# Positive Solutions for a Class of Singular Boundary Value Problems with Fractionalq-Difference Equations

Authors: Jufang Wang; Changlong Yu; Yanping Guo
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We discuss a class of singular
boundary value problems of fractionalq-difference equations. Some
existence and uniqueness results are obtained by a fixed point
theorem in partially ordered sets. Finally, we give an example to
illustrate the results.


---

## Body

## 1. Introduction

In recent years, many papers on fractional differential equations have appeared, because of their demonstrated applications in various fields of science and engineering; see [1–11] and the references therein. Based on the increasingly extensive application of discrete fractional calculus and the development of q-difference calculus or quantum calculus (see [12–19] and the references therein), fractional q-difference equations have attracted the attention of researchers for the numerous applications in a number of fields such as physics, chemistry, aerodynamics, biology, economics, control theory, mechanics, electricity, signal and image processing, biophysics, blood flow phenomena, aerodynamics, and fitting of experimental data; see [20–23]. Some recent work on the existence theory of fractional q-difference equations can be found in [24–29]. However, the study of singular boundary value problems (BVPs) with fractional q-difference equations is at its infancy and lots of work on the topic should be done.Recently, in [25], Ferreira has investigated the existence of positive solution for the following fractional q-difference equations BVP (1)
D
q
α
y
x
+
f
x
,
y
t
=
0
,
0
<
x
<
1
,
y
0
=
D
q
y
0
=
0
,
D
q
y
1
=
β
≥
0
,by applying a fixed point theorem in cones.More recently, in [30], Caballero et al. have studied positive solutions for the following BVP: (2)
D
0
+
α
u
t
+
f
t
,
u
t
=
0
,
0
<
t
<
1
,
u
(
0
)
=
u
(
1
)
=
0
,by fixed point theorem in partially ordered sets.Motivated by the work above, in this paper, we discuss the existence and uniqueness of solutions for the singular BVPs of factionalq-difference equations given by (3)
(
D
q
α
u
)
(
t
)
+
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
u
(
1
)
=
0
,
(
D
q
u
)
(
0
)
=
0
,where 2
<
α
≤
3 and f
:
[
0,1
]
×
[
0
,
∞
)
→
[
0
,
∞
) is continuous with lim
t
→
0
+
⁡
f
(
t
,
·
)
=
∞ (i.e., f is singular at t
=
0).


## 2. Preliminary Results

For convenience, we present some definitions and lemmas which will be used in the proofs of our results.Letq
∈
(
0,1
) and define (4)
a
q
=
1
-
q
a
1
-
q
,
a
∈
R
.Theq-analogue of the power function (
a
-
b
)
n with n
∈
N
0 is (5)
a
-
b
0
=
1
,
a
-
b
n
=
∏
k
=
0
n
-
1
‍
a
-
b
q
k
,
n
∈
N
,
a
,
b
∈
R
.More generally, if α
∈
R, then (6)
a
-
b
(
α
)
=
a
α
∏
n
=
0
n
-
1
‍
a
-
b
q
n
a
-
b
q
α
+
n
.Note that if b
=
0 then a
(
α
)
=
a
α. The q-gamma function is define by (7)
Γ
q
x
=
1
-
q
(
x
-
1
)
1
-
q
x
-
1
,
x
∈
R
∖
{
0
,
-
1
,
-
2
,
…
}
,and it satisfies Γ
q
(
x
+
1
)
=
[
x
]
q
Γ
q
(
x
).Following, let us recall some basic concepts ofq-calculus [12].Definition 1.
For0
<
q
<
1, we define the q-derivative of a real-value function f as (8)
D
q
f
x
=
f
x
-
f
q
x
1
-
q
x
,
D
q
f
0
=
lim
x
→
0
D
q
f
x
.Note that lim
q
→
1
-
⁡
D
q
f
(
x
)
=
f
′
(
x
).Definition 2.
The higher orderq-derivatives are defined inductively as (9)
(
D
q
0
f
)
(
x
)
=
f
(
x
)
,
(
D
q
n
f
)
(
t
)
=
D
q
(
D
q
n
-
1
f
)
(
t
)
,
n
∈
N
.
For example,D
q
(
t
k
)
=
[
k
]
q
t
k
-
1, where k is a positive integer and the bracket [
k
]
q
=
(
q
k
-
1
)
/
(
q
-
1
). In particular, D
q
(
t
2
)
=
(
1
+
q
)
t.Definition 3.
Theq-integral of a function f in the interval [
0
,
b
] is given by (10)
I
q
f
x
=
∫
0
x
‍
f
t
d
q
t
=
x
1
-
q
∑
n
=
0
∞
‍
f
x
q
n
q
n
,
x
∈
0
,
b
.If a
∈
[
0
,
b
] and f is defined in the interval [
0
,
b
], its integral from a to b is define by (11)
∫
a
b
‍
f
(
t
)
d
q
t
=
∫
0
b
‍
f
(
t
)
d
q
t
-
∫
0
a
‍
f
(
t
)
d
q
t
.Similarly as done for derivatives, an operator I
q
n can be define, namely, (12)
(
I
q
0
f
)
(
x
)
=
f
(
x
)
,
(
I
q
n
f
)
(
x
)
=
I
q
(
I
q
n
-
1
f
)
(
x
)
,
n
∈
N
.
Observe that(13)
D
q
I
q
f
(
x
)
=
f
(
x
)
,and if f is continuous at x
=
0, then I
q
D
q
f
(
x
)
=
f
(
x
)
-
f
(
0
).
We now point out three formulas (i
D
q denotes the derivative with respect to variable i) (14)
a
t
-
s
(
α
)
=
a
α
t
-
s
(
α
)
,
(15)
t
D
q
t
-
s
(
α
)
=
α
q
t
-
s
(
α
-
1
)
,
D
x
q
∫
0
x
‍
f
x
,
t
d
q
t
=
∫
0
x
x
D
q
f
x
,
t
d
q
t
+
f
q
x
,
x
.Remark 4.
We note that ifα
≥
0 and a
≤
b
≤
t, then (
t
-
a
)
(
α
)
≥
(
t
-
b
)
(
α
) [24].Definition 5 (see [21]).
Letα
≥
0 and f be a function defined on [
0,1
]. The fractional q-integral of the Riemann-Liuville type is (
I
q
0
f
)
(
x
)
=
f
(
x
) and (16)
I
q
α
f
x
=
1
Γ
q
α
∫
0
x
‍
x
-
q
t
α
-
1
f
t
d
q
t
,
α
>
0
,
x
∈
0,1
.Definition 6 (see [23]).
The fractionalq-derivative of the Riemann-Liuville type of α
≥
0 is defined by (
D
q
0
f
)
(
x
)
=
f
(
x
) and (17)
(
D
q
α
f
)
(
x
)
=
(
D
q
m
I
q
m
-
α
f
)
(
x
)
,
α
>
0
,where m is the smallest integer greater than or equal to α.Lemma 7 (see [21, 23]).
Letα
,
β
≥
0 and let f be a function define on [
0,1
]. Then, the next formulas hold:(1)
(
I
q
β
I
q
α
f
)
(
x
)
=
(
I
q
α
+
β
f
)
(
x
),
(2)
(
D
q
α
I
q
α
f
)
(
x
)
=
f
(
x
).Lemma 8 (see [24]).
Letα
>
0 and let p be a positive integer. Then, the following equality holds: (18)
I
q
α
D
q
p
f
x
=
D
q
p
I
q
α
f
x
-
∑
k
=
0
p
-
1
‍
x
α
-
p
+
k
Γ
q
α
+
k
-
p
+
1
D
q
k
f
0
.Lemma 9.
Lety
(
t
)
∈
C
[
0,1
]
⋂
‍
L
1
[
0,1
] and 2
<
α
≤
3; then the BVP (19)
D
q
α
u
t
+
y
t
=
0
,
0
<
t
<
1
,
u
0
=
u
1
=
0
,
D
q
u
0
=
0
,has a unique solution(20)
u
(
t
)
=
∫
0
1
‍
G
(
t
,
q
s
)
y
(
s
)
d
q
s
,where(21)
G
t
,
s
=
1
Γ
q
(
α
)
1
-
s
(
α
-
1
)
t
α
-
1
-
t
-
s
(
α
-
1
)
,
0
≤
s
≤
t
≤
1
,
1
-
s
(
α
-
1
)
t
α
-
1
,
0
≤
t
≤
s
≤
1
.Proof.
By Lemmas7 and 8, we see that(22)
D
q
α
u
t
=
-
y
t
D
q
α
u
t
⟺
I
q
α
D
q
3
I
q
3
-
α
u
t
=
-
I
q
α
y
t
D
q
α
u
t
⟺
u
t
=
c
1
t
α
-
1
+
c
2
t
α
-
2
D
q
α
u
t
⟺
u
t
=
+
c
3
t
α
-
3
-
1
Γ
q
α
∫
0
t
t
-
q
s
α
-
1
D
q
α
u
t
⟺
u
t
=
c
3
t
α
-
3
1
Γ
q
α
×
y
s
d
q
s
,where c
1
,
c
2, and c
3 are some constants to be determined. Since u
(
0
)
=
0, we must have c
3
=
0. Now, differentiating both sides of (22) and using (15), we obtain (23)
D
q
u
t
=
α
-
1
q
c
1
t
α
-
2
+
α
-
2
q
c
2
t
α
-
3
D
q
u
t
-
1
Γ
q
(
α
)
∫
0
t
‍
α
-
1
q
t
-
q
s
(
α
-
2
)
y
(
s
)
d
q
s
.Using (
D
q
u
)
(
0
)
=
0 and u
(
1
)
=
0, we must set c
2
=
0, and (24)
c
1
=
1
Γ
q
(
α
)
∫
0
1
‍
1
-
q
s
(
α
-
1
)
y
(
s
)
d
q
s
.
Finally, we obtain(25)
u
t
=
t
α
-
1
Γ
q
α
∫
0
1
‍
1
-
q
s
α
-
1
y
s
d
q
s
u
t
-
1
Γ
q
α
∫
0
t
‍
t
-
q
s
α
-
1
y
s
d
q
s
u
t
=
∫
0
1
‍
G
(
t
,
q
s
)
y
(
s
)
d
q
s
.The proof is complete.Lemma 10.
FunctionG defined above satisfies the following conditions:(i)
G
(
t
,
q
s
) is a continuous function on [
0,1
]
×
[
0,1
];
(ii)
G
(
t
,
q
s
)
≥
0 for t
,
s
∈
[
0,1
].Proof.
(i) Obviously,G
(
t
,
q
s
) is continuous on [
0,1
]
×
[
0,1
].
(ii) Let(26)
g
1
t
,
s
=
1
-
s
α
-
1
t
α
-
1
-
t
-
s
α
-
1
,
0
≤
s
≤
t
≤
1
,
g
2
(
t
,
s
)
=
1
-
s
(
α
-
1
)
t
α
-
1
,
0
≤
t
≤
s
≤
1
.It is clear that g
2
(
t
,
q
s
)
≥
0, for t
,
s
∈
[
0,1
]. Now, in view of Remark 4, for t
≠
0
(27)
g
1
t
,
q
s
=
1
-
q
s
α
-
1
t
α
-
1
-
t
-
q
s
α
-
1
g
1
t
,
q
s
=
t
α
-
1
1
-
q
s
α
-
1
-
1
-
q
s
t
α
-
1
g
1
t
,
q
s
≥
t
α
-
1
1
-
q
s
α
-
1
-
1
-
q
s
α
-
1
=
0
.Therefore, G
(
t
,
q
s
)
≥
0. This proof is complete.ByJ we denote the class of those functions β
:
[
0
,
∞
)
→
[
0,1
) satisfying the following condition; β
(
t
n
)
→
1 implies t
n
→
0.Theorem 11 (see [31]).
Let(
X
,
≤
) be a partially ordered set and suppose that there exists a metric d in X such that (
X
,
d
) is a complete metric space. Let T
:
X
→
X be a nondecreasing mapping such that there exists an element x
0
∈
X with x
0
≤
T
x
0. Suppose that there exists β
∈
J such that(28)
d
T
x
,
T
y
≤
β
d
x
,
y
·
d
x
,
y
f
o
r
x
,
y
∈
X
w
i
t
h
x
≥
y
.Assume that either T is continuous or X is such that (29)
i
f
x
n
i
s
a
n
o
n
d
e
c
r
e
a
s
i
n
g
s
e
q
u
e
n
c
e
i
n
X
s
u
c
h
t
h
a
t
x
n
⟶
x
t
h
e
n
x
n
≤
x
∀
n
∈
N
.Besides if (30)
f
o
r
e
a
c
h
x
,
y
∈
X
t
h
e
r
e
e
x
i
s
t
s
z
∈
X
w
h
i
c
h
i
s
c
o
m
p
a
r
a
b
l
e
t
o
x
a
n
d
y
,then T has a unique fixed point.LetC
[
0,1
]
=
{
x
:
[
0,1
]
→
R
,
continuous
} be the Banach space with the classic metric given by d
(
x
,
y
)
=
sup
⁡
0
≤
t
≤
1
{
|
x
(
t
)
-
y
(
t
)
|
}.Notice that this space can be equipped with a partial order given by(31)
x
,
y
∈
C
[
0,1
]
,
x
≤
y
⟺
x
(
t
)
≤
y
(
t
)
,
for
t
∈
[
0,1
]
.In [32], it is proved that (
C
[
0,1
]
,
≤
) satisfies condition (29) of Theorem 11. Moreover, for x
,
y
∈
C
[
0,1
], as the function max
⁡
(
x
,
y
)
∈
C
[
0,1
]
,
(
C
[
0,1
]
,
≤
) satisfies condition (30).


## 3. Main Result

In this section, we will consider the question of positive solutions for BVP (3). At first, we prove some lemmas required for the main result.Lemma 12.
Let0
<
σ
<
1, 2
<
α
≤
3 and F
:
(
0,1
]
→
R is a continuous function with lim
t
→
0
+
F
(
t
)
=
∞. Suppose that t
σ
F
(
t
) is a continuous function on [
0,1
]. Then the function defined by (32)
H
(
t
)
=
∫
0
1
G
(
t
,
q
s
)
F
(
s
)
d
q
sis continuous on [
0,1
], where G
(
t
,
s
) is Green function be given in Lemma 9.Proof.
We will divide the proof into three parts.
Case 1 (t
0
=
0). First, H
(
0
)
=
0. Since t
σ
F
(
t
) is continuous on [
0,1
], we can find a positive constant M such that |
t
σ
F
(
t
)
|
≤
M for any t
∈
[
0,1
]. Thus, (33)
H
t
-
H
0
=
H
t
=
∫
0
1
‍
G
t
,
q
s
F
s
d
q
s
=
∫
0
1
‍
G
t
,
q
s
s
-
σ
s
σ
F
s
d
q
s
=
∫
0
t
‍
1
-
q
s
α
-
1
t
α
-
1
-
t
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
+
∫
t
1
‍
1
-
q
s
α
-
1
t
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
=
∫
0
1
‍
1
-
q
s
α
-
1
t
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
-
∫
0
t
‍
t
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
≤
M
∫
0
1
‍
1
-
q
s
α
-
1
t
α
-
1
Γ
q
α
s
-
σ
d
q
s
+
M
∫
0
t
‍
t
-
q
s
α
-
1
Γ
q
α
s
-
σ
d
q
s
=
M
t
α
-
1
Γ
q
α
∫
0
1
‍
1
-
q
s
α
-
1
s
-
σ
d
q
s
+
∫
0
t
‍
1
-
q
s
t
(
α
-
1
)
s
-
σ
d
q
s
.For ∫
0
t
‍
(
1
-
(
q
s
/
t
)
)
(
α
-
1
)
s
-
σ
d
q
s, let u
=
s
/
t; then we obtain(34)
∫
0
t
‍
1
-
q
s
t
(
α
-
1
)
s
-
σ
d
q
s
=
t
1
-
σ
∫
0
1
‍
1
-
q
u
(
α
-
1
)
u
-
σ
d
q
u
.Hence,(35)
H
t
≤
M
t
α
-
1
Γ
q
α
∫
0
1
‍
1
-
q
s
α
-
1
s
-
σ
d
q
s
H
t
+
M
t
α
-
σ
Γ
q
α
∫
0
1
‍
1
-
q
u
α
-
1
u
-
σ
d
q
u
H
t
=
M
t
α
-
1
Γ
q
α
+
M
t
α
-
σ
Γ
q
α
β
q
(
1
-
σ
,
α
)
,where β
q denotes the q-beta function.
Whent
→
0, we see that H
(
t
)
→
H
(
0
); that is H
(
t
) is continuous at t
0
=
0.
Case 2 (t
0
∈
(
0,1
)). We should prove H
(
t
n
)
→
H
(
t
0
) when t
n
→
t
0. Without loss of generality, we consider t
n
>
t
0 (it is the same argument for t
n
<
t
0). In fact, (36)
H
t
n
-
H
t
0
=
∫
0
1
‍
t
n
α
-
1
1
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
-
∫
0
t
n
‍
t
n
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
s
d
q
s
-
∫
0
1
‍
t
0
α
-
1
1
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
s
d
q
s
+
∫
0
t
0
‍
t
0
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
=
∫
0
1
‍
(
t
n
α
-
1
-
t
0
α
-
1
)
1
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
-
∫
0
t
0
‍
t
n
-
q
s
α
-
1
-
t
0
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
s
d
q
s
-
∫
t
0
t
n
‍
t
n
-
q
s
α
-
1
Γ
q
α
s
-
σ
s
σ
F
(
s
)
d
q
s
≤
M
t
n
α
-
1
-
t
0
α
-
1
Γ
q
α
∫
0
1
‍
1
-
q
s
α
-
1
s
-
σ
d
q
s
+
M
Γ
q
α
∫
0
t
0
‍
t
n
-
q
s
α
-
1
-
t
0
-
q
s
α
-
1
s
-
σ
d
q
s
+
M
Γ
q
α
∫
t
0
t
n
‍
t
n
-
q
s
α
-
1
s
-
σ
d
q
s
=
M
Γ
q
(
α
)
β
q
(
1
-
σ
,
α
)
(
t
n
α
-
1
-
t
0
α
-
1
)
+
M
Γ
q
(
α
)
(
a
n
+
b
n
)
,where (37)
a
n
=
∫
0
t
0
t
n
-
q
s
α
-
1
-
t
0
-
q
s
α
-
1
s
-
σ
d
q
s
,
b
n
=
∫
t
0
t
n
t
n
-
q
s
α
-
1
s
-
σ
d
q
s
.When n
→
∞, we verify a
n
→
0.
Ast
n
→
t
0, then (
(
t
n
-
q
s
)
(
α
-
1
)
-
(
t
0
-
q
s
)
(
α
-
1
)
)
s
-
σ
→
0, when n
→
∞. Moreover, (38)
t
n
-
q
s
α
-
1
-
t
0
-
q
s
α
-
1
s
-
σ
≤
2
s
-
σ
,
∫
0
1
2
s
-
σ
d
q
s
=
2
1
-
σ
q
s
1
-
σ
0
1
=
2
1
-
σ
q
<
∞
.
We have(
(
t
n
-
q
s
)
(
α
-
1
)
-
(
t
0
-
q
s
)
(
α
-
1
)
)
s
-
σ converges pointwise to the zero function and |
(
t
n
-
q
s
)
(
α
-
1
)
-
(
t
0
-
q
s
)
(
α
-
1
)
|
s
-
σ is bounded by a function belonging to L
1
[
0,1
], by Lebesgue’s dominated convergence theorem a
n
→
0 when n
→
∞.
Now, we proveb
n
→
0 when n
→
∞.
In fact, as(39)
b
n
=
∫
t
0
t
n
t
n
-
q
s
α
-
1
s
-
σ
d
q
s
b
n
≤
∫
t
0
t
n
s
-
σ
d
q
s
=
s
1
-
σ
1
-
σ
q
t
0
t
n
b
n
=
1
1
-
σ
q
t
n
1
-
σ
-
t
0
1
-
σ
,and taking into account that t
n
→
t
0, we get b
n
→
0 when n
→
∞.
Above all, we obtain|
H
(
t
n
)
-
H
(
t
0
)
|
→
0, when n
→
∞.
Case 3 (t
0
=
1). It is easy to check that H
(
1
)
=
0 and H
(
t
) is continuous at t
0
=
1. The proof is the same as the proof of Case 1.Lemma 13.
Suppose that0
<
σ
<
1. Then,(40)
max
⁡
0
≤
t
≤
1
∫
0
1
‍
G
(
t
,
q
s
)
s
-
σ
d
q
s
=
A
α
-
1
-
A
α
-
σ
Γ
q
(
α
)
β
q
(
1
-
σ
,
α
)
,where A
=
(
(
α
-
1
)
/
(
α
-
σ
)
)
1
/
(
1
-
σ
).Proof.
(41)
∫
0
1
G
t
,
q
s
s
-
σ
d
q
s
=
∫
0
t
1
-
q
s
α
-
1
t
α
-
1
-
t
-
q
s
α
-
1
Γ
q
α
s
-
σ
d
q
s
+
∫
t
1
1
-
q
s
α
-
1
t
α
-
1
Γ
q
α
s
-
σ
d
q
s
=
∫
0
1
1
-
q
s
α
-
1
t
α
-
1
Γ
q
α
s
-
σ
d
q
s
-
∫
0
t
t
-
q
s
α
-
1
Γ
q
α
s
-
σ
d
q
s
=
t
α
-
1
Γ
q
α
∫
0
1
1
-
q
s
α
-
1
s
-
σ
d
q
s
-
t
α
-
σ
Γ
q
α
∫
0
1
1
-
q
s
α
-
1
s
-
σ
d
q
u
=
t
α
-
1
-
t
α
-
σ
Γ
q
α
β
q
1
-
σ
,
α
.
Letg
(
t
)
=
t
α
-
1
-
t
α
-
σ, t
∈
[
0,1
].
Sinceg
′
(
t
)
=
(
α
-
1
)
t
α
-
2
-
(
α
-
σ
)
t
α
-
σ
-
1, let g
′
(
t
)
=
0; we can get g
(
t
) has a maximum at the point t
0
=
A
=
(
(
α
-
1
)
/
(
α
-
σ
)
)
1
/
(
1
-
σ
).
Hence,(42)
max
⁡
0
≤
t
≤
1
∫
0
1
‍
G
(
t
,
q
s
)
s
-
σ
d
q
s
=
A
α
-
1
-
A
α
-
σ
Γ
q
(
α
)
β
q
(
1
-
σ
,
α
)
.For the convenience, we denotemax
⁡
0
≤
t
≤
1
⁡
∫
0
1
G
(
t
,
q
s
)
s
-
σ
d
q
s by K.Next, we denote the class of functionsϕ
:
[
0
,
∞
)
→
[
0
,
∞
) by A satisfying(i)
ϕ is nondecreasing;
(ii)
ϕ
(
x
)
<
x for any x
>
0;
(iii)
β
(
x
)
=
ϕ
(
x
)
/
x
∈
J, where J is the class of functions appearing in Theorem 11.We give our main result as follows.Theorem 14.
Let0
<
σ
<
1, 2
<
α
≤
3
,
f
:
[
0,1
]
×
[
0
,
∞
)
→
[
0
,
∞
) is continuous and lim
t
→
0
+
⁡
f
(
t
,
·
)
=
∞, and t
σ
f
(
t
,
y
) is a continuous function on [
0,1
]
×
[
0
,
∞
). Assume that there exists 0
<
λ
≤
1
/
K such that for x
,
y
∈
[
0
,
∞
) with y
≥
x and t
∈
[
0,1
], (43)
0
≤
t
σ
(
f
(
t
,
y
)
-
f
(
t
,
x
)
)
≤
λ
ϕ
(
y
-
x
)
,where ϕ
∈
A. Then the BVP (3) has a unique positive solution (i.e., x
(
t
)
>
0 for t
∈
(
0,1
)).Proof.
We define the coneP by (44)
P
=
{
u
∈
C
[
0,1
]
:
u
(
t
)
≥
0
}
.It is clear that P is a complete metric space as P is a closed set of C
[
0,1
]. It is also easy to check that P satisfies conditions (29) and (30) of Theorem 11.
We define the operatorT by (45)
T
u
t
=
∫
0
1
‍
G
t
,
q
s
f
s
,
u
s
d
q
s
T
u
t
=
∫
0
1
‍
G
(
t
,
q
s
)
s
-
σ
s
σ
f
(
s
,
u
(
s
)
)
d
q
s
.In view of Lemma 12, T
u
∈
C
[
0,1
]. Moreover, it follows from the nonnegativeness of G
(
t
,
q
s
) and t
σ
f
(
t
,
y
) that T
u
∈
P for u
∈
P. Thus, T
:
P
→
P.
Next, we will prove that assumptions in Theorem11 are satisfied.
First, foru
≥
v, we have (46)
T
u
t
=
∫
0
1
‍
G
t
,
q
s
f
s
,
u
s
d
q
s
T
u
t
=
∫
0
1
‍
G
t
,
q
s
s
-
σ
s
σ
f
s
,
u
s
d
q
s
T
u
t
≥
∫
0
1
‍
G
t
,
q
s
s
-
σ
s
σ
f
s
,
v
s
d
q
s
T
u
t
=
(
T
v
)
(
t
)
.Hence, the operator T is nondecreasing. Besides, for u
≥
v and u
≠
v, (47)
d
T
u
,
T
v
=
max
⁡
t
∈
0,1
T
u
t
-
T
v
t
=
max
⁡
t
∈
0,1
T
u
t
-
T
v
t
=
max
⁡
t
∈
0,1
∫
0
1
‍
G
t
,
q
s
f
s
,
u
s
-
f
s
,
v
s
d
q
s
=
max
⁡
t
∈
0,1
∫
0
1
‍
G
t
,
q
s
s
-
σ
s
σ
f
s
,
u
s
-
f
s
,
v
s
d
q
s
≤
max
⁡
t
∈
[
0,1
]
∫
0
1
‍
G
(
t
,
q
s
)
s
-
σ
λ
ϕ
(
u
(
s
)
-
v
(
s
)
)
d
q
s
.Since ϕ is nondecreasing and u
(
s
)
-
v
(
s
)
≤
d
(
u
,
v
), (48)
d
T
u
,
T
v
≤
max
⁡
t
∈
0,1
∫
0
1
‍
G
t
,
q
s
s
-
σ
λ
ϕ
d
u
,
v
d
q
s
d
T
u
,
T
v
=
λ
ϕ
d
u
,
v
max
⁡
t
∈
0,1
∫
0
1
‍
G
t
,
q
s
s
-
σ
d
q
s
d
T
u
,
T
v
=
λ
ϕ
(
d
(
u
,
v
)
)
K
.Moreover, when 0
<
λ
≤
1
/
K, we get (49)
d
T
u
,
T
v
≤
ϕ
d
u
,
v
d
T
u
,
T
v
=
ϕ
d
u
,
v
d
u
,
v
·
d
u
,
v
d
T
u
,
T
v
=
β
(
d
(
u
,
v
)
)
·
d
(
u
,
v
)
.Obviously, the last inequality is satisfied for u
=
v.
Taking into account that the zero function satisfies0
≤
T
0, in view of Theorem 11, the operator T has a unique fixed point x
(
t
) in P.
At last, we will provex
(
t
) is a positive solution. We assume that there exists 0
<
t
1
<
1 such that x
(
t
1
)
=
0. Since x
(
t
) of problem (3) is a fixed point of the operator T, we have (50)
x
(
t
)
=
∫
0
1
G
(
t
,
q
s
)
f
(
s
,
x
(
s
)
)
d
q
s
,
for
0
<
t
<
1
,
x
(
t
1
)
=
∫
0
1
G
(
t
1
,
q
s
)
f
(
s
,
x
(
s
)
)
d
q
s
=
0
.For the nonnegative character of G
(
t
,
q
s
) and f
(
s
,
x
), the last relation gives (51)
G
t
1
,
q
s
f
s
,
x
s
=
0
a
.
e
.
(
s
)
.
f is continuous and lim
t
→
0
+
⁡
f
(
t
,
·
)
=
∞; then for M
>
0, we can find δ
>
0, and, for s
∈
[
0,1
]
∩
(
0
,
δ
), we have f
(
s
,
0
)
>
M. It is clear that [
0,1
]
∩
(
0
,
δ
)
⊂
{
s
∈
[
0,1
]
:
f
(
s
,
x
(
s
)
)
>
M
} and μ
(
[
0,1
]
∩
(
0
,
δ
)
)
>
0, where μ is the Lebesgue measure on [
0,1
]. That is to say, G
(
t
1
,
q
s
)
f
(
s
,
x
(
s
)
)
=
0 a.e. (
s
). This is a contradiction because G
(
t
1
,
q
s
) is a rational function in s.
Therefore,x
(
t
)
>
0 for t
∈
(
0,1
).
The proof is complete.


## 4. Example

Consider the following singular BVP:(52)
D
1
/
2
5
/
2
u
(
t
)
+
λ
(
t
2
+
1
)
ln
⁡
(
1
+
u
(
t
)
)
t
1
/
2
=
0
,
0
<
t
<
1
,
λ
>
0
,
u
(
0
)
=
u
(
1
)
=
0
,
(
D
1
/
2
u
)
(
0
)
=
0
.Here, α
=
2.5, q
=
1
/
2, σ
=
1
/
2, and f
(
t
,
u
)
=
λ
(
t
2
+
1
)
ln
⁡
(
1
+
u
(
t
)
)
/
t
1
/
2 for (
t
,
u
)
∈
[
0,1
]
×
[
0
,
∞
). Notice that f is continuous in [
0,1
]
×
[
0
,
∞
) and lim
t
→
0
+
⁡
f
(
t
,
·
)
=
∞.At first, we defineϕ by (53)
ϕ
:
[
0
,
∞
)
⟶
[
0
,
∞
)
,
ϕ
(
x
)
=
ln
⁡
(
1
+
x
)
.It is clear that ϕ
(
x
)
=
ln
⁡
(
1
+
x
) is a nondecreasing function; for u
≥
v, we can get (54)
ϕ
(
u
)
-
ϕ
(
v
)
≥
0
.Moreover, foru
≥
v, ϕ also satisfies (55)
ϕ
(
u
)
-
ϕ
(
v
)
≤
ϕ
(
u
-
v
)
.In fact, when u
≥
v, (56)
ϕ
u
-
v
-
ϕ
u
-
ϕ
v
=
ln
⁡
⁡
1
+
u
-
v
ϕ
u
-
v
ϕ
u
-
ϕ
v
-
ln
⁡
⁡
1
+
u
-
ln
⁡
⁡
1
+
v
ϕ
u
-
v
ϕ
u
-
ϕ
v
=
ln
⁡
1
+
u
-
v
1
+
v
1
+
u
ϕ
u
-
v
ϕ
u
-
ϕ
v
=
ln
⁡
(
1
+
(
u
-
v
)
v
1
+
u
)
≥
0
,equivalently (57)
ϕ
(
u
)
-
ϕ
(
v
)
≤
ϕ
(
u
-
v
)
.Above all, 0
≤
ϕ
(
u
)
-
ϕ
(
v
)
≤
ϕ
(
u
-
v
) for u
≥
v.Second, foru
≥
v and t
∈
[
0,1
], we have (58)
0
≤
t
1
/
2
f
t
,
u
-
f
t
,
v
0
=
λ
t
2
+
1
ln
⁡
⁡
1
+
u
-
ln
⁡
⁡
1
+
v
0
≤
λ
t
2
+
1
ln
⁡
⁡
1
+
u
-
v
0
≤
2
λ
ln
⁡
(
1
+
u
-
v
)
;that is, f satisfies assumptions of Theorem 14.Third, we should proveϕ
(
x
) belongs to A. By elemental calculus, it is easy to check that ϕ is nondecreasing and ϕ
(
x
)
<
x, for x
>
0.In order to proveβ
(
x
)
=
ϕ
(
x
)
/
x
∈
J, we notice that if β
(
t
n
)
→
1, then the sequence (
t
n
) is a bounded sequence because in contrary case, that is, t
n
→
∞, we get (59)
β
(
t
n
)
=
ln
⁡
(
1
+
t
n
)
t
n
⟶
0
.Now, we assume that t
n
↛
0, and then we find ε
>
0 such that for each n
∈
N there exists ρ
n
≥
n with t
ρ
n
≥
ε.Since(
t
n
) is a bounded sequence, we can find a subsequence (
t
k
n
) of (
t
ρ
n
) with t
k
n
→
a, for certain a
∈
[
0,1
). When β
(
t
n
)
→
1, it implies that (60)
β
(
t
k
n
)
=
ln
⁡
(
1
+
t
k
n
)
t
k
n
⟶
1
.and, as the unique solution of ln
⁡
(
1
+
x
)
=
x is x
0
=
0, we deduce that a
=
0. Therefore, t
k
n
→
0 and this contradicts the fact that t
k
n
≥
ε for every n
∈
N.Thus,t
n
→
0 and this proves that β
∈
J.Finally, in view of Theorem14, (61)
2
λ
≤
1
K
=
1
1
/
4
3
/
2
-
1
/
4
1
/
2
/
Γ
1
/
2
3
/
2
·
β
1
/
2
1
/
2
,
3
/
2
2
λ
≈
10.96511985
;that is, when λ
≤
5.48256, boundary value problem (52) has a unique positive solution.



---

*Source: 418569-2015-04-06.xml*