2 Preliminaries and lemmas
In this section, we shall state some necessary definitions and preliminaries results.
Definition 2.1 Let E be a real Banach space. A nonempty closed set
P
⊂
E
is called a cone if it satisfies the following two conditions:
(1)
x
∈
P
,
λ
>
0
implies
λ
x
∈
P
;
(2)
x
∈
P
,
−
x
∈
P
implies
x
=
0
.
A cone is said solid if it contains interior points,
P
∘
≠
θ
. A cone P is called to be generating if
E
=
P
−
P
, i.e., every element
y
∈
E
can be represented in the form
y
=
x
−
z
, where
x
,
z
∈
P
. A cone P in E induces a partial ordering in E given by
u
≤
v
if
v
−
u
∈
P
. If
u
≤
v
and
u
≠
v
, we write
u
<
v
; if cone P is solid and
v
−
u
∈
P
∘
, we write
u
≪
v
.
Definition 2.2 A cone
P
⊂
E
is said to be normal if there exists a positive constant N such that
∥
x
+
y
∥
≥
N
,
∀
x
,
y
∈
P
,
∥
x
∥
=
1
,
∥
y
∥
=
1
.
Definition 2.3 Let E be a metric space and S be a bounded subset of E. The measure of non-compactness
ϒ
(
S
)
of S is defined by
ϒ
(
S
)
=
inf
{
δ
>
0
:
S
admits a finite cover by sets of diameter
≤
δ
}
.
Definition 2.4 An operator
B
:
D
⟶
E
is said to be completely continuous if it is continuous and compact. B is called a k-set-contraction (
k
≥
0
) if it is continuous, bounded and
ϒ
(
B
(
S
)
)
≤
k
ϒ
(
S
)
for any bounded set
S
⊂
D
, where
ϒ
(
S
)
denotes the measure of noncompactness of S.
A k-set-contraction is called a strict-set contraction if
k
<
1
. An operator B is said to be condensing if it is continuous, bounded, and
ϒ
(
B
(
S
)
)
<
ϒ
(
S
)
for any bounded set
S
⊂
D
with
ϒ
(
S
)
>
0
.
Obviously, if B is a strict-set contraction, then B is a condensing mapping, and if operator B is completely continuous, then B is a strict-set contraction.
It is well known that
y
∈
C
2
(
0
,
1
)
∩
C
[
0
,
1
]
is a solution of the problem (1.1) if and only if
x
∈
C
[
0
,
1
]
is a solution of the following nonlinear integral equation:
y
(
t
)
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
{
∑
k
=
1
m
[
γ
(
1
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
+
δ
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
}
,
where
G
(
t
,
s
)
=
1
ρ
{
(
γ
+
δ
−
γ
t
)
(
β
+
α
s
)
,
0
≤
s
≤
t
≤
1
,
(
β
+
α
t
)
(
γ
+
δ
−
γ
s
)
,
0
≤
t
≤
s
≤
1
,
where
ρ
=
γ
β
+
α
γ
+
α
δ
>
0
. In what follows, we write
J
1
=
[
0
,
t
1
]
,
J
k
=
(
t
k
−
1
,
t
k
]
(
k
=
1
,
2
,
…
,
m
),
J
m
=
(
t
m
,
1
]
. By making use of (2.1), we can prove that
G
(
t
,
s
)
has the following properties.
Proposition 2.1
0
≤
G
(
t
,
s
)
≤
G
(
s
,
s
)
≤
1
ρ
(
α
+
δ
)
(
α
+
β
)
,
t
,
s
∈
[
0
,
1
]
.
Proposition 2.2
0
≤
σ
G
(
t
,
s
)
≤
G
(
t
,
s
)
,
t
,
s
∈
[
a
,
b
]
⊂
[
0
,
1
]
, where
a
∈
(
0
,
t
1
]
,
b
∈
[
t
m
,
1
)
and
0
<
σ
=
min
{
β
+
α
a
α
+
β
,
δ
+
(
1
−
b
)
γ
γ
+
δ
}
<
1
.
Let
P
C
[
J
,
E
]
=
{
y
:
y
is a map from
J
into
E
such that
y
(
t
)
is continuous at
t
≠
t
k
, left
continuous at
t
=
t
k
and
y
(
t
k
+
)
exists for
k
=
1
,
…
,
m
}
and
P
C
[
J
,
P
]
=
{
y
∈
P
C
[
J
,
E
]
:
y
(
t
)
≥
θ
}
. It is easy to verify
P
C
[
J
,
E
]
is a Banach space with norm
∥
y
∥
P
C
=
sup
t
∈
J
∥
y
(
t
)
∥
. Obviously,
P
C
[
J
,
P
]
is a cone in Banach space
P
C
[
J
,
E
]
.
Let
P
C
1
[
J
,
E
]
=
{
y
:
y
is a map from
J
into
E
such that
y
′
(
t
)
exist and is continuous at
t
≠
t
k
,
y
(
t
)
left continuous at
t
=
t
k
, and
y
′
(
t
k
+
)
,
y
′
(
t
k
−
)
exist for
k
=
1
,
…
,
m
}
,
P
C
1
[
J
,
P
]
=
{
y
∈
P
C
1
[
J
,
E
]
:
y
(
t
)
≥
θ
,
y
′
(
t
)
≥
θ
}
. It is easy to see that
P
C
1
[
J
,
E
]
is a Banach space with the norm
∥
y
∥
1
=
max
{
∥
y
∥
P
C
,
∥
y
′
∥
P
C
}
. Evidently,
∥
y
∥
1
≤
∥
y
∥
P
C
+
∥
y
′
∥
P
C
and
P
C
1
[
J
,
P
]
is a cone in Banach space
P
C
1
[
J
,
E
]
. For any
y
∈
P
C
1
[
J
,
E
]
, by making use of the mean value theorem
y
(
t
k
)
−
y
(
t
k
−
h
)
∈
h
c
o
¯
{
y
′
(
t
)
:
t
k
−
h
<
t
<
t
k
}
(
h
>
0
), obviously we see that
y
′
(
t
k
−
)
exists and
y
′
(
t
k
−
)
=
lim
h
→
0
−
y
(
t
k
)
−
y
(
t
k
−
h
)
h
.
Let
K
=
{
y
∈
P
C
[
J
¯
,
P
]
:
y
≥
σ
∥
y
∥
,
t
∈
[
a
,
b
]
}
. For any
0
<
r
<
R
<
+
∞
, let
K
r
=
{
y
∈
K
:
∥
y
∥
<
r
}
,
∂
K
r
=
{
y
∈
K
:
∥
y
∥
=
r
}
,
∂
K
¯
r
,
R
=
{
y
∈
K
:
r
≤
∥
y
∥
≤
R
}
.
A map
y
∈
P
C
1
[
J
,
E
]
∩
C
2
[
J
′
,
E
]
is called a nonnegative solution of problem (1.1) if
y
(
t
)
≥
θ
,
y
′
(
t
)
≥
θ
for
t
∈
J
and
y
(
t
)
satisfies problem (1.1). An operator
y
∈
P
C
1
[
J
,
E
]
∩
C
2
[
J
′
,
E
]
is called a positive solution of problem (1.1) if y is a nonnegative solution of problem (1.1) and
y
(
t
)
≢
θ
.
For convenience and simplicity in the following discussion, we denote
![]()
where ν denote 0 or ∞.
To establish the existence of multiple positive solutions in E of problem (1.1), let us list the following assumptions, which will stand throughout the paper:
(H1)
f
∈
C
(
J
×
P
4
,
P
)
,
h
∈
C
(
J
,
P
)
and
∥
f
(
t
,
y
1
,
y
2
,
y
3
,
y
4
)
∥
≤
a
(
t
)
+
∑
i
=
1
4
b
i
(
t
)
∥
y
i
∥
,
where
a
(
t
)
and
b
i
(
t
)
are Lebesgue integrable functionals on J (
i
=
1
,
2
,
3
,
4
) and satisfying
![]()
(H2)
I
k
∈
C
(
P
,
P
)
,
I
¯
k
∈
C
(
P
,
P
)
and there exist positive constants
c
k
,
c
k
∗
and
c
¯
k
(
k
=
1
,
…
,
m
) satisfying
∑
k
=
1
m
(
c
k
+
c
k
∗
+
c
¯
k
)
+
1
ρ
(
α
+
β
)
∑
k
=
1
m
[
(
γ
+
δ
)
(
c
k
∗
+
c
¯
k
)
+
γ
c
k
]
<
1
4
such that
∥
I
k
(
y
)
∥
≤
c
k
∥
y
∥
,
∥
I
¯
k
(
y
1
,
y
2
)
∥
≤
c
k
∗
∥
y
1
∥
+
c
¯
k
∥
y
2
∥
,
∀
t
∈
J
,
(H3) for any bounded set
B
i
⊂
E
,
i
=
1
,
2
,
3
,
4
,
f
(
t
,
B
1
,
B
2
,
B
3
,
B
4
)
and
I
k
(
B
1
)
together with
I
¯
k
(
B
1
,
B
2
)
are relatively compact sets,
(H4)
f
0
>
m
,
(H5)
f
∞
<
m
, where
m
=
max
{
(
σ
∫
a
b
G
(
s
,
s
)
h
(
s
)
d
s
)
−
1
,
(
∫
a
b
G
t
′
(
t
,
s
)
h
(
s
)
d
s
)
−
1
,
1
ρ
(
α
+
β
)
(
γ
+
δ
)
∫
0
1
h
(
s
)
a
(
s
)
d
s
}
.
We shall reduce problem (1.1) to an integral equation in E. To this end, we first consider operator
A
:
K
⟶
P
C
[
J
¯
,
E
]
defined by
(
A
y
)
(
t
)
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
{
∑
k
=
1
m
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
∑
k
=
1
m
I
k
(
y
(
t
k
)
)
}
.
Lemma 2.1
y
∈
P
C
1
[
J
¯
,
E
]
∩
C
2
[
J
′
,
E
]
is a solution of problem (1.1) if and only if
y
∈
P
C
1
[
J
¯
,
E
]
is a solution of the following impulsive integral equation:
y
(
t
)
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
i.e., y
is a fixed point of operator
A
defined by (2.6) in
P
C
1
[
J
,
E
]
.
Proof First suppose that
y
∈
P
C
1
[
J
,
E
]
is a solution of problem (1.1). It is easy to see by the integration of problem (1.1) that
y
′
(
t
)
=
y
′
(
0
)
−
∫
0
t
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
y
′
(
t
k
+
)
−
y
′
(
t
k
−
)
]
=
y
′
(
0
)
−
∫
0
t
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
−
∑
0
<
t
k
<
t
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
,
∀
t
∈
J
.
Integrate again, we get
y
(
t
)
=
y
(
0
)
+
y
′
(
0
)
t
−
∫
0
t
(
t
−
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
,
∀
t
∈
J
.
Letting
t
=
1
in (2.8) and (2.9), we find that
y
′
(
1
)
=
y
′
(
0
)
−
∫
0
1
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
−
∑
k
=
1
m
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
.
y
(
1
)
=
y
(
0
)
+
y
′
(
0
)
−
∫
0
1
(
1
−
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
k
=
1
m
[
I
k
(
y
(
t
k
)
)
−
(
1
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
.
Since
y
(
0
)
=
β
α
y
′
(
0
)
,
y
(
1
)
=
−
δ
γ
{
y
′
(
0
)
−
∫
0
1
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
−
∑
k
=
1
m
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
}
.
We get
y
′
(
0
)
=
α
ρ
{
δ
∫
0
1
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
γ
∫
0
1
(
1
−
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
}
,
∀
t
∈
J
.
Substituting (2.12) and (2.13) into (2.9), we obtain
y
(
t
)
=
1
ρ
(
α
t
+
β
)
{
∫
0
1
(
γ
(
1
−
s
)
+
δ
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
−
∑
k
=
1
m
[
γ
I
k
(
y
(
t
k
)
)
−
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
}
−
∫
0
t
γ
(
t
−
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
=
1
ρ
∫
0
t
[
(
α
t
+
β
)
(
γ
(
1
−
s
)
+
δ
)
−
ρ
(
t
−
s
)
]
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
1
ρ
(
α
t
+
β
)
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
.
Conversely, if
y
∈
P
C
1
[
J
,
E
]
is a solution of the integral equation (2.7). Evidently,
Δ
y
|
t
=
t
k
=
I
k
(
y
(
t
k
)
)
(
k
=
1
,
…
,
m
). For
t
=
t
k
, direct differentiation of the integral equation (2.7) implies
y
′
(
t
)
=
1
ρ
{
−
γ
∫
0
t
(
α
s
+
β
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
α
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
−
∑
0
<
t
k
<
t
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
}
−
α
ρ
∑
k
=
1
m
[
γ
I
k
(
y
(
t
k
)
)
−
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
and
y
′
′
=
−
f
(
t
,
y
(
t
)
,
y
′
(
t
)
,
(
T
y
)
(
t
)
,
(
S
y
)
(
t
)
)
. So
y
∈
C
2
[
J
′
,
E
]
and
−
Δ
y
′
|
t
=
t
k
=
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
(
k
=
1
,
…
,
m
). It is easy to verify that
α
y
(
0
)
−
β
y
′
(
0
)
=
0
and
γ
y
(
1
)
+
δ
y
′
(
1
)
=
0
. The proof is complete. □
Thanks to (2.1), we know that
G
t
′
(
t
,
s
)
=
1
ρ
{
−
γ
(
β
+
α
s
)
,
0
≤
s
≤
t
≤
1
,
α
(
γ
+
δ
−
γ
s
)
,
0
≤
t
≤
s
≤
1
.
In the following, let
w
′
=
sup
t
,
s
∈
J
,
t
≠
s
|
G
t
′
(
t
,
s
)
|
. For
B
⊂
P
C
1
[
J
,
E
]
, we denote
B
′
=
{
y
′
:
y
′
∈
B
}
⊂
P
C
[
J
,
E
]
,
B
(
t
)
=
{
y
(
t
)
:
y
∈
B
}
⊂
E
and
B
′
(
t
)
=
{
y
′
(
t
)
:
y
′
∈
B
}
⊂
E
(
t
∈
J
).
Lemma 2.2 (
12
)
Let
D
⊂
P
C
1
[
J
,
E
]
be a bounded set. Suppose that
D
′
(
t
)
is equi-continuous on each
J
k
(
k
=
1
,
…
,
m
). Then
ϒ
P
C
1
(
D
)
=
max
{
sup
t
∈
J
ϒ
(
D
(
t
)
)
,
sup
t
∈
J
ϒ
(
D
′
(
t
)
)
}
,
where ϒ and
ϒ
P
C
1
denote the Kuratowski measures of noncompactness of bounded sets in
E
and
P
C
1
[
J
,
E
]
, respectively.
Lemma 2.3 (
15
)
Let
H
⊂
P
C
[
J
,
E
]
be bounded equicontinuous, then
ϒ
(
H
(
t
)
)
is continuous on
J
and
ϒ
(
{
∫
J
y
(
t
)
d
t
:
y
∈
H
}
)
≤
∫
J
ϒ
(
H
(
t
)
)
d
t
.
Lemma 2.4 (
15
)
H
⊂
P
C
1
[
J
,
E
]
is relatively compact if and only if each element
y
(
t
)
∈
H
and
y
′
(
t
)
∈
H
are uniformly bounded and equicontinuous on each
J
k
(
k
=
1
,
…
,
m
).
Lemma 2.5 (
15
)
Let
E
be a Banach space and
H
⊂
C
[
J
,
E
]
if
H
is countable and there exists
φ
∈
L
[
J
,
R
+
]
such that
∥
y
(
t
)
∥
≤
φ
(
t
)
,
t
∈
J
,
y
∈
H
. Then
ϒ
(
{
y
(
t
)
:
y
∈
H
}
)
is integrable on
J, and
ϒ
(
{
∫
J
y
(
t
)
d
t
:
y
∈
H
}
)
≤
2
∫
J
ϒ
{
y
(
t
)
:
y
∈
H
}
d
t
.
Lemma 2.6
A
(
K
)
⊂
K
.
Proof For any
y
∈
K
, from Proposition 2.1 and (2.6), we obtain
∥
A
y
∥
≤
∫
0
1
G
(
s
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
{
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
−
∑
k
=
1
m
γ
I
k
(
y
(
t
k
)
)
}
.
On the other hand, for any
t
∈
[
a
,
b
]
, by (2.6) and Proposition 2.2, we know that
A
y
(
t
)
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
≥
σ
∫
0
1
G
(
s
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
{
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
}
≥
σ
∥
A
y
∥
.
Hence,
A
(
K
)
⊂
K
. □
Lemma 2.7
Suppose that (H
1) and (H
3) hold. Then
A
:
K
⟶
K
is completely continuous.
Proof Firstly, we show that
A
:
K
⟶
K
is continuous. Assume that
y
n
,
y
0
∈
K
and
∥
y
n
−
y
0
∥
⟶
0
,
∥
y
n
′
−
y
0
′
∥
⟶
0
(
n
⟶
∞
). Since
f
∈
C
(
J
¯
×
P
×
P
×
P
×
P
,
P
)
,
I
k
∈
C
(
J
,
P
)
and
I
¯
k
∈
C
(
J
,
P
)
, then
![]()
and
lim
n
⟶
∞
∥
I
¯
k
(
y
n
(
t
k
)
,
y
n
′
(
t
k
)
)
−
I
¯
k
(
y
0
(
t
k
)
,
y
0
′
(
t
k
)
)
∥
=
0
,
k
=
1
,
…
,
m
.
Thus, for any
t
∈
J
, from the Lebesgue dominated convergence theorem together with (2.14) and (2.15), we know that
![]()
Hence,
A
:
K
⟶
K
is continuous.
Let
B
⊂
K
be any bounded set, then there exists a positive constant
R
0
∗
such that
∥
y
∥
1
≤
R
0
∗
. Thus, for any
y
∈
B
,
t
∈
(
0
,
1
)
, we know that
|
(
A
y
)
′
(
t
)
|
=
1
ρ
|
{
−
γ
∫
0
t
(
α
s
+
β
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
α
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
}
−
∑
0
<
t
k
<
t
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
+
α
ρ
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
|
.
Therefore,
|
(
A
y
)
′
(
t
)
|
≤
1
ρ
{
−
γ
∫
0
t
(
α
s
+
β
)
h
(
s
)
|
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
|
d
s
+
α
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
h
(
s
)
|
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
|
d
s
}
+
∑
0
<
t
k
<
t
|
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
|
+
α
ρ
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
+
γ
|
I
k
(
y
(
t
k
)
)
|
]
≤
1
ρ
{
−
γ
∫
0
t
(
α
s
+
β
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
∥
y
∥
1
)
d
s
+
α
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
∥
y
(
s
)
∥
1
)
d
s
}
+
∑
k
=
1
m
(
c
k
∗
+
c
¯
k
)
∥
y
(
s
)
∥
1
+
α
ρ
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
(
c
k
∗
+
c
¯
k
)
+
γ
c
k
]
∥
y
(
s
)
∥
1
≤
1
ρ
{
−
γ
∫
0
t
(
α
s
+
β
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
+
α
∫
t
1
(
γ
(
1
−
s
)
+
δ
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
}
+
∑
k
=
1
m
(
c
k
∗
+
c
¯
k
)
R
0
∗
+
α
ρ
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
(
c
k
∗
+
c
¯
k
)
+
γ
c
k
]
R
0
∗
,
∀
t
∈
J
,
t
≠
t
k
,
k
=
1
,
…
,
m
. Let
![]()
Integrating
ψ
(
t
)
from 0 to 1 and exchanging integral sequence, then
∫
0
1
ψ
(
t
)
d
t
=
∫
0
1
∫
0
s
α
(
γ
+
δ
−
γ
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
t
d
s
−
γ
∫
0
1
∫
s
1
(
β
+
α
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
t
d
s
≤
α
∫
0
1
s
(
γ
+
δ
−
γ
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
−
γ
∫
0
1
(
β
+
α
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
≤
ρ
∫
0
1
G
(
s
,
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
<
+
∞
.
Thus, by (H1) and (2.17), we have
ψ
(
t
)
∈
L
1
(
J
)
. Hence, for any
0
≤
t
1
≤
t
2
≤
1
and for all
y
∈
E
, from (2.16), we know that
∥
(
A
y
)
(
t
1
)
−
(
A
y
)
(
t
2
)
∥
=
|
∫
t
1
t
2
(
A
y
)
′
(
t
)
d
t
|
≤
∫
t
1
t
2
(
ψ
(
t
)
+
M
1
)
d
t
.
From (2.17), (2.18), and the absolutely continuity of integral function, we see that
A
(
B
)
is equicontinuous.
On the other hand, for any
y
∈
B
and
t
∈
J
, we know that
|
(
A
y
)
(
t
)
|
=
|
∫
0
1
G
(
t
,
s
)
h
(
s
)
f
(
s
,
y
(
s
)
,
y
′
(
s
)
,
(
T
y
)
(
s
)
,
(
S
y
)
(
s
)
)
d
s
+
∑
0
<
t
k
<
t
[
I
k
(
y
(
t
k
)
)
−
(
t
−
t
k
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
]
+
1
ρ
(
α
t
+
β
)
{
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
I
¯
k
(
y
(
t
k
)
,
y
′
(
t
k
)
)
−
γ
I
k
(
y
(
t
k
)
)
]
}
|
≤
∫
0
1
G
(
s
,
s
)
h
(
s
)
(
a
(
s
)
+
(
∑
i
=
1
2
b
i
(
s
)
+
K
0
b
3
(
s
)
+
H
0
b
4
(
s
)
)
R
0
∗
)
d
s
+
∑
k
=
1
m
(
c
k
∗
+
c
¯
k
)
R
0
∗
)
+
α
ρ
∑
k
=
1
m
[
(
γ
(
1
−
t
k
)
+
δ
)
(
c
k
∗
+
c
¯
k
)
+
γ
c
k
]
R
0
∗
<
+
∞
.
Therefore,
A
(
B
)
is uniformly bounded. By virtue of Lemma 2.3 and (H3), we know that
![]()
So,
ϒ
(
A
(
B
)
)
=
0
. Therefore, A is compact. To sum up, the conclusion of Lemma 2.7 follows. □
The main tools of the paper are the following well-known fixed-point index theorems (see
2
3
4
).
Lemma 2.8
Let
A
:
K
→
K
be a completely continuous mapping and
A
y
≠
y
for
y
∈
∂
K
r
. Thus, we have the following conclusions:
(i) If
∥
y
∥
≤
∥
A
y
∥
for
y
∈
∂
K
r
, then
i
(
A
,
K
r
,
K
)
=
0
.
(ii) If
∥
y
∥
≥
∥
A
y
∥
for
y
∈
∂
K
r
, then
i
(
A
,
K
r
,
K
)
=
1
.