# Investigating a Coupled Hybrid System of Nonlinear Fractional Differential Equations

Authors: Wiyada Kumam; Mian Bahadur Zada; Kamal Shah; Rahmat Ali Khan
Journal: Discrete Dynamics in Nature and Society
 (2018)

---

## Abstract

We study sufficient conditions for existence of solutions to the coupled systems of higher order hybrid fractional differential equations with three-point boundary conditions. For this motive, we apply the coupled fixed point theorem of Krasnoselskii type to form adequate conditions for existence of solutions to the proposed system. We finish the paper with suitable illustrative example.


---

## Body

## 1. Introduction

Fractional calculus is found to be more practical and effective than the classical calculus in the mathematical modeling of several phenomena. Fractional differential equations are very important and significant part of the mathematics and have various applications in viscoelasticity, electroanalytical chemistry, and many physical problems [1–6]. A systematic presentation of the applications of fractional differential equations can be found in the book of Balachandran and Park [7]. In recent years, many works have been devoted to the study of the mathematical aspects of fractional order differential equations [8–12]. There are numerous advanced and efficient methods, which have been focusing on the existence of solution to fractional differential equations. One of the powerful tools for obtaining the existence of solutions to such equations is the fixed point methods. Many authors use fixed point theorems to prove the existence and uniqueness of solution to nonlinear fractional differential equations; see, for example, [13–17].On the other hand, the study for coupled systems of fractional differential equations is also important as such systems occur in various problems of applied nature, for instance, [18–25]. Additionally, fixed point theory can be used to develop the existence theory for the coupled systems of fractional hybrid differential equations [13, 16, 17]. Bashiri et al. [17] discussed the existence of solution to the following system of fractional hybrid differential equations of order p
∈
(
0,1
):(1)
D
p
x
t
-
f
t
,
x
t
=
g
t
,
y
t
,
I
α
y
t
,
a
.
e
.
t
∈
0
,
T
,
T
>
0
,
D
p
y
t
-
f
t
,
y
t
=
g
t
,
x
t
,
I
α
x
t
,
a
.
e
.
t
∈
0
,
T
,
T
>
0
,
x
0
=
0
,
y
0
=
0
,where α
>
0, and the functions f
:
[
0
,
T
]
×
R
→
R
,
f
(
0,0
)
=
0 and g
:
[
0
,
T
]
×
R
×
R
→
R satisfy certain conditions. D
p is the R-L fractional derivative of order p.Recently, the existence of solutions for fractional differential equations involving the Caputo fractional derivative was studied in [13, 26–28]. Motivated by the work of Bashiri et al. [17], in this paper we are concerned with the existence of solutions to three-point boundary value problem for a coupled system of hybrid fractional differential equations of order p
∈
[
n
-
1
,
n
) given by(2)
D
p
x
t
-
f
t
,
x
t
=
g
t
,
y
t
,
I
α
y
t
,
a
.
e
.
t
∈
0,1
,
D
p
y
t
-
f
t
,
y
t
=
g
t
,
x
t
,
I
α
x
t
,
a
.
e
.
t
∈
0,1
,
x
0
=
δ
1
x
η
1
,
y
0
=
δ
1
y
η
1
,
x
1
=
δ
2
x
η
2
,
y
1
=
δ
2
y
η
2
,
x
i
0
=
y
i
0
=
0
,
for
i
=
1,2
,
3
,
…
,
n
-
2
,where α
>
0, 0
<
η
1
<
η
2
<
1, f
:
[
0,1
]
×
R
→
R, f
(
0
,
x
(
0
)
)
=
0, and g
:
[
0,1
]
×
R
×
R
→
R. D
p is the Caputo fractional derivative of order p. Moreover, an example is given to illustrate the validity of the existence result.


## 2. Preliminaries

Throughout this manuscriptΦ
=
{
φ
:
R
+
→
R
+
such  that
φ
(
r
)
<
r
for
r
>
0
and
φ
(
0
)
=
0
}, C
(
[
0
,
T
]
×
R
,
R
) denote the class of continuous functions f
:
[
0
,
T
]
×
R
→
R, and C
(
[
0
,
T
]
×
R
×
R
,
R
) denote the class of functions g
:
[
0
,
T
]
×
R
×
R
→
R such that(i)
the mapt
→
g
(
t
,
x
,
y
) is measurable for each x
,
y
∈
R,
(ii)
the mapx
→
g
(
t
,
x
,
y
) is continuous for each x
∈
R,
(iii)
the mapy
→
g
(
t
,
x
,
y
) is continuous for each y
∈
R.We need the following definitions which can be found in [9].Definition 1.
The Riemann-Liouville fractional integral of orderα
>
0 of function f
∈
L
1
(
R
+
) is defined as(3)
I
α
f
t
=
1
Γ
α
∫
0
t
t
-
s
α
-
1
f
s
d
s
,provided that the right side is pointwise defined on (
0
,
∞
).Definition 2.
Letα be a positive real number, such that m
-
1
≤
α
<
m, m
∈
N, and f
m
(
x
) exists, a function of class C. Then Caputo fractional derivative of f is defined as(4)
D
c
α
f
t
=
1
Γ
m
-
α
∫
0
t
t
-
s
m
-
α
-
1
f
m
s
d
s
,provided that the right side is pointwise defined on (
0
,
∞
), where m
=
[
α
]
+
1 and [
α
] represents the integer part of α.Definition 3 (see [29]).
The mappingF
:
X
×
X
→
X has a coupled fixed point (
x
,
y
)
∈
X
×
X if F
(
x
,
y
)
=
x and F
(
y
,
x
)
=
y.Theorem 4 (see [17]).
LetS be a nonempty, closed, convex, and bounded subset of the Banach space X and S
~
=
S
×
S. Suppose that A
:
X
→
X and B
:
S
→
X are two operators such that (
C
1
)
there existsφ
A
∈
Φ such that, for all x
,
y
∈
X, one has (5)
A
x
-
A
y
≤
σ
φ
A
x
-
y
,
for
some  constant
σ
>
0
,
(
C
2
)
B is completely continuous,
(
C
3
)
x
=
A
x
+
B
y
⇒
x
∈
S, for all y
∈
S.
Then the operatorT
(
x
,
y
)
=
A
x
+
B
y has at least a coupled fixed point in S
~ whenever σ
<
1.Lemma 5.
The following result holds for fractional differential equations:(6)
I
α
D
c
α
h
t
=
h
t
+
c
0
+
c
1
t
+
c
2
t
2
+
⋯
+
c
m
-
1
t
m
-
1
,for arbitrary c
i
∈
R, i
=
0,1
,
2
,
…
,
m
-
1, where m
=
[
α
]
+
1 and [
α
] represents the integer part of α.


## 3. Existence Results

Let us set the following notations for convenience:(7)
λ
=
1
-
δ
1
1
-
δ
2
η
2
n
-
1
+
1
-
δ
2
δ
1
η
1
n
-
1
,
(8)
F
0
=
2
λ
f
1,0
+
1
2
m
a
x
t
∈
0,1
f
t
,
0
+
m
a
x
η
1
∈
0,1
f
η
1
,
0
+
m
a
x
η
2
∈
0,1
f
η
2
,
0
.For the forthcoming analysis, we assume that (
A
1
)
the functionx
→
x
-
f
(
t
,
x
) is increasing in R for all t
∈
[
0,1
];
(
A
2
)
there existsM
≥
L
>
0 such that (9)
f
t
,
x
t
-
f
t
,
y
t
≤
L
x
t
-
y
t
8
M
+
x
t
-
y
t
,
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
≤
sup
⁡
L
δ
2
x
t
-
y
t
16
M
+
δ
2
x
t
-
y
t
,
f
η
1
,
x
η
1
-
f
η
1
,
y
η
1
λ
≤
sup
⁡
L
x
t
-
y
t
16
M
+
x
t
-
y
t
,
f
η
2
,
x
η
2
-
f
η
2
,
y
η
2
λ
≤
sup
⁡
L
x
t
-
y
t
16
M
+
x
t
-
y
t
;
(
A
3
)
there exists a continuous functionh
∈
C
(
[
0,1
]
,
R
) such that (10)
g
t
,
x
t
,
y
t
≤
h
t
,
x
,
y
∈
R
,
t
∈
0,1
.Lemma 6.
Iff
(
0
,
x
(
0
)
)
=
0 and (
∂
i
f
t
,
x
t
/
∂
t
i
)
|
t
=
0
=
0 for i
=
1,2
,
…
,
(
n
-
2
), then integral representation of the system (2) is given by (11)
x
t
=
f
t
,
x
t
+
1
λ
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
δ
1
η
1
n
-
1
f
1
,
x
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
+
1
λ
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
1
-
δ
1
f
1
,
x
1
-
1
-
δ
2
δ
1
f
η
1
,
x
η
1
t
n
-
1
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
,
(12)
y
t
=
f
t
,
y
t
+
1
λ
δ
2
δ
1
η
1
n
-
1
f
η
2
,
y
η
2
-
δ
1
η
1
n
-
1
f
1
,
y
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
y
η
1
+
1
λ
1
-
δ
1
δ
2
f
η
2
,
y
η
2
-
1
-
δ
1
f
1
,
y
1
-
1
-
δ
2
δ
1
f
η
1
,
y
η
1
t
n
-
1
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
.Proof.
Applying the operatorI
p on the first equation of system (2) and using Lemma 5, we obtain(13)
x
t
=
f
t
,
x
t
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
C
0
+
C
1
t
+
C
2
t
2
+
⋯
+
C
n
-
1
t
n
-
1
.Applying the initial conditions x
(
i
)
(
0
)
=
0, for i
=
1,2
,
3
,
…
,
(
n
-
2
), we conclude that C
1
=
C
2
=
⋯
=
C
n
-
2
=
0. Therefore (13) becomes(14)
x
t
=
f
t
,
x
t
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
C
0
+
C
n
-
1
t
n
-
1
.Now, to find the values of C
0 and C
n
-
1, since x
(
0
)
=
δ
1
x
(
η
1
), from (14) we have (15)
f
0
,
x
0
+
I
p
g
0
,
y
0
,
I
α
y
0
+
C
0
=
δ
1
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
C
0
+
C
n
-
1
η
1
n
-
1
⟹
1
-
δ
1
C
0
-
δ
1
C
n
-
1
η
1
n
-
1
-
δ
1
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
=
0
.Also, since x
(
1
)
=
δ
2
x
(
η
2
), from (14), we have (16)
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
C
0
+
C
n
-
1
=
δ
2
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
C
0
+
C
n
-
1
η
2
n
-
1
⟹
1
-
δ
2
C
0
+
1
-
δ
2
η
2
n
-
1
C
n
-
1
+
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
2
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
=
0
.Solving (15) and (16) for C
0, we get (17)
1
-
δ
2
η
2
n
-
1
1
-
δ
1
+
δ
1
η
1
n
-
1
1
-
δ
2
C
0
+
δ
1
η
1
n
-
1
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
=
0
,and using (7), we can write (18)
C
0
=
δ
1
1
-
δ
2
η
2
n
-
1
λ
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
1
η
1
n
-
1
λ
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
.Similarly, solving (15) and (16) and using (7), we obtained that(19)
C
n
-
1
=
-
1
-
δ
1
λ
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
1
-
δ
2
δ
1
λ
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
1
-
δ
1
δ
2
λ
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
.Substituting the values of C
0 and C
1 in (14), we can write (20)
x
t
=
f
t
,
x
t
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
1
η
1
n
-
1
λ
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
1
-
δ
1
λ
f
1
,
x
1
+
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
t
n
-
1
-
1
-
δ
2
δ
1
λ
f
η
1
,
x
η
1
+
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
t
n
-
1
+
1
-
δ
1
δ
2
λ
f
η
2
,
x
η
2
+
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
t
n
-
1
,which implies that (21)
x
t
=
f
t
,
x
t
+
1
λ
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
δ
1
η
1
n
-
1
f
1
,
x
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
+
1
λ
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
1
-
δ
1
f
1
,
x
1
-
1
-
δ
2
δ
1
f
η
1
,
x
η
1
t
n
-
1
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
y
s
,
I
α
y
s
d
s
.Similarly, repeating the above process with the second equation of system (2), we obtain integral equation (12).Now, we are in a position to present the existence theorem for the system (2).Theorem 7.
Assume that hypotheses(
A
1
)–(
A
3
) hold. Then there exists a solution for coupled systems (2) of higher order hybrid FHDEs with three-point boundary conditions.Proof.
SetX
=
C
(
[
0,1
]
,
R
) and a subset S of X defined by (22)
S
=
x
∈
X
:
x
≤
N
,
where
N
≥
7
L
+
F
0
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
.Clearly S is a closed, convex, and bounded subset of the Banach space X. Now, since x
(
t
) is a solution of the FHDEs system (2) if and only if x
(
t
) satisfies the system of integral equations in Lemma 6, to show the existence solution of system (2) it is enough to show the existence solution of the integral equations in Lemma 6. For this, define two operators A
:
X
→
X and B
:
S
→
X by (23)
A
x
t
=
f
t
,
x
t
+
1
λ
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
δ
1
η
1
n
-
1
f
1
,
x
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
+
1
λ
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
1
-
δ
1
f
1
,
x
1
-
1
-
δ
2
δ
1
f
η
1
,
x
η
1
t
n
-
1
,
B
x
t
=
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
.Then the operators form of system (2) is (24)
x
t
=
A
x
t
+
B
y
t
,
t
∈
0,1
,
y
t
=
A
y
t
+
B
x
t
,
t
∈
0,1
.We have to show that the operators A and B satisfy all the conditions of Theorem 4. For this, let x
,
y
∈
X; then we have (25)
A
x
t
-
A
y
t
≤
f
t
,
x
t
-
f
t
,
y
t
+
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
δ
1
η
1
n
-
1
f
1
,
x
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
λ
-
δ
2
δ
1
η
1
n
-
1
f
η
2
,
y
η
2
-
δ
1
η
1
n
-
1
f
1
,
y
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
y
η
1
λ
+
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
1
-
δ
1
f
1
,
x
1
-
1
-
δ
2
δ
1
f
η
1
,
x
η
1
λ
-
1
-
δ
1
δ
2
f
η
2
,
y
η
2
-
1
-
δ
1
f
1
,
y
1
-
1
-
δ
2
δ
1
f
η
1
,
y
η
1
λ
t
n
-
1
≤
f
t
,
x
t
-
f
t
,
y
t
+
δ
1
η
1
n
-
1
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
-
f
η
1
,
y
η
1
λ
+
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
f
η
2
,
y
η
2
λ
+
1
-
δ
1
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
+
1
-
δ
2
δ
1
f
η
1
,
x
η
1
-
f
η
1
,
y
η
1
λ
+
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
f
η
2
,
y
η
2
λ
<
f
t
,
x
t
-
f
t
,
y
t
+
2
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
+
2
f
η
1
,
x
η
1
-
f
η
1
,
y
η
1
λ
+
2
f
η
2
,
x
η
2
-
f
η
2
,
y
η
2
λ
.
Taking the supremum overt and using (7), we get (26)
A
x
t
-
A
y
t
≤
1
2
L
x
t
-
y
t
4
M
+
x
t
-
y
t
+
L
δ
2
x
t
-
y
t
4
M
+
δ
2
x
t
-
y
t
+
L
x
t
-
y
t
4
M
+
x
t
-
y
t
+
L
x
t
-
y
t
4
M
+
x
t
-
y
t
=
1
2
3
L
x
t
-
y
t
4
M
+
x
t
-
y
t
+
L
δ
2
x
t
-
y
t
4
M
+
δ
2
x
t
-
y
t
≤
1
2
φ
x
t
-
y
t
.Thus A satisfies condition (
C
1
) of Theorem 4 with σ
=
1
/
2 and φ
(
r
)
=
3
L
r
/
4
(
M
+
r
)
+
L
δ
2
r
/
4
(
M
+
δ
2
r
).
Next, we show thatB is compact and continuous operator on S. Let {
x
m
} be a sequence in S such that {
x
m
}
→
x
∈
S. Then for all t
∈
[
0,1
], we have (27)
l
i
m
m
→
∞
B
x
m
t
=
1
Γ
p
l
i
m
m
→
∞
∫
0
t
t
-
s
p
-
1
g
s
,
x
m
s
,
I
α
x
m
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
l
i
m
m
→
∞
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
m
s
,
I
α
x
m
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
l
i
m
m
→
∞
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
m
s
,
I
α
x
m
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
l
i
m
m
→
∞
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
m
s
,
I
α
x
m
s
d
s
=
1
Γ
p
∫
0
t
t
-
s
p
-
1
l
i
m
m
→
∞
g
s
,
x
m
s
,
I
α
x
m
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
l
i
m
m
→
∞
g
s
,
x
m
s
,
I
α
x
m
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
l
i
m
m
→
∞
g
s
,
x
m
s
,
I
α
x
m
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
l
i
m
m
→
∞
g
s
,
x
m
s
,
I
α
x
m
s
d
s
=
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
=
B
x
t
.Thus the map B is continuous on S.
Letx
∈
S; then we have (28)
B
x
t
≤
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
≤
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
+
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
≤
1
Γ
p
∫
0
t
t
-
s
p
-
1
h
s
d
s
+
2
λ
Γ
p
∫
0
1
1
-
s
p
-
1
h
s
d
s
+
∫
0
η
1
η
1
-
s
p
-
1
h
s
d
s
+
∫
0
η
2
η
2
-
s
p
-
1
h
s
d
s
≤
h
L
1
Γ
p
∫
0
t
t
-
s
p
-
1
d
s
+
2
λ
∫
0
1
1
-
s
p
-
1
d
s
+
∫
0
η
1
η
1
-
s
p
-
1
d
s
+
∫
0
η
2
η
2
-
s
p
-
1
d
s
=
h
L
1
Γ
p
t
-
s
p
-
p
0
t
+
2
λ
1
-
s
p
-
p
0
1
+
η
1
-
s
p
-
p
0
η
1
+
η
2
-
s
p
-
p
0
η
2
=
h
L
1
Γ
p
t
p
p
+
2
λ
1
p
+
η
1
p
p
+
η
2
p
p
=
h
L
1
Γ
p
+
1
t
p
+
2
λ
1
+
η
1
p
+
η
2
p
≤
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
.Taking the supremum over t, we get (29)
B
x
≤
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
,
∀
x
∈
S
.Thus B is uniformly bounded on S.
Now, lett
1
,
t
2
∈
[
0,1
] such that t
1
≠
t
2; then for any x
∈
S, we have (30)
B
x
t
1
-
B
x
t
2
=
1
Γ
p
∫
0
t
1
t
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
1
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
1
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
1
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
1
Γ
p
∫
0
t
2
t
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
1
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
1
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
1
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
≤
1
Γ
p
∫
0
t
1
t
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
-
∫
0
t
2
t
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
≤
1
Γ
p
∫
0
t
1
t
1
-
s
p
-
1
h
s
d
s
-
∫
0
t
1
t
2
-
s
p
-
1
h
s
d
s
+
∫
0
t
1
t
2
-
s
p
-
1
h
s
d
s
-
∫
0
t
2
t
2
-
s
p
-
1
h
s
d
s
≤
h
L
1
Γ
p
∫
0
t
1
t
1
-
s
p
-
1
-
t
2
-
s
p
-
1
d
s
+
∫
t
1
t
2
t
2
-
s
p
-
1
d
s
=
h
L
1
Γ
p
t
1
-
s
p
-
p
-
t
2
-
s
p
-
p
0
t
1
+
t
2
-
s
p
-
p
t
1
t
2
=
h
L
1
Γ
p
t
2
-
t
1
p
p
+
t
1
p
p
-
t
2
p
p
+
t
2
-
t
1
p
p
≤
h
L
1
Γ
p
+
1
t
2
-
t
1
p
+
t
1
p
-
t
2
p
+
t
2
-
t
1
p
.Since t
p is uniformly continuous on [
0,1
] for n
-
1
<
p
<
n, for any ε
>
0 we can find δ
1
∗
>
0 such that (31)
t
1
p
-
t
2
p
<
Γ
p
+
1
3
h
L
1
ε
,
whenever
t
1
-
t
2
<
δ
1
∗
.Let δ
∗
=
min
⁡
δ
1
∗
,
Γ
(
p
+
1
)
/
3
h
L
1
1
/
p; then, we have (32)
B
x
t
1
-
B
x
t
2
≤
h
L
1
Γ
p
+
1
Γ
p
+
1
3
h
L
1
ε
+
Γ
p
+
1
3
h
L
1
ε
+
Γ
p
+
1
3
h
L
1
ε
=
ε
,
whenever
t
1
-
t
2
<
δ
∗
.Thus B
(
S
) is equicontinuous and hence B is completely continuous on S.
To prove hypothesis(
C
3
) of Theorem 4, let x
∈
X and y
∈
S such that x
=
A
x
+
B
y; using (7) and (8), we have (33)
x
t
≤
A
x
t
+
B
x
t
=
f
t
,
x
t
+
δ
2
δ
1
η
1
n
-
1
f
η
2
,
x
η
2
-
δ
1
η
1
n
-
1
f
1
,
x
1
+
δ
1
1
-
δ
2
η
2
n
-
1
f
η
1
,
x
η
1
λ
+
1
-
δ
1
δ
2
f
η
2
,
x
η
2
-
1
-
δ
1
f
1
,
x
1
-
1
-
δ
2
δ
1
f
η
1
,
x
η
1
λ
t
n
-
1
+
1
Γ
p
∫
0
t
t
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
1
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
1
1
-
δ
2
η
2
n
-
1
λ
-
1
-
δ
2
δ
1
t
n
-
1
λ
1
Γ
p
∫
0
η
1
η
1
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
+
δ
2
δ
1
η
1
n
-
1
λ
+
1
-
δ
1
δ
2
t
n
-
1
λ
1
Γ
p
∫
0
η
2
η
2
-
s
p
-
1
g
s
,
x
s
,
I
α
x
s
d
s
≤
f
t
,
x
t
+
f
η
2
,
x
η
2
+
f
1
,
x
1
+
f
η
1
,
x
η
1
λ
+
f
η
2
,
x
η
2
+
f
1
,
x
1
+
f
η
1
,
x
η
1
λ
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
.This implies (34)
x
t
≤
f
t
,
x
t
-
f
t
,
0
+
f
t
,
0
+
2
f
1,0
λ
+
f
η
1
,
0
λ
+
f
η
2
,
0
λ
+
2
f
1
,
δ
2
x
η
2
-
f
1,0
λ
+
f
η
1
,
x
η
1
-
f
η
1
,
0
λ
+
f
η
2
,
x
η
2
-
f
η
2
,
0
λ
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
≤
L
+
2
L
+
L
+
L
+
2
λ
1
2
f
t
,
0
+
f
1,0
+
f
η
1
,
0
+
f
η
2
,
0
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
≤
7
L
+
F
0
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
.Taking the supremum over t on [
0,1
], we can write (35)
x
t
≤
7
L
+
F
0
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
≤
N
.That is, x
∈
S. Thus condition (
C
3
) of Theorem 4 holds. Therefore, all the conditions of Theorem 4 are satisfied; hence the operator T
(
x
,
y
)
=
A
x
+
B
y has a coupled fixed point on S
~. Consequently, system (2) has a solution defined on [
0,1
].To illustrate Theorem7, we construct the following example.Example 8.
We discuss the following hybrid fractional differential equations with three-point boundary conditions:(36)
D
5
/
2
x
t
-
e
-
t
x
t
11
+
x
t
=
t
3
3
-
cos
⁡
y
t
+
cos
⁡
I
3
/
2
y
t
,
D
5
/
2
y
t
-
e
-
t
y
t
11
+
y
t
=
t
3
3
-
c
o
s
x
t
+
cos
⁡
I
3
/
2
x
t
,
x
0
=
1
4
x
1
4
,
y
0
=
1
4
y
1
4
,
x
1
=
1
2
x
1
2
,
y
1
=
1
2
y
1
2
,
x
′
0
=
y
′
0
=
0
,where t
∈
[
0,1
], f
:
[
0,1
]
×
R
→
R
,
f
(
0
,
x
(
0
)
)
=
0, and g
:
[
0,1
]
×
R
×
R
→
R. D
5
/
2 is the Caputo fractional derivative of order 5
/
2.
Here(37)
δ
1
=
η
1
=
1
4
,
δ
2
=
η
2
=
1
2
,
f
t
,
x
t
=
e
-
t
x
t
11
+
x
t
,
g
t
,
x
t
,
I
α
x
t
=
t
3
3
-
c
o
s
x
t
+
c
o
s
I
3
/
2
x
t
.
Therefore,(38)
λ
=
19
32
,
F
0
=
2
λ
f
1,0
+
1
2
m
a
x
t
∈
0,1
f
t
,
0
+
m
a
x
η
1
∈
0,1
f
η
1
,
0
+
m
a
x
η
2
∈
0,1
f
η
2
,
0
=
0
.Now, for M
=
11 and L
=
512
/
19
e
<
M, we have (39)
f
t
,
x
t
-
f
t
,
y
t
=
e
-
t
x
t
11
+
x
t
-
e
-
t
y
t
11
+
y
t
≤
x
t
-
y
t
+
y
t
11
+
x
t
-
y
t
+
y
t
-
y
t
11
+
x
t
-
y
t
+
y
t
≤
x
t
-
y
t
11
+
x
t
-
y
t
+
y
t
≤
x
t
-
y
t
11
+
x
t
-
y
t
≤
512
/
19
e
x
t
-
y
t
16
11
+
x
t
-
y
t
,
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
=
f
1
,
1
/
2
x
1
/
2
-
f
1
,
1
/
2
y
1
/
2
19
/
32
=
32
19
e
-
1
1
/
2
x
1
/
2
11
+
1
/
2
x
1
/
2
-
e
-
1
1
/
2
y
1
/
2
11
+
1
/
2
y
1
/
2
,
f
1
,
δ
2
x
η
2
-
f
1
,
δ
2
y
η
2
λ
≤
32
19
e
δ
2
x
1
/
2
-
y
1
/
2
+
δ
2
y
1
/
2
11
+
δ
2
x
1
/
2
-
y
1
/
2
+
δ
2
y
1
/
2
-
δ
2
y
1
/
2
11
+
δ
2
x
1
/
2
-
y
1
/
2
+
δ
2
y
1
/
2
≤
32
19
e
δ
2
x
1
/
2
-
y
1
/
2
11
+
δ
2
x
1
/
2
-
y
1
/
2
+
δ
2
y
1
/
2
≤
32
19
e
×
δ
2
x
1
/
2
-
y
1
/
2
11
+
δ
2
x
1
/
2
-
y
1
/
2
≤
sup
⁡
512
/
19
e
δ
2
x
t
-
y
t
16
11
+
δ
2
x
t
-
y
t
.Similarly, (40)
f
η
1
,
x
η
1
-
f
η
1
,
y
η
1
λ
≤
s
u
p
512
/
19
e
δ
2
x
t
-
y
t
16
11
+
δ
2
x
t
-
y
t
,
f
η
2
,
x
η
2
-
f
η
2
,
y
η
2
λ
≤
sup
⁡
512
/
19
e
δ
2
x
t
-
y
t
16
11
+
δ
2
x
t
-
y
t
.Next, (41)
g
t
,
x
t
,
I
α
x
t
=
t
3
3
-
cos
⁡
x
t
+
cos
⁡
I
3
/
2
x
t
≤
t
3
3
=
h
t
.That is, there exists a continuous function h
∈
C
(
[
0,1
]
,
R
) such that (42)
g
t
,
x
t
,
y
t
≤
h
t
,
x
,
y
∈
R
,
t
∈
0,1
.Finally, since (43)
h
t
=
s
u
p
∫
0
1
t
3
3
d
t
=
1
12
,
L
=
512
19
e
,
F
0
=
0
,
λ
=
19
32
,we can write (44)
7
L
+
F
0
+
h
L
1
Γ
p
+
1
1
+
2
λ
1
+
η
1
p
+
η
2
p
<
70
.Thus N
≥
70. It follows that assumptions (
A
1
) and (
A
2
) are satisfied. Therefore, by Theorem 7 we conclude that problem (36) has a solution.


## 4. Conclusion

We have successfully developed appropriate conditions for existence of at least one solution to a complicated higher order coupled system of nonlinear hybrid fractional differential equations. The respective conditions have been derived by using coupled fixed point theorem of Krasnoselskii type. The obtained results were also demonstrated by a suitable example.



---

*Source: 5937572-2018-03-20.xml*