# The Unique Positive Solution for Singular Hadamard Fractional Boundary Value Problems

Authors: Jinxiu Mao; Zengqin Zhao; Chenguang Wang
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In this paper, we investigate singular Hadamard fractional boundary value problems. The existence and uniqueness of the exact iterative solution are established only by using an iterative algorithm. The iterative sequences have been proved to converge uniformly to the exact solution, and estimation of the approximation error and the convergence rate have also been derived.


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## Body

## 1. Introduction

Fractional differential operators play an important role in describing phenomenons in many fields such as physics, chemistry, control, and electromagnetism [1–9]. They have many applications in fractional differential equations and fractional integral equations. In [10], authors investigate a class of fractional integral equations arising from a symmetric transition model(1)
d
d
t
1
2
D
0
t
-
β
u
′
t
+
1
2
D
t
T
-
β
u
′
t
+
∇
F
t
,
u
t
=
0
,
a
.
e
.
t
∈
0
,
T
u
0
=
u
T
=
0 where 0
D
t
-
β and t
D
T
-
β are the left and right Riemann-Liouville fractional integrals of order 0
≤
β
<
1, respectively, and ∇
F
(
t
,
x
) is the gradient of F at x.The fractional order diffusion equation(2)
∂
α
ϕ
∂
t
α
=
K
∂
β
ϕ
∂
x
β where 0
<
α
≤
1 and 1
<
β
≤
2, contains fractional differential operators which can describe different diffusion process.In the past ten years, fractional differential equations have been considered in many papers (see [11–22]). Most of the works on the topic have been based on Riemann-Liouville type and Caputo type fractional differential equations. By means of fixed point theorems and variational methods, authors obtain at least one or multiple positive solutions for boundary value problems of fractional differential equations. Very recently, more studies have been carried out on the boundary value problems of nonlinear Hadamard fractional differential equations. An important characteristic of Hadamard fractional derivative is that it contains logarithmic function of arbitrary exponent. However, there are few results about this topic (see [23–28]).By using the Krasnoselskii-Zabreiko fixed point theorem, Yang [26] obtained at least one positive solution for the boundary value problem (3)
D
q
u
t
+
f
t
,
u
t
=
0
,
t
∈
1
,
e
,
u
m
1
=
0
,
u
e
=
∫
1
e
g
t
u
t
d
t
t
, where f
∈
C
(
[
1
,
e
]
×
R
+
,
R
), g
∈
C
(
[
1
,
e
]
,
R
+
). D
q was the Hadamard fractional derivative of order q. 0
≤
m
≤
n
-
2
,
n
∈
N
,
n
≥
3, n
-
1
<
q
≤
n.In [28], the authors studied the Hadamard fractional differential equation (4)
D
q
H
x
t
+
σ
t
f
t
,
x
t
=
0
,
2
<
q
≤
3
,
t
∈
1
,
+
∞
, with boundary conditions (5)
x
1
=
x
′
1
=
0
,
D
q
-
1
H
x
∞
=
a
H
I
β
x
ξ
+
b
∑
i
=
1
m
-
2
α
i
x
η
i
,
1
<
ξ
<
η
1
<
η
2
<
⋯
<
η
m
-
2
<
+
∞
, where D
q
H denoted Hadamard fractional derivative of order q and H
I
β denoted Hadamard fractional integral of order β
.
f
∈
C
(
[
1
,
∞
)
×
[
0
,
∞
)
,
[
0
,
∞
)
), σ
:
[
1
,
∞
)
→
[
0
,
∞
). By employing the complete continuity of the associated integral operator T and the monotone iterative method, the authors obtained twin positive solutions and the unique positive solution.Most of the above works required the associated integral operators to be completely continuous because fixed point theorems could be applied. Furthermore, the uniqueness of positive solutions was rarely investigated while the existence and multiplicity of positive solutions were investigated widely.Inspired by the above results, in this work, we study the existence and uniqueness of positive solutions for the following boundary value problem:(6)
D
q
H
x
t
+
f
t
,
x
t
=
0
,
2
<
q
≤
3
,
t
∈
1
,
e
,
x
1
=
x
′
1
=
0
,
x
e
=
0
,where D
q
H denotes Hadamard fractional derivative of order q; f
:
(
1
,
e
)
×
[
0
,
∞
)
→
[
0
,
∞
) is continuous.In this work, only by using the monotone iterative technique, we aim to establish the unique positive solution for problem (6). The main contributions of this work are as follows: (a) the nonlinear term f
(
t
,
x
) can be singular at t
=
1 and t
=
e; (b) we do not need the continuity and complete continuity of the associated integral operator; (c) we get the unique positive solution.Throughout this work, we assume that the following conditions hold without further mention.(
H
1
)
:
f
:
(
1
,
e
)
×
[
0
,
∞
)
→
[
0
,
∞
) is continuous.(
H
2
): For (
t
,
x
)
∈
(
1
,
e
)
×
[
0
,
∞
), f
(
t
,
x
) is nondecreasing in x and there exists a constant k
∈
(
0,1
) such that, for ∀
σ
∈
(
0,1
],(7)
f
t
,
σ
x
≥
σ
k
f
t
,
x
.It is easy to verify that ifσ
∈
(
1
,
+
∞
), then f
(
t
,
σ
x
)
≤
σ
k
f
(
t
,
x
)
.


## 2. Preliminaries

The way to attack this new problem follows a scheme similar to that used in [21], with the necessary adaptations that Hadamard fractional derivative contains logarithmic function of arbitrary exponent.In this section, we present some basic concepts and conclusions needed in the proof of our main results.Definition 1.
The Hadamard fractional integral of orderβ
>
0 of a function x
:
(
1
,
∞
)
→
R is given by (8)
I
H
β
x
t
=
1
Γ
β
∫
1
t
l
o
g
t
s
β
-
1
x
s
s
d
s
.Definition 2.
The Hadamard fractional derivative of orderq is defined by (9)
D
q
H
x
t
=
1
Γ
n
-
q
t
d
d
t
n
∫
1
t
l
o
g
t
s
n
-
q
-
1
x
s
s
d
s
, where n
-
1
<
q
≤
n, n
=
[
q
]
+
1
,
q
>
0.
Specifically,H
D
n
x
(
t
)
=
x
(
n
)
(
t
)
,
n
=
1,2
,
3
,
…
.Lemma 3.
Suppose thath
(
t
)
∈
C
(
1
,
e
), 0
<
∫
1
e
‍
h
(
s
)
d
s
/
s
<
+
∞
. Then the Hadamard type fractional differential equation(10)
D
q
H
x
t
+
h
t
=
0
,
2
<
q
≤
3
,
t
∈
1
,
e
,
x
1
=
x
′
1
=
0
,
x
e
=
0has a unique solution (11)
x
t
=
∫
1
e
G
t
,
s
h
s
d
s
swhere(12)
G
t
,
s
=
1
Γ
q
l
o
g
t
q
-
1
1
-
l
o
g
s
q
-
1
-
l
o
g
t
s
q
-
1
,
1
≤
s
≤
t
≤
e
;
l
o
g
t
q
-
1
1
-
l
o
g
s
q
-
1
,
1
≤
t
≤
s
≤
e
.Proof.
As argued in [9], the solution of the Hadamard differential equation in (10) can be written as the equivalent integral equation(13)
x
t
=
c
1
l
o
g
t
q
-
1
+
c
2
l
o
g
t
q
-
2
+
c
3
l
o
g
t
q
-
3
-
1
Γ
q
∫
1
t
l
o
g
t
s
q
-
1
h
s
d
s
s
.From x
(
1
)
=
x
′
(
1
)
=
0, we have c
3
=
c
2
=
0
. Thus (47) reduces to(14)
x
t
=
c
1
l
o
g
t
q
-
1
-
1
Γ
q
∫
1
t
l
o
g
t
s
q
-
1
h
s
d
s
s
.Using x
(
e
)
=
0 we obtain(15)
c
1
=
1
Γ
q
∫
1
e
l
o
g
e
s
q
-
1
h
s
d
s
s
.Substituting (15) into (14), we obtain (16)
x
t
=
l
o
g
t
q
-
1
.
1
Γ
q
∫
1
e
l
o
g
e
s
q
-
1
h
s
d
s
s
-
1
Γ
q
∫
1
t
l
o
g
t
s
q
-
1
h
s
d
s
s
=
∫
1
e
G
t
,
s
h
s
d
s
s
.Takeρ
(
t
)
=
(
l
o
g
t
)
q
-
1
(
1
-
l
o
g
t
) and ρ
^
(
t
)
=
(
1
-
l
o
g
t
)
q
-
1
l
o
g
t for q
>
2, t
∈
[
1
,
e
]. We can prove that G
(
t
,
s
) have the following properties.Lemma 4 (see [26]).
Fort
,
s
∈
[
1
,
e
], Green’s function G
(
t
,
s
) satisfies the following properties:
(i)G
(
t
,
s
) is continuous on [
1
,
e
]
×
[
1
,
e
] and G
(
t
,
s
)
≥
0,
(ii)ρ
(
t
)
ρ
^
(
s
)
≤
Γ
(
q
)
G
(
t
,
s
)
≤
(
q
-
1
)
ρ
(
t
),
(iii)ρ
(
t
)
ρ
^
(
s
)
≤
Γ
(
q
)
G
(
t
,
s
)
≤
(
q
-
1
)
ρ
^
(
s
),
(iv)G
(
t
,
s
)
=
G
(
e
/
t
,
e
/
s
)
.
FromΓ
(
q
+
1
)
=
q
Γ
(
q
)
,
q
>
0, and (ii) we get(17)
ρ
t
1
Γ
q
ρ
^
s
≤
G
t
,
s
≤
1
Γ
q
-
1
ρ
t
.In this paper, we will work in the Banach spaceE
=
C
[
1
,
e
] with the norm x
=
m
a
x
t
∈
[
1
,
e
]
|
x
(
t
)
|
.Define a setP
⊂
E as follows.P
=
{
x
∈
E
∣ there are constants 0
<
l
x
<
1
<
L
x such that l
x
ρ
(
t
)
≤
x
(
t
)
≤
L
x
ρ
(
t
)
,
t
∈
[
1
,
e
]
}
. Evidently ρ
(
t
)
∈
P
. Therefore, P is not empty.


## 3. The Main Results

Theorem 5.
Assume(
H
1
)
,
(
H
2
) hold. And (18)
0
<
∫
1
e
f
t
,
ρ
t
d
t
t
<
∞
. Then problem (6) has at least one positive solution x
∗
(
t
)
.Proof.
Define the operatorT
:
E
→
E by(19)
T
x
t
=
∫
1
e
G
t
,
s
f
s
,
x
s
d
s
s
.We can see easily the equivalence between x is a solution of (6) and x is a fixed point of T.
Claim 1. The operator T
:
P
→
P is nondecreasing.
In fact, forx
∈
P, it is obvious that T
x
∈
E, T
x
(
1
)
=
T
x
(
e
)
=
0
. For any x
∈
P and t
∈
[
1
,
e
], from (17), (20)
T
x
t
=
∫
1
e
G
t
,
s
f
s
,
x
s
d
s
s
.
≤
∫
1
e
1
Γ
q
-
1
ρ
t
f
s
,
L
x
ρ
s
d
s
s
≤
ρ
t
L
x
k
1
Γ
q
-
1
∫
1
e
f
s
,
ρ
s
d
s
s
≤
L
T
x
ρ
t and (21)
T
x
t
=
∫
1
e
G
t
,
s
f
s
,
x
s
d
s
s
.
≥
∫
1
e
1
Γ
q
ρ
^
s
ρ
t
f
s
,
l
x
ρ
s
d
s
s
≥
ρ
t
l
x
k
1
Γ
q
∫
1
e
ρ
^
s
f
s
,
ρ
s
d
s
s
=
l
T
x
ρ
t
, where L
T
x and l
T
x are positive constants satisfying (22)
L
T
x
>
m
a
x
1
,
L
x
k
1
Γ
q
-
1
∫
1
e
f
s
,
ρ
s
d
s
s
,
0
<
l
T
x
<
m
i
n
1
,
l
x
k
1
Γ
q
∫
1
e
ρ
^
s
f
s
,
ρ
s
d
s
s
. Thus, it follows that there are constants 0
<
l
T
x
<
1
<
L
T
x such that, for t
∈
[
1
,
e
], (23)
l
T
x
ρ
t
≤
T
x
t
≤
L
T
x
ρ
t
. Therefore, for any x
∈
P, T
x
∈
P, T is the operator P
→
P
. From (19), it is easy to see that T is nondecreasing with respect to x. Hence, Claim 1 holds.
Claim 2. There exist a nondecreasing sequence {
u
n
} and a nonincreasing sequence {
v
n
} and there exists x
∗
∈
P such that(24)
u
n
t
→
x
∗
t
,
v
n
t
→
x
∗
t
,uniformly on [
1
,
e
]
.
First, there exist two constantsl
T
ρ
,
L
T
ρ with 0
<
l
T
ρ
<
1
<
L
T
ρ since T
ρ
∈
P
. Take δ and γ to be fixed numbers satisfying(25)
0
<
δ
≤
l
T
ρ
1
/
1
-
k
,
γ
≥
L
T
ρ
1
/
1
-
k
.Obviously, 0
<
δ
<
1
<
γ
.
We construct two iterative sequences as follows:(26)
u
0
t
=
δ
ρ
t
,
v
0
t
=
γ
ρ
t
,
(27)
u
n
=
T
u
n
-
1
,
v
n
=
T
v
n
-
1
,
n
=
1,2
,
3
,
…
.Then(28)
u
0
≤
u
1
≤
…
≤
u
n
≤
…
v
n
≤
…
≤
v
1
≤
v
0
.In fact, from (26), we have u
0
,
v
0
∈
P and u
0
≤
v
0
. Furthermore, (29)
u
1
t
=
T
u
0
t
=
∫
1
e
G
t
,
s
f
s
,
δ
ρ
s
d
s
s
≥
δ
k
∫
1
e
G
t
,
s
f
s
,
ρ
s
d
s
s
=
δ
k
T
ρ
≥
δ
k
l
T
ρ
ρ
t
≥
δ
k
δ
1
-
k
ρ
t
=
u
0
t
,
v
1
t
=
T
v
0
t
=
∫
1
e
G
t
,
s
f
s
,
γ
ρ
s
d
s
s
≤
γ
k
∫
1
e
G
t
,
s
f
s
,
ρ
s
d
s
s
=
γ
k
T
ρ
≤
γ
k
L
T
ρ
ρ
t
≤
γ
k
γ
1
-
k
ρ
t
=
v
0
t
. From u
0
≤
v
0 and T nondecreasing, (28) holds. Let c
0
=
δ
/
γ; then 0
<
c
0
<
1
. It follows from (7) that (30)
T
c
0
u
≥
c
0
k
T
u
. And, for any natural number n, (31)
u
n
=
T
u
n
-
1
=
T
n
u
0
=
T
n
δ
ρ
=
T
n
c
0
γ
ρ
≥
c
0
k
n
T
n
γ
ρ
=
c
0
k
n
v
n
. Thus, for any natural number n and p
∗, we have(32)
0
≤
u
n
+
p
∗
-
u
n
≤
v
n
-
u
n
≤
1
-
c
0
k
n
v
n
≤
1
-
c
0
k
n
γ
ρ
,which implies that {
u
n
} is a cauchy sequence in [
u
0
,
v
0
]
. So there exists x
∗
∈
[
u
0
,
v
0
]
⊂
P such that(33)
u
n
t
→
x
∗
t
,and from (32)(34)
v
n
t
→
x
∗
t
.
Fromx
∗
∈
[
u
0
,
v
0
], we have T
x
∗
∈
[
u
0
,
v
0
]
. Combining with T nondecreasing on x,(35)
u
n
≤
T
x
∗
≤
v
n
. Let n
→
∞,(36)
x
∗
t
=
T
x
∗
t
, which implies x
∗ is a positive solution of problem (6).Theorem 6.
Assume(
H
1
)
,
(
H
2
) hold. Then, we have the following:
(i)Problem (6) has unique positive solution x
∗
(
t
) in P and there exist constants l
,
L with 0
<
l
<
1
<
L such that (37)
l
ρ
t
≤
x
∗
t
≤
L
ρ
t
,
t
∈
1
,
e
.
(ii)For any initial valuex
0
(
t
)
∈
P, there exists a sequence {
x
n
(
t
)
} that uniformly converges to the unique positive solution x
∗
(
t
), and we have the error estimation(38)
m
a
x
x
n
t
-
x
∗
t
=
o
1
-
λ
k
n
,where λ is a constant with 0
<
λ
<
1 and determined by x
0; o
(
1
-
λ
k
n
) represents the same order infinitesimal of (
1
-
λ
k
n
)
.Proof.
Letu
0
,
v
0
,
u
n
,
v
n be defined in (26) and (27).
(i) It follows from Theorem5 that problem (6) has a positive solution x
∗
(
t
)
∈
P, which implies that there exists constants l and L with 0
<
l
<
L
<
1 such that (39)
l
ρ
t
≤
x
∗
t
≤
L
ρ
t
,
t
∈
1
,
e
. Let x
¯
(
t
)
∈
P be another positive solution of problem (6). Then there exist constants c
1 and c
2 with 0
<
c
1
<
1
<
c
2 such that (40)
c
1
ρ
t
≤
x
¯
t
≤
c
2
ρ
t
,
t
∈
1
,
e
. Let δ defined in (25) be small enough so that δ
<
c
1 and γ defined in (25) be large enough so that γ
>
c
2
. Then (41)
u
0
t
≤
x
¯
t
≤
v
0
t
,
t
∈
1
,
e
. Note that T
x
¯
(
t
)
=
x
¯
(
t
) and T is nondecreasing; we have (42)
u
n
t
≤
x
¯
t
≤
v
n
t
,
t
∈
1
,
e
. Letting n
→
∞, we obtain that x
∗
(
t
)
=
x
¯
(
t
)
. Hence, the positive solution of problem (6) is unique.
(ii) For anyx
0
(
t
)
∈
P, there exist constants l
0 and L
0 with 0
<
l
0
<
1
<
L
0 such that (43)
l
0
ρ
t
≤
x
0
t
≤
L
0
ρ
t
,
t
∈
1
,
e
. Similar to (i), take δ and γ to be defined by (25) satisfying δ
<
l
0 and γ
>
L
0
. Then (44)
u
0
t
≤
x
0
t
≤
v
0
t
,
t
∈
1
,
e
. Let (45)
x
n
t
=
T
x
n
-
1
t
=
∫
1
e
G
t
,
s
f
s
,
x
n
-
1
s
d
s
s
,
n
=
1,2
,
…
. Note that T is nondecreasing, (46)
u
n
t
≤
x
n
t
≤
v
n
t
,
t
∈
1
,
e
. Letting n
→
∞, it follows from (33) and (34) that x
n
(
t
)
→
x
∗
(
t
) uniformly on [
1
,
e
].
At the same time, (38) follows from (32).Remark 7.
We just investigate a simple form of boundary value problems for Hadamard differential equations. We can easily apply the monotone iterative technique to multipoint or multistrip boundary value problems.Remark 8.
Suppose thatβ
i
(
t
)
(
i
=
0,1
,
2
,
…
m
) are nonnegative continuous functions on (
1
,
e
), which may be unbounded at the end points of (
1
,
e
). Ω is the set of functions f
(
t
,
x
) which satisfy the conditions (
H
1
) and (
H
2
). Then we have the following conclusions:
(1)β
i
(
t
)
∈
Ω
,
x
b
∈
Ω, where 0
<
b
<
1.
(2) If0
<
b
i
<
+
∞
(
i
=
1,2
,
…
m
) and b
>
m
a
x
1
≤
i
≤
m
{
b
i
}, then [
β
0
(
t
)
+
∑
i
=
1
m
‍
β
i
(
t
)
x
b
i
]
1
/
b
∈
Ω.
(3) Iff
(
t
,
x
)
∈
Ω, then β
i
(
t
)
f
(
t
,
x
)
∈
Ω.
(4) Iff
i
(
t
,
x
)
∈
Ω
(
i
=
1,2
,
…
m
), then m
a
x
1
≤
i
≤
m
{
f
i
(
t
,
x
)
}
∈
Ω
,
m
i
n
1
≤
i
≤
m
{
f
i
(
t
,
x
)
}
∈
Ω.
The above four facts can be verified directly. This indicates that there are many kinds of functions which satisfy the conditions(
H
1
) and (
H
2
).


## 4. An Example

Consider the following boundary value problem:(47)
D
5
/
2
H
u
t
+
a
t
u
1
/
4
+
b
t
u
1
/
3
=
0
,
t
∈
1
,
e
,
u
1
=
u
′
1
=
0
,
u
e
=
0
,where q
=
5
/
2, f
(
t
,
u
)
=
a
(
t
)
u
1
/
4
+
b
(
t
)
u
1
/
3, a
(
t
)
,
b
(
t
)
∈
C
(
(
1
,
e
)
,
(
0
,
+
∞
)
)
.Analysis 1.
First,f
∈
C
(
(
1
,
e
)
×
[
0
,
∞
)
,
[
0
,
∞
)
) and so (
H
1
) holds.
For anyσ
∈
(
0,1
), we take k
=
1
/
2 and have (48)
f
t
,
σ
u
≥
σ
k
f
t
,
u
. Then (
H
2
) holds.
Obviously(49)
0
<
∫
1
e
f
t
,
ρ
t
d
t
t
<
∞
, where ρ
(
t
)
=
(
l
o
g
t
)
5
/
2
-
1
(
1
-
l
o
g
t
). Hence all conditions of Theorem 5 are satisfied, and consequently we have the following corollary.Corollary 9.
Problem (47) has unique positive solution u
∗
(
t
). For any initial value x
0
∈
P, the successive iterative sequence {
x
n
(
t
)
} generated by (50)
x
n
t
=
∫
0
1
G
t
,
s
a
s
u
1
/
4
+
b
s
u
1
/
3
d
s
s
,
n
=
1,2
,
… uniformly converges to the unique positive solution u
∗
(
t
) on [
1
,
e
]. We have the error estimation (51)
m
a
x
x
n
t
-
u
∗
t
=
∘
1
-
λ
1
/
2
n
, where λ is a constant with 0
<
λ
<
1 and determined by the initial value x
0
. And there are constants l
,
L with 0
<
l
<
1
<
L such that (52)
l
ρ
t
≤
u
∗
t
≤
L
ρ
t
,
t
∈
1
,
e
.



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*Source: 5923490-2019-06-20.xml*