Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 420514, 6 pages
http://dx.doi.org/10.1155/2013/420514
Research Article
Existence and Exact Asymptotic Behavior of Positive Solutions
for a Fractional Boundary Value Problem
Habib Mâagli, Noureddine Mhadhebi, and Noureddine Zeddini
King Abdulaziz University, Rabigh Campus, College of Sciences and Arts, Department of Mathematics, P.O. Box 344,
Rabigh 21911, Saudi Arabia
Correspondence should be addressed to Noureddine Zeddini; noureddine.zeddini@ipein.rnu.tn
Received 4 November 2012; Accepted 25 December 2012
Academic Editor: Chuanzhi Bai
Copyright © 2013 Habib Mâagli et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We establish the existence and uniqueness of a positive solution 𝑢 for the fractional boundary value problem 𝐷𝛼 𝑢(𝑥) = −𝑎(𝑥)𝑢𝜎 (𝑥),
𝑥 ∈ (0, 1) with the condition lim𝑥 → 0 𝐷𝛼−1 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.
1. Introduction
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[1–3]). In
consequence, the subject of fractional differential equations
is gaining much importance and attention. Motivated by the
surge in the development of this subject, we consider the
following problem:
𝐷𝛼 𝑢 (𝑥) = −𝑎 (𝑥) 𝑢𝜎 (𝑥) ,
lim+ 𝐷𝛼−1 𝑢 (𝑥) = 0,
𝑥→0
𝑥 ∈ (0, 1) ,
𝑢 (1) = 0,
(1)
where 1 < 𝛼 ≤ 2, −1 < 𝜎 < 1, 𝑎 is a nonnegative continuous
function on (0, 1) that may be singular at 𝑥 = 0 or 𝑥 = 1 and
𝐷𝛼 is the Riemann-Liouville fractional derivative. Then we
study the existence and exact asymptotic behavior of positive
solutions for this problem.
We recall that for a measurable function V, the RiemannLiouville fractional integral 𝐼𝛽 V and the Riemann-Liouville
derivative 𝐷𝛽 V of order 𝛽 > 0 are, respectively, defined
by
𝐼𝛽 V (𝑥) =
𝐷𝛽 V (𝑥) =
𝑥
1
∫ (𝑥 − 𝑡)𝛽−1 V (𝑡) 𝑑𝑡,
Γ (𝛽) 0
1
𝑑 𝑛 𝑥
( ) ∫ (𝑥 − 𝑡)𝑛−𝛽−1 V (𝑡) 𝑑𝑡
Γ (𝑛 − 𝛽) 𝑑𝑥
0
=(
(2)
𝑑 𝑛
) 𝐼𝑛−𝛽 V (𝑥) ,
𝑑𝑥
provided that the right-hand sides are pointwise defined on
(0, 1]. Here 𝑛 = [𝛽] + 1 and [𝛽] means the integral part of the
number 𝛽 and Γ is the Euler Gamma function.
Moreover, we have the following well-known properties
(see [2, 4]):
(i) 𝐼𝛽 𝐼𝛾 V(𝑥) = 𝐼𝛽+𝛾 V(𝑥) for 𝑥 ∈ [0, 1], V ∈ 𝐿1 ((0, 1]), 𝛽 +
𝛾 ≥ 1,
(ii) 𝐷𝛽 𝐼𝛽 V(𝑥) = V(𝑥) for a.e. 𝑥 ∈ [0, 1], where V ∈
𝐿1 ((0, 1]), 𝛽 > 0,
(iii) 𝐷𝛽 V(𝑥) = 0 if and only if V(𝑥) = ∑𝑛𝑗=1 𝑐𝑗 𝑡𝛽−𝑗 , where
𝑛 = [𝛽] + 1 and (𝑐1 , 𝑐2 , . . . , 𝑐𝑛 ) ∈ R𝑛 .
Several results are obtained for fractional differential equation with different boundary conditions (see [5–15] and the
2
Abstract and Applied Analysis
references therein), but none of them deals with the existence
of a positive solution for problem (1).
Our aim in this paper is to establish the existence and
uniqueness of a positive solution 𝑢 ∈ 𝐶2−𝛼 ([0, 1]) for problem
(1) with a precise asymptotic behavior, where 𝐶2−𝛼 ([0, 1]) is
the set of all functions 𝑓 such that 𝑡 → 𝑡2−𝛼 𝑓(𝑡) is continuous
on [0, 1]. Note that for 0 < 𝛼 < 2, the solution 𝑢 for problem
(1) blows up at 𝑥 = 0.
To state our result, we need some notations. We will use
K to denote the set of Karamata functions 𝐿 defined on (0, 𝜂]
by
2. Technical Lemmas
To keep the paper self-contained, we begin this section by
recapitulating some properties of Karamata regular variation
theory. The following is due to [16, 17].
Lemma 2. The following hold.
(i) Let 𝐿 ∈ K and 𝜀 > 0, then one has
lim 𝑡𝜀 𝐿 (𝑡) = 0.
(9)
𝑡 → 0+
(3)
(ii) Let 𝐿 1 , 𝐿 2 ∈ K and 𝑝 ∈ R. Then one has 𝐿 1 +𝐿 2 ∈ K,
𝑝
𝐿 1 𝐿 2 ∈ K, and 𝐿 1 ∈ K.
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 𝐶1 ((0, 𝜂]) such that
Example 3. Let 𝑚 ∈ N∗ . Let 𝑐 > 0, (𝜇1 , 𝜇2 , . . . , 𝜇𝑚 ) ∈ R𝑚 and
let 𝑑 be a sufficiently large positive real number such that the
function
𝐿 (𝑡) := 𝑐 exp (∫
𝜂
𝑡
lim+
𝑡→0
𝑧 (𝑠)
𝑑𝑠) ,
𝑠
𝑡𝐿󸀠 (𝑡)
= 0.
𝐿 (𝑡)
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) ≤ 𝑐𝑓(𝑥), for all 𝑥 ∈ 𝑆. We denote
also 𝑥+ = max(𝑥, 0) for 𝑥 ∈ R.
Throughout this paper we assume that 𝑎 is nonnegative
on (0, 1) and satisfies the following condition.
(𝐻0 ) 𝑎 ∈ 𝐶((0, 1)) such that
−𝜆
−𝜇
𝑎 (𝑡) ≈ 𝑡 𝐿 1 (𝑡) (1 − 𝑡) 𝐿 2 (1 − 𝑡) ,
𝑡 ∈ (0, 1) ,
(5)
∫
0
𝐿 1 (𝑡)
𝑑𝑡 < ∞,
𝑡𝜆+(2−𝛼)𝜎
∫
𝜂
0
𝐿 2 (𝑡)
𝑑𝑡 < ∞.
𝑡𝜇−𝛼+1
(6)
𝜃 (𝑥) .
<
𝜂
−1, then ∫0 𝑠𝜇 𝐿(𝑠)𝑑𝑠 diverges and
𝜂
> −1, then ∫0 𝑠𝜇 𝐿(𝑠)𝑑𝑠
𝑡 𝜇
∫0 𝑠 𝐿(𝑠)𝑑𝑠∼𝑡 → 0+ (𝑡1+𝜇 𝐿(𝑡)/(𝜇 + 1)).
(ii) If 𝜇
converges and
Lemma 5. Let 𝐿 ∈ K be defined on (0, 𝜂]. Then one has
lim 𝜂
𝑡 → 0+ ∫
𝑡
(7)
𝐿 (𝑡)
(𝐿 (𝑠) /𝑠) 𝑑𝑠
(11)
𝜂
𝐿 (𝑡)
(𝐿 (𝑠) /𝑠) 𝑑𝑠
= 0.
(12)
Proof. We distinguish two cases.
𝜂
Case 1. We suppose that ∫0 (𝐿(𝑠)/𝑠)𝑑𝑠 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
𝐿 (𝑡) ≤ ∫
𝑡
0
(8)
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 Section 3,
we give the proof of Theorem 1.
= 0.
If further ∫0 (𝐿(𝑠)/𝑠)𝑑𝑠 converges, then one has
lim 𝑡
𝑡 → 0+ ∫
0
Theorem 1. Let 𝜎 ∈ (−1, 1) and assume that 𝑎 satisfies
(H0 ). Then problem (1) has a unique positive solution 𝑢 ∈
𝐶2−𝛼 ([0, 1]) satisfying for 𝑥 ∈ (0, 1)
𝑢 (𝑥) ≈ 𝑥
Lemma 4. Let 𝜇 ∈ R and let 𝐿 be a function in K defined on
(0, 𝜂]. One has the following.
𝜂
Our main result is the following.
𝛼−2
Applying Karamata’s theorem (see [16, 17]), we get the
following.
∫𝑡 𝑠𝜇 𝐿(𝑠)𝑑𝑠∼𝑡 → 0+ − (𝑡1+𝜇 𝐿(𝑡)/(𝜇 + 1)).
In the sequel, we introduce the function 𝜃 defined on (0, 1) by
1 − 𝑥 if 𝜇 < 𝜎 + 𝛼 − 1,
{
{
{
1/(1−𝜎)
𝜂
{
{
{(1 − 𝑥) (∫ 𝐿 2 (𝑠) 𝑑𝑠)
{
{
{
𝑠
{
1−𝑥
{
{
{
if 𝜇 = 𝜎 + 𝛼 − 1,
{
{
{
1/(1−𝜎)
𝜃 (𝑥) = {(1 − 𝑥)(𝛼−𝜇)/(1−𝜎) (𝐿 2 (1 − 𝑥))
{
{
{
if 𝜎 + 𝛼 − 1 < 𝜇 < 𝛼,
{
{
{
1/(1−𝜎)
{
1−𝑥
{
𝐿 2 (𝑠)
{
{
if 𝜇 = 𝛼.
𝑑𝑠)
{(∫
{
{
𝑠
{
{ 0
{
(10)
is defined and positive on (0, 𝜂], for some 𝜂 > 1, where
log𝑘 𝑥 = log ∘ log ∘ ⋅ ⋅ ⋅ ∘ log 𝑥 (𝑘 times). Then 𝐿 ∈ K.
(i) If 𝜇
where 𝜆 + (2 − 𝛼)𝜎 ≤ 1, 𝜇 ≤ 𝛼, 𝐿 1 , 𝐿 2 ∈ K satisfying
𝜂
𝑚
𝑑 −𝜇𝑘
𝐿 (𝑡) = 𝑐∏(log𝑘 ( ))
𝑡
𝑘=1
(4)
𝐿 (𝑠)
𝑑𝑠.
𝑠
(13)
It follows that lim𝑡 → 0+ 𝐿(𝑡) = 0. So we deduce (11).
Now put
𝜑 (𝑡) =
𝐿 (𝑡)
,
𝑡
for 𝑡 ∈ (0, 𝜂) .
(14)
Abstract and Applied Analysis
3
Using that lim𝑡 → 0+ (𝑡𝜑󸀠 (𝑡)/𝜑(𝑡)) = −1, we obtain
𝑡
𝑡
𝑡
0
0
0
Using the fact that lim𝑥 → 0 𝐷𝛼−1 𝑢(𝑥) = 0 and 𝑢(1) = 0, we
obtain 𝑐1 = 0 and 𝑐2 = 𝐼𝛼 𝑓(1). So
∫ 𝜑 (𝑠) 𝑑𝑠∼𝑡 → 0+ − ∫ 𝑠𝜑󸀠 (𝑠) 𝑑𝑠 = −𝑡𝜑 (𝑡) + ∫ 𝜑 (𝑠) 𝑑𝑠.
𝑢 (𝑥) =
(15)
𝑡
0
𝑡
𝐿 (𝑠)
𝐿 (𝑠)
𝑑𝑠∼𝑡 → 0+ − 𝐿 (𝑡) + ∫
𝑑𝑠.
𝑠
𝑠
0
𝜔
𝜔
𝑡
𝑡
(17)
This implies that
𝑡
𝜔
𝐿 (𝑠)
𝐿 (𝑠)
𝑑𝑠∼𝑡 → 0+ 𝐿 (𝑡) − 𝜔𝜑 (𝜔) + ∫
𝑑𝑠.
𝑠
𝑠
𝑡
(18)
This proves (11) and completes the proof.
𝑡 󳨀→ ∫
𝑡
𝐿 (𝑠)
𝑑𝑠 ∈ K.
𝑠
(19)
𝜂
If further ∫0 (𝐿(𝑠)/𝑠)𝑑𝑠 converges, we have by (11) that
𝑡 󳨀→ ∫
𝑡
0
𝐿 (𝑠)
𝑑𝑠 ∈ K.
𝑠
𝜇
𝜇
min (1, ) (1 − 𝑎𝑡𝜆 ) ≤ 1 − 𝑎𝑡𝜇 ≤ max (1, ) (1 − 𝑎𝑡𝜆 ) .
𝜆
𝜆
(25)
Proposition 9. On (0, 1) × (0, 1), one has
𝐺𝛼 (𝑥, 𝑡) ≈ 𝑥𝛼−2 (1 − 𝑡)𝛼−2 (1 − max (𝑥, 𝑡)) .
lim 𝐷𝛼−1 𝑢 (𝑥) = 0,
(21)
0
].
(27)
Since (𝑥 − 𝑡)+ /𝑥(1 − 𝑡) ∈ (0, 1) for 𝑡 ∈ (0, 1), then, by applying
Lemma 8 with 𝜇 = 𝛼 − 1 and 𝜆 = 1, we obtain
=𝑥
has a unique solution given by
𝑢 (𝑥) = 𝐺𝛼 𝑓 (𝑥) := ∫ 𝐺𝛼 (𝑥, 𝑡) 𝑓 (𝑡) 𝑑𝑡,
(𝑥 − 𝑡)+
(1 − 𝑡)𝛼−1 𝑥𝛼−2
[1 − 𝑥(
)
Γ (𝛼)
𝑥 (1 − 𝑡)
(20)
𝑢 (1) = 0,
1
(26)
𝛼−1
𝐺𝛼 (𝑥, 𝑡) =
𝛼−2
(1 − 𝑡)
𝛼−2
(𝑥 − 𝑡)+
)
1−𝑡
(28)
(1 − max (𝑥, 𝑡)) ,
which completes the proof.
In the sequel we put
𝑡 ∈ (0, 1) ,
𝑥→0
𝑏 (𝑡) = 𝑡−𝛽 𝐿 3 (𝑡) (1 − 𝑡)−𝛾 𝐿 4 (1 − 𝑡) ,
(29)
where 𝐿 3 , 𝐿 4 ∈ K and we aim to give some estimates on
𝑥2−𝛼 𝐺𝛼 𝑏(𝑥).
Proposition 10. Assume that 𝐿 3 , 𝐿 4 ∈ K, 𝛽 ≤ 1, 𝛾 ≤ 𝛼 with
(22)
𝜂
∫ 𝑡−𝛽 𝐿 3 (𝑡) 𝑑𝑡 < ∞,
0
where
𝐺𝛼 (𝑥, 𝑡) =
Lemma 8. For 𝜆, 𝜇 ∈ (0, ∞), and 𝑎, 𝑡 ∈ [0, 1], one has
𝐺𝛼 (𝑥, 𝑡) ≈ 𝑥𝛼−2 (1 − 𝑡)𝛼−1 (1 −
Lemma 7. Given 1 < 𝛼 ≤ 2 and 𝑓 is such that the function
𝑡 → (1 − 𝑡)𝛼−1 𝑓(𝑡) is continuous and integrable on (0, 1), then
the boundary value problem
𝐷𝛼 𝑢 (𝑡) = −𝑓 (𝑡) ,
In the following, we give some estimates on the function 𝐺𝛼 .
So, we need the following lemma.
Proof. For 𝑥, 𝑡 ∈ (0, 1) × (0, 1), we have
Remark 6. Let 𝐿 ∈ K defined on (0, 𝜂]; then, using (4) and
(11), we deduce that
𝜂
= ∫ 𝐺𝛼 (𝑥, 𝑡) 𝑓 (𝑡) 𝑑𝑡.
0
∫ 𝜑 (𝑠) 𝑑𝑠∼𝑡 → 0+ 𝑡𝜑 (𝑡) − 𝜔𝜑 (𝜔) + ∫ 𝜑 (𝑠) 𝑑𝑠.
𝜔
(24)
1
(16)
So (12) holds.
𝜂
Case 2. We suppose that ∫0 (𝐿(𝑠)/𝑠)𝑑𝑠 diverges. We have,
for some 𝜔 < 𝜂,
∫
𝑥
1
∫ (𝑥 − 𝑡)𝛼−1 𝑓 (𝑡) 𝑑𝑡
Γ (𝛼) 0
−
This implies that
∫
1 𝛼−2 1
𝑥 ∫ (1 − 𝑡)𝛼−1 𝑓 (𝑡) 𝑑𝑡
Γ (𝛼)
0
𝜂
∫ 𝑡𝛼−1−𝛾 𝐿 4 (𝑡) 𝑑𝑡 < ∞.
0
(30)
Then for 𝑥 ∈ (0, 1),
1
𝛼−1
[𝑥𝛼−2 (1 − 𝑡)𝛼−1 − ((𝑥 − 𝑡)+ ) ] ,
Γ (𝛼)
(23)
is the Green function for the boundary value problem (21).
Proof. Since 𝑢0 = −𝐼𝛼 𝑓 is a solution of the equation 𝐷𝛼 𝑢 =
−𝑓, then 𝐷𝛼 (𝑢 + 𝐼𝛼 𝑓) = 0. Consequently there exist two
constants 𝑐1 , 𝑐2 ∈ R such that 𝑢(𝑥) + 𝐼𝛼 𝑓(𝑥) = 𝑐1 𝑥𝛼−1 + 𝑐2 𝑥𝛼−2 .
1−𝑥
{
{
𝜂
{
𝐿 4 (𝑡)
{
{
𝑑𝑡
(1 − 𝑥) ∫
{
{
2−𝛼
𝑡
1−𝑥
𝑥 𝐺𝛼 𝑏 (𝑥) ≈ {
{(1 − 𝑥)𝛼−𝛾 𝐿 4 (1 − 𝑥)
{
{
{
{ 1−𝑥 𝐿 4 (𝑡)
{
𝑑𝑡
∫
{ 0
𝑡
if 𝛾 < 𝛼 − 1,
if 𝛾 = 𝛼 − 1,
if 𝛼 − 1 < 𝛾 < 𝛼,
if 𝛾 = 𝛼.
(31)
4
Abstract and Applied Analysis
This together with (34) implies that (36) holds on (0, 1).
Proof. Using Proposition 9, we have
𝑥
𝑥2−𝛼 𝐺𝛼 𝑏 (𝑥) ≈ (1 − 𝑥) ∫ (1 − 𝑡)𝛼−2−𝛾 𝑡−𝛽 𝐿 3 (𝑡) 𝐿 4 (1 − 𝑡) 𝑑𝑡
0
1
3. Proof of Theorem 1
+ ∫ (1 − 𝑡)
𝛼−1−𝛾 −𝛽
𝑡 𝐿 3 (𝑡) 𝐿 4 (1 − 𝑡) 𝑑𝑡
𝑥
We begin this section by giving a preliminary result that will
play a crucial role in the proof of Theorem 1.
= (1 − 𝑥) 𝐼 (𝑥) + 𝐽 (𝑥) .
(32)
For 0 < 𝑥 ≤ 1/2, we use Lemma 4 and hypotheses (30) to
deduce that
1−𝛽
{𝑥 𝑥 𝐿 3 (𝑥)
𝐼 (𝑥) ≈ { 𝐿 3 (𝑡)
𝑑𝑡
∫
𝑡
{ 0
𝐽 (𝑥) ≈ ∫
1
1/2
(1 − 𝑡)
≈1+∫
1/2
𝑥
𝛼−1−𝛾
𝑥2−𝛼 𝐺𝛼 𝜔 (𝑥) ≈ 𝜃 (𝑥) .
if 𝛽 < 1,
𝐿 4 (1 − 𝑡) 𝑑𝑡 + ∫
1/2
𝑥
−𝛽
𝑡 𝐿 3 (𝑡) 𝑑𝑡
𝑡−𝛽 𝐿 3 (𝑡) 𝑑𝑡
(33)
Hence, it follows from Lemma 2 and hypothesis (30) that, for
0 < 𝑥 ≤ 1/2, we have
𝑥2−𝛼 𝐺𝛼 𝑏 (𝑥) ≈ 1.
(34)
Now, for 1/2 ≤ 𝑥 < 1, we use again Lemma 4 and hypothesis
(30) to deduce that
1/2
0
𝑥
𝑡−𝛽 𝐿 3 (𝑡) 𝑑𝑡 + ∫
1/2
≈1+∫
1/2
1−𝑥
1−𝑥
0
𝑡−𝜆−(2−𝛼)𝜎 (1 − 𝑡)−𝜇+𝜎 𝐿 1 (𝑡) 𝐿 2 (1 − 𝑡)
{
{
{
{
if 𝜇 < 𝜎 + 𝛼 − 1,
{
{
{
−𝜇+𝜎
{
−𝜆−(2−𝛼)𝜎
{
𝐿 1 (𝑡) 𝐿 2 (1 − 𝑡)
(1 − 𝑡)
{
{𝑡
{
𝜎/(1−𝜎)
𝜂
{
{
𝐿
(𝑠)
{
2
{
if 𝜇 = 𝜎 + 𝛼 − 1,
{
{ ×(∫1−𝑡 𝑠 𝑑𝑠)
= { −𝜆−(2−𝛼)𝜎
−(𝜇−𝜎𝛼)/(1−𝜎)
𝐿 1 (𝑡)
𝑡
(1 − 𝑡)
{
{
{
{
𝜎/(1−𝜎)
{
if 𝜎 + 𝛼 − 1 < 𝜇 < 𝛼,
{
{ ×(𝐿 2 (1 − 𝑡))
{
{
−𝜇
−𝜆−(2−𝛼)𝜎
{
𝑡
𝐿
−
𝑡)
𝐿 2 (1 − 𝑡)
(1
(𝑡)
{
1
{
{
𝜎/(1−𝜎)
{
1−𝑡
{
{
{ ×(∫ 𝐿 2 (𝑠) 𝑑𝑠)
if 𝜇 = 𝛼.
𝑠
{
0
̃ 2 (1 − 𝑡) ,
̃ 1 (𝑡) 𝐿
𝜔 (𝑡) = 𝑡−𝛽 (1 − 𝑡)−𝛾 𝐿
𝑡𝛼−2−𝛾 𝐿 4 (𝑡) 𝑑𝑡
(35)
𝑡𝛼−1−𝛾 𝐿 4 (𝑡) 𝑑𝑡
(38)
(39)
where 𝛽 ≤ 1, 𝛾 ≤ 𝛼 and, according to Lemma 2, the functions
̃ 1 (𝑡) and 𝑡 → 𝐿
̃ 2 (𝑡) are in K. Moreover, using
𝑡 → 𝐿
𝜂 −𝛽
𝜂
̃
̃ 2 (𝑡)𝑑𝑡 <
Lemma 4, we have ∫0 𝑡 𝐿1 (𝑡)𝑑𝑡 < ∞ and ∫0 𝑡−𝛾 𝐿
∞. So the result follows from Proposition 10.
Proof of Theorem 1. From Proposition 11, there exists 𝑀 > 1
such that for each 𝑥 ∈ (0, 1)
1
𝜃 (𝑥) ≤ 𝑥2−𝛼 𝐺𝛼 𝜔 (𝑥) ≤ 𝑀𝜃 (𝑥) ,
𝑀
𝛼−𝛾
{(1 − 𝑥) 𝐿 4 (1 − 𝑥) if 𝛾 < 𝛼,
1−𝑥
≈{
𝐿 4 (𝑡)
𝑑𝑡
if 𝛾 = 𝛼.
∫
{ 0
𝑡
(40)
where 𝜔(𝑡) = 𝑎(𝑡)𝑡(𝛼−2)𝜎 𝜃𝜎 (𝑡).
Put 𝑐0 = 𝑀1/(1−|𝜎|) and let
Hence, it follows from Lemmas 2 and 5 that, for 𝑥 ∈ [1/2, 1),
we have
1−𝑥
{
{
𝜂
{
𝐿 4 (𝑡)
{
{
𝑑𝑡
(1 − 𝑥) ∫
{
{
2−𝛼
𝑡
1−𝑥
𝑥 𝐺𝛼 𝑏 (𝑥) ≈ {
{(1 − 𝑥)𝛼−𝛾 𝐿 4 (1 − 𝑥)
{
{
{
{ 1−𝑥 𝐿 4 (𝑡)
{
𝑑𝑡
∫
{ 0
𝑡
𝜔 (𝑡) = 𝑎 (𝑡) 𝑡(𝛼−2)𝜎 𝜃𝜎 (𝑡)
So, we can see that
(1 − 𝑡)𝛼−2−𝛾 𝐿 4 (1 − 𝑡) 𝑑𝑡
1
if 𝛾 < 𝛼 − 1,
{
{
{ 𝜂 𝐿 4 (𝑡)
≈ {∫
𝑑𝑡
if 𝛾 = 𝛼 − 1,
{
𝑡
{ 1−𝑥 𝛼−1−𝛾
𝐿 4 (1 − 𝑥) if 𝛾 > 𝛼 − 1,
{(1 − 𝑥)
𝐽 (𝑥) ≈ ∫
(37)
Proof. For 𝑡 ∈ (0, 1), we have
if 𝛽 = 1,
≈ 1.
𝐼 (𝑥) ≈ ∫
Proposition 11. Assume that the function 𝑎 satisfies (𝐻0 ) and
put 𝜔(𝑡) = 𝑎(𝑡)𝑡(𝛼−2)𝜎 𝜃𝜎 (𝑡) for 𝑡 ∈ (0, 1). Then one has, for
𝑥 ∈ (0, 1),
if 𝛾 < 𝛼 − 1,
Λ = {V ∈ 𝐶 ([0, 1]) :
1
𝜃 ≤ V ≤ 𝑐0 𝜃} .
𝑐0
(41)
In order to use a fixed point theorem, we denote 𝑎̃(𝑡) =
𝑎(𝑡)𝑡(𝛼−2)𝜎 and we define the operator 𝑇 on Λ by
if 𝛾 = 𝛼 − 1,
if 𝛼 − 1 < 𝛾 < 𝛼,
if 𝛾 = 𝛼.
(36)
𝑇V (𝑥) = 𝑥2−𝛼 𝐺𝛼 (̃𝑎V𝜎 ) (𝑥) .
(42)
For this choice of 𝑐0 , we can easily prove that for V ∈ Λ, we
have 𝑇V ≤ 𝑐0 𝜃 and 𝑇V ≥ (1/𝑐0 )𝜃.
Abstract and Applied Analysis
5
(iii) If 𝜇 = 𝜎 + 𝛼 − 1 and 𝛽 < 1, then for 𝑥 ∈ (0, 1),
Now, we have
𝑇V (𝑥) =
=
𝑥2−𝛼 1
∫ 𝐺 (𝑥, 𝑡) 𝑎̃ (𝑡) V𝜎 (𝑡) 𝑑𝑡
Γ (𝛼) 0 𝛼
1
1
𝛼−1
∫ [(1 − 𝑡)𝛼−1 − 𝑥2−𝛼 ((𝑥 − 𝑡)+ ) ]
Γ (𝛼) 0
𝑢 (𝑥) ≈ 𝑥𝛼−2 (1 − 𝑥) [log (
(43)
Since the function (𝑥, 𝑡) → (1 − 𝑡)𝛼−1 − 𝑥2−𝛼 ((𝑥 − 𝑡)+ )𝛼−1
is continuous on [0, 1] × [0, 1] and the function 𝑡 → (1 −
𝑡)𝛼−1 𝑎̃(𝑡)𝜃𝜎 (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 𝑇V = V. Put 𝑢(𝑥) = 𝑥𝛼−2 V(𝑥). Then 𝑢 ∈ 𝐶2−𝛼 ([0, 1]) and
𝑢 satisfies the equation
(44)
Since the function 𝑡 → (1 − 𝑡)𝛼−1 𝑎(𝑡)𝑢𝜎 (𝑡) is continuous
and integrable on (0, 1), then by Lemma 7 the function 𝑢 is
a positive continuous solution of problem (1).
Finally, let us prove that 𝑢 is the unique positive continuous solution satisfying (8). To this aim, we assume that (1)
has two positive solutions 𝑢, V ∈ 𝐶2−𝛼 ([0, 1]) satisfying (8)
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 ≤ 𝑐V. It
follows that 𝑢𝜎 ≤ 𝑐|𝜎| V𝜎 and consequently
−𝐷𝛼 (𝑐|𝜎| V − 𝑢) = 𝑎 (𝑐|𝜎| V𝜎 − 𝑢𝜎 ) ≥ 0,
lim+ 𝐷𝛼−1 (𝑐|𝜎| V − 𝑢) (𝑡) = 0,
𝑡→0
(𝑐|𝜎| V − 𝑢) (1) = 0.
(45)
Which implies by Lemma 7 that 𝑐|𝜎| V − 𝑢 = 𝐺𝛼 (𝑎(𝑐|𝜎| V𝜎 −
𝑢𝜎 )) ≥ 0. By symmetry, we obtain also that V ≤ 𝑐|𝜎| 𝑢. Hence
𝑐|𝜎| ∈ 𝐽 and 𝑐 ≤ 𝑐|𝜎| . Since |𝜎| < 1, then 𝑐 = 1 and
consequently 𝑢 = V.
Example 12. Let 𝜎 ∈ (−1, 1) and 𝑎 be a positive continuous
function on (0, 1) such that
𝑎 (𝑡) ≈ (1 − 𝑡)−𝜇 log (
3 −𝛽
) ,
1−𝑡
(46)
where 𝜇 < 𝛼 and 𝛽 ∈ R or 𝜇 = 𝛼 and 𝛽 > 1. Then, using
Theorem 1, problem (1) has a unique positive continuous
solution 𝑢 satisfying the following estimates.
(i) If 𝜇 < 𝜎 + 𝛼 − 1 or 𝜇 = 𝜎 + 𝛼 − 1 and 𝛽 > 1, then for
𝑥 ∈ (0, 1),
𝑢 (𝑥) ≈ 𝑥𝛼−2 (1 − 𝑥) .
(47)
(ii) If 𝜇 = 𝜎 + 𝛼 − 1 and 𝛽 = 1, then for 𝑥 ∈ (0, 1),
𝑢 (𝑥) ≈ 𝑥𝛼−2 (1 − 𝑥) [log (log (
1/(1−𝜎)
3
.
))]
1−𝑥
(48)
(49)
(iv) If 𝜎 + 𝛼 − 1 < 𝜇 < 𝛼, then for 𝑥 ∈ (0, 1),
𝑢 (𝑥) ≈ 𝑥𝛼−2 (1 − 𝑥)(𝛼−𝜇)/(1−𝜎) [log (
× 𝑎̃ (𝑡) V𝜎 (𝑡) 𝑑𝑡.
𝑢 (𝑥) = 𝐺𝛼 (𝑎𝑢𝜎 ) (𝑥) .
(1−𝛽)/(1−𝜎)
3
.
)]
1−𝑥
−𝛽/(1−𝜎)
3
. (50)
)]
1−𝑥
(v) If 𝜇 = 𝛼 and 𝛽 > 1, then for 𝑥 ∈ (0, 1),
𝑢 (𝑥) ≈ 𝑥𝛼−2 [log (
(1−𝛽)/(1−𝜎)
3
)]
.
1−𝑥
(51)
Acknowledgment
The authors thank the anonymous referees for a careful
reading of the paper and for their helpful suggestions.
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