2 Preliminaries
In this section, we recall some basic definitions and properties from fractional calculus theory. For more details about fractional calculus, we refer the readers to 1310.
Definition 2.1 The Riemann-Liouville fractional derivative of order α
>
0
of a continuous function φ
:
(
0
,
+
∞
)
→
R
is given by
D
0
+
α
φ
(
t
)
=
1
Γ
(
n
−
α
)
(
d
d
t
)
(
n
)
∫
0
t
φ
(
s
)
(
t
−
s
)
α
−
n
+
1
d
s
,
where n
=
[
α
]
+
1
, [
α
]
denotes the integer part of number α, provided that the right side is pointwise defined on (
0
,
+
∞
)
.
Definition 2.2 The Riemann-Liouville fractional integral of order α>0 of a function φ:(0,+∞)→R is given by
I
0
+
α
φ
(
t
)
=
1
Γ
(
α
)
∫
0
t
(
t
−
s
)
α
−
1
φ
(
s
)
d
s
,
provided that the right side is pointwise defined on (0,+∞).
From the definition of the Riemann-Liouville derivative, we can obtain the following statement.
Lemma 2.1 (see 10)
Letα>0. If we assumeu
∈
C
(
0
,
1
)
∩
L
(
0
,
1
)
, then the fractional differential equation
D
0
+
α
u
(
t
)
=
0
hasu
(
t
)
=
c
1
t
α
−
1
+
c
2
t
α
−
2
+
⋯
+
c
N
t
α
−
N
,
c
i
∈
R
, i
=
1
,
2
,
…
,
N
as unique solutions, whereNis the smallest integer greater than or equal toα.
Lemma 2.2 (see 10)
Assume thatu∈C(0,1)∩L(0,1)with a fractional derivative of orderα>0that belongs toC
(
0
,
1
)
∩
L
(
0
,
1
)
. Then
I
0
+
α
D
0
+
α
u
(
t
)
=
u
(
t
)
+
c
1
t
α
−
1
+
c
2
t
α
−
2
+
⋯
+
c
N
t
α
−
N
,
for someci∈R, i=1,2,…,N, whereNis the smallest integer greater than or equal toα.
The Green function of fractional differential equation boundary value problem is given by
Lemma 2.3 (see 10)
Leth
∈
C
(
[
0
,
1
]
)
and3<α≤4. The unique solution to
D
0
+
α
u
(
t
)
=
h
(
t
)
,
0
<
t
<
1
,
u
(
0
)
=
u
(
1
)
=
u
′
(
0
)
=
u
′
(
1
)
=
0
,
is
u
(
t
)
=
∫
0
1
G
(
t
,
s
)
h
(
s
)
d
s
,
where
G
(
t
,
s
)
=
{
(
t
−
s
)
α
−
1
+
(
1
−
s
)
α
−
2
t
α
−
2
[
(
s
−
t
)
+
(
α
−
2
)
(
1
−
t
)
s
]
Γ
(
α
)
,
0
≤
s
≤
t
≤
1
,
t
α
−
2
(
1
−
s
)
α
−
2
[
(
s
−
t
)
+
(
α
−
2
)
(
1
−
t
)
s
]
Γ
(
α
)
,
0
≤
t
≤
s
≤
1
.
HereG
(
t
,
s
)
is called the Green function of boundary value problem (3)-(4).
The following properties of the Green function will be used later.
Lemma 2.4 (see 10)
The following properties hold:
(i) G
(
t
,
s
)
=
G
(
1
−
s
,
1
−
t
)
fort
,
s
∈
(
0
,
1
)
;
(ii) (
α
−
2
)
t
α
−
2
(
1
−
t
)
2
s
2
(
1
−
s
)
α
−
2
≤
Γ
(
α
)
G
(
t
,
s
)
≤
M
0
s
2
(
1
−
s
)
α
−
2
, t,s∈(0,1);
(iii) G
(
t
,
s
)
>
0
, fort,s∈(0,1);
(iv) (
α
−
2
)
s
2
(
1
−
s
)
α
−
2
t
α
−
2
(
1
−
t
)
2
≤
Γ
(
α
)
G
(
t
,
s
)
≤
M
0
t
α
−
2
(
1
−
t
)
2
, t,s∈(0,1), where
M
0
=
max
{
α
−
1
,
(
α
−
2
)
2
}
.
Let (
X
,
⪯
)
be a partially ordered set endowed with a metric d such that (
X
,
d
)
is complete metric space. Let F
:
X
×
X
→
X
be a given mapping.
Definition 2.3 We say that (X,⪯) is directed if for every (
x
,
y
)
∈
X
2
there exists z
∈
X
such that x
⪯
z
and y
⪯
z
.
Definition 2.4 We say that (
X
,
⪯
,
d
)
is regular if the following conditions hold: (
c
1
) = if {
x
n
}
is a nondecreasing sequence in X such that
x
n
→
x
∈
X
, then
x
n
⪯
x
for all n;; (
c
2
) = if {
y
n
}
is a decreasing sequence in X such that
y
n
→
y
∈
X
, then
y
n
⪰
y
for all n..
Example 2.1 Let X
=
C
(
[
0
,
T
]
)
, T
>
0
, be the set of real continuous functions on [
0
,
T
]
. We endow X with the standard metric d given by
d
(
u
,
v
)
=
max
0
≤
t
≤
T
|
u
(
t
)
−
v
(
t
)
|
,
u
,
v
∈
X
.
We define the partial order ⪯ on X by
u
,
v
∈
X
,
u
⪯
v
⟺
u
(
t
)
≤
v
(
t
)
for all
t
∈
[
0
,
T
]
.
Let u
,
v
∈
X
. For w
=
max
{
u
,
v
}
, that is, w
(
t
)
=
max
{
u
(
t
)
,
v
(
t
)
}
for all t
∈
[
0
,
T
]
, we have u
⪯
w
and v
⪯
w
. This implies that (X,⪯) is directed. Now, let {xn} be a nondecreasing sequence in X such that d
(
x
n
,
x
)
→
0
as n
→
∞
, for some x
∈
X
. Then, for all t∈[0,T], {
x
n
(
t
)
}
is a nondecreasing sequence of real numbers converging to x
(
t
)
. Thus we have
x
n
(
t
)
≤
x
(
t
)
for all n, that is, xn⪯x for all n. Similarly, if {yn} is a decreasing sequence in X such that d
(
y
n
,
y
)
→
0
as n→∞, for some y
∈
X
, we get that yn⪰y for all n. Then we proved that (
X
,
⪯
,
d
)
is regular.
Definition 2.5 (see 15)
An element (x,y)∈X2 is called a coupled fixed point of F if F
(
x
,
y
)
=
x
and F
(
y
,
x
)
=
y
.
Definition 2.6 (see 15)
We say that F has the mixed monotone property if for all (
x
,
y
)
, (
u
,
v
)
∈
X
2
, we have
x
⪯
u
,
y
⪰
v
⟹
F
(
x
,
y
)
⪯
F
(
u
,
v
)
.
Denote by Φ the set of functions φ
:
[
0
,
+
∞
)
→
[
0
,
+
∞
)
satisfying: (
φ
1
) = φ is continuous;; (
φ
2
) = φ is nondecreasing;; (
φ
3
) =
φ
−
1
(
{
0
}
)
=
{
0
}
..
The following two lemmas are fundamental in the proofs of our main results.
Lemma 2.5 (see 17)
Let(X,⪯)be a partially ordered set and suppose that there exists a metricdonXsuch that(X,d)is a complete metric space. LetF:X×X→Xbe a mapping having the mixed monotone property onXsuch that
ψ
(
d
(
F
(
x
,
y
)
,
F
(
u
,
v
)
)
)
≤
ψ
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
−
φ
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
for allx
,
y
,
u
,
v
∈
X
withx
⪰
u
andy
⪯
v
, whereψ
,
φ
∈
Φ
. Suppose also that(X,⪯,d)is regular and there exist
x
0
,
y
0
∈
X
such that
x
0
⪯
F
(
x
0
,
y
0
)
,
y
0
⪰
F
(
y
0
,
x
0
)
.
ThenFhas a coupled fixed point(
x
∗
,
y
∗
)
∈
X
2
. Moreover, if{
x
n
}
and{yn}are the sequences inXdefined by
x
n
+
1
=
F
(
x
n
,
y
n
)
,
y
n
+
1
=
F
(
y
n
,
x
n
)
,
n
=
0
,
1
,
…
,
then
lim
n
→
∞
d
(
x
n
,
x
∗
)
=
lim
n
→
∞
d
(
y
n
,
y
∗
)
=
0
.
Lemma 2.6 (see 17)
Adding to the hypotheses of Lemma 2.5 the condition(X,⪯)is regular, we obtain the uniqueness of the coupled fixed point. Moreover, we have the equality
x
∗
=
y
∗
.
3 Main result
Let Banach space E
=
C
(
[
0
,
1
]
)
be endowed with the norm
∥
u
∥
∞
=
max
0
≤
t
≤
1
|
u
(
t
)
|
. We define the partial order ⪯ on E by
u
,
v
∈
E
,
u
⪯
v
⟺
u
(
t
)
≤
v
(
t
)
for all
t
∈
[
0
,
1
]
.
In Example 2.1, we proved that (
E
,
⪯
)
with the classic metric given by
d
(
u
,
v
)
=
max
0
≤
t
≤
1
|
u
(
t
)
−
v
(
t
)
|
,
u
,
v
∈
E
satisfies the following properties: (E,⪯) is directed and (
E
,
⪯
,
d
)
is regular.
Define the closed cone P
⊂
E
by
P
=
{
u
∈
E
:
u
⪰
0
}
,
where 0 denotes the zero function.
Definition 3.1 (see 15)
We say that (
u
−
,
u
+
)
∈
C
(
[
0
,
1
]
)
×
C
(
[
0
,
1
]
)
is a coupled lower and upper solution to (1)-(2) if
u
−
(
t
)
≤
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
−
(
s
)
,
u
+
(
s
)
)
d
s
,
for all
0
≤
t
≤
1
and
u
+
(
t
)
≥
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
+
(
s
)
,
u
−
(
s
)
)
d
s
,
for all
0
≤
t
≤
1
.
Our main result is the following.
Theorem 3.1 Let0
<
σ
<
1
, 3<α≤4, f
:
(
0
,
1
)
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
→
[
0
,
+
∞
)
is continuous,
lim
t
→
0
+
f
(
t
,
⋅
,
⋅
)
=
+
∞
andt
↦
t
σ
f
(
t
,
x
,
y
)
is continuous on[
0
,
1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
. Assume that there exists0
<
λ
≤
(
1
−
σ
)
Γ
(
α
−
σ
+
1
)
/
(
2
Γ
(
3
−
σ
)
)
such that forx
,
y
,
z
,
w
∈
[
0
,
+
∞
)
withx
≥
z
, y
≤
w
andt
∈
[
0
,
1
]
,
0
≤
t
σ
(
f
(
t
,
x
,
y
)
−
f
(
t
,
z
,
w
)
)
≤
λ
η
(
max
{
(
x
−
z
)
,
(
w
−
y
)
}
)
,
whereη
:
[
0
,
+
∞
)
→
[
0
,
+
∞
)
, β
:
u
↦
u
−
η
(
u
)
∈
Φ
. Suppose also that (1)-(2) has a coupled lower and upper solution(
u
−
,
u
+
)
∈
P
×
P
. Then the boundary value problem (1)-(2) has a unique positive solution
u
∗
∈
C
(
[
0
,
1
]
)
. The sequences{
u
n
}
and{
v
n
}
defined by
u
0
=
u
−
,
u
n
+
1
=
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
n
(
s
)
,
v
n
(
s
)
)
d
s
,
n
=
0
,
1
,
…
,
v
0
=
u
+
,
v
n
+
1
=
∫
0
1
G
(
t
,
s
)
f
(
s
,
v
n
(
s
)
,
u
n
(
s
)
)
d
s
,
n
=
0
,
1
,
…
converge uniformly to
u
∗
.
Proof Suppose that u is a solution of boundary value problem (1)-(2). Then
u
(
t
)
=
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
(
s
)
,
u
(
s
)
)
d
s
,
t
∈
[
0
,
1
]
.
We define the operator F
:
P
×
P
→
E
by
F
(
u
,
v
)
(
t
)
=
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
,
t
∈
[
0
,
1
]
.
• Step 1. We shall prove that
F
(
P
×
P
)
⊆
P
.
Let u
,
v
∈
P
. Let us prove that F
(
u
,
v
)
∈
C
(
[
0
,
1
]
)
. We have
F
(
u
,
v
)
(
t
)
=
∫
0
1
G
(
t
,
s
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
,
t
∈
[
0
,
1
]
.
By the continuity of
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
in [
0
,
1
]
, it is easy to check that F
(
u
,
v
)
(
0
)
=
0
. Now, let
t
0
∈
[
0
,
1
]
. We have to prove that
|
F
(
u
,
v
)
(
t
)
−
F
(
u
,
v
)
(
t
0
)
|
→
0
as
t
→
t
0
.
We distinguish three cases:
t
0
=
0
,
t
0
∈
(
0
,
1
]
and
t
∈
(
t
0
,
1
]
,
t
0
∈
(
0
,
1
]
and
t
∈
[
0
,
t
0
)
.
Case 1.
t
0
=
0
. Since s
↦
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
is continuous on [0,1], there exists a constant M
>
0
such that |
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
|
≤
M
for all s
∈
[
0
,
1
]
. We have
|
F
(
u
,
v
)
(
t
)
−
F
(
u
,
v
)
(
t
0
)
|
=
|
∫
0
1
G
(
t
,
s
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
|
.
Using Lemma 2.3, we have
|
F
(
u
,
v
)
(
t
)
−
F
(
u
,
v
)
(
t
0
)
|
=
|
∫
0
1
t
α
−
2
(
1
−
s
)
α
−
2
[
(
s
−
t
)
+
(
α
−
2
)
(
1
−
t
)
s
]
Γ
(
α
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
+
∫
0
t
(
t
−
s
)
α
−
1
Γ
(
α
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
|
≤
|
∫
0
1
t
α
−
2
(
1
−
s
)
α
−
2
[
(
s
−
t
)
+
(
α
−
2
)
(
1
−
t
)
s
]
Γ
(
α
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
|
+
|
∫
0
t
(
t
−
s
)
α
−
1
Γ
(
α
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
|
≤
M
∫
0
1
(
α
−
1
)
t
α
−
2
(
1
−
s
)
α
−
2
Γ
(
α
)
s
1
−
σ
d
s
+
M
∫
0
1
t
α
−
1
(
1
−
s
)
α
−
2
Γ
(
α
)
s
−
σ
d
s
+
M
∫
0
1
(
α
−
2
)
t
α
−
1
(
1
−
s
)
α
−
2
Γ
(
α
)
s
1
−
σ
d
s
+
M
∫
0
t
(
t
−
s
)
α
−
1
Γ
(
α
)
s
−
σ
d
s
=
M
t
α
−
2
Γ
(
α
−
1
)
B
(
2
−
σ
,
α
−
1
)
+
M
t
α
−
1
Γ
(
α
)
B
(
1
−
σ
,
α
−
1
)
+
M
(
α
−
2
)
t
α
−
1
Γ
(
α
)
B
(
2
−
σ
,
α
−
1
)
+
M
t
α
−
σ
Γ
(
α
)
B
(
1
−
σ
,
α
)
→
0
as
t
→
0
+
,
where B
(
⋅
,
⋅
)
denotes the beta function.
Case 2.
t
0
∈
(
0
,
1
)
and t
∈
(
t
0
,
1
]
. In this case,
|
F
(
u
,
v
)
(
t
)
−
F
(
u
,
v
)
(
t
0
)
|
=
|
∫
0
1
(
α
−
1
)
(
t
α
−
2
−
t
0
α
−
2
)
(
1
−
s
)
α
−
2
s
1
−
σ
Γ
(
α
)
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
+
∫
0
1
(
t
0
α
−
1
−
t
α
−
1
)
(
1
−
s
)
α
−
2
s
−
σ
Γ
(
α
)
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
+
∫
0
1
(
α
−
2
)
(
t
0
α
−
1
−
t
α
−
1
)
(
1
−
s
)
α
−
2
s
1
−
σ
Γ
(
α
)
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
+
∫
0
t
0
[
(
t
−
s
)
α
−
1
−
(
t
0
−
s
)
α
−
1
]
s
−
σ
Γ
(
α
)
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
+
∫
t
0
t
(
t
−
s
)
α
−
1
s
−
σ
Γ
(
α
)
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
|
.
Now, we have
|
F
(
u
,
v
)
(
t
)
−
F
(
u
,
v
)
(
t
0
)
|
≤
M
(
α
−
1
)
(
t
α
−
2
−
t
0
α
−
2
)
Γ
(
α
)
∫
0
1
(
1
−
s
)
α
−
2
s
1
−
σ
d
s
+
M
(
t
α
−
1
−
t
0
α
−
1
)
Γ
(
α
)
∫
0
1
(
1
−
s
)
α
−
2
s
−
σ
d
s
+
M
(
α
−
2
)
(
t
α
−
1
−
t
0
α
−
1
)
Γ
(
α
)
∫
0
1
(
1
−
s
)
α
−
2
s
1
−
σ
d
s
+
M
Γ
(
α
)
∫
0
t
0
[
(
t
−
s
)
α
−
1
−
(
t
0
−
s
)
α
−
1
]
s
−
σ
d
s
+
M
Γ
(
α
)
∫
t
0
t
(
t
−
s
)
α
−
1
s
−
σ
d
s
=
M
(
t
α
−
2
−
t
0
α
−
2
)
Γ
(
α
−
1
)
B
(
2
−
σ
,
α
−
1
)
+
M
(
t
α
−
1
−
t
0
α
−
1
)
Γ
(
α
)
[
B
(
1
−
σ
,
α
−
1
)
+
(
α
−
2
)
B
(
2
−
σ
,
α
−
1
)
]
+
M
t
α
−
σ
Γ
(
α
)
B
(
1
−
σ
,
α
)
−
M
t
0
α
−
σ
Γ
(
α
)
B
(
1
−
σ
,
α
)
→
0
as
t
→
t
0
+
.
Case 3.
t
0
∈
(
0
,
1
]
and t
∈
[
0
,
t
0
)
. The proof is similar to that of Case 2, so we omit it.
Thus we proved that F
(
u
,
v
)
is continuous on [0,1] for all u
,
v
∈
C
(
[
0
,
1
]
)
. Moreover, taking into account Lemma 2.4 and as
t
σ
f
(
t
,
x
,
y
)
≥
0
for all t∈[0,1], x
,
y
≥
0
, our claim (10) is proved. Now the mapping
F
:
P
×
P
→
P
is well defined.
• Step 2. We shall prove that F has the mixed monotone property with respect to the partial order ⪯ given by (7).
Let (
x
,
y
)
,
(
u
,
v
)
∈
P
×
P
such that x
⪯
u
and y
⪰
v
. From (8), we have
s
σ
f
(
s
,
x
(
s
)
,
y
(
s
)
)
≤
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
,
for all s∈[0,1]. This implies that
∫
0
1
G
(
t
,
s
)
s
−
σ
s
σ
f
(
s
,
x
(
s
)
,
y
(
s
)
)
d
s
≤
∫
0
1
G
(
t
,
s
)
s
−
σ
s
σ
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
,
that is,
∫
0
1
G
(
t
,
s
)
f
(
s
,
x
(
s
)
,
y
(
s
)
)
d
s
≤
∫
0
1
G
(
t
,
s
)
f
(
s
,
u
(
s
)
,
v
(
s
)
)
d
s
,
which gives us that
F
(
x
,
y
)
(
t
)
≤
F
(
u
,
v
)
(
t
)
,
for all t∈[0,1], and then we have
F
(
x
,
y
)
⪯
F
(
u
,
v
)
.
Then F has the mixed monotone property.
• Step 3. We shall prove that F satisfies the contractive condition (5) for some ψ
,
φ
∈
Φ
.
Let (
x
,
y
)
,
(
u
,
v
)
∈
P
×
P
such that x
⪰
u
and y
⪯
v
. For all t∈[0,1], using (8), we have
|
F
(
x
,
y
)
(
t
)
−
F
(
u
,
v
)
(
t
)
|
=
∫
0
1
G
(
t
,
s
)
[
f
(
s
,
x
(
s
)
,
y
(
s
)
)
−
f
(
s
,
u
(
s
)
,
v
(
s
)
)
]
d
s
=
∫
0
1
G
(
t
,
s
)
s
−
σ
s
σ
[
f
(
s
,
x
(
s
)
,
y
(
s
)
)
−
f
(
s
,
u
(
s
)
,
v
(
s
)
)
]
d
s
≤
∫
0
1
G
(
t
,
s
)
s
−
σ
λ
η
(
max
{
x
(
s
)
−
u
(
s
)
,
v
(
s
)
−
y
(
s
)
}
)
d
s
≤
λ
η
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
∫
0
1
G
(
t
,
s
)
s
−
σ
d
s
≤
λ
η
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
max
z
∈
[
0
,
1
]
∫
0
1
G
(
z
,
s
)
s
−
σ
d
s
.
Thus we have
d
(
F
(
x
,
y
)
,
F
(
u
,
v
)
)
≤
λ
η
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
max
z
∈
[
0
,
1
]
∫
0
1
G
(
z
,
s
)
s
−
σ
d
s
.
Now, let z
∈
[
0
,
1
]
. We have
∫
0
1
G
(
z
,
s
)
s
−
σ
d
s
=
∫
0
1
z
α
−
2
(
1
−
s
)
α
−
2
[
(
s
−
z
)
+
(
α
−
2
)
(
1
−
z
)
s
]
Γ
(
α
)
s
−
σ
d
s
+
∫
0
z
(
z
−
s
)
α
−
1
Γ
(
α
)
s
−
σ
d
s
≤
∫
0
1
(
α
−
1
)
z
α
−
2
(
1
−
s
)
α
−
2
Γ
(
α
)
s
1
−
σ
d
s
+
∫
0
1
z
α
−
1
(
1
−
s
)
α
−
2
Γ
(
α
)
s
−
σ
d
s
+
∫
0
1
(
α
−
2
)
z
α
−
1
(
1
−
s
)
α
−
2
Γ
(
α
)
s
1
−
σ
d
s
+
∫
0
z
(
z
−
s
)
α
−
1
Γ
(
α
)
s
−
σ
d
s
≤
z
α
−
2
Γ
(
α
−
1
)
B
(
2
−
σ
,
α
−
1
)
+
z
α
−
1
Γ
(
α
)
B
(
1
−
σ
,
α
−
1
)
+
(
α
−
2
)
z
α
−
1
Γ
(
α
)
B
(
2
−
σ
,
α
−
1
)
+
z
α
−
σ
Γ
(
α
)
B
(
1
−
σ
,
α
)
≤
1
Γ
(
α
−
1
)
B
(
2
−
σ
,
α
−
1
)
+
1
Γ
(
α
)
B
(
1
−
σ
,
α
−
1
)
+
(
α
−
2
)
Γ
(
α
)
B
(
2
−
σ
,
α
−
1
)
+
1
Γ
(
α
)
B
(
1
−
σ
,
α
)
=
2
Γ
(
3
−
σ
)
(
1
−
σ
)
Γ
(
α
−
σ
+
1
)
,
which implies that
max
z
∈
[
0
,
1
]
∫
0
1
G
(
z
,
s
)
s
−
σ
d
s
≤
2
Γ
(
3
−
σ
)
(
1
−
σ
)
Γ
(
α
−
σ
+
1
)
.
Now, using the above inequality, (11) and the fact that λ
<
(
1
−
σ
)
Γ
(
α
−
σ
+
1
)
2
Γ
(
3
−
σ
)
, we get
d
(
F
(
x
,
y
)
,
F
(
u
,
v
)
)
≤
η
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
=
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
−
β
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
.
Thus we proved that for all (x,y),(u,v)∈P×P such that x⪰u and y⪯v, we have
ψ
(
d
(
F
(
x
,
y
)
,
F
(
u
,
v
)
)
)
≤
ψ
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
−
φ
(
max
{
d
(
x
,
u
)
,
d
(
y
,
v
)
}
)
,
where ψ
(
t
)
=
t
and φ
(
t
)
=
β
(
t
)
.
• Step 4. Existence of (
x
0
,
y
0
)
∈
P
×
P
such that
x
0
⪯
F
(
x
0
,
y
0
)
and
y
0
⪰
F
(
y
0
,
x
0
)
.
We take (
x
0
,
y
0
)
=
(
u
−
,
u
+
)
, the coupled lower and upper solution to (1)-(2).
Now, from Lemmas 2.5 and 2.6, there exists a unique
u
∗
∈
P
such that
u
∗
=
F
(
u
∗
,
u
∗
)
, that is u∗ is the unique positive solution to (1)-(2). The convergence of the sequences {un} and {
v
n
}
to u∗ follows immediately from (6). □
Now, we end this paper with the following example.
Example 3.1 Consider the boundary value problem
D
0
+
7
/
2
u
(
t
)
=
(
t
−
1
/
2
)
2
2
t
(
u
(
t
)
+
1
u
(
t
)
+
1
)
,
0
<
t
<
1
,
u
(
0
)
=
u
(
1
)
=
u
′
(
0
)
=
u
′
(
1
)
=
0
.
In this case, f
(
t
,
x
,
y
)
=
(
t
−
1
/
2
)
2
2
t
(
x
+
1
y
+
1
)
, for (
t
,
x
,
y
)
∈
(
0
,
1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
. Note that f is continuous on (
0
,
1
]
×
[
0
,
+
∞
)
×
[
0
,
+
∞
)
and
lim
t
→
0
+
f
(
t
,
⋅
,
⋅
)
=
+
∞
. Let σ
=
λ
=
1
/
2
and η
(
t
)
=
(
1
/
2
)
t
. For all x,y,z,w∈[0,+∞) with x≥z, y≤w and t∈[0,1], we have
0
≤
t
1
/
2
(
f
(
t
,
x
,
y
)
−
f
(
t
,
z
,
w
)
)
=
(
t
−
1
/
2
)
2
2
[
(
x
−
z
)
+
(
w
−
y
)
]
≤
(
t
−
1
/
2
)
2
max
{
x
−
z
,
w
−
y
}
≤
1
4
max
{
x
−
z
,
w
−
y
}
=
λ
η
(
max
{
x
−
z
,
w
−
y
}
)
.
On the other hand,
(
1
−
σ
)
Γ
(
α
−
σ
+
1
)
2
Γ
(
3
−
σ
)
=
3
2
Γ
(
5
/
2
)
=
2
π
>
λ
=
1
2
.
Consider now, the pair (u−,u+)∈P×P defined by
u
−
≡
0
and
u
+
≡
7
/
17
. Using Lemma 2.4(iv), one can show easily that (
u
−
,
u
+
)
is a coupled lower and upper solution to (12)-(13).
Finally, applying Theorem 3.1, we deduce that (12)-(13) has one and only one positive solution u∗∈C([0,1]).