# Three-Point Boundary Value Problems for Conformable Fractional Differential Equations

Authors: H. Batarfi; Jorge Losada; Juan J. Nieto; W. Shammakh
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We study a fractional differential equation using a recent novel concept of fractional derivative with initial and three-point boundary conditions. We first obtain Green's function for the linear problem and then we study the nonlinear differential equation.


---

## Body

## 1. Introduction

In this paper, we study a class of differential equations supplemented with three-point boundary conditions. Precisely, we consider the following problem:(1)
D
α
D
+
λ
x
t
=
f
t
,
x
t
,
t
∈
[
0,1
]
,
(2)
x
0
=
0
,
x
′
0
=
0
,
x
1
=
β
x
η
,where D
α is the conformable fractional derivative of order α
∈
(
1,2
], D is the ordinary derivative, f
:
[
0,1
]
×
R
→
R is a known continuous function, λ and β are real numbers, λ
>
0, and η
∈
(
0,1
).Fractional calculus and fractional differential equations are relevant areas of research. There are several concepts of fractional derivatives, some classical, such as Riemann-Liouville or Caputo definitions, and some novel, such as conformable fractional derivative [1], β-derivative [2], or a new definition [3, 4]. The relation between these definitions and their potential applications needs further study.The conformable fractional derivative aims at extending the usual derivative satisfying some natural properties (see [1]) and gives a new solution for some fractional differential equations. In this paper we present a boundary value problem involving this fractional derivative.Sequential fractional differential equations have been considered for other types of fractional derivatives, see, for example, [5, 6].The paper is organized as follows. In Section2 we recall some concepts relative to the conformable fractional calculus. In Section 3 we solve the corresponding linear problem and obtain Green's function. In Section 4 we study the nonlinear problem and finally, in Section 5, we present an example to illustrate the applicability of our results.


## 2. Preliminaries

We recall some definitions and results concerning conformable fractional derivative.Definition 1 (see [1]).
Given a functionx
:
[
0
,
+
∞
)
→
R, the conformable fractional derivative of order α
∈
(
0,1
] of x at t is defined by (3)
D
α
x
t
=
lim
ε
→
0
x
t
+
ε
t
1
-
α
-
x
t
ε
,for all t
>
0. If f is α-differentiable in some interval (
0
,
a
) with a
>
0, then we define (4)
D
α
x
(
0
)
=
lim
t
→
0
+
D
α
x
(
t
)
,whenever the limit of the right hand side exists.We remark that ifx is differentiable, then (5)
D
α
x
(
t
)
=
t
1
-
α
x
′
t
.Reciprocally, if D
α
x
(
t
) exists, then for t
≠
0 we have (6)
x
′
t
=
lim
δ
→
0
x
(
t
+
δ
)
-
x
(
t
)
δ
=
lim
ε
→
0
x
t
+
ε
t
1
-
α
-
x
t
ε
t
1
-
α
=
t
α
-
1
D
α
x
(
t
)
.Hence, D
α
x
(
t
)
=
t
1
-
α
x
′
(
t
). Of course, for t
=
0 this is not valid and it would be useful to deal with equations and solutions with singularities.Definition 2.
Letα
∈
(
n
,
n
+
1
] and let x be an n-differentiable function at t
>
0; the fractional conformable derivative of order α at t
>
0 is given by (7)
D
α
x
(
t
)
=
lim
ε
→
0
x
(
α
-
1
)
(
t
+
ε
t
α
-
α
)
-
x
(
α
-
1
)
(
t
)
ε
,where α is the smallest integer greater than or equal to α. For t
=
0 we proceed in a similar way as in Definition 1.Definition 3.
Givenα
∈
(
0,1
], the fractional integral of order α at t
≥
0 is given by (8)
D
-
α
x
(
t
)
≡
I
α
x
(
t
)
=
I
1
(
t
α
-
1
x
)
(
t
)
=
∫
0
t
‍
x
(
s
)
s
1
-
α
d
s
.Definition 4.
Givenα
∈
(
n
-
1
,
n
], the fractional integral of order α is given by (9)
D
-
α
x
(
t
)
≡
I
α
x
(
t
)
=
I
n
(
t
α
-
α
x
)
(
t
)
,where I
n denotes the operator I
1 (usual integration) of order n.Remark 5.
Some authors (see [7, 8]) have argued that conformable fractional derivative is not a truly fractional operator. This question seems today to still be open and perhaps it is a philosophical issue. However, in any case, the study of boundary value problems involving this new derivative has, in our opinion, a point of interest and deserves to be researched in more detail.In [1, Theorem 3.1], authors have proved that for α
∈
(
0,1
] and x a given continuous function, D
α
I
α
x
(
t
)
=
x
(
t
) for t
≥
0. In this paper, we consider α
∈
(
1,2
], so we need the following results.Lemma 6.
Givenα
∈
(
1,2
] and x a continuous function defined in the domain of I
α, one has that D
α
I
α
x
(
t
)
=
x
(
t
) for t
≥
0.Proof.
Sincex is continuous, then I
α
x
(
t
) is twice differentiable. In view of [1, Remark 2.1] we have (10)
D
α
I
α
x
t
=
t
2
-
α
d
2
d
t
2
∫
0
t
‍
∫
0
t
1
‍
x
(
s
)
s
α
-
2
d
s
d
t
1
=
t
2
-
α
d
d
t
∫
0
t
‍
x
(
s
)
s
α
-
2
d
s
=
t
2
-
α
x
(
t
)
t
α
-
2
=
x
(
t
)
.
Thus, statement of Lemma6 has been proved.Lemma 7.
Givenα
∈
(
1,2
] and x
:
[
0
,
+
∞
)
→
R an α-differentiable function, one has that D
α
x
(
t
)
=
0 if and only if x
(
t
)
=
c
1
t
+
c
2, where c
1, c
2
∈
R.Proof.
This fact follows easily in view of the mean value theorem for conformable fractional differentiable functions (see [1, Theorem 2.4]).


## 3. Linear Boundary Problem

In order to study boundary value problem (1)-(2), we consider now the linear equation (11)
D
α
D
+
λ
x
t
=
σ
t
,
t
∈
[
0,1
]
,where 1
<
α
≤
2 and σ
∈
C
[
0,1
].Lemma 8.
Consider(12)
β
≠
λ
+
e
-
λ
-
1
λ
η
+
e
-
λ
η
-
1
.Then, the unique solution of (11) subject to the boundary conditions (2) is given by (13)
x
t
=
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
+
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
s
s
s
s
s
s
s
s
s
-
∫
0
1
‍
e
-
λ
1
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
,where (14)
A
t
=
1
Δ
λ
t
+
e
-
λ
t
-
1
,
Δ
=
λ
+
e
-
λ
-
1
-
β
λ
η
+
e
-
λ
η
-
1
≠
0
.Proof.
Integrating (11) we obtain(15)
D
+
λ
x
(
t
)
=
I
α
σ
(
t
)
+
c
1
t
+
c
2
.Thus, in view of Lemmas 6 and 7, every solution of (15) is a solution for (11).
Lety
(
t
)
=
e
λ
t
x
(
t
). Equation (15) can be rewritten as (16)
D
y
(
t
)
=
I
α
σ
(
t
)
+
c
1
t
+
c
2
e
λ
t
.Integrating from 0 to t, we obtain(17)
x
t
=
c
1
λ
2
λ
t
-
1
+
e
-
λ
t
+
c
2
λ
1
-
e
-
λ
t
+
c
3
+
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
.Imposing boundary conditions (2), we conclude that c
3
=
0, c
2
=
0, and (18)
c
1
=
λ
2
Δ
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
s
s
s
s
s
-
∫
0
1
‍
e
-
λ
(
1
-
s
)
(
∫
0
s
‍
σ
(
u
)
u
α
-
2
(
s
-
u
)
d
u
)
d
s
.Substituting these values of c
1, c
2, and c
3 in (17), we finally obtain (13) and that expression gives the unique solution.We now obtain Green’s function corresponding to the fractional differential equations (11) of order α
+
1 with 1
<
α
≤
2 subject to boundary conditions (2).By changing the order of integration, we note that(19)
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
σ
u
u
α
-
2
s
-
u
d
u
d
s
=
∫
0
t
‍
∫
u
t
‍
e
λ
s
-
t
s
-
u
d
s
σ
(
u
)
u
α
-
2
d
u
.Hence, solution (17) with c
2
=
c
3
=
0 takes the form (20)
x
(
t
)
=
c
1
λ
2
λ
t
-
1
+
e
-
λ
t
+
∫
0
t
‍
k
(
t
,
s
)
σ
(
s
)
s
α
-
2
d
s
,where (21)
k
(
t
,
s
)
=
∫
s
t
‍
e
λ
(
u
-
t
)
(
u
-
s
)
d
u
=
e
λ
(
s
-
t
)
-
λ
s
-
1
+
λ
t
λ
2
.Now, using the boundary condition x
(
1
)
=
β
x
(
η
), we get (22)
c
1
=
λ
2
Δ
β
∫
0
η
‍
k
η
,
s
σ
s
s
α
-
s
d
s
-
∫
0
1
‍
k
1
,
s
σ
s
s
α
-
2
d
s
.Therefore, we finally conclude the following: (23)
x
t
=
A
(
t
)
Δ
β
∫
0
η
‍
k
(
η
,
s
)
σ
(
s
)
s
α
-
s
d
s
s
s
s
s
s
s
s
s
-
∫
0
1
‍
k
(
1
,
s
)
σ
(
s
)
s
α
-
2
d
s
+
∫
0
t
‍
k
(
t
,
s
)
σ
(
s
)
s
α
-
2
d
s
;so, we deduce the following result.Theorem 9.
The unique solution of (11) subject to boundary conditions (2) is given by (24)
x
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
σ
(
s
)
s
α
-
2
d
s
,where (25)
G
(
t
,
s
)
=
-
k
(
1
,
s
)
ψ
(
t
)
,
i
f
0
≤
max
⁡
η
,
t
<
s
≤
1
;
-
k
(
1
,
s
)
ψ
(
t
)
+
k
(
t
,
s
)
,
i
f
0
≤
η
<
s
<
t
≤
1
;
(
β
k
(
η
,
s
)
-
k
(
1
,
s
)
)
ψ
(
t
)
,
i
f
0
≤
t
<
s
≤
η
≤
1
;
(
β
k
(
η
,
s
)
-
k
(
1
,
s
)
)
ψ
(
t
)
+
k
(
t
,
s
)
,
i
f
0
≤
s
<
min
⁡
{
η
,
t
}
≤
1
,with (26)
ψ
t
=
A
t
Δ
,
t
∈
[
0,1
]
.Remark 10.
In other words, corresponding Green's function for the homogeneous problem (11) satisfying the boundary conditions (2) is given by (25).Remark 11.
Note thatG
(
t
,
s
) is independent of α, but the solution depends, of course, on α.


## 4. Nonlinear Problem

LetC
=
C
[
0,1
] be the Banach space of all continuous functions defined in [
0,1
] endowed with the usual supremum norm defined by x
=
sup
⁡
{
|
x
(
t
)
|
,
t
∈
[
0,1
]
}.For the sake of convenience, we set(27)
B
=
1
+
A
1
β
η
α
1
-
e
-
λ
η
+
1
-
e
-
λ
λ
α
(
α
-
1
)
>
0
,with (28)
A
1
=
sup
⁡
t
∈
[
0,1
]
A
t
,where A
(
t
) is given by (14).In view of Lemma8, we transform boundary value problem (1)-(2) into (29)
x
=
T
x
,
x
∈
C
,where T
:
C
→
C is defined by (30)
T
x
t
=
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
+
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
dddddidd
-
∫
0
1
‍
e
-
λ
1
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
.Observe that problem (1)-(2) has solutions if the operator (30) has fixed points.Theorem 12.
Letf
:
[
0,1
]
×
R
→
R be a jointly continuous function satisfying the condition (31)
f
(
t
,
v
)
-
f
(
t
,
w
)
≤
L
v
-
w
∀
t
∈
0,1
,
v
,
w
∈
R
,where L
>
0 is the Lipschitz constant. Then, boundary value problem (1)-(2) has a unique solution if B
<
1
/
L, where B is given by (27).Proof.
First, forT defined by (30), we show that T
B
r
⊂
B
r, where B
r is the closed ball of radius r
>
0 in C; that is, B
r
=
{
x
∈
C
:
x
≤
r
}. Now we consider M
>
sup
⁡
{
f
t
,
0
:
t
∈
[
0,1
]
} and we choose (32)
r
>
M
B
1
-
L
B
.For x
∈
B
r, we have (33)
T
x
=
sup
⁡
t
∈
[
0,1
]
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
ssssssssss
+
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
ssssssssssssssssssssssss
·
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
ssssssssssssssssssss
-
∫
0
1
‍
e
-
λ
(
1
-
s
)
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
f
f
f
f
s
s
s
s

·


∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
≤
sup
⁡
t
∈
[
0,1
]
∫
0
t
‍
e
-
λ
(
t
-
s
)
s
s
s
s
s
s
s
s
s
s
s
·
∫
0
s
‍
f
u
,
x
u
-
f
u
,
0
+
f
u
,
0
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
∫
0
s
‍
·
u
α
-
2
s
-
u
d
u
d
s
+
sup
⁡
t
∈
[
0,1
]
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
·
∫
0
s
‍
f
u
,
x
u
-
f
u
,
0
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
i
i
s
s
s
s
∫
0
s
‍
+
f
u
,
0
u
α
-
2
s
-
u
d
u
d
s
ddddddddddddd
+
∫
0
1
‍
e
-
λ
(
1
-
s
)
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
·
∫
0
s
‍
f
u
,
x
u
-
f
u
,
0
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
+
f
u
,
0
∫
0
s
‍
u
α
-
2
s
-
u
d
u


d
s
≤
sup
⁡
t
∈
0,1
∫
0
t
‍
e
-
λ
t
-
s
s
s
s
s
s
s
s
s
s
s
s
·
∫
0
s
‍
L
x
u
+
f
u
,
0
u
α
-
2
s
-
u
d
u
d
s
+
A
1
β
∫
0
η
‍
e
-
λ
η
-
s
d
d
s
d
d
d
i
d
d
d
·
∫
0
s
‍
L
x
u
+
f
u
,
0
u
α
-
2
s
-
u
d
u
d
s
ssssssss
+
∫
0
1
‍
e
-
λ
1
-
s
∫
0
s
‍
L
x
u
+
f
u
,
0
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
·
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
≤
L
r
+
M
sup
⁡
t
∈
0,1
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
ssssssssssssss
+
A
1
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
sup
⁡
t
∈
0,1
+
∫
0
1
‍
e
-
λ
1
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
≤
L
r
+
M
α
(
α
-
1
)
sup
⁡
t
∈
[
0,1
]
∫
0
t
‍
e
-
λ
(
t
-
s
)
s
α
d
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
sup
⁡
t
∈
0,1
+
A
1
β
∫
0
η
‍
e
-
λ
η
-
s
s
α
d
s
+
∫
0
1
‍
e
-
λ
1
-
s
s
α
d
s
≤
L
r
+
M
1
+
A
1
β
η
α
1
-
e
-
λ
η
+
1
-
e
-
λ
λ
α
α
-
1
=
(
L
r
+
M
)
B
≤
r
.Now, for x, y
∈
C and for each t
∈
[
0,1
], we obtain (34)
|
(
T
x
)
(
t
)
-
(
T
y
)
(
t
)
|
=
sup
⁡
t
∈
[
0,1
]
(
T
x
)
(
t
)
-
(
T
y
)
(
t
)
≤
sup
⁡
t
∈
0,1
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
f
u
,
x
u
-
f
u
,
y
u
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
∫
0
s
‍
·
u
α
-
2
s
-
u
d
u
d
s
sssssssssss
+
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
·
∫
0
s
‍
f
u
,
x
u
-
f
u
,
y
u
u
α
-
2
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
∫
0
s
‍
·
s
-
u
d
u
d
s
ssssssssssssssssssssss
+
∫
0
1
‍
e
-
λ
1
-
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
d
d
s
s
s
s
·
∫
0
s
‍
f
u
,
x
u
-
f
u
,
y
u
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
∫
0
s
‍
·
u
α
-
2
s
-
u
d
u
d
s
≤
L
x
-
y
·
sup
⁡
t
∈
[
0,1
]
∫
0
t
‍
e
-
λ
(
t
-
s
)
∫
0
s
‍
u
α
-
2
(
s
-
u
)
d
u
d
s
ssssss
+
sup
⁡
t
∈
[
0,1
]
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
z
z
z
z
z
z
z
z
s
s
s
s
s
z
z
z
z
sup
⁡
t
∈
[
0,1
]
+
∫
0
1
‍
e
-
λ
1
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
≤
L
1
+
A
1
β
η
α
1
-
e
-
λ
η
+
1
-
e
-
λ
λ
α
(
α
-
1
)
x
-
y
=
B
L
x
-
y
.As B
<
1
/
L, we conclude that T is a contraction. Thus, the statement of the theorem follows by the classical Banach fixed point theorem. This concludes the proof.Now we recall a known result due to Krasnoselskii (see [9, Theorem 4.4
.
1]) which we will use to prove existence of at least one solution to (1)-(2).Theorem 13.
LetN be a closed, convex, and nonempty subset of a Banach space X. Let T
1 and T
2 be operators such that (i)
T
1
x
+
T
2
y
∈
N whenever x, y
∈
N,
(ii)
T
1 is compact and continuous,
(iii)
T
2 is a contraction mapping. Then there exists z
∈
N such that z
=
T
1
z
+
T
2
z.Theorem 14.
Letf
:
[
0,1
]
×
R
→
R be a jointly continuous function satisfying the following conditions: (H1)
|
f
(
t
,
v
)
-
f
(
t
,
w
)
|
≤
L
|
v
-
w
| for all t
∈
[
0,1
], v, w
∈
R,
(H2)
|
f
(
t
,
v
)
|
≤
μ
(
t
) for all (
t
,
v
)
∈
[
0,1
]
×
R with μ
∈
C. Then, boundary value problem (1)-(2) has at least one solution on C if (35)
A
1
|
β
|
η
α
(
1
-
e
-
λ
η
)
+
(
1
+
e
-
λ
)
λ
α
(
α
-
1
)
<
1
.Proof.
Lettingμ
=
sup
⁡
t
∈
[
0,1
]
|
μ
(
t
)
|, we fix (36)
r
≥
1
+
A
1
β
η
α
1
-
e
-
λ
η
+
1
-
e
-
λ
λ
α
α
-
1
μ
,and we consider B
r as in Theorem 12.
Define the operatorsT
1 and T
2 as (37)
(
T
1
x
)
(
t
)
=
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
,
(
T
2
y
)
(
t
)
=
A
(
t
)
β
∫
0
η
‍
e
-
λ
η
-
s
∫
0
s
‍
f
u
,
y
u
u
α
-
2
s
-
u
d
u
d
s
s
s
s
s
s
s
i
i
s
s
+
∫
0
1
‍
e
-
λ
1
-
s

∫
0
s
‍
f
u
,
y
u
u
α
-
2
s
-
u
d
u
d
s
.For x, y
∈
B
r, it follows by (36) that (38)
T
1
x
+
T
2
y
≤
1
+
A
1
|
β
|
η
α
(
1
-
e
-
λ
η
)
+
(
1
-
e
-
λ
)
λ
α
(
α
-
1
)
μ
≤
r
.Thus, T
1
x
+
T
2
y
∈
B
r. In view of condition (35), we have that T
2 is a contraction mapping.
Now we show thatT
1 is compact and continuous. The continuity of f implies that the operator T
1 is continuous. In addition, T
1 is uniformly bounded on B
r as (39)
T
1
x
=
sup
⁡
t
∈
[
0,1
]
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
≤
μ
sup
⁡
t
∈
0,1
∫
0
t
‍
e
-
λ
t
-
s
∫
0
s
‍
u
α
-
2
s
-
u
d
u
d
s
=
μ
α
(
α
-
1
)
∫
0
t
‍
e
-
λ
(
t
-
s
)
s
α
d
s
≤
μ
1
-
e
-
λ
λ
α
(
α
-
1
)
.Setting Ω
=
[
0,1
]
×
B
r, we define M
r
=
sup
⁡
(
t
,
x
)
∈
Ω
|
f
(
t
,
x
)
|, and consequently we have that, for 0
≤
t
2
<
t
1
≤
1, (40)
|
(
T
1
x
)
(
t
1
)
-
(
T
1
x
)
(
t
2
)
|
=
∫
0
t
1
‍
e
-
λ
t
1
-
s
∫
0
s
‍
f
u
,
x
u
u
α
-
2
s
-
u
d
u
d
s
d
d
d
-
∫
0
t
2
‍
e
-
λ
(
t
2
-
s
)
(
∫
0
s
‍
f
(
u
,
x
(
u
)
)
u
α
-
2
(
s
-
u
)
d
u
)
d
s
≤
M
r
α
α
-
1
e
λ
t
1
-
e
λ
t
2
e
-
λ
t
1
t
1
α
-
e
-
λ
t
2
t
2
α
,which is independent of x and tends to zero as t
2
→
t
1. This shows that T
1 is relatively compact on B
r. Hence, by the Ascoli-Arzelá Theorem, T
1 is compact on B
r. Thus, all the hypotheses of Theorem 13 are satisfied and the conclusion of Theorem 13 implies that the boundary value problem (1)-(2) has at least one solution on B
r
⊂
C, with r satisfying (36). This completes the proof.


## 5. Example

Consider the boundary value problem over the interval[
0,1
], given by:(41)
D
3
/
2
D
+
4
x
t
=
L
t
2
+
cos
⁡
t
+
arctan
x
t
,
x
(
0
)
=
0
,
x
′
0
=
0
,
x
(
1
)
=
x
1
2
.Here, f
(
t
,
v
)
=
L
t
2
+
cos
⁡
t
+
arctan
v, λ
=
4, β
=
1, η
=
1
/
2. Clearly, (42)
f
t
,
v
-
f
t
,
w
≤
L
arctan
v
-
arctan
w
≤
L
v
-
w
,
A
1
=
4
+
e
-
4
-
1
4
+
e
-
4
-
1
-
(
2
+
e
-
2
-
1
)
≈
1.6029
,
B
=
1
+
A
1
(
1
/
2
3
/
2
(
1
-
e
-
2
)
+
1
-
e
-
4
)
3
≈
1.0212
.For L
<
0.9792, it follows, by Theorem 12, that boundary value problem (41) has a unique solution.Remark 15.
Authors of [5] obtained a similar result considering fractional Caputo derivative instead of fractional conformable derivative in (41).



---

*Source: 706383-2015-03-18.xml*