# Existence and Estimates of  Positive Solutions for Some Singular Fractional Boundary Value Problems

Authors: Habib Mâagli; Noureddine Mhadhebi; Noureddine Zeddini
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We establish the existence and uniqueness of a positive solutionu for the following
fractional boundary value problem:D
α
u
(
x
)
=
−
a
(
x
)
u
σ
(
x
), x
∈
(
0
,
1
) with the conditions lim
x
→
0
+
⁡
x
2
−
α
u
(
x
)
=
0, u
(
1
)
=
0, where 1
<
α
≤
2, σ
∈
(
−
1
,
1
), and a is a nonnegative continuous function on (
0
,
1
) that may be singular at x
=
0 or x
=
1. We also give the global behavior of such a solution.


---

## Body

## 1. Introduction

Recently, the theory of fractional differential equations has been developed very quickly and the investigation for the existence of solutions of these differential equations has attracted considerable attention of researchers in the last few years (see [1–11] and the references therein).Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetism. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see [12–14]). In consequence, the subject of fractional differential equations is gaining much importance. Motivated by the surge in the development of this subject, we consider the following singular Dirichlet problem:
(1)
D
α
u
(
x
)
=
-
a
(
x
)
u
σ
(
x
)
,
x
∈
(
0,1
)
,
lim
⁡
x
→
0
+
x
2
-
α
u
(
x
)
=
0
,
u
(
1
)
=
0
,
where 1
<
α
≤
2, -
1
<
σ
<
1, and a is a nonnegative continuous function on (
0,1
) that may be singular at x
=
0 or x
=
1. Then we study the existence and exact asymptotic behavior of positive solutions for this problem.We recall that, for a measurable functionv, the Riemann-Liouville fractional integral I
β
v and the Riemann-Liouville derivative D
β
v of order β
>
0 are, respectively, defined by
(2)
I
β
v
(
x
)
=
1
Γ
(
β
)
∫
0
x
‍
(
x
-
t
)
β
-
1
v
(
t
)
d
t
,
D
β
v
(
x
)
=
1
Γ
(
n
-
β
)
(
d
d
x
)
n
∫
0
x
‍
(
x
-
t
)
n
-
β
-
1
v
(
t
)
d
t
=
(
d
d
x
)
n
I
n
-
β
v
(
x
)
,
provided that the right hand sides are pointwise defined on (
0,1
]. Here n
=
[
β
]
+
1 and [
β
] means the integer part of the number β and Γ is the Euler Gamma function.Moreover, we have the following well-known properties (see [3, 13, 15]):(i)
I
β
I
γ
v
(
x
)
=
I
β
+
γ
v
(
x
) for x
∈
[
0,1
], v
∈
L
1
(
(
0,1
]
), β
+
γ
≥
1;
(ii)
D
β
I
β
v
(
x
)
=
v
(
x
) for a.e. x
∈
[
0,1
], where v
∈
L
1
(
(
0,1
]
), β
>
0;
(iii)
ifv
∈
C
(
(
0,1
)
)
∩
L
1
(
(
0,1
)
) and D
β
v
(
x
)
=
0, then v
(
x
)
=
∑
j
=
1
n
‍
c
j
t
β
-
j, where (
c
1
,
c
2
,
…
,
c
n
)
∈
ℝ
n and n is the smallest integer greater than or equal to β.Several results are obtained for fractional differential equations with different boundary conditions, but none of them deal with the existence of a positive solution to problem (1).Our aim in this paper is to establish the existence and uniqueness of a positive solutionu
∈
C
2
-
α
(
[
0,1
]
) for problem (1) with a precise asymptotic behavior, where C
2
-
α
(
[
0,1
]
) is the set of all functions f such that t
→
t
2
-
α
f
(
t
) is continuous on [
0,1
].To state our result, we need some notations. We will use𝒦 to denote the set of Karamata functions L defined on (
0
,
η
] by
(3)
L
(
t
)
∶
=
c
exp
⁡
(
∫
t
η
‍
z
(
s
)
s
d
s
)
,
for some η
>
1, where c
>
0 and z
∈
C
(
[
0
,
η
]
) such that z
(
0
)
=
0. It is clear that a function L is in 𝒦 if and only if L is a positive function in C
1
(
(
0
,
η
]
) such that
(4)
lim
⁡
t
→
0
+
t
L
′
(
t
)
L
(
t
)
=
0
.
For two nonnegative functions f and g defined on a set S, the notation f
(
x
)
≈
g
(
x
), x
∈
S, means that there exists c
>
0 such that (
1
/
c
)
f
(
x
)
≤
g
(
x
)
≤
c
f
(
x
) for all x
∈
S. We denote by x
+
=
max
⁡
(
x
,
0
) for x
∈
ℝ and by B
+
(
(
0,1
)
) the set of all nonnegative measurable functions on (
0,1
).Throughout this paper, we assume thata is nonnegative on (
0,1
) and satisfies the following condition:(
H
0
)
a
∈
C
(
(
0,1
)
) such that for t
∈
(
0,1
)
(5)
a
(
t
)
≈
t
-
λ
L
1
(
t
)
(
1
-
t
)
-
μ
L
2
(
1
-
t
)
,
where λ
≤
α
+
(
2
-
α
)
(
1
-
σ
), μ
≤
α, L
1
,
L
2
∈
𝒦 satisfying
(6)
∫
0
η
‍
L
1
(
t
)
t
λ
+
(
2
-
α
)
σ
-
1
d
t
<
∞
,
∫
0
η
‍
L
2
(
t
)
t
μ
-
α
+
1
d
t
<
∞
.
In the sequel, we introduce the function θ defined on (
0,1
) by
(7)
θ
(
x
)
=
x
min
⁡
⁡
(
1
,
(
2
-
λ
+
(
α
-
2
)
σ
)
/
(
1
-
σ
)
)
(
L
~
1
(
x
)
)
1
/
(
1
-
σ
)
×
(
1
-
x
)
min
⁡
⁡
(
1
,
(
α
-
μ
)
/
(
1
-
σ
)
)
(
L
~
2
(
1
-
x
)
)
1
/
(
1
-
σ
)
,
where
(8)
L
~
1
(
x
)
=
{
1
,
if
λ
<
α
-
(
α
-
1
)
×
(
1
-
σ
)
,
∫
x
η
‍
L
1
(
s
)
s
d
s
,
if
λ
=
α
-
(
α
-
1
)
×
(
1
-
σ
)
,
L
1
(
x
)
,
if
α
-
(
α
-
1
)
(
1
-
σ
)
<
λ
<
α
+
(
2
-
α
)
(
1
-
σ
)
,
∫
0
x
‍
L
1
(
s
)
s
d
s
,
if
λ
=
α
+
(
2
-
α
)
×
(
1
-
σ
)
,
L
~
2
(
x
)
=
{
1
,
if
μ
<
α
+
σ
-
1
,
∫
x
η
‍
L
2
(
s
)
s
d
s
,
if
μ
=
α
+
σ
-
1
,
L
2
(
x
)
,
if
α
+
σ
-
1
<
μ
<
α
,
∫
0
x
‍
L
2
(
s
)
s
d
s
,
if
μ
=
α
.
Our main result is the following.Theorem 1.
Letσ
∈
(
-
1,1
) and assume that a satisfies (
H
0
). Then problem (1) has a unique positive solution u
∈
C
2
-
α
(
[
0,1
]
) satisfying for x
∈
(
0,1
),
(9)
u
(
x
)
≈
x
α
-
2
θ
(
x
)
.Remark 2.
Note that, forx
∈
(
0,1
), we have
(10)
x
α
-
2
θ
(
x
)
≈
x
min
⁡
(
α
-
1
,
(
α
-
λ
)
/
(
1
-
σ
)
)
×
(
L
~
1
(
x
)
)
1
/
(
1
-
σ
)
(
1
-
x
)
min
⁡
(
1
,
(
α
-
μ
)
/
(
1
-
σ
)
)
×
(
L
~
2
(
1
-
x
)
)
1
/
(
1
-
σ
)
.
This implies in particular that, for 1
<
α
<
2 and α
<
λ
≤
α
+
(
2
-
α
)
(
1
-
σ
), the solution u blows up at x
=
0 and for λ
<
α, lim
⁡
x
→
0
+
u
(
x
)
=
0.This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section3, we give the proof of Theorem 1.


## 2. Technical Lemmas

To let the paper be self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].Lemma 3.
The following hold.(i)
LettingL
∈
𝒦 and ɛ
>
0, then one has
(11)
lim
⁡
t
→
0
+
t
ɛ
L
(
t
)
=
0
.
(ii)
LetL
1
,
L
2
∈
𝒦 and p
∈
ℝ. Then one has L
1
+
L
2
∈
𝒦, L
1
L
2
∈
𝒦, and L
1
p
∈
𝒦.Example 4.
Letm be a positive integer. Let c
>
0, (
μ
1
,
μ
2
,
…
,
μ
m
)
∈
ℝ
m, and d be a sufficiently large positive real number such that the function
(12)
L
(
t
)
=
c
∏
k
=
1
m
‍
(
log
⁡
k
(
d
t
)
)
-
μ
k
is defined and positive on (
0
,
η
], for some η
>
1, where log
⁡
k
x
=
log
⁡
∘
log
⁡
∘
⋯
∘
log
⁡
x
(
k times). Then L
∈
𝒦.Applying Karamata’s theorem (see [16, 17]), we get the following.Lemma 5.
Letμ
∈
ℝ and L be a function in 𝒦 defined on (
0
,
η
]. One has the following:(i)
ifμ
<
-
1, then ∫
0
η
‍
s
μ
L
(
s
)
d
s diverges and ∫
t
η
‍
s
μ
L
(
s
)
d
s
~
t
→
0
+
-
t
1
+
μ
L
(
t
)
/
(
μ
+
1
);
(ii)
ifμ
>
-
1, then ∫
0
η
‍
s
μ
L
(
s
)
d
s converges and ∫
0
t
‍
s
μ
L
(
s
)
d
s
~
t
→
0
+
t
1
+
μ
L
(
t
)
/
(
μ
+
1
).Lemma 6.
LetL
∈
𝒦 be defined on (
0
,
η
]. Then one has
(13)
lim
⁡
t
→
0
+
L
(
t
)
∫
t
η
‍
(
L
(
s
)
/
s
)
d
s
=
0
.
If further ∫
0
η
‍
(
L
(
s
)
/
s
)
d
s converges, then one has
(14)
lim
⁡
t
→
0
+
L
(
t
)
∫
0
t
‍
(
L
(
s
)
/
s
)
d
s
=
0
.Proof.
We distinguish two cases.
Case  1. We suppose that ∫
0
η
‍
(
L
(
s
)
/
s
)
d
s converges. Since the function t
→
L
(
t
)
/
t is nonincreasing in (
0
,
ω
], for some ω
<
η, it follows that, for each t
≤
ω, we have
(15)
L
(
t
)
≤
∫
0
t
‍
L
(
s
)
s
d
s
.
It follows that lim
⁡
t
→
0
+
L
(
t
)
=
0. So we deduce (13).
Now put(16)
φ
(
t
)
=
L
(
t
)
t
,
for
t
∈
(
0
,
η
)
.
Using that lim
⁡
t
→
0
+
(
t
φ
′
(
t
)
/
φ
(
t
)
)
=
-
1, we obtain
(17)
∫
0
t
‍
φ
(
s
)
d
s
~
t
→
0
+
-
∫
0
t
‍
s
φ
′
(
s
)
d
s
=
-
t
φ
(
t
)
+
∫
0
t
‍
φ
(
s
)
d
s
.
This implies that
(18)
∫
0
t
‍
L
(
s
)
s
d
s
~
t
→
0
+
-
L
(
t
)
+
∫
0
t
‍
L
(
s
)
s
d
s
.
So (14) holds.
Case 2. We suppose that ∫
0
η
‍
(
L
(
s
)
/
s
)
d
s diverges. We have, for some ω
<
η,
(19)
∫
t
ω
‍
φ
(
s
)
d
s
~
t
→
0
+
t
φ
(
t
)
-
ω
φ
(
ω
)
+
∫
t
ω
‍
φ
(
s
)
d
s
.
This implies that
(20)
∫
t
ω
‍
L
(
s
)
s
d
s
~
t
→
0
+
L
(
t
)
-
ω
φ
(
ω
)
+
∫
t
ω
‍
L
(
s
)
s
d
s
.
This proves (13) and completes the proof.Remark 7.
LetL
∈
𝒦 be defined on (
0
,
η
]; then using (4) and (13), we deduce that
(21)
t
⟶
∫
t
η
‍
L
(
s
)
s
d
s
∈
𝒦
.
If further ∫
0
η
‍
(
L
(
s
)
/
s
)
d
s converges, we have by (13) that
(22)
t
⟶
∫
0
t
‍
L
(
s
)
s
d
s
∈
𝒦
.Lemma 8.
Given1
<
α
≤
2 and φ
∈
C
(
[
0,1
]
), then the unique continuous solution of
(23)
D
α
u
(
x
)
=
-
φ
(
x
)
,
x
∈
(
0,1
)
,
lim
⁡
x
→
0
x
2
-
α
u
(
x
)
=
0
,
u
(
1
)
=
0
is given by
(24)
u
(
x
)
=
G
α
φ
(
x
)
∶
=
∫
0
1
‍
G
α
(
x
,
t
)
φ
(
t
)
d
t
,
where
(25)
G
α
(
x
,
t
)
=
1
Γ
(
α
)
[
x
α
-
1
(
1
-
t
)
α
-
1
-
(
(
x
-
t
)
+
)
α
-
1
]
is Green's function for the boundary value problem (23).Proof.
Sinceφ
∈
C
(
[
0,1
]
), then u
0
=
-
I
α
φ is a solution of the equation D
α
u
=
-
φ. Hence D
α
(
u
+
I
α
φ
)
=
0. Consequently there exist two constants c
1
,
c
2
∈
ℝ such that u
(
x
)
+
I
α
φ
(
x
)
=
c
1
x
α
-
1
+
c
2
x
α
-
2. Using the fact that lim
⁡
x
→
0
x
2
-
α
u
(
x
)
=
0 and u
(
1
)
=
0, we obtain c
2
=
0 and c
1
=
I
α
φ
(
1
). So
(26)
u
(
x
)
=
1
Γ
(
α
)
x
α
-
1
∫
0
1
‍
(
1
-
t
)
α
-
1
φ
(
t
)
d
t
-
1
Γ
(
α
)
∫
0
x
‍
(
x
-
t
)
α
-
1
φ
(
t
)
d
t
=
∫
0
1
‍
G
α
(
x
,
t
)
φ
(
t
)
d
t
.In the following, we give some estimates on the Green functionG
α
(
x
,
y
). So, we need the following lemma.Lemma 9.
Forλ
,
μ
∈
(
0
,
∞
) and t
∈
[
0,1
] one has
(27)
min
⁡
(
1
,
μ
λ
)
(
1
-
t
λ
)
≤
1
-
t
μ
≤
max
⁡
(
1
,
μ
λ
)
(
1
-
t
λ
)
.Proposition 10.
On(
0,1
)
×
(
0,1
), one has(i)
G
α
(
x
,
t
)
≈
x
α
-
2
(
1
-
t
)
α
-
2
min
⁡
(
x
,
t
)
(
1
-
max
⁡
(
x
,
t
)
);
(ii)
there exist two constants
c
1
,
c
2
>
0
s
u
c
h
t
h
a
t
(28)
c
1
x
α
-
1
t
(
1
-
t
)
α
-
1
(
1
-
x
)
≤
G
α
(
x
,
t
)
≤
c
2
x
α
-
2
t
(
1
-
t
)
α
-
1
.Proof.
(i) Forx
,
t
∈
(
0,1
)
×
(
0,1
) we have
(29)
G
α
(
x
,
t
)
=
(
1
-
t
)
α
-
1
x
α
-
1
Γ
(
α
)
[
1
-
(
(
x
-
t
)
+
x
(
1
-
t
)
)
α
-
1
]
.
Since (
x
-
t
)
+
/
x
(
1
-
t
)
∈
(
0,1
) for x
,
t
∈
(
0,1
), then by applying Lemma 9 with μ
=
α
-
1 and λ
=
1, we obtain
(30)
G
α
(
x
,
t
)
≈
x
α
-
1
(
1
-
t
)
α
-
1
(
1
-
(
x
-
t
)
+
x
(
1
-
t
)
)
=
x
α
-
2
(
1
-
t
)
α
-
2
min
⁡
(
x
,
t
)
(
1
-
max
⁡
⁡
(
x
,
t
)
)
.
(ii) Using the following inequalities forx
,
t
∈
[
0,1
],
(31)
x
(
1
-
x
)
t
(
1
-
t
)
≤
min
⁡
(
x
,
t
)
(
1
-
max
⁡
(
x
,
t
)
)
≤
t
(
1
-
t
)
,
we deduce the result from (i).As a consequence of Proposition10, we obtain the following.Corollary 11.
Letf
∈
B
+
(
(
0,1
)
) and put G
α
f
(
x
)
:
=
∫
0
1
‍
G
α
(
x
,
t
)
f
(
t
)
d
t for x
∈
(
0,1
]. Then
(32)
G
α
f
(
x
)
<
∞
for
x
∈
(
0,1
)
i
f
f
∫
0
1
‍
t
(
1
-
t
)
α
-
1
f
(
t
)
d
t
<
∞
.Proposition 12.
Given1
<
α
<
2 and f such that the function t
→
t
(
1
-
t
)
α
-
1
f
(
t
) is continuous and integrable on (
0,1
), then G
α
f is the unique solution in C
2
-
α
(
[
0,1
]
) of the following boundary value problem:
(33)
D
α
u
(
x
)
=
-
f
(
x
)
,
x
∈
(
0,1
)
,
lim
⁡
x
→
0
+
x
2
-
α
u
(
x
)
=
0
,
u
(
1
)
=
0
.Proof.
From Corollary11, the function G
α
f is defined on (
0,1
) and by Proposition 10, we have
(34)
G
α
|
f
|
(
x
)
≤
c
2
x
α
-
2
∫
0
1
‍
t
(
1
-
t
)
α
-
1
|
f
(
t
)
|
d
t
,
which implies that I
2
-
α
(
G
α
|
f
|
) is bounded on (
0,1
). Now, using Fubini's theorem, we have
(35)
I
2
-
α
(
G
α
f
)
(
x
)
=
1
Γ
(
2
-
α
)
∫
0
x
‍
(
x
-
t
)
1
-
α
G
α
f
(
t
)
d
t
=
1
Γ
(
2
-
α
)
∫
0
1
‍
(
∫
0
x
‍
(
x
-
t
)
1
-
α
G
α
(
t
,
s
)
d
t
)
f
(
s
)
d
s
.
On the other hand, we have
(36)
1
Γ
(
2
-
α
)
∫
0
x
‍
(
x
-
t
)
1
-
α
G
α
(
t
,
s
)
d
t
=
1
Γ
(
2
-
α
)
Γ
(
α
)
[
(
1
-
s
)
α
-
1
∫
0
x
‍
(
x
-
t
)
1
-
α
t
α
-
1
d
t
-
∫
0
x
‍
(
x
-
t
)
1
-
α
(
(
t
-
s
)
+
)
α
-
1
d
t
]
=
x
(
1
-
s
)
α
-
1
-
1
Γ
(
2
-
α
)
Γ
(
α
)
∫
0
x
‍
(
x
-
t
)
1
-
α
×
(
(
t
-
s
)
+
)
α
-
1
d
t
.
Now, suppose that s
≤
x; then we have
(37)
∫
0
x
‍
(
x
-
t
)
1
-
α
(
(
t
-
s
)
+
)
α
-
1
d
t
=
∫
s
x
‍
(
x
-
t
)
1
-
α
(
t
-
s
)
α
-
1
d
t
.
By considering the substitution t
=
s
+
θ
(
x
-
s
), we obtain
(38)
∫
s
x
‍
(
x
-
t
)
1
-
α
(
t
-
s
)
α
-
1
d
t
=
Γ
(
α
)
Γ
(
2
-
α
)
(
x
-
s
)
.
Moreover if x
≤
s and t
∈
(
0
,
x
), we have ∫
0
x
‍
(
x
-
t
)
1
-
α
(
(
t
-
s
)
+
)
α
-
1
d
t
=
0.
So, it follows that(39)
1
Γ
(
2
-
α
)
∫
0
x
‍
(
x
-
t
)
1
-
α
G
α
(
t
,
s
)
d
t
=
x
(
1
-
s
)
α
-
1
-
(
x
-
s
)
+
.
This implies that
(40)
I
2
-
α
(
G
α
f
)
(
x
)
=
∫
0
1
‍
[
x
(
1
-
s
)
α
-
1
-
(
x
-
s
)
+
]
f
(
s
)
d
s
=
x
∫
0
x
‍
(
(
1
-
s
)
α
-
1
-
1
)
f
(
s
)
d
s
+
∫
0
x
‍
s
f
(
s
)
d
s
+
x
∫
x
1
‍
(
1
-
s
)
α
-
1
f
(
s
)
d
s
,
D
α
(
G
α
f
)
(
x
)
=
d
2
d
x
2
(
I
2
-
α
(
G
α
f
)
)
(
x
)
=
-
f
(
x
)
,
for
x
∈
(
0,1
)
.
Moreover, using part (
i
) of Proposition 10 and the dominated convergence theorem, we conclude that lim
⁡
x
→
0
+
⁡
x
2
-
α
G
α
f
(
x
)
=
0 and G
α
f
(
1
)
=
0.
Finally, we prove the uniqueness. Letu
,
v
∈
C
2
-
α
(
[
0,1
]
) be two solutions of (33) and put w
=
v
-
u. Then w
∈
C
2
-
α
(
[
0,1
]
)
⊂
L
1
(
(
0,1
)
)
∩
C
(
(
0,1
)
) and D
α
w
=
0. Hence, it follows that w
(
x
)
=
c
1
x
α
-
1
+
c
2
x
α
-
2. Using the fact that lim
⁡
x
→
0
+
x
2
-
α
w
(
x
)
=
w
(
1
)
=
0, we conclude that w
=
0 and so u
=
v.
In the sequel, we assume thatβ
≤
2 and γ
≤
α and we put
(41)
b
(
t
)
=
t
-
β
L
3
(
t
)
(
1
-
t
)
-
γ
L
4
(
1
-
t
)
,
where L
3
,
L
4
∈
𝒦 satisfy
(42)
∫
0
η
‍
L
3
(
t
)
t
β
-
1
d
t
<
∞
,
∫
0
η
‍
t
α
-
1
-
γ
L
4
(
t
)
d
t
<
∞
.
So, we aim to give some estimates on the potential function G
α
b
(
x
).
We define the Karamata functionsψ
β, ϕ
γ by
(43)
ψ
β
(
x
)
=
{
∫
0
x
‍
L
3
(
t
)
t
d
t
,
if
β
=
2
,
L
3
(
x
)
,
if
1
<
β
<
2
,
∫
x
η
‍
L
3
(
t
)
t
d
t
,
if
β
=
1
,
1
,
if
β
<
1
,
(44)
ϕ
γ
(
x
)
=
{
∫
0
x
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
,
L
4
(
x
)
,
if
α
-
1
<
γ
<
α
,
∫
x
η
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
-
1
,
1
,
if
γ
<
α
-
1
.Then, we have the following.Proposition 13.
Forx
∈
(
0,1
),
(45)
G
α
b
(
x
)
≈
x
min
⁡
⁡
(
α
-
1
,
α
-
β
)
(
1
-
x
)
min
⁡
⁡
(
1
,
α
-
γ
)
ψ
β
(
x
)
ϕ
γ
(
1
-
x
)
.Proof.
Using Proposition10, we have
(46)
x
2
-
α
G
α
b
(
x
)
≈
∫
0
1
‍
(
1
-
t
)
α
-
2
-
γ
t
-
β
min
⁡
(
x
,
t
)
×
(
1
-
max
⁡
(
x
,
t
)
)
L
3
(
t
)
L
4
(
1
-
t
)
d
t
≈
(
1
-
x
)
∫
0
x
‍
(
1
-
t
)
α
-
2
-
γ
t
1
-
β
×
L
3
(
t
)
L
4
(
1
-
t
)
d
t
+
x
∫
x
1
‍
(
1
-
t
)
α
-
1
-
γ
t
-
β
×
L
3
(
t
)
L
4
(
1
-
t
)
d
t
=
(
1
-
x
)
I
(
x
)
+
x
J
(
x
)
.
For 0
<
x
≤
1
/
2, we have I
(
x
)
≈
∫
0
x
‍
t
1
-
β
L
3
(
t
)
d
t. So, using Lemma 5 and hypothesis (42), we deduce that
(47)
I
(
x
)
≈
{
x
2
-
β
L
3
(
x
)
,
if
β
<
2
,
∫
0
x
‍
L
3
(
t
)
t
d
t
,
if
β
=
2
.
Now, we have
(48)
J
(
x
)
≈
∫
x
1
/
2
‍
t
-
β
L
3
(
t
)
d
t
+
∫
1
/
2
1
‍
(
1
-
t
)
α
-
1
-
γ
L
4
(
1
-
t
)
d
t
≈
1
+
∫
x
1
/
2
‍
t
-
β
L
3
(
t
)
d
t
,
which implies by Lemma 5 that
(49)
J
(
x
)
≈
{
x
1
-
β
L
3
(
x
)
,
if
1
<
β
≤
2
,
∫
x
η
‍
L
3
(
t
)
t
d
t
,
if
β
=
1
,
1
,
if
β
<
1
.
Hence, it follows by Lemma 6 and hypothesis (42) that, for 0
<
x
≤
1
/
2, we get
(50)
x
2
-
α
G
α
b
(
x
)
≈
{
∫
0
x
‍
L
3
(
t
)
t
d
t
,
if
β
=
2
,
x
2
-
β
L
3
(
x
)
,
if
1
<
β
<
2
,
x
∫
x
η
‍
L
3
(
t
)
t
d
t
,
if
β
=
1
,
x
,
if
β
<
1
.
That is
(51)
G
α
b
(
x
)
≈
x
min
⁡
⁡
(
α
-
1
,
α
-
β
)
ψ
β
(
x
)
.
Now, for 1
/
2
≤
x
<
1, we use again Lemma 5 and hypothesis (42) to deduce that
(52)
I
(
x
)
≈
∫
0
1
/
2
‍
t
1
-
β
L
3
(
t
)
d
t
+
∫
1
/
2
x
‍
(
1
-
t
)
α
-
2
-
γ
L
4
(
1
-
t
)
d
t
≈
1
+
∫
1
/
2
x
‍
(
1
-
t
)
α
-
2
-
γ
L
4
(
1
-
t
)
d
t
≈
{
(
1
-
x
)
α
-
1
-
γ
L
4
(
1
-
x
)
,
if
α
-
1
<
γ
≤
α
,
∫
1
-
x
η
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
-
1
,
1
,
if
γ
<
α
-
1
,
J
(
x
)
≈
∫
0
1
-
x
‍
t
α
-
1
-
γ
L
4
(
t
)
d
t
≈
{
(
1
-
x
)
α
-
γ
L
4
(
1
-
x
)
,
if
γ
<
α
,
∫
0
1
-
x
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
.
Hence, it follows from Lemma 3 that, for x
∈
[
1
/
2
,
1
), we get
(53)
x
2
-
α
G
α
b
(
x
)
≈
{
∫
0
1
-
x
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
,
(
1
-
x
)
α
-
γ
L
4
(
1
-
x
)
,
if
α
-
1
<
γ
<
α
,
(
1
-
x
)
∫
1
-
x
η
‍
L
4
(
t
)
t
d
t
,
if
γ
=
α
-
1
,
1
-
x
,
if
γ
<
α
-
1
.
That is
(54)
x
2
-
α
G
α
b
(
x
)
≈
(
1
-
x
)
min
⁡
⁡
(
1
,
α
-
γ
)
ϕ
γ
(
1
-
x
)
.
This together with (51) implies that, for x
∈
(
0,1
), we have
(55)
G
α
b
(
x
)
≈
x
min
⁡
⁡
(
α
-
1
,
α
-
β
)
(
1
-
x
)
min
⁡
⁡
(
1
,
α
-
γ
)
ψ
β
(
x
)
ϕ
γ
(
1
-
x
)
.


## 3. Proof of Theorem1

In order to prove Theorem1, we need the following Lemma.Lemma 14.
Assume that the functiona satisfies (
H
0
) and put ω
(
t
)
=
a
(
t
)
t
(
α
-
2
)
σ
(
θ
(
t
)
)
σ for t
∈
(
0,1
). Then one has, for x
∈
(
0,1
),
(56)
G
α
ω
(
x
)
≈
x
α
-
2
θ
(
x
)
.Proof.
Putr
=
min
⁡
(
α
-
1
,
(
α
-
λ
)
/
(
1
-
σ
)
) and s
=
min
⁡
(
1
,
(
α
-
μ
)
/
(
1
-
σ
)
). Then for t
∈
(
0,1
), we have
(57)
ω
(
t
)
=
t
-
λ
+
r
σ
L
1
(
t
)
(
L
~
1
(
t
)
)
σ
/
(
1
-
σ
)
×
(
1
-
t
)
-
μ
+
s
σ
L
2
(
1
-
t
)
(
L
~
2
(
1
-
t
)
)
σ
/
(
1
-
σ
)
.
Let β
=
λ
-
r
σ, γ
=
μ
-
s
σ, L
3
(
t
)
=
L
1
(
t
)
(
L
~
1
(
t
)
)
σ
/
(
1
-
σ
), and L
4
(
t
)
=
L
2
(
t
)
(
L
~
2
(
t
)
)
σ
/
(
1
-
σ
). Then, using Proposition 13, we obtain by a simple computation that
(58)
x
2
-
α
G
α
(
ω
)
(
x
)
≈
θ
(
x
)
.Proof of Theorem1.
From Lemma14, there exists M
>
1 such that, for each x
∈
(
0,1
),
(59)
1
M
θ
(
x
)
≤
x
2
-
α
G
α
ω
(
x
)
≤
M
θ
(
x
)
,
where ω
(
t
)
=
a
(
t
)
t
(
α
-
2
)
σ
θ
σ
(
t
).
Putc
0
=
M
1
/
(
1
-
|
σ
|
) and let
(60)
Λ
=
{
v
∈
C
(
[
0,1
]
)
:
1
c
0
θ
≤
v
≤
c
0
θ
}
.
In order to use a fixed point theorem, we denote a
~
(
t
)
=
a
(
t
)
t
(
α
-
2
)
σ and we define the operator T on Λ by
(61)
T
v
(
x
)
=
x
2
-
α
G
α
(
a
~
v
σ
)
(
x
)
.
For this choice of c
0, we can easily prove that, for v
∈
Λ, we have T
v
≤
c
0
θ and T
v
≥
(
1
/
c
0
)
θ.
Now, we have(62)
T
v
(
x
)
=
x
2
-
α
Γ
(
α
)
∫
0
1
‍
G
α
(
x
,
t
)
a
~
(
t
)
v
σ
(
t
)
d
t
=
x
2
-
α
Γ
(
α
)
∫
0
1
‍
[
-
(
(
x
-
t
)
+
)
α
-
1
x
α
-
1
(
1
-
t
)
α
-
1
-
(
(
x
-
t
)
+
)
α
-
1
]
a
~
(
t
)
v
σ
(
t
)
d
t
.
Since the function (
x
,
t
)
→
x
α
-
1
(
1
-
t
)
α
-
1
-
(
(
x
-
t
)
+
)
α
-
1 is continuous on [
0,1
]
×
[
0,1
] and by Proposition 10, Corollary 11, and Lemma 14, the function t
→
t
(
1
-
t
)
α
-
1
a
~
(
t
)
θ
σ
(
t
) is integrable on (
0,1
), we deduce that the operator T is compact from Λ to itself. It follows by the Schauder fixed point theorem that there exists v
∈
Λ such that T
v
=
v. Put u
(
x
)
=
x
α
-
2
v
(
x
). Then u
∈
C
2
-
α
(
[
0,1
]
) and u satisfies the equation
(63)
u
(
x
)
=
G
α
(
a
u
σ
)
(
x
)
.
Since the function t
→
t
(
1
-
t
)
α
-
1
a
(
t
)
u
σ
(
t
) is continuous and integrable on (
0,1
), then by Proposition 12, the function u is a positive continuous solution of problem (1).
Finally, let us prove thatu is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions u
,
v
∈
C
2
-
α
(
[
0,1
]
) satisfying (9) and consider the nonempty set J
=
{
m
≥
1
:
1
/
m
≤
u
/
v
≤
m
} and put c
=
inf
⁡
J. Then c
≥
1 and we have (
1
/
c
)
v
≤
u
≤
c
v. It follows that u
σ
≤
c
|
σ
|
v
σ and consequently
(64)
-
D
α
(
c
|
σ
|
v
-
u
)
=
a
(
c
|
σ
|
v
σ
-
u
σ
)
≥
0
,
lim
⁡
t
→
0
+
x
2
-
α
(
c
|
σ
|
v
-
u
)
(
t
)
=
0
,
(
c
|
σ
|
v
-
u
)
(
1
)
=
0
,
which implies by Proposition 12 that c
|
σ
|
v
-
u
=
G
α
(
a
(
c
|
σ
|
v
σ
-
u
σ
)
)
≥
0. By symmetry, we also obtain that v
≤
c
|
σ
|
u. Hence, c
|
σ
|
∈
J and c
≤
c
|
σ
|. Since |
σ
|
<
1, then c
=
1 and consequently u
=
v.Example 15.
Letσ
∈
(
-
1,1
) and a be a positive continuous function on (
0,1
) such that
(65)
a
(
t
)
≈
t
-
λ
(
1
-
t
)
-
μ
log
⁡
(
2
1
-
t
)
,
where λ
<
α
+
(
2
-
α
)
(
1
-
σ
) and μ
<
α. Then, using Theorem 1, problem (1) has a unique positive continuous solution u satisfying the following estimates:
(66)
u
(
x
)
≈
x
min
⁡
(
α
-
1
,
(
α
-
λ
)
/
(
1
-
σ
)
)
(
L
~
1
(
x
)
)
1
/
(
1
-
σ
)
×
(
1
-
x
)
min
⁡
⁡
(
1
,
(
α
-
μ
)
/
(
1
-
σ
)
)
(
L
~
2
(
1
-
x
)
)
1
/
(
1
-
σ
)
,
where
(67)
L
~
1
(
x
)
=
{
1
,
if
λ
≠
α
-
(
α
-
1
)
(
1
-
σ
)
,
log
⁡
(
2
x
)
,
if
λ
=
α
-
(
α
-
1
)
(
1
-
σ
)
,
L
~
2
(
x
)
=
{
1
,
if
μ
<
α
+
σ
-
1
,
(
log
⁡
(
2
x
)
)
2
,
if
μ
=
α
+
σ
-
1
,
log
⁡
(
2
x
)
,
if
α
+
σ
-
1
<
μ
<
α
.



---

*Source: 120781-2014-04-01.xml*