for some n, since α
(
t
)
≤
z
(
t
)
≤
β
(
t
). From property (b), we can get
(48)
u
n
+
1
(
t
)
≤
z
(
t
)
≤
v
n
+
1
(
t
)
,
t
∈
J
.
Since u
0
(
t
)
=
α
(
t
)
≤
z
(
t
)
≤
β
(
t
)
=
v
0
(
t
), t
∈
J, by induction, we can conclude that
(49)
u
n
(
t
)
≤
z
(
t
)
≤
v
n
(
t
)
,
∀
n
.
Passing to the limit as n
→
∞, we obtain
(50)
u
(
t
)
≤
z
(
t
)
≤
v
(
t
)
,
t
∈
J
.
This ends the proof.



---

*Source: 657629-2014-02-04.xml*