International Journal of Analysis and Applications

Volume 17, Number 1 (2019), 47-63

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-47

NONLINEAR SEQUENTIAL RIEMANN-LIOUVILLE AND CAPUTO FRACTIONAL

DIFFERENTIAL EQUATIONS WITH NONLOCAL AND INTEGRAL BOUNDARY

CONDITIONS

SUPHAWAT ASAWASAMRIT1, NAWAPOL PHUANGTHONG1, SOTIRIS K.

NTOUYAS2,3 AND JESSADA TARIBOON1,∗

1Intelligent and Nonlinear Dynamic Innovations Research Center, Department of Mathematics, Faculty of

Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand

2Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece

3Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics,

Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

∗Corresponding author: jessada.t@sci.kmutnb.ac.th

Abstract. In this paper, we discuss the existence and uniqueness of solutions for a new class of sequential

fractional differential equations of Riemann-Liouville and Caputo types with nonlocal integral boundary

conditions, by using standard fixed point theorems. We also demonstrate the application of the obtained

results with the aid of examples.

Received 2018-09-10; accepted 2018-10-24; published 2019-01-04.

2010 Mathematics Subject Classification. 26A33, 34A08, 34B15.

Key words and phrases. fractional derivatives; fractional integral; boundary value problems; existence; uniqueness; fixed

point theorems.

c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

47

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-47


Int. J. Anal. Appl. 17 (1) (2019) 48

1. Introduction

Fractional differential equations have gained considerable importance due to their widespread applica-

tions in various disciplines of social and natural sciences, and engineering. In recent years, there has been

a significant development in fractional calculus and fractional differential equations, for instance, see the

monographs by Kilbas et al. [12], Lakshmikantham et al. [14], Miller and Ross [15], Podlubny [16], Samko

et al. [18], Diethelm [9], Ahmad et al. [7] and the papers [1, 4–6, 8, 10, 17, 20, 21].

Recently in [2] the authors studied a class of nonlinear differential equations with multiple fractional

derivatives and Caputo type integro-differential boundary conditions


Dα[Dβx(t) −g(t,x(t))] = f(t,x(t)), t ∈ J := [0,T],

x(0) = 0, (Dγx)(T) = λ(Iδx)(T),

(1.1)

where Dχ is Caputo fractional derivative of order χ ∈{α,β,γ}, 0 < α,β,γ < 1, Iδ is the Riemann-Liouville

fractional integral of order δ > 0, f,g : J × R → R are given functions and λ 6=
Γ(β + δ + 1)

Tγ+δΓ(β −γ + 1)
. The

existence of solutions for the problem (1.1) is established by applying Leray-Schauder nonlinear alternative

[11] and Krasnoselskii’s fixed point theorem [13]. The uniqueness result for the problem (1.1) is obtained by

means of a celebrated fixed point theorem due to Banach.

In [3] existence criteria are developed for the solutions of Caputo-Hadamard type fractional neutral dif-

ferential equations supplemented with Dirichlet boundary conditions


Dω[Dκx(t) −h(t,x(t))] = f(t,x(t)), t ∈ [1,T], T > 1,

x(1) = 0, x(T) = 0,

(1.2)

where Dρ denotes the Caputo-Hadamard fractional derivatives of order ρ ∈ (0, 1) with ρ ∈ {ω,κ} and

f,h : [1,T] ×R → R are appropriate functions.

Very recently in [19], the authors discussed existence and uniqueness of solutions for two sequential

Caputo-Hadamard and Hadamard-Caputo fractional differential equations subject to separated boundary

conditions as 


CDp(HDqx)(t) = f(t,x(t)), t ∈ (a,b),

α1x(a) + α2(
HDqx)(a) = 0, β1x(b) + β2(

HDqx)(b) = 0,

(1.3)

and 


HDq(CDpx)(t) = f(t,x(t)), t ∈ (a,b),

α1x(a) + α2(
CDpx)(a) = 0, β1x(b) + β2(

CDpx)(b) = 0,

(1.4)

where CDp and HDq are the Caputo and Hadamard fractional derivatives of orders p and q, respectively,

0 < p,q ≤ 1, f : [a,b] ×R → R is a continuous function, a > 0 and αi,βi ∈ R, i = 1, 2.



Int. J. Anal. Appl. 17 (1) (2019) 49

Motivated by the above papers, we consider in the present paper the following boundary value problem


RLDq[CDrx(t) −g(t,x(t))] = f(t,x(t)), 0 < t < T,

x(η) = φ(x), Ipx(T) = h(x),

(1.5)

where RLDq, CDr are Riemann-Liouville and Caputo fractional derivatives of orders q,r ∈ (0, 1), respectively,

Ip is the Riemann-Liouville fractional integral of order p > 0, f,g : J×R → R are given continuous functions

and φ,h : C(J,R) → R are two given functionals.

The rest of the paper is arranged as follows. In Section 2, we establish basic results that lays the

foundation for defining a fixed point problem equivalent to the given problem (1.5). The main results, based

on Banach’s contraction mapping principle, Krasnoselskii’s fixed point theorem and nonlinear alternative of

Leray-Schauder type, are obtained in Section 3. Examples illustrating the obtained results are also included.

2. Preliminaries

In this section, we recall some basic concepts of fractional calculus [12, 16] and present known results

needed in our forthcoming analysis.

Definition 2.1. The Riemann-Liouville fractional derivative of order q for a function f : (0,∞) → R is

defined by

RLDqf(t) =
1

Γ(n−q)

(
d

dt

)n ∫ t
0+

(t−s)n−q−1 f(s)ds, q > 0, n = [q] + 1,

where [q] denotes the integer part of the real number q, provided the right-hand side is pointwise defined on

(0,∞).

Definition 2.2. The Riemann-Liouville fractional integral of order q for a function f : (0,∞) → R is

defined by

Iqf(t) =
1

Γ(q)

∫ t
0+

(t−s)q−1 f(s)ds, q > 0,

provided the right-hand side is pointwise defined on (0,∞).

Definition 2.3. The Caputo derivative of fractional order q for a n-times derivative function f : (0,∞) → R

is defined as

CDqf(t) =
1

Γ(n−q)

∫ t
0+

(t−s)n−q−1
(
d

ds

)n
f(s)ds, q > 0, n = [q] + 1.

Lemma 2.1. If α + β > 0, then the equation (IαIβu)(t) = (Iα+βu)(t), t ∈ J is satisfied for u ∈ L1(J,R).

Lemma 2.2. Let β > α. Then the equation (DαIβu)(t) = (Iβ−αu)(t), t ∈ J is satisfied for u ∈ C(J,R).

Lemma 2.3. Let n = [α] + 1 if α 6∈ N and n = α if α ∈ N. Then the following relations hold:

(i) for k ∈{0, 1, 2, . . . ,n− 1}, Dαtk = 0;



Int. J. Anal. Appl. 17 (1) (2019) 50

(ii) if β > n then Dαtβ−1 =
Γ(β)

Γ(β −α)
tβ−α−1;

(iii) Iαtβ−1 =
Γ(β)

Γ(β + α)
tβ+α−1.

Lemma 2.4. Let q > 0. Then for y ∈ C(0,T) ∩L(0,T) it holds

RLIq
(
RLDqy

)
(t) = y(t) + c1t

q−1 + c2t
q−2 + · · · + cntq−n,

where ci ∈ R, i = 1, 2, . . . ,n and n− 1 < q < n.

Lemma 2.5. Let q > 0. Then for y ∈ C(0,T) ∩L(0,T) it holds

RLIq
(
CDqy

)
(t) = y(t) + c0 + c1t + c2t

2 + · · · + cn−1tn−1,

where ci ∈ R, i = 0, 1, 2, . . . ,n− 1 and n = [q] + 1.

In the following, for simplicity, we use the notation Iq for RLIq.

Lemma 2.6. Let p > 0, 0 < q,r ≤ 1, with q + r > 1,

Λ =
Γ(q)

Γ(q + r)

Tp

Γ(p + 1)
ηq+r−1 −

Γ(q)

Γ(p + q + r)
Tp+q+r−1 6= 0, (2.1)

and ĝ,y ∈ C(J,R) and two functionals φ,h : C(J,R) → R. The unique solution of the linear problem


RLDq[CDrx(t) − ĝ(t)] = y(t), 0 < t < T,

x(η) = φ(x), Ipx(T) = h(x),

(2.2)

is given by

x(t) = Irĝ(t) + Iq+ry(t)

+
tq+r−1

Λ

Γ(q)

Γ(q + r)

[(
φ(x) − Irĝ(η) − Iq+ry(η)

) Tp
Γ(p + 1)

−
(
h(x) − Ip+rĝ(T) − Ip+q+ry(T)

)]
+

1

Λ

[
Γ(q)

Γ(q + r)
ηq+r−1

(
h(x) − Ip+rĝ(T) − Ip+q+ry(T)

)
−
(
φ(x) − Irĝ(η) − Iq+ry(η)

) Γ(q)
Γ(p + q + r)

Tp+q+r−1
]
.

Proof. Firstly, we apply the Riemann-Liouville fractional integral of order q to both sides of equation (2.2),

and using Lemma 2.4, we have

CDrx(t) = ĝ(t) + Iqy(t) + c1t
q−1, (2.3)



Int. J. Anal. Appl. 17 (1) (2019) 51

where a constant c1 ∈ R. After that, using Riemann-Liouville fractional integral of order r to both sides the

above equation and applying Lemma 2.5, we get

x(t) = Irĝ(t) + Iq+ry(t) + c1
Γ(q)

Γ(q + r)
tq+r−1 + c2, (2.4)

where a constant c2 ∈ R. Observe that the equation (2.4) is well defined as q + r > 1.

Using nonlocal boundary condition of problem (2.2) to the above equation, we obtain the linear system

c1
Γ(q)

Γ(q + 1)
ηq+r−1 + c2 = φ(x) − Irĝ(η) − Iq+ry(η),

c1
Γ(q)

Γ(p + q + 1)
Tp+q+r−1 + c2

Tp

Γ(p + 1)
= h(x) − Ip+rĝ(T) − Ip+q+ry(T).

Note that the two functionals φ(x) and h(x) are constants. Solving the system of linear equations for

constants c1, c2, we have

c1 =
1

Λ

[
Tp

Γ(p + 1)

(
φ(x) − Irĝ(η) − Iq+ry(η)

)
−
(
h(x) − Ip+rĝ(T) − Ip+q+ry(T)

)]
,

c2 =
1

Λ

[
Γ(q)

Γ(q + r)
ηq+r−1

(
h(x) − Ip+rĝ(T) − Ip+q+ry(T)

)
−

Γ(q)

Γ(p + q + r)
Tp+q+r−1

(
φ(x) − Irĝ(η) − Iq+ry(η)

)]
.

Substituting two constants c1 and c2 into equation (2.4), we obtain the required solution. The converse

follows by direct computation. The proof is completed. �

3. Main results

Let J = [0,T] and C = C(J,R) denotes the Banach space of all continuous functions from J to R endowed

with the norm defined by ‖x‖ = supt∈J |x(t)|. By Lemma 2.6, we define an operator A : C →C by

(Ax)(t)

= Irg(s,x(s))(t) + Iq+rf(s,x(s))(t)

+
tq+r−1

Λ

Γ(q)

Γ(q + r)

[(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

) Tp
Γ(p + 1)

−
(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)]
(3.1)

+
1

Λ

[
Γ(q)

Γ(q + r)
ηq+r−1

(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)
−
(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

) Γ(q)
Γ(p + q + r)

Tp+q+r−1
]
,

with Λ 6= 0. It should be noticed that problem (1.5) can be transformed into a fixed point equation x = Ax.



Int. J. Anal. Appl. 17 (1) (2019) 52

To accomplish of the study, we will use fixed point theorems to prove that the operator A has fixed points.

For the sake of convenience, we define four constants by

Φ1 =

[
Tr

Γ(r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηr

Γ(r + 1)
+

Tp+r

Γ(p + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+r

Γ(p + r + 1)
ηq+r−1 +

Γ(q)

Γ(p + q + 1)

ηr

Γ(r + 1)
Tp+q+r−1

)]
,

Φ2 =

[
Tq+r

Γ(q + r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηq+r

Γ(q + r + 1)

+
Tp+q+r

Γ(p + q + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+q+r

Γ(p + q + r + 1)
ηq+r−1 +

+
Γ(q)

Γ(p + q + 1)

ηq+r

Γ(q + r + 1)
Tp+q+r−1

)]
,

Φ3 =

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1

Γ(p + 1)
+

Γ(q)

|Λ|Γ(p + q + 1)
Tp+q+r−1

]
,

Φ4 =

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1 +

Γ(q)

|Λ|Γ(q + 1)
ηq+r−1

]
.

The first existence and uniqueness result is obtained by using Banach contraction mapping principle.

Theorem 3.1. Let g,f : J × R → R, be continuous functions and φ,h : C(J,R) → R be two functionals

satisfying the assumption:

(H1) there exist positive constants Li, i = 1, 2, 3, 4 such that:

|g(t,x) −g(t,y)| ≤ L1|x−y|, |f(t,x) −f(t,y)| ≤ L2|x−y|, t ∈ J,x,y ∈ R,

|φ(u) −φ(v)| ≤ L3|u−v| and |h(u) −h(v)| ≤ L4|u−v|, u,v ∈ C(J,R).

If the inequality

Ω1 := L1Φ1 + L2Φ2 + L3Φ3 + L4Φ4 < 1, (3.2)

holds, then the boundary value problem (1.5) has a unique solution on J.

Proof. By using the Banach’s contraction mapping principle, we shall show that A of a fixed point problem,

x = Ax, has a unique fixed point which is the unique solution of problem (1.5).

To prove the embedding property, we set

sup
t∈[0,T]

|g(t, 0)| = M1 < ∞, sup
t∈[0,T]

|f(t, 0)| = M2 < ∞, |φ(0)| = M3, |h(0)| = M4,

and choose

r ≥
M1Φ1 + M2Φ2 + M3Φ3 + M4Φ4

1 − Ω1
.



Int. J. Anal. Appl. 17 (1) (2019) 53

Now, we show that ABr ⊂ Br, where Br = {x ∈ C : ‖x‖ ≤ r}. For any x ∈ Br, and taking into account

assumption (H1), we obtain

|Ax(t)|

≤ sup
t∈[0,T]

{
Irg(s,x(s))(t) + Iq+rf(s,x(s))(t)

+
tq+r−1

Λ

Γ(q)

Γ(q + r)

[(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

)
Tp

Γ(p + 1)

−
(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)]
+

1

Λ

[
Γ(q)

Γ(q + r)
ηq+r−1

(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)
−
(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

)
Γ(q)

Γ(p + q + r)
Tp+q+r−1

]}
≤ Ir(|g(s,x(s)) −g(s, 0)| + |g(s, 0)|)(T)

+Iq+r(|f(s,x(s)) −f(s, 0)| + |f(s, 0)|)(T)

+
Tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[
Tp

Γ(p + 1)

(
(|φ(x) −φ(0)| + |φ(0)|) (T)

+Ir
(
|g(s,x(s)) −g(s, 0)| + |g(s, 0)|

)
(η)

+Iq+r
(
|f(s,x(s)) −f(s, 0)| + |f(s, 0)|

)
(η)

)
+

(
(|h(x) −h(0)| + |h(0)|) (T)

+Ip+r
(
|g(s,x(s)) −g(s, 0)| + |g(s, 0)|

)
(T)

+Ip+q+r
(
|f(s,x(s)) −f(s, 0)| + |f(s, 0)|

)
(T)

)]
+

1

|Λ|

[
Γ(q)

Γ(q + 1)
ηq+r−1

(
(|h(x) −h(0)| + |h(0)|) (T)

+Ip+r
(
|g(s,x(s)) −g(s, 0)| + |g(s, 0)|

)
(T)

+Ip+q+r
(
|f(s,x(s)) −f(s, 0)| + |f(s, 0)|

)
(T)

)
+

Γ(q)

Γ(p + q + 1)
Tp+q+r−1

(
(|φ(x) −φ(0)| + |φ(0)|) (T)

+Ir
(
|g(s,x(s)) −g(s, 0)| + |g(s, 0)|

)
(η)

+Iq+r
(
|f(s,x(s)) −f(s, 0)| + |f(s, 0)|

)
(η)

)]
≤ (L1r + M1)

Tr

Γ(r + 1)
+ (L2r + M2)

Tq+r

Γ(q + r + 1)

+
Tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[
Tp

Γ(p + 1)

(
(L3r + M3) + (L1r + M1)

ηr

Γ(r + 1)

+ (L2r + M2)
ηq+r

Γ(q + r + 1)

)
+

(
(L4r + M4) + (L1r + M1)

Tp+r

Γ(p + r + 1)

+ (L2r + M2)
Tp+q+r

Γ(p + q + r + 1)

)]



Int. J. Anal. Appl. 17 (1) (2019) 54

+
1

|Λ|

[
Γ(q)

Γ(q + 1)
ηq+r−1

(
(L4r + M4) + (L1r + M1)

Tp+r

Γ(p + r + 1)

+ (L2r + M2)
Tp+q+r

Γ(p + q + r + 1)

)
+

Γ(q)

Γ(p + q + r)

(
(L3r + M3)

+ (L1r + M1)
ηr

Γ(r + 1)
+ (L2r + M2)

ηq+r

Γ(q + r + 1)

)]
≤

[
Tr

Γ(r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηr

Γ(r + 1)
+

Tp+r

Γ(p + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+r

Γ(p + r + 1)
ηq+r−1 +

Γ(q)

Γ(p + q + 1)

ηr

Γ(r + 1)
Tp+q+r−1

)]
×(L1r + M1) +

[
Tq+r

Γ(q + r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηq+r

Γ(q + r + 1)

+
Tp+q+r

Γ(p + q + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+q+r

Γ(p + q + r + 1)
ηq+r−1 +

+
Γ(q)

Γ(p + q + 1)

ηq+r

Γ(q + r + 1)
Tp+q+r−1

)]
(L2r + M2)

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1

Γ(p + 1)
+

Γ(q)

|Λ|Γ(p + q + 1)
Tp+q+r−1

]
(L3r + M3)

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1 +

Γ(q)

|Λ|Γ(q + 1)
ηq+r−1

]
(L4r + M4)

= Φ1(L1r + M1) + Φ2(L2r + M2) + Φ3(L3r + M3) + Φ4(L4r + M4)

= Ω1r + (M1Φ1 + M2Φ2 + M3Φ3 + M4Φ4)

≤ r.

This mean that ‖Ax‖≤ r which yields ABr ⊂ Br. For all t ∈ [0,T] and for each x,y ∈C, we have

|Ax(t) −Ay(t)|

≤ Ir(|g(s,x(s)) −g(s,y(s))|)(T) + Iq+r(|f(s,x(s)) −f(s,y(s))|)(T)

+
Tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[
Tp

Γ(p + 1)

(
(|φ(x) −φ(y)|) (T)

+Ir (|g(s,x(s)) −g(s,y(s))|) (η) + Iq+r (|f(s,x(s)) −f(s,y(s))|) (η)
)

+

(
(|h(x) −h(y)|) (T) + Ip+r (|g(s,x(s)) −g(s,y(s))|) (T)

+Ip+q+r (|f(s,x(s)) −f(s,y(s))|) (T)
)]

+
1

|Λ|

[
Γ(q)

Γ(q + 1)
ηq+r−1

(
(|h(x) −h(y)|) (T)

+Ip+r (|g(s,x(s)) −g(s,y(s))|) (T) + Ip+q+r (|f(s,x(s)) −f(s,y(s))|) (T)
)

+
Γ(q)

Γ(p + q + 1)
Tp+q+r−1

(
(|φ(x) −φ(y)|) (T) + Ir (|g(s,x(s)) −g(s,x(s))|) (η)



Int. J. Anal. Appl. 17 (1) (2019) 55

+Iq+r (|f(s,x(s)) −f(s,y(s))|) (η)
)]

≤ (L1|x−y|)
Tr

Γ(r + 1)
+ (L2|x−y|)

Tq+r

Γ(q + r + 1)

+
Tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[
Tp

Γ(p + 1)

(
(L3|x−y|)

+ (L1|x−y|)
ηr

Γ(r + 1)
+ (L2|x−y|)

ηq+r

Γ(q + r + 1)

)
+

(
(L4|x−y|) + (L1|x−y|)

Tp+r

Γ(p + r + 1)
+ (L2|x−y|)

Tp+q+r

Γ(p + q + r + 1)

)]
+

1

|Λ|

[
Γ(q)

Γ(q + 1)
ηq+r−1

(
(L4|x−y|) + (L1|x−y|)

Tp+r

Γ(p + r + 1)

+ (L2|x−y|)
Tp+q+r

Γ(p + q + r + 1)

)
+

Γ(q)

Γ(p + q + r)

(
(L3|x−y|)

+ (L1|x−y|)
ηr

Γ(r + 1)
+ (L2|x−y|)

ηq+r

Γ(q + r + 1)

)]
≤

[
Tr

Γ(r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηr

Γ(r + 1)
+

Tp+r

Γ(p + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+r

Γ(p + r + 1)
ηq+r−1 +

Γ(q)

Γ(p + q + 1)

ηr

Γ(r + 1)
Tp+q+r−1

)]
×(L1|x−y|) +

[
Tq+r

Γ(q + r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηq+r

Γ(q + r + 1)

+
Tp+q+r

Γ(p + q + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+q+r

Γ(p + q + r + 1)
ηq+r−1

+
Γ(q)

Γ(p + q + 1)

ηq+r

Γ(q + r + 1)
Tp+q+r−1

)]
L2|x−y|

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1

Γ(p + 1)
+

Γ(q)

|Λ|Γ(p + q + 1)
Tp+q+r−1

]
L3|x−y|

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1 +

Γ(q)

|Λ|Γ(q + 1)
ηq+r−1

]
L4|x−y|

= Φ1L1|x−y|) + Φ2L2|x−y|) + Φ3L3|x−y|) + Φ4L4|x−y|)

= Ω1|x−y|.

The above result implies that ‖Ax−Ay‖ ≤ Ω1‖x−y‖. As Ω1 < 1, therefore A is a contraction operator.

Hence, by the Banach contraction mapping principle, we obtain that A has a unique fixed point which is

the unique solution of the problem (1.5). The proof is completed. �

Example 3.1. Consider the following nonlinear sequential Riemann-Liouville and Caputo fractional differ-

ential equation with nonlocal integral boundary conditions

RLD
4
5

(
CD

1
2 x(t) −

et

(t2 + 40) + 20

|x(t)|
|x(t)| + 1

)



Int. J. Anal. Appl. 17 (1) (2019) 56

=
cos2(2πt)

(t + 10)
2

+ 50
·
(
x2(t) + 2|x(t)|
|x(t)| + 1

)
+ et, 0 < t < 3, (3.3)

x

(
1

2

)
=
x2(2) + 2|x(2)|
60(|x(2)| + 1)

+ 30, I
2
3 x(3) =

|x(1)|
25(|x(1)| + 1)

.

Setting constants q = 4/5, r = 1/2, p = 2/3, η = 1/2, T = 3, then we can compute constants as

Φ1 = 12.42305820, Φ2 = 14.24066077, Φ3 = 6.845515569, Φ4 = 4.835810257. Setting functions

g(t,x) =
et

(t2 + 40) + 20

|x|
|x| + 1

, f(t,x) =
cos2(2πt)

(t + 10)
2

+ 50
·
(
x2 + 2|x|
|x| + 1

)
+ et

φ(x) =
x2 + 2|x|

60(|x| + 1)
+ 30. h(x) =

|x|
25(|x| + 1)

,

so we get |g(t,x) −g(t,y)| ≤ (1/60)|x−y|, |f(t,x) −f(t,y)| ≤ (1/75)|x−y|, |φ(x) −φ(y)| ≤ (1/30)|x−y|

and |h(x) − h(y)| ≤ (1/25)|x − y|. Therefore the condition (H1) is satisfied with L1 = 1/60, L2 = 1/75,

L3 = 1/30 and L4 = 1/25. We can show that

Ω1 = 0.8185425995 < 1.

Hence, by Theorem 3.1, the boundary value problem (3.3) has a unique solution on [0, 3].

The second existence result will be proved by using the following Krasnoselskii’s fixed point theorem.

Lemma 3.1. (Krasnoselskii’s fixed point theorem) [13]. Let M be a closed, bounded, convex and nonempty

subset of a Banach space X. Let A, B be the operators such that (a) Ax+By ∈ M whenever x,y ∈ M; (b) A

is compact and continuous; (c) B is a contraction mapping. Then there exists z ∈ M such that z = Az +Bz.

Theorem 3.2. Assume that g,f : J×R → R, are continuous functions and two functionals φ,h : C(J×R) →

R satisfying the assumption (H1). In addition we suppose that:

(H2) |g(t,x)| ≤ δ1(t), |f(t,x)| ≤ δ2(t), ∀(t,x) ∈ J ×R and δ1,δ2 ∈ C(J,R+),

|φ(u)| ≤ δ3, |h(u)| ≤ δ4, ∀u ∈ C(J ×R) and δ3,δ4 ∈ R+.

If the inequality

Ω2 := L1

(
Φ1 −

Tq

Γ(r + 1)

)
+ L2

(
Φ2 −

Tq+r

Γ(q + r + 1)

)
+ L3Φ3 + L4Φ4 < 1, (3.4)

then the boundary value problem (1.5) has at least one solution on J.

Proof. To applied Lemma 3.1, we let supt∈J |δ1(t)| = ‖δ1‖, supt∈J |δ2(t)| = ‖δ2‖, and a positive constant r

as

r ≥‖δ1‖Φ1 + ‖δ2‖Φ2 + δ3Φ3 + δ4Φ4.



Int. J. Anal. Appl. 17 (1) (2019) 57

Define a ball Br by Br = {x ∈ C : ‖x‖ ≤ r} which is closed, bounded, convex and nonempty subset of a

Banach space C. In addition, we define the operators P and Q on Br as

(Px)(t) = Irg(s,x(s))(t) + Iq+rf(s,x(s))(t), t ∈ [0,T],

(Qx)(t)

=
tq+r−1

Λ

Γ(q)

Γ(q + r)

[(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

) Tp
Γ(p + 1)

−
(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)]
+

1

Λ

[
Γ(q)

Γ(q + r)
ηq+r−1

(
h(x(t)) − Ip+rg(s,x(s))(T) − Ip+q+rf(s,x(s))(T)

)
−
(
φ(x(t)) − Irg(s,x(s))(η) − Iq+rf(s,x(s))(η)

) Γ(q)
Γ(p + q + r)

Tp+q+r−1
]
,

t ∈ [0,T].

Obvious that Ax = Px + Qx. To prove that P and Q satisfy (a) of Lemma 3.1, for x,y ∈ Br, we have

‖Px + Qy‖

≤ ‖δ1‖
[

Tr

Γ(r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηr

Γ(r + 1)
+

Tp+r

Γ(p + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+r

Γ(p + r + 1)
ηq+r−1 +

Γ(q)

Γ(p + q + 1)

ηr

Γ(r + 1)
Tp+q+r−1

)]
+‖δ2‖

[
Tq+r

Γ(q + r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

|Λ|

(
Tp

Γ(p + 1)

ηq+r

Γ(q + r + 1)

+
Tp+q+r

Γ(p + q + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+q+r

Γ(p + q + r + 1)
ηq+r−1 +

+
Γ(q)

Γ(p + q + 1)

ηq+r

Γ(q + r + 1)
Tp+q+r−1

)]
+ δ3

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1

Γ(p + 1)
+

Γ(q)

|Λ|Γ(p + q + 1)
Tp+q+r−1

]
+ δ4

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1 +

Γ(q)

|Λ|Γ(q + 1)
ηq+r−1

]
= ‖δ1‖Φ1 + ‖δ2‖Φ2 + δ3Φ3 + δ4Φ4

≤ r.

This shows that Px + Qy ∈ Br.

The operator Q satisfies the condition (c) of Lemma 3.1 from assumption (H1) together with (3.4). The

final step is to show that the operator P is satisfied condition (b) of Lemma 3.1. Since the functions f,g are

continuous, we get that the operator P is continuous. Now we will show that the operator P is compact.



Int. J. Anal. Appl. 17 (1) (2019) 58

For any x ∈ Br, we obtain

‖Px‖≤‖δ1‖
Tr

Γ(q + 1)
+ ‖δ2‖

Tq+r

Γ(q + r + 1)
.

Therefore, the set P(Br) is uniformly bounded. Let us let sup(t,x)∈J×Br |g(t,x)| = g < ∞ and sup(t,x)∈J×Br |f(t,x)|

= f < ∞. Let t1, t2 ∈ J with t1 < t2. Then we have

|(Px)(t2) − (Px)(t1)|

≤
g

Γ(r)

∣∣∣∣
∫ t1
0

[
(t2 −s)r−1 − (t1 −s)r−1

]
ds

∣∣∣∣ + gΓ(r)
∣∣∣∣
∫ t2
t1

(t2 −s)r−1ds
∣∣∣∣

+
f

Γ(q + r)

∣∣∣∣
∫ t1
0

[
(t2 −s)q+r−1 − (t1 −s)q+r−1

]
ds

∣∣∣∣
+

f

Γ(q + r)

∣∣∣∣
∫ t2
t1

(t2 −s)q+r−1ds
∣∣∣∣

≤
g

Γ(r + 1)
[|tr2 − t

r
2| + 2(t2 − t1)

r] +
f

Γ(q + r + 1)

[∣∣tq+r2 − tq+r1 ∣∣ + 2(t2 − t1)r] ,
which is independent of x and tends to zero as t1 → t2. Thus, the set P(Br) is equicontinuous. Hence,

by the Arzelá-Ascoli theorem, the set P(Br) is relatively compact. Therefore, the operator P is compact

which is satisfied condition (b) of Lemma 3.1. Thus all the assumptions of Lemma 3.1 are satisfied. So the

boundary value problem (1.5) has at least one solution on J. The proof is completed. �

Remark 3.1. In the above theorem we can interchange the roles of the operators P and Q to obtain a second

result replacing (3.4) by the following condition:

Ω3 := L1
Tr

Γ(r + 1)
+ L2

Tq+r

Γ(q + r + 1)
< 1. (3.5)

Remark 3.2. Since Ω2 < Ω1 and Ω3 < Ω1, the condition (3.2) can be relaxed by (3.4) and (3.5). However,

the conclusion of both theorems has different mentions between uniqueness and multiplicity of solutions.

Example 3.2. Consider the following nonlinear sequential Riemann-Liouville and Caputo fractional differ-

ential equation with nonlocal integral boundary conditions

RLD
1
2

(
CD

2
3 x(t) −

e2t

(t2 + 100)2 + 19300
·
|x(t)|
|x(t)| + 1

)

=
cos2(2πt)

t2 + 28000
·
(
|x(t)|
|x(t)| + 1

)
+ cos(πt), 0 < t < 4, (3.6)

x(2) =
|x(3)|

9990(|x(3)| + 1)
, I

2
3 x(4) =

|x(2)|
9840(|x(2)| + 1)

+ 35.

Setting constants q = 1/2, r = 2/3, p = 2/3, η = 2, T = 4, then we can fine that Φ1 = 6717.422119,

Φ2 = 6652.469591, Φ3 = 3119.677669, Φ4 = 2175.349828. Next we set the following functions

g(t,x) =
e2t

(t2 + 100)2 + 19300
·
|x|
|x| + 1

,



Int. J. Anal. Appl. 17 (1) (2019) 59

f(t,x) =
cos2(2πt)

t2 + 28000
·
(
|x|
|x| + 1

)
+ cos(πt)

φ(x) =
|x|

9990(|x| + 1)
, h(x) =

|x|
9840(|x| + 1)

+ 35.

Since |g(t,x)−g(t,y)| ≤ (1/29300)|x−y|, |f(t,x)−f(t,y)| ≤ (1/28000)|x−y|, |φ(x)−φ(t,y)| ≤ (1/9990)|x−y|

and |h(x) −h(y)| ≤ (1/9840)|x−y|, the condition (H1) fulfilled. It is obvious that

|g(t,x)| ≤
e2t

29300
, |f(t,x)| ≤ 1 + cos(πt), |φ(x)| ≤ 1, h(x) ≤ 36.

Then the condition (H2) is satisfied. In addition we have

Ω2 = 0.999918 < 1.

Hence, by Theorem 3.2, the boundary value problem (3.6) has at least one solution on [0, 4].

Remark 3.3. The problem (3.6) can not be applied by Theorem 3.1 since Ω1 = 1.000204 > 1.

Now, our third existence result is based on Leray-Schauder’s Nonlinear Alternative.

Lemma 3.2. (Nonlinear alternative for single-valued maps) [11]. Let E be a Banach space, C be a closed,

convex subset of E, U be an open subset of C and 0 ∈ U. Suppose that A : U → C is a continuous, compact

(that is, A(U) is a relatively compact subset of C) map. Then either

(i) A has a fixed point in U, or

(ii) there is a u ∈ ∂U (the boundary of U in C) and λ ∈ (0, 1) with u = λA(u).

Theorem 3.3. Assume that g,f : J×R → R are continuous functions and two functionals φ,h : C(J×R) →

R. In addition we suppose that:

(H3) there exist continuous nondecreasing functions ψ1,ψ2 : [0,∞) → (0,∞) and functions p1,p2 ∈

C(J,R+) such that

|g(t,x)| ≤ p1(t)ψ1(‖x‖), |f(t,x)| ≤ p2(t)ψ2(‖x‖) for each (t,x) ∈ J ×R;

(H4) there exists a constant N > 0 such that

N

Φ1‖p1‖ψ1(N) + Φ2‖p2‖ψ2(N) + Φ3|φ(N)| + Φ4|h(N)|
> 1.

Then the boundary value problem (1.5) has at least one solution on J.

Proof. Let us define a positive number R and let a ball BR = {x ∈C : ‖x‖≤ R} be a closed, convex subset

of C. Next, we will prove that the operator A, defined by (3.1), maps bounded sets (balls) into bounded sets

in C. For any t ∈ J and x ∈ BR, we have

|Ax(t)|



Int. J. Anal. Appl. 17 (1) (2019) 60

≤ Ir|g(s,x(s))|(t) + Iq+r|f(s,x(s))|(t)

+
tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[(
|φ(x(t))| + Ir|g(s,x(s))|(η) + Iq+r|f(s,x(s))|(η)

) Tp
Γ(p + 1)

+
(
|h(x(t))| + Ip+r|g(s,x(s))|(T) + Ip+q+r|f(s,x(s))|(T)

)]
+

1

|Λ|

[
Γ(q)

Γ(q + r)
ηq+r−1

(
|h(x(t))| + Ip+r|g(s,x(s))|(T) + Ip+q+r|f(s,x(s))|(T)

)
+
(
|φ(x(t))| + Ir|g(s,x(s))|(η) + Iq+r|f(s,x(s))|(η)

) Γ(q)
Γ(p + q + r)

Tp+q+r−1
]

≤ ‖p1‖ψ1(‖x‖)
Tr

Γ(r + 1)
+ ‖p2‖ψ2(‖x‖)

Tq+r

Γ(q + r + 1)

+
Tq+r−1

|Λ|
Γ(q)

Γ(q + r)

[
Tp

Γ(p + 1)

(
|φ(‖x‖)| + (‖p1‖ψ1(‖x‖))

ηr

Γ(r + 1)

+ (‖p2‖ψ2(‖x‖))
ηq+r

Γ(q + r + 1)

)
+

(
|h(‖x‖)| + (‖p1‖ψ1(‖x‖))

Tp+r

Γ(p + r + 1)

+ (‖p2‖ψ2(‖x‖))
Tp+q+r

Γ(p + q + r + 1)

)]
+

1

|Λ|

[
Γ(q)

Γ(q + 1)
ηq+r−1

(
|h(‖x‖)| + (‖p1‖ψ1(‖x‖))

Tp+r

Γ(p + r + 1)

+ (‖p2‖ψ2(‖x‖))
Tp+q+r

Γ(p + q + r + 1)

)
+

Γ(q)

Γ(p + q + r)

(
|φ(‖x‖)|

+ (‖p1‖ψ1(‖x‖))
ηr

Γ(r + 1)
+ (‖p2‖ψ2(‖x‖))

ηq+r

Γ(q + r + 1)

)]
≤

[
Tr

Γ(r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

||Λ||

(
Tp

Γ(p + 1)

ηr

Γ(r + 1)
+

Tp+r

Γ(p + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+r

Γ(p + r + 1)
ηq+r−1 +

Γ(q)

Γ(p + q + 1)

ηr

Γ(r + 1)
Tp+q+r−1

)]
×(‖p1‖ψ1(‖x‖)) +

[
Tq+r

Γ(q + r + 1)
+

Γ(q)

Γ(q + r)

Tq+r−1

||Λ||

(
Tp

Γ(p + 1)

ηq+r

Γ(q + r + 1)

+
Tp+q+r

Γ(p + q + r + 1)

)
+

1

|Λ|

(
Γ(q)

Γ(q + 1)

Tp+q+r

Γ(p + q + r + 1)
ηq+r−1 +

+
Γ(q)

Γ(p + q + 1)

ηq+r

Γ(q + r + 1)
Tp+q+r−1

)]
‖p2‖ψ2(‖x‖)

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1

Γ(p + 1)
+

Γ(q)

|Λ|Γ(p + q + 1)
Tp+q+r−1

]
|φ(‖x‖)|

+

[
Γ(q)

|Λ|Γ(q + r)
Tp+q+r−1 +

Γ(q)

|Λ|Γ(q + 1)
ηq+r−1

]
|h(‖x‖)|

= Φ1‖p1‖ψ1(‖x‖) + Φ2‖p2‖ψ2(‖x‖) + Φ3|φ(‖x‖)| + Φ4|h(‖x‖)|

≤ Φ1‖p1‖ψ1(R) + Φ2‖p2‖ψ2(R) + Φ3|φ(R)| + Φ4|h(R)|.

Therefore, from the above result, we conclude that

‖Ax‖≤ Φ1‖p1‖ψ1(R) + Φ2‖p2‖ψ2(R) + Φ3|φ(R)| + Φ4|h(R)|.



Int. J. Anal. Appl. 17 (1) (2019) 61

Then the set A(BR) is uniformly bounded. Next, we show that the operator A maps bounded sets into

equicontinuous sets of C. Let ν1,ν2 ∈ J with ν1 < ν2 and for any x ∈ BR, then we have

|(Ax)(ν2) − (Ax)(ν1)|

≤ Ir|g(s,x(s))(ν2) −g(s,x(s))(ν1)| + Iq+r(|f(s,x(s))(ν2) −f(s,x(s))(ν1)|)

+

∣∣∣νq+r−12 −νq+r−11 ∣∣∣
|Λ|

Γ(q)

Γ(q + r)

[
(|φ(x(ν2)) −φ(x(ν1)))

Tp

Γ(p + 1)

+ (|h(x(ν2)) −h(x(ν1))|)
]

+
1

|Λ|

[
Γ(q)

Γ(q + r)
ηq+r−1 (|h(x(ν2)) −h(x(ν1))|)

+ (|φ(x(ν2)) −φ(x(ν1))|)
Γ(q)

Γ(p + q + r)
Tp+q+r−1

]
≤
‖p1‖ψ1(R)
Γ(r + 1)

[|tr2 − t
r
2| + 2(t2 − t1)

r] +
‖p2‖ψ2(R)

Γ(q + r + 1)

[∣∣tq+r2 − tq+r1 ∣∣ + 2(t2 − t1)r]
+

∣∣∣νq+r−12 −νq+r−11 ∣∣∣
|Λ|

Γ(q)

Γ(q + r)

[
(|φ(x(ν2)) −φ(x(ν1))|)

Tp

Γ(p + 1)

+ (|h(x(ν2)) −h(x(ν1))|)
]

+
1

|Λ|

[
Γ(q)

Γ(q + r)
ηq+r−1 (|h(x(ν2)) −h(x(ν1))|)

+ (|φ(x(ν2)) −φ(x(ν1))|)
Γ(q)

Γ(p + q + r)
Tp+q+r−1

]
.

Obviously the right hand side of the above inequality tends to zero independently of x ∈ BR as ν1 → ν2,

which implies that the set A(BR) is equicontinuous. Therefore it follows by the Arzelá-Ascoli theorem that

the set A(BR) is relative compact. Then the operator A is compact.

Let x(t) be a solution of problem (1.5). Then, for t ∈ J and x ∈ BR, we have

‖x‖≤ Φ1‖p1‖ψ1(‖x‖) + Φ2‖p2‖ψ2(‖x‖) + Φ3|φ(‖x‖)| + Φ4|h(‖x‖)|.

Consequently, we have

‖x‖
Φ1‖p1‖ψ1(‖x‖) + Φ2‖p2‖ψ2(‖x‖) + Φ3|φ(‖x‖)| + Φ4|h(‖x‖)|

≤ 1.

Let us define a subset of BR as

U = {x ∈C : ‖x‖ < N}, (3.7)

where N is satisfied the condition (H4). Note that the operator A : U → C is continuous and completely

continuous. From the choice of U, there is no x ∈ ∂U such that x = θAx for some θ ∈ (0, 1). Then, by

nonlinear alternative of Leray-Schauder type, Lemma 3.2, we get that the operator A has a fixed point in

U, which is a solution of the boundary value problem (1.5). This completes the proof. �



Int. J. Anal. Appl. 17 (1) (2019) 62

Example 3.3. Consider the following nonlinear sequential Riemann-Liouville and Caputo fractional differ-

ential equation with nonlocal integral boundary conditions

RLD
4
5

(
CD

2
5 x(t) −

2e−t cos2 t

1000

(
|x|5

x4 + 1
+ 1

))
=

2 sin4 t

1000

(
x8

|x|7 + 1
+ 1

)
, 0 < t < 5, (3.8)

x(2) =
x(4)

500
, I

3
5 x(5) =

x(3)

200
.

Setting constants q = 4/5, r = 2/5, p = 3/5, η = 2, T = 5, then we get Φ1 = 72.200440, Φ2 = 129.62057,

Φ3 = 34.389063 and Φ4 = 2.841029. Let the following functions

g(t,x) =
2e−t cos2 t

1000

(
|x|5

x4 + 1
+ 1

)
, f(t,x) =

2 sin4 t

1000

(
x8

|x|7 + 1
+ 1

)
,

φ(x) =
x

500
, h(x) =

x

200
.

It follows that

|g(t,x)| ≤ 2 cos2 t
(
|x| + 1
1000

)
and |f(t,x)| ≤ 2 sin4 t

(
|x| + 1
1000

)
.

Hence, we choose p1(t) = 2 cos
2 t, ψ1(|x|) = (|x| + 1)/(1000), p2(t) = 2 sin4 t, ψ2(|x|) = (|x| + 1)/(1000).

Then there exists a constant N > 0.97645553 satisfying inequality

N

(72.200440)(2)
(
N+1
1000

)
+ (129.62057)(2)

(
N+1
1000

)
+ (34.389063)

∣∣ N
500

∣∣ + (22.841029) ∣∣ N
200

∣∣ > 1.
Thus, by Theorem 3.3, the boundary value problem (3.8) has at least one solution on [0, 5].

The following result can be obtained by substituting p1(t),p2(t) ≡ 1 and linear functions ψ1(|x|) =

M1|x| + K1 and ψ2(|x|) = M2|x| + K2 in Theorem 3.3.

Corollary 3.1. Assume that the continuous functions g,f : J×R → R and two functionals φ,h : C(J×R) →

R are satisfied

|g(t,x)| ≤ M1|x| + K1, |f(t,x)| ≤ M2|x| + K2 for each (t,x) ∈ J ×R,

|φ(x)| ≤ M3|x| + K4, |h(x)| ≤ M4|x| + K4 for each x ∈ C(J,R),

where M1,M2,M3,M4 > 0 and K1,K2,K3,K4 ≥ 0. If M1Φ1 + M2Φ2 + M3Φ3 + M4Φ4 < 1, then boundary

value problem (1.5) has at least one solution on [0,T].

Acknowledgements:

This research was funded by Faculty of Applied Science, King Mongkut’s University of Technology North

Bangkok, Thailand. Contract no. 6042102.



Int. J. Anal. Appl. 17 (1) (2019) 63

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	1. Introduction
	2. Preliminaries
	3. Main results
	Acknowledgements:
	References