EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 3687-3707
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Gronwall-Type Inequalities and Qualitative Studies on
Higher-Variable Orders of Atangana-Baleanu Fractional

Operators via Increasing Functions

Hasanen A. Hammad1,2,∗, Manuel De la Sen3

1 Department of Mathematics, College of Science, Qassim University, Buraydah 51452,
Saudi Arabia
2 Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt
3 Institute of Research and Development of Processes, Department of Electricity and
Electronics, Faculty of Science and Technology, Univesity of the Basque Country,
48940-Leioa (Bizkaia), Spain

Abstract. This paper introduces a novel extension of Caputo-Atangana-Baleanu and Riemann-
Atangana-Baleanu fractional derivatives from constant to increasing variable order. We generalize
the fractional order from a fixed value in (0, 1] to a time-dependent function in (k, k + 1], where
k ≥ 0. The corresponding Atangana-Baleanu fractional integral is also extended. Key properties of
these new definitions are explored, including a generalized Gronwall inequality. We then delve into
the analysis of higher-variable initial fractional differential equations using the Caputo-Atangana-
Baleanu operator with an increasing function, establishing existence and uniqueness results via
Picard’s iterative method. The findings presented in this work are expected to stimulate further
research on inequalities and fractional differential equations related to Atangana-Baleanu fractional
calculus with respect to increasing functions. Concrete examples are provided to illustrate the
practical applications of our results.

2020 Mathematics Subject Classifications: 74H10, 54H25, 34A08, 34A12

Key Words and Phrases: Fractional derivatives, variable order derivatives, fixed point tech-
niques, existence results, differential equations

1. Introduction

For a better explanation of chaotic complex systems, fractional calculus has drawn the
attention of numerous authors in a variety of fields over the past three decades. These
fields have many applications in qualitative theories, electrical networks, etc. For more
information, see [16, 18, 22]. The reason why this trend has so many readers is that the
fractional differentiation of the function produces its complete spectrum which includes

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5592

Email addresses: h.abdelwareth@qu.edu.sa (H.A. Hammad), manuel.delasen@ehu.eus (M. De la Sen)

https://www.ejpam.com 3687 Copyright: © 2024 The Author(s). (CC BY-NC 4.0)



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3688

the corresponding integer-order counterpart as a special case. In addition, the use of
these equations and formulas in its mathematical models contributes fundamentally to
real-world applications because it generates full dynamics of the topic under study and
benefits from higher degrees of freedom. Regarding applications of viscoelasticity, physics,
and dynamics, reputable results can be found in [1, 7, 13, 21].

Some scientists and engineers have adapted fractional calculus to singular and nonsin-
gular kernels in order to recognize and explain the genuine phenomena in their respective
domains. A novel definition of a fractional operator with an exponential kernel was stud-
ied by Caputo and Fabrizio [8]. The Atangana-Baleanu (AB) fractional operator was
introduced by Atangana and Baleanu [5] and has a fresh and intriguing definition of a
Mittag-Leffler (ML) kernel. The AB fractional operator was extended to higher arbitrary
orders by Abdeljawad [2]. Following that, a number of researchers examined the qualita-
tive characteristics and approximate solutions of fractional differential equations (FDEs)
utilizing Atangana-Baleanu-Caputo (ABC) fractional operators, Caputo-Fabrizio deriva-
tives, and others applied the technique of FP theory to find the existence solutions to
theses operators; for more information, see [6, 9, 10, 12, 14, 15, 23–26, 28].

Recently, a fractional derivative of a function with respect to (w.r.t.) another function
with a ML kernel was proposed by Fernandez and Baleanu [11], and it is actually thought
of as a generalized AB fractional operator. By establishing the appropriate AB-fractional
integral of a function w.r.t. another function, authors [20] established a link between the
AB fractional operator and the Riemann-Liouville (RL) fractional integral w.r.t. another
function. Following that, Kashuri [17] introduced a fractional integral operator known as
the Atangana-Baleanu-Kashuri (ABK) fractional integral.

Inspired of the above works, in this article, we increase the fractional derivatives of
ABC and RAB with respect to an increasing function from a fractional order ϖ ∈ (0, 1]
to an arbitrary variable order ϖ(τ) ∈ (k, k+ 1], k ≥ 0. Several characteristics and uses of
these concepts are also studied. Further, in the framework of the AB fractional integrals,
a brand-new generalized Gronwall inequality is also demonstrated. Moreover, Picard’s
iterative approach is used to establish the existence and uniqueness results of a higher-
variable order ABC fractional issue under initial boundary constraints our paper extends
and generalizes the results of [3]. Finally, illustrative examples are provided to support
our results.

2. Preliminaries

This part is devoted to present some crucial foundational material for fractional calcu-
lus. Let us denote by Ck(ℑ,R) the BS of all the kth continuously differentiable functions
κ equipped with usual norm ∥κ∥ = sup {|κ(r)| : r ∈ ℑ = [κ, ϱ]}.

Definition 1. [4] Let ψ : ℑ → R be an increasing and differentiable function. For the
integrable function ξ : ℑ → R, the ϖth left-sided ψ−RL fractional integral w.r.t. another



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3689

function ψ(z), is described as

RLℜϖ,ψκ ξ (z) =
1

Γ (ϖ)

z∫
κ

(ψ(z)− ψ(r))ψ′(r)ξ (r) dr,

for all z ∈ ℑ = [κ, ϱ], where Γ (ϖ) =
∞∫
0

e−rrϖ−1dr, ϖ > 0.

Definition 2. [5] For the function ξ ∈ H1(κ, ϱ) and ϖ ∈ (0, 1], the ϖth left-sided RL-AB
fractional derivative is defined by

(
RLABDϖ

κ ξ
)
z =

Λ(ϖ)

1−ϖ

d

dz

z∫
κ

Lϖ

(
−ϖ
1−ϖ

(z − r)ϖ
)
ξ (r) dr, z ∈ ℑ,

where Λ(ϖ) is the normalization function with Λ(0) = Λ(1) = 1, and Lϖ is the ML
function given by

Lϖ(s) =
∞∑
j=0

sj

Γ (1 +ϖj)
, Re(ϖ) > 0, s ∈ C.

Definition 3. [5] For the function ξ ∈ H1(κ, ϱ) and ϖ ∈ (0, 1], the ϖth left-sided ABC
fractional derivative is proposed by

(
CABDϖ

κ ξ
)
z =

Λ(ϖ)

1−ϖ

z∫
κ

Lϖ

(
−ϖ
1−ϖ

(z − r)ϖ
)
ξ′ (r) dr, z ∈ ℑ.

Definition 4. [5] For the function ξ ∈ H1(κ, ϱ) and ϖ ∈ (0, 1], the ϖth left-sided RL-AB
fractional integral is formed as(

RLABℜϖκ ξ
)
z =

1−ϖ

Λ(ϖ)
ξ(z) +

ϖ

Λ(ϖ)

RL
ℜϖκ ξ (z) , z ∈ ℑ.

Definition 5. [17] For the function ξ ∈ Hq
t (κ, ϱ), (where 1 ≤ q <∞, t ∈ R) , and ϖ ∈

(0, 1], the ϖth left-sided KAB fractional integral is written as

(
KAB
κ ℜϖ,ηξ

)
z =

1−ϖ

Λ(ϖ)
ξ(z) +

ϖ

Λ(ϖ)

1

Γ (ϖ)

z∫
κ

rη−1

(
zη − rη

η

)ϖ−1

ξ (r) dr, z ∈ ℑ, η > 0.

Definition 6. [8] Assume that ϖ ∈ (0, 1]. For the function ξ ∈ H1(κ, ϱ), the ϖth left-sided
ψ−RL-AB fractional derivative under an increasing differentiable function ψ : ℑ → R with
ψ′(z) ̸= 0, for all z ∈ ℑ is given by

(
RLABDϖ,ψ

κ ξ
)
(z) =

Λ(ϖ)

(1−ϖ)ψ′(z)

d

dz

z∫
κ

ψ′(r)

(
−ϖ
1−ϖ

(ψ(z)− ψ(r))ϖ
)
ξ (r) dr, z ∈ ℑ.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3690

Definition 7. [8] Assume that ϖ ∈ (0, 1]. For the function ξ ∈ H1(κ, ϱ), the ϖth left-
sided ψ−ABC fractional derivative under an increasing differentiable function ψ : ℑ → R
with ψ′(z) ̸= 0, for all z ∈ ℑ is defined by(

CABDϖ,ψ
κ ξ

)
(z) =

Λ(ϖ)

(1−ϖ)

z∫
κ

ψ′(r)Lϖ

(
−ϖ
1−ϖ

(ψ(z)− ψ(r))ϖ
)
ξ′ψ (r) dr, z ∈ ℑ,

where ξ′ψ (r) =
ξ′(z)
ψ′(z) .

Definition 8. [20] Assume that ϖ ∈ (0, 1]. For the function ξ ∈ H1(κ, ϱ), the ϖth left-
sided ψ−RL-AB fractional integral under an increasing differentiable function ψ : ℑ → R
with ψ′(z) ̸= 0, for all z ∈ ℑ is described as(

RLABℜϖ,ψκ ξ
)
(z) =

1−ϖ

Λ(ϖ)
ξ(z) +

ϖ

Λ(ϖ)

(
RLℜϖ,ψκ ξ

)
(z) , z ∈ ℑ.

Remark 1. It should be noted that

(i) Definitions 6, 7 and 8 reduce to Definitions 2, 3 and 4, respectively, by taking ψ(z) =
z.

(ii) Definition 8 follows immediately from Definition 5, by considering ψ(z) = zη

η .

Lemma 1. [4] Assume that ϖ, ν > 0 and ξ : ℑ → R. Then

(1) RLℜϖ,ψκ (ξ (z)− ξ (κ))ν−1 = Γ(ν)
Γ(ϖ+ν) (ξ (z)− ξ (κ))ϖ+ν−1 ;

(2) RLℜϖ,ψκ
RLℜν,ψκ ξ (z) =RL ℜϖ+ν,ψ

κ ξ (z) ;

(3)

((
1

ψ(z)
d
dz

)k
RLℜk,ψκ ξ

)
(z) = ξ (z) , k ∈ N.

Lemma 2. [20] For ϖ ∈ (0, 1] and ξ : ℑ → R, the relations below are satisfied:

(i)
(
RLABℜϖκ RLABDϖ,ψ

κ ξ
)
(z) = ξ (z) ;

(ii)
(
RLABDϖ,ψ

κ
RLABℜϖκ ξ

)
(z) = ξ (z) .

Definition 9. [22] For the function ξ ∈ L([0, T ]),

(a) the variable order of the RL fractional integral is remembered as

ℜϖ(z)
+0 ξ(z) =

1

Γ(ϖ(z))

∫ z

0
(z − r)ϖ(r)−1ξ(r)dr;

(b) the variable order of the Caputo fractional derivative is given by

CD
ϖ(z)
0 ξ(z) =

1

Γ(k −ϖ(z))

∫ z

0
(z − r)k−ϖ(r)−1ξ(k)(r)dr.

where ϖ : [0, T ] → (0, 1], T > 0 is a continuous function.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3691

3. Derivatives of higher-variable orders

In this section, we consider ψ : ℑ → R to be an increasing function with ψ′(z) ̸= 0
to investigate the definitions of higher-variables order fractional derivatives and integrals
within the AB framework with regard to a function ψ.

Consider a partition of ℑ = [κ, ϱ] as

{ℑ1 = [κ, τ1], ℑ2 = (τ1, τ2], ℑ3 = (τ2, τ3], · · · ,ℑk = (τk−1, ϱ]},

and assume that ϖ : ℑ → (k, k + 1] is a piecewise function such that

ϖ(τ) =
k∑

u=1

ϖu(τ)Iu(τ) =


ϖ1, if τ ∈ ℑ1

ϖ2, if τ ∈ ℑ2
...

ϖk, if τ ∈ ℑk,

where Iu is the indicator function of ℑu = (τu−1, τk] and k < ϖu < k + 1 are constants
with u = 1, 2, · · · , k, such that τ0 = κ and τk = ϱ and

Iu(τ) =

{
1 for τ ∈ ℑu,
0 otherwise.

Assume that Ck(ℑu,R) refers to the space of all kth continuously differentiable func-
tions ξ. Clearly, it is a Banach space under the norm ∥ξ∥ = sup {|ξ (z)| : z ∈ ℑ = [κ, ϱ]} .
Here, we shall write for simplicity ϖu(τ) = ϖu and θu(τ) = θu for all τ ∈ ℑu.

Definition 10. Let ϖu ∈ (k, k + 1] and θu = ϖu − k, for k ≥ 0, u ≥ 1. For the function
℘ ∈ H1(κ, ϱ), the ϖuth left-sided ψ−RL-AB fractional derivative under the function ψ
with ψ′(z) ̸= 0, for all z ∈ ℑ is defined by(

RLABDϖu,ψ
κ ℘

)
(z)

=

(
1

ψ′(z)

d

dz

)k (
RLABDθu,ψ

κ ℘(z)
)

=

(
1

ψ′(z)

d

dz

)k Λ(θu)

(1− θu)ψ′(z)

d

dz

z∫
κ

ψ′(r)Lθu

(
−θu
1− θu

(ψ(z)− ψ(r))θu
)
℘ (r) dr

=

(
1

ψ′(z)

d

dz

)k+1 Λ(ϖu − k)

(k + 1−ϖu)ψ′(z)

×
z∫

κ

ψ′(r)Lϖu−k

(
− (ϖu − k)

k + 1−ϖu
(ψ(z)− ψ(r))ϖu−k

)
℘ (r) dr.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3692

Definition 11. Let ϖu ∈ (k, k + 1] and θu = ϖu − k, for k ≥ 0, u ≥ 1. For the function
℘(k) ∈ H1(κ, ϱ), the ϖuth left-sided ψ−ABC fractional derivative under the function ψ
with ψ′(z) ̸= 0, for all z ∈ ℑ is described as(

CABDϖu,ψ
κ ℘

)
(z) =

(
CABDθu,ψ

κ ℘
(k)
ψ

)
(z)

=
Λ(θu)

(1− θu)

z∫
κ

ψ′(r)Lθu

(
−θu
1− θu

(ψ(z)− ψ(r))θu
)
℘
(k+1)
ψ (r) dr

=
Λ(ϖu − k)

(k + 1−ϖu)

×
z∫

κ

ψ′(r)Lϖu−k

(
− (ϖu − k)

k + 1−ϖu
(ψ(z)− ψ(r))ϖu−k

)
℘
(k+1)
ψ (r) dr,

where ℘
(k)
ψ (z) =

(
1

ψ′(z)
d
dz

)k
℘(z) and ℘

(0)
ψ (z) = ℘(z). If ϖu = n ∈ N, then

(
CABDϖu,ψ

κ ℘
)
(z) =

℘
(n)
ψ (z).

Definition 12. Let ϖu ∈ (k, k + 1] and θu = ϖu − k, for k ≥ 0, u ≥ 1. For the function
℘ ∈ H1(κ, ϱ), the ϖuth left-sided ψ−RL-AB fractional integral under the function ψ with
ψ′(z) ̸= 0, for all z ∈ ℑ is defined by(

RLABℜϖu,ψκ ℘
)
(z) =

(
RLℜk,ψκ

ABℜθu,ψκ ℘
)
(z) =

(
ABℜθu,ψκ

RLℜk,ψκ ℘
)
(z)

=
k + 1−ϖu

Λ(ϖu − k)

RL

ℜk,ψκ ℘ (z) +
ϖu − k

Λ(ϖu − k)

RL

ℜϖu,ψκ ℘ (z) ,

where ℜk,ψκ takes the form

ℜk,ψκ ℘ (z) =
1

Γ (k)

z∫
κ

ψ′(r) (ψ(z)− ψ(r))k−1 ξ (r) dr.

Remark 2. For u ≥ 1, it is clear that

(i) if we take ϖu = ϖ ∈ (0, 1] in Definitions 10, 11 and 12, then we have Definitions 6,
7 and 8, respectively.

(ii) if we put ϖu = ϖ = k + 1, then ϖu = 1 and hence the following is true for our
generalization to the higher-variable order cases:(

RLABDϖu,ψ
κ ℘

)
(z) =

(
1

ψ′(z)

d

dz

)k (
RLABD1,ψ

κ ℘(z)
)
= ℘

(k+1)
ψ (z) ,(

CABDϖu,ψ
κ ℘

)
(z) =

(
CABD1,ψ

κ ℘
(k)
ψ

)
(z) = ℘

(k+1)
ψ (z) ,(

RLABℜϖu,ψκ ℘
)
(z) =

(
RLℜk,ψκ

ABℜ1,ψ
κ ℘

)
(z) =

(
RLℜk+1,ψ

κ ℘
)
(z) .



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3693

Lemma 3. For ϖu = ϖ ∈ (0, 1], u ≥ 1, the equations below hold

(i)
(
RLABℜϖ,ψκ

CABDϖ,ψ
κ ℘

)
(z) = ℘(z)− ℘(κ).

(ii)
(
CABDϖ,ψ

κ
RLABℜϖ,ψκ ℘

)
(z) = ℘(z)− ℘(κ)Lϖ

(
−ϖ
1−ϖ (ψ(z)− ψ(κ))ϖ

)
.

Proof. (i) Utilizing Definitions 7 and 8, we get(
RLABℜϖ,ψκ

CABDϖ,ψ
κ ℘

)
(z)

=
Λ(ϖ)

(1−ϖ)

(
CABDϖ,ψ

κ ℘
)
(z)

+
ϖ

Λ(ϖ)

(
RLℜϖ,ψκ

CABDϖ,ψ
κ ℘

)
(z)

=
∞∑
j=0

(
−ϖ
1−ϖ

)j z∫
κ

ψ′(r)

(
ψ(z)− ψ(r)

Γ(1 + jϖ)

)jϖ
℘′
ψ (r) dr

+
ϖ

1−ϖ

RL
ℜϖ,ψκ

∞∑
j=0

(
−ϖ
1−ϖ

)j z∫
κ

ψ′(r)
[ψ(z)− ψ(r)]

Γ(1 + jϖ)

jϖ

℘′
ψ (r) dr

=

∞∑
j=0

(
−ϖ
1−ϖ

)j
RLℜjϖ+1,ψ

κ
℘′(z)

ψ′(z)
+

ϖ

1−ϖ

RL
ℜϖ,ψκ

∞∑
j=0

(
−ϖ
1−ϖ

)j
RLℜjϖ+1,ψ

κ
℘′(z)

ψ′(z)

=

∞∑
j=0

(
−ϖ
1−ϖ

)j
RLℜjϖ+1,ψ

κ
℘′(z)

ψ′(z)
−

∞∑
j=0

(
−ϖ
1−ϖ

)j+1
RLℜjϖ+ϖ+1,ψ

κ
℘′(z)

ψ′(z)

= RLℜ1,ψ
κ

℘′(z)

ψ′(z)
=

z∫
κ

℘′(r)dr = ℘(z)− ℘(κ).

(ii) Again, utilizing Definitions 7, 8 and the identity

RLℜµ+1,ψ
κ

(
1

ψ′(z)

d

dz

)
℘(z) =RL ℜµ,ψκ ℘(z)− ℘(κ)

(ψ(z)− ψ(κ))µ

Γ(1 + µ)
, Re(µ) > 0,

one has(
CABDϖ,ψ

κ
RLABℜϖ,ψκ ℘

)
(z)

= CABDϖ,ψ
κ

(
1−ϖ

Λ(ϖ)
℘(z) +

ϖ

Λ(ϖ)

(
RLℜϖ,ψκ ℘

)
(z)

)
=

1−ϖ

Λ(ϖ)

(
CABDϖ,ψ

κ ℘
)
(z) +

ϖ

Λ(ϖ)

CAB
Dϖ,ψ

κ

(
RLℜϖ,ψκ ℘

)
(z)

=

∞∑
j=0

(
−ϖ
1−ϖ

)j
RLℜjϖ+1,ψ

κ

(
1

ψ′(z)

d

dz

)
℘(z)



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3694

+
ϖ

1−ϖ

∞∑
j=0

(
−ϖ
1−ϖ

)j
RLℜjϖ+1,ψ

κ

(
1

ψ′(z)

d

dz

)(
RLℜϖ,ψκ ℘

)
(z)

=
∞∑
j=0

(
−ϖ
1−ϖ

)j {
RLℜjϖ,ψκ ℘(z)− ℘(κ) [ψ(z)− ψ(κ)]

Γ(1 + jϖ)

jϖ
}

−
∞∑
j=0

(
−ϖ
1−ϖ

)j+1
RLℜjϖ+ϖ,ψ

κ ℘(z)

= ℘(z)−
∞∑
j=0

(
−ϖ
1−ϖ

)j ℘(κ) [ψ(z)− ψ(κ)]
Γ(1 + jϖ)

jϖ

= ℘(z)− ℘(κ)Lϖ
(

−ϖ
1−ϖ

(ψ(z)− ψ(κ))ϖ
)
.

Lemma 4. Assume that ℘ ∈ Ck[ℑu,R] and ψ ∈ Ck[ℑu,R+]. For ϖu ∈ (k, k + 1] and
θu = ϖu − k, for k ≥ 0, u ≥ 1 and all τ ∈ ℑu, the following equations are true:

(i)
(
RLABDϖu,ψ

κ
RLABℜϖu,ψκ ℘

)
(z) = ℘(z).

(ii)
(
RLABℜϖu,ψκ

RLABDϖu,ψ
κ ℘

)
(z) = ℘(z).

(iii)
(
CABDϖu,ψ

κ
RLABℜϖu,ψκ ℘

)
(z) = ℘(z)−℘(κ)Lϖu−k

(
−(ϖu−k)
1−(ϖu−k)(ψ(z)− ψ(κ))ϖu−k

)
.

(iv)
(
RLABℜϖu,ψκ

CABDϖu,ψ
κ ℘

)
(z) = ℘(z)−

k∑
m=0

℘
(m)
ψ (κ)
m! (ψ(z)− ψ(κ))m.

Proof. (i) In light of Definitions 10 and 12 and using Lemmas 1 and 2, for u ≥ 1, we
can write(
RLABDϖu,ψ

κ
RLABℜϖu,ψκ ℘

)
(z) =

((
1

ψ(z)

d

dz

)k
RLABDθu,ψ

κ
RLABℜθu,ψκ

RLℜk,ψκ ℘

)
(z)

=

((
1

ψ(z)

d

dz

)k
RLℜk,ψκ ℘

)
(z) = ℘(z).

(ii) According to Definitions 10 and 12, for u ≥ 1, we have(
RLABℜϖu,ψκ

RLABDϖu,ψ
κ ℘

)
(z)

=
k + 1−ϖu

Λ(ϖu − k)

RL

ℜk,ψκ

(
RLABDϖu,ψ

κ ℘ (z)
)
+

ϖu − k

Λ(ϖu − k)

RL

ℜϖu,ψκ

(
RLABDϖu,ψ

κ ℘ (z)
)

= ℜk,ψκ

(
1

ψ′(z)

d

dz

)k+1 ∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j z∫
κ

ψ′(r)

(
ψ(z)− ψ(r)

Γ(1 + j (ϖu − k))

)j(ϖu−k)
℘ (r) dr

+
(ϖu − k)

(k + 1−ϖu)

RL

ℜϖu,ψκ

(
1

ψ′(z)

d

dz

)k+1 ∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3695

×
z∫

κ

ψ′(r)

(
ψ(z)− ψ(r)

Γ(1 + j (ϖu − k))

)j(ϖu−k)
℘ (r) dr,

which implies that(
RLABℜϖu,ψκ

RLABDϖu,ψ
κ ℘

)
(z)

=
∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j
ℜk,ψκ

(
1

ψ′(z)

d

dz

)k+1
RLℜj(ϖu−k)+1,ψ

κ ℘ (z)

−
∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j+1

ℜϖu,ψκ

(
1

ψ′(z)

d

dz

)k+1
RLℜj(ϖu−k)+1,ψ

κ ℘ (z)

=
∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j
RLℜj(ϖu−k)+1,ψ

κ ℘ (z)

−
∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j+1
RLℜj(ϖu−k)+(ϖu−k),ψ

κ ℘ (z)

= ℘(z).

(iii) Using Definitions 10, 12, Lemmas 1 and 3, for u ≥ 1, one has(
CABDϖu,ψ

κ
RLABℜϖu,ψκ ℘

)
(z)

=

(
CABDθu,ψ

κ

(
1

ψ(z)

d

dz

)k
RLℜk,ψκ

RLABℜθu,ψκ ℘

)
(z)

=
(
CABDθu,ψ

κ
RLABℜθu,ψκ ℘

)
(z)

= ℘(z)− ℘(κ)Lθu
(

−θu
1− θu

(ψ(z)− ψ(κ))θu
)

= ℘(z)− ℘(κ)Lϖu−k
(

−(ϖu − k)

1− (ϖu − k)
(ψ(z)− ψ(κ))ϖu−k

)
.

(iv) Based on Definitions 10, 12 and Lemma 3, for u ≥ 1, we get(
RLABℜϖu,ψκ

CABDϖu,ψ
κ ℘

)
(z)

=
(
RLℜk,ψκ

RLABℜθu,ψκ
CABDθu,ψ

κ ℘
(k)
ψ

)
(z) =RL ℜk,ψκ

(
℘
(k)
ψ (z)− ℘

(k)
ψ (κ)

)
= ℘(z)−

k−1∑
m=0

℘
(m)
ψ (κ)
m!

(ψ(z)− ψ(r))m −
℘
(k)
ψ (κ)
k!

(ψ(z)− ψ(r))k

= ℘(z)−
k∑

m=0

℘
(m)
ψ (κ)
m!

(ψ(z)− ψ(κ))m.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3696

Lemma 5. Assume that ℘ ∈ Ck(ℑu,R) and ψ ∈ Ck(ℑu,R+) with ψ′(z) ̸= 0. For ϖu ∈
(k, k+1], θu = ϖu−k, λ ≥ k+1 and ζ ≥ 0, for k ≥ 0, u ≥ 1, the relations below are true:

(i) RLABℜϖu,ψκ (℘ (z)− ℘ (κ))ζ = (k+1−ϖu)Γ(1+ζ)(℘(z)−℘(κ))ζ+k
Λ(ϖu−k)Γ(1+ζ+k) + (ϖu−k)Γ(1+ζ)(℘(z)−℘(κ))ζ+ϖu

Λ(ϖu−k)Γ(1+ζ+k) .

(ii) CABDϖu,ψ
κ (℘ (z)− ℘ (κ))λ = Λ(ϖu−k)

k+1−ϖu

∞∑
j=0

(
−(ϖu−k)
k+1−ϖu

)j
Γ(1+λ)(℘(z)−℘(κ))j(ϖu−k)+λ−k

Γ(j(ϖu−k)+λ−k+1) .

(iii) CABDϖu,ψ
κ (℘ (z)− ℘ (κ))σ = 0, σ = 0, 1, ..., k.

(iv)
(
RLABℜϖu,ψκ 1

)
(z) = (k+1−ϖu)(℘(z)−℘(κ))k

Λ(ϖu−k)Γ(k+1) + (ϖu−k)(℘(z)−℘(κ))ϖu
Λ(ϖu−k)Γ(1+k) .

(v)
(
CABDϖu,ψ

κ 1
)
(z) = 0.

Proof. (i) Using Definition 12 and Lemma 3, for u ≥ 1, we have

RLABℜϖu,ψκ (℘ (z)− ℘ (κ))ζ

=
k + 1−ϖu

Λ(ϖu − k)

RL

ℜk,ψκ (℘ (z)− ℘ (κ))ζ +
ϖu − k

Λ(ϖu − k)

RL

ℜϖu,ψκ (℘ (z)− ℘ (κ))ζ

=
k + 1−ϖu

Λ(ϖu − k)

Γ(1 + ζ)

Γ(1 + ζ + k)
(℘ (z)− ℘ (κ))ζ+k

+
ϖu − k

Λ(ϖu − k)

Γ(1 + ζ)

Γ(1 + ζ + k)
(℘ (z)− ℘ (κ))ζ+ϖu .

(ii) From Definitions 3, 11 and Lemma 3, it follows that for u ≥ 1,

CABDϖu,ψ
κ (℘ (z)− ℘ (κ))λ

= CABDθu,ψ
κ

(
1

ψ′(z)

d

dz

)k
(℘ (z)− ℘ (κ))λ

= CABDθu,ψ
κ

Γ(1 + λ)

Γ(λ− k + 1)
(℘ (z)− ℘ (κ))λ−k

=
Λ(θu)

1− θu

z∫
κ

ψ′(r)
∞∑
j=0

(
−θu
1− θu

)j Γ(1 + λ) (℘ (r)− ℘ (κ))λ−(k+1)

Γ(λ− k)Γ(jθu + 1)
(℘ (z)− ℘ (r))jθu dr

=
Γ(1 + λ)Λ(θu)

Γ(λ− k) (1− θu)

∞∑
j=0

(
−θu
1− θu

)j
RLℜjθu+1,ψ

κ (℘ (z)− ℘ (κ))λ−(k+1)

=
Λ(θu)

1− θu

∞∑
j=0

(
−θu
1− θu

)j Γ(1 + λ)

Γ(jθu + λ− k + 1)
(℘ (z)− ℘ (κ))jθu+λ−k

=
Λ(ϖu − k)

k + 1−ϖu

∞∑
j=0

(
−(ϖu − k)

k + 1−ϖu

)j Γ(1 + λ) (℘ (z)− ℘ (κ))j(ϖu−k)+λ−k

Γ(j(ϖu − k) + λ− k + 1)
.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3697

(iii) From Definitions 7 and 11, one has

CABDϖu,ψ
κ (℘ (z)− ℘ (κ))σ

= CABDθu,ψ
κ

(
1

ψ′(z)

d

dz

)k
(℘ (z)− ℘ (κ))σ

= CABDθu,ψ
κ

Γ(1 + σ)

Γ(λ− σ + 1)
(℘ (z)− ℘ (κ))σ−k

=
Λ(θu)

1− θu

z∫
κ

Lθu

(
−θu
1− θu

(℘ (z)− ℘ (r))θu
)

Γ(1 + σ)

Γ(λ− σ + 1)
(℘ (r)− ℘ (κ))σ−k dr

= 0.

Taking ζ = σ = 0 in portions (i) and (iii), we conclude (iv) and (v), respectively.

4. Generalizing Gronwall’s inequality

This part will begin with the following generalization of Gronwall’s inequality.

Lemma 6. [27] Assume that the function ψ ∈ C1(ℑu,R+) is increasing with ψ′(z) ̸= 0,
for each z ∈ ℑ and ϖ > 0. Let ℓ(z) be a nonnegative and nondecreasing function (NNF,
for abbreviate), ℏ(z) be a nonnegative function locally integrable (NFLI, for short) on ℑ
and ϱ be a NFLI on ℑ. If the inequality

ϱ (z) ≤ ℏ(z) + ℓ(z)

z∫
κ

Lθuψ
′(r) (ψ (z)− ψ (r))ϖ−1 ϱ (r) dr, z ∈ ℑ

holds, then

ϱ (z) ≤ ℏ(z) +
z∫

κ

∞∑
j=1

[ℓ(z)Γ(ϖ)]j

Γ(jϖ)
ψ′(r) (ψ (z)− ψ (r))jϖ−1 ℏ (r) dr,

for every z ∈ ℑ.

Lemma 7. [27] Assume that all requirements of Lemma 6 are true, if the function ℏ(z)
is nondecreasing on ℑ, one has

ϱ (z) ≤ ℏ(z)Lϖ [ℓ(z)Γ(ϖ) (ψ (z)− ψ (κ))ϖ] , z ∈ ℑ.

In this role, we will present a novel Gronwall inequality within the context of the
ψ −RL−AB fractional operator.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3698

Lemma 8. Suppose that the function ψ ∈ C1(ℑu,R+) is increasing with ψ′(z) ̸= 0, for

each z ∈ ℑ and ϖu = ϖ ∈ (0, 1], for u ≥ 1. Assume that Θ(z) = U(z)Λ(ϖ)
Λ(ϖ)−(1−ϖ)G(z) is a NFLI

on ℑ, Ξ(z) = ϖG(z)
Λ(ϖ)−(1−ϖ)G(z) is a NNF and ϱ is a NFLI on ℑ, such that

ϱ (z) ≤ U(z) +G(z)RLABℜϖ,ψκ ϱ (z) , z ∈ ℑ. (1)

Then, for each z ∈ ℑ, we have

ϱ (z) ≤ Θ(z) +

z∫
κ

∞∑
j=1

[Ξ(z))]j

Γ(jϖ)
ψ′(r) (ψ (z)− ψ (r))jϖ−1Θ(r) dr.

Proof. Utilizing (1), Definitions 1 and 8, one has

ϱ (z) ≤ U(z) +G(z)
(
RLABℜϖ,ψκ ϱ

)
(z)

≤ U(z) +G(z)

1−ϖ

Λ(ϖ)
ϱ(z) +

ϖ

Λ(ϖ)

1

Γ (ϖ)

z∫
κ

(ψ(z)− ψ(r))ω−1 ψ′(r)ϱ (r) dr

 .
Therefore,

ϱ (z) ≤ U(z)Λ(ϖ)

Λ(ϖ)− (1−ϖ)G(z)

+
ϖG(z)

Λ(ϖ)− (1−ϖ)G(z)

1

Γ (ϖ)

z∫
κ

(ψ(z)− ψ(r))ϖ−1 ψ′(r)ϱ (r) dr

Lemma 6 allows us to obtain

ϱ (z) ≤ U(z)Λ(ϖ)

Λ(ϖ)− (1−ϖ)G(z)

+

z∫
κ

∞∑
j=1

1

Γ(jϖ)

(
ϖG(z)

Λ(ϖ)− (1−ϖ)G(z)

)j U(r)Λ(ϖ) (ψ(z)− ψ(r))jϖ−1 ψ′(r)

Λ(ϖ)− (1−ϖ)G(z)
dr

≤ Θ(z) +

z∫
κ

∞∑
j=1

[Ξ(z))]j

Γ(jϖ)
ψ′(r) (ψ (z)− ψ (r))jϖ−1Θ(r) dr.

Corollary 1. In light of assumptions of Lemma 8, if the function Θ(z) is nondecreasing
on ℑ, then, we get

ϱ (z) ≤ Θ(z)Lϖ [Ξ(z) (ψ (z)− ψ (κ))ϖ] , z ∈ ℑ.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3699

Proof. Based on Lemma 8, one can write

ϱ (z) ≤ U(z)Λ(ϖ)

Λ(ϖ)− (1−ϖ)G(z)
Lϖ

(
ϖG(z) (ψ(z)− ψ(r))ϖ

Λ(ϖ)− (1−ϖ)G(z)

)
≤ Θ(z)Lϖ [Ξ(z) (ψ (z)− ψ (κ))ϖ] .

A novel Gronwall inequality in the context of the ψ−KAB fractional operator will be
concluded here.

Corollary 2. Let ϖu = ϖ > 0, for u ≥ 1. Assume that Θ(z) = U(z)Λ(ϖ)
Λ(ϖ)−(1−ϖ)G(z) is a NFLI

on ℑ, Ξ(z) = ϖG(z)
Λ(ϖ)−(1−ϖ)G(z) is a NNF and ϱ is a NFLI on ℑ, such that

ϱ (z) ≤ U(z) +G(z)KABℜϖ,ηκ ϱ (z) , z ∈ ℑ,

Then, for each z ∈ ℑ, we get

ϱ (z) ≤ Θ(z) +

z∫
κ

∞∑
j=1

η1−jϖrη−1 [Ξ(z))]j

Γ(jϖ)
(zη − rη)jϖ−1Θ(r) dr.

Proof. The proof follows immediately by taking ψ(z) = zη

η in Lemma 8.

Corollary 3. Via the assumptions of Corollary 2, if the function Θ(z) is nondecreasing
on ℑ, then, we get

ϱ (z) ≤ Θ(z)Lϖ

[
Ξ(z)

(
zη − rη

η

)ϖ]
, z ∈ ℑ.

Proof. Taking ψ(z) = zη

η in Corollary 1, we get the proof.

5. Solving a fractional differential equation

This part is devoted to presenting the existence and uniqueness of solution to the initial
FDE below: {

CABDϖu,ψ
κ ϱ(z) = ϕ(z, ϱ(z)), z ∈ ℑ, u ≥ 1,

ϱ
(i)
ψ (κ) = γi, i = 0, 1, ..., k,

(2)

where CABDϖu,ψ
κ is the ϖuth left-sided ψ−ABC fractional derivative such that ϖu ∈

(k, k+1], γi ∈ R (i ≥ 0) are constants, ϕ : ℑ×R → R is a continuous function, ψ : ℑ → R
is an increasing function with ψ′(z) ∈ Ck(ℑu,R+) and ψ′(z) ̸= 0, for all z ∈ ℑ and

ϱ(z) ∈ Ck(ℑu,R) is a recognized function in which ϱ
(i)
ψ (z) =

(
1

ψ′(z)
d
dz

)i
ϱ(z) such that

ϱ
(0)
ψ (z) = ϱ(z).



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3700

In fact, by utilizing Lemma 4 in conjunction with the initial FDE (2) and the ϖuth
left-sided ψ–RL-AB fractional integral operator on both sides of (2), we get

ϱ (z) =
k∑
i=0

γi
i!

(ψ(z)− ψ(κ))i +RLAB ℜϖu,ψκ ϕ(z, ϱ(z)). (3)

Now, we shall apply Picard’s iterative technique [19] to demonstrate the existence and
uniqueness of the solution to Problem (2).

Theorem 1. Assume that the assertions below hold:

(i) there exists a constant T > 0 such that

sup
z∈ℑ

|ϕ(z, ϱ0(z))| ≤ T,

(ii) there exists a constant P > 0 such that

|ϕ(z, ϱ1)− ϕ(z, ϱ2)| ≤ P |ϱ1 − ϱ2| , for all z ∈ ℑ, ϱ1, ϱ2 ∈ Ck(ℑu,R).

(iii) we have the inequality

P

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
< 1. (4)

Then, the problem (2) has a unique solution on ℑ.

Proof. It is evident that the solution to system (2) is the same as the solution to the
FIE (3). Set

ϱ0 (z) =
k∑
i=0

γi
i!

(ψ(z)− ψ(κ))i , (5)

and

ϱs (z) =
k∑
i=0

γi
i!

(ψ(z)− ψ(κ))i +RLAB ℜϖu,ψκ ϕ(z, ϱs−1(z)), s ∈ N. (6)

Clearly, the series ϱ0 (z) +
∞∑
m=0

(ϱm − ϱm−1) has a partial sum

ϱs (z) = ϱ0 (z) +
s∑

m=0

(ϱm − ϱm−1).

We want to show that the sequence {ϱs (z)} converges to ϱ (z) . By a mathematical induc-
tion, for all z ∈ [κ, υ], we can write

∥ϱs (z)− ϱs−1 (z)∥



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3701

≤ TP s−1

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)s
, s ∈ N.(7)

Based on (5) and (6) and Lemma 5 (iv), one has

∥ϱ1 − ϱ0∥ = sup
z∈ℑ

∣∣∣RLABℜϖu,ψκ ϕ(z, ϱ0(z))
∣∣∣

≤ T

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
.

Hence, for s = 1, the inequality (7) is true. After that, consider the inequality (7) is
satisfied when s = n. Therefore,

∥ϱn+1 − ϱn∥ = sup
z∈ℑ

∣∣∣RLABℜϖu,ψκ ϕ(z, ϱn(z))−RLAB ℜϖu,ψκ ϕ(z, ϱn−1(z))
∣∣∣

= sup
z∈ℑ

∣∣∣RLABℜϖu,ψκ [ϕ(z, ϱn(z))− ϕ(z, ϱn−1(z))]
∣∣∣

≤ RLABℜϖu,ψκ (P ∥ϱn(z)− ϱn−1(z)∥)

≤ RLABℜϖu,ψκ

(
TPn

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)n)

≤ TP (s+1)−1

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)n+1

.

Hence, the inequality (7) is fulfilled for s = n + 1. Then inequality (7) is true for every
s ∈ N and all z ∈ [κ, υ]. Thus, one can write

∞∑
s=1

∥ϱs (z)− ϱs−1 (z)∥ ≤
∞∑
s=1

TP s−1

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)s
.

The series on the right side of the aforementioned inequality is convergent as a result of as-

sumption (4), and so
∞∑
s=1

∥ϱs − ϱs−1∥ is also convergent that shows that ϱ0+
∞∑
s=1

∥ϱs − ϱs−1∥
converges.

Put

ϱ = ϱ0 +
∞∑
s=1

∥ϱs − ϱs−1∥ ,

it follows that
∥ϱs − ϱ∥ → 0, as s→ ∞. (8)

This indicates that the solution to problem (2) exists. From (8), we get

∥ϕ(., ϱs−1(.))− ϕ(., ϱ(.))∥ ≤ P |ϱs−1 − ϱ| → 0, as s→ ∞.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3702

Thus,
lim
s→∞

ϕ(z, ϱs−1(z)) = ϕ(z, ϱ(z)). (9)

As s→ ∞ in (6) and applying (9), we have

ϱ (z) =
k∑
i=0

γi
i!

(ψ(z)− ψ(κ))i +RLAB ℜϖu,ψκ ϕ(z, ϱ(z)),

which is a solution of the initial FDE (2).
Finally, for the uniqueness, assume that ϱ̂ is another solution to Problem (2). Thus,

we get

∥ϱ− ϱ̂∥ = sup
z∈ℑ

∣∣∣RLABℜϖu,ψκ ϕ(z, ϱ(z))−RLAB ℜϖu,ψκ ϕ(z, ϱ̂(z))
∣∣∣

= sup
z∈ℑ

∣∣∣RLABℜϖu,ψκ [ϕ(z, ϱ(z))− ϕ(z, ϱ̂(z))]
∣∣∣

≤ RLABℜϖu,ψκ (P ∥ϱ(z)− ϱ̂(z)∥)

≤ p

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
∥ϱ− ϱ̂∥ .

In light of (4), we conclude that ∥ϱ− ϱ̂∥ , that is, ϱ(z) = ϱ̂(z). This completes the proof.

6. Supportive examples

In this part, we support our results by the following examples:

Example 1. Consider the following initial FDE:{
CABDϖu,ψ

1 ϱ(z) = cos
(
z2
)
− ϱ(z)

1
4
−ϱ(z) , z ∈ [1, 3],

ϱ(1) = 1, ϱ′ψ(1) = 1,
(10)

where ψ(z) = ln(z) and

ϖu(τ) =

{
0.98, τ ∈ [1, 2],
1.21, τ ∈ (2, 3],

for u ≥ 1.

The requirements of Theorem 1 shall be examined as follows:

|ϕ(z, ϱ1)− ϕ(z, ϱ2)| =

∣∣∣∣∣cos (z2)− ϱ1(z)
1
4 − ϱ1(z)

− cos
(
z2
)
+

ϱ2(z)
1
4 − ϱ(z)

∣∣∣∣∣
=

∣∣∣∣∣ ϱ1(z)
1
4 − ϱ1(z)

− ϱ2(z)
1
4 − ϱ(z)

∣∣∣∣∣
≤ 1

4
|ϱ1(z)− ϱ2(z)| .

Then, P = 1
4 > 0. If we take Λ(ϖu − k) = 1, then, we have the following cases:



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3703

(a) If τ ∈ [1, 2], one has k = 1, and

P

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
≈ 0.171925 < 1.

(b) If τ ∈ (2, 3], we have k = 3, and

P

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
≈ 0.568509 < 1.

Therefore, all axioms of Theorem 1 are fulfilled. Hence, there exists a unique solution
to Problem (10).

Example 2. Consider the following initial FDE:{
CABDϖu,ψ

0 ϱ(z) = z3 − ϱ(z), z ∈ [0, 2],

ϱ(0) = 0, ϱ′ψ(0) = 0, ϱ
′′
ψ(0) = 2

(11)

where ψ(z) = z and

ϖu(τ) =

{
0.2, τ ∈ [12 , 1],
0.4, τ ∈ (1, 2],

for u ≥ 1.

Assume that problem (11) has an exact solution ϱ(z) = z3.
The conditions of Theorem 1 shall be checked as follows:

|ϕ(z, ϱ1)− ϕ(z, ϱ2)| =
∣∣z3 − ϱ1(z)− z3 + ϱ2(z)

∣∣
≤ |ϱ1(z)− ϱ2(z)| .

Then, P = 1 > 0. If we take Λ(ϖu − k) = 1, then, we have the following cases:

(a) If τ ∈ [12 , 1], one has k = 1, and

P

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
≈ 0.645318 < 1.

Example 3. If τ ∈ (1, 2], we have k = 2, and

P

(
(k + 1−ϖu) (ψ(υ)− ψ(κ))k

Λ(ϖu − k)Γ(k + 1)
+

(ϖu − k) (ψ(υ)− ψ(κ))ϖu

Λ(ϖu − k)Γ(ϖu + 1)

)
≈ 0.735214 < 1.

Hence, all assumptions of Theorem 1 are fulfilled. Therefore, the problem (11) possesses
a unique solution.

Next, using Picard’s iterative method, we will compute the solution of system (11) as
follows:

ϱs (z) = ϱ0 (z) +
CAB Dϖu,ψ

0

(
z3 − ϱs−1(z)

)
, ϱ0 (z) = z3.



H.A. Hammad, M. De la Sen / Eur. J. Pure Appl. Math, 17 (4) (2024), 3687-3707 3704

Then

ϱ1 (z) = z3 + CABDϖu,ψ
0

(
z3 − ϱ0(z)

)
= z3,

ϱ2 (z) = z3 + CABDϖu,ψ
0

(
z3 − ϱ1(z)

)
= z3,

ϱ3 (z) = z3,

...

ϱk (z) = z3.

This corresponds to the exact solution.

7. Conclusion and open problems

This paper delves into the application of AB fractional operators with higher-variable
orders using increasing functions. We explore the qualitative properties of these operators
and demonstrate the existence of a unique solution for an initial FDE using Picard’s itera-
tion method. Our findings are supported by two illustrative examples. We also introduce
a generalized Gronwall inequality within the framework of AB fractional integrals. This
study contributes to the advancement of fractional calculus and its potential applications.
Future research will focus on applying these extended operators to real-world dynamic sys-
tems, investigating new properties and inequalities associated with them, and exploring
the corresponding right-sided fractional operators for RL-AB, ABC, and KAB.

8. Abbreviations

AB→Atangana-Baleanu
ML→Mittag-Leffler
FDE→fractional differential equations
ABC→Atangana-Baleanu-Caputo
w.r.t.→with respect to
RL→Riemann-Liouville
ABK→Atangana-Baleanu-Kashuri
RAB→Riemann-Atangana-Baleanu
RL-AB→Riemann-Liouville-tangana-Baleanu
NNF→nonnegative and nondecreasing function
NFLI→nonnegative function locally integrable
FIE→fractional integral equation

Conflicts of interest

The authors declare that they have no conflicts of interest.



REFERENCES 3705

Author’s contributions

All authors contributed equally and significantly in writing this article.

Funding

This work was supported in part by the Basque Government under Grant IT1555-22.

Acknowledgments

The authors thank the Basque Government for Grant IT1555-22 and to MICIU/AEI/
10.13039/501100011033 and ERDF/E for Grants PID2021-123543OB-C21 and PID2021-
123543OB-C22.

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