EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 2, 2022, 557-571
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The fractional differential equations with uncertainty by
conformable derivative

Atimad Harir1,∗, Said Melliani1, Lalla Saadia Chadli1
1 Laboratory of Applied Mathematics and Scientific Computing,
Sultan Moulay Slimane University, P.O. Box 523, Beni Mellal, 23000, Morocco

Abstract. We provide a fractional order fuzzy fractional differential equation q ∈ (0, 1]. A fuzzy
fractional integral and a fuzzy conformable derivative are shown and proved. To prove fuzzy
solutions for fractional differential equations with fuzzy beginning values and deterministic or
fuzzy functions, two alternative techniques are used. The application has been submitted.
2020 Mathematics Subject Classifications: 34K36, 34K37, 46S40
Key Words and Phrases: Fuzzy fractional differential equation, conformable derivative, fuzzy
number

1. Introduction

In this paper we will consider the fractional differential equation

y(q)(t) = F (t, y, k), q ∈ (0, 1] (1)
y(0) = c

where k = (k1, . . . , kn) is a vector of constants, t ∈ (0, a) and y(q) is the conformable
derivative of y of order q ∈ (0, 1] see [12]. We suppose the existence of imprecise parameters
kj and c in Eq (1). Because fuzzy sets theory is a valuable tool for representing imprecision
and processing vagueness in mathematical models [10, 12], the goal of this paper is to
solve Eq (1) with fuzzy parameters using fuzzy conformable derivative using the same
technique as Buckley and Feuring [10]. As a result, the first part of this work proposes
a new solution to the q ∈ (0, 1] order fuzzy fractional initial value problem. This new
solution’s basic attributes are listed. [8] developed the concept of the fuzzy conformable
derivative, which is the most natural and efficient definition of the conformable derivative
of order q ∈ (0, 1]. The following are the key advantages of this derivative:

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4299

Email addresses: atimad.harir@gmail.com (A. Harir), s.melliani@usms.ma (S. Melliani) ,
sa.chadli@yahoo.fr (L. S. Chadli )

https://www.ejpam.com 557 © 2022 EJPAM All rights reserved.



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 558

Many applications and phenomena can be modeled using conformable derivatives and
need to be solved (physical applications see[2, 4, 5, 13–17].)
It can be extended to solve exactly and numerically fractional differential equations and
systems easily and efficiently. It creates new comparisons of conformable derivatives and
other previous fractional definitions in many applications. This paper initially proposes
a new solution to the q ∈ (0, 1] order fuzzy fractional initial value problem. This new
solution’s basic qualities are listed below.
The following is a breakdown of the paper’s sections: section 2 covers the fundamental
ideas of fuzzy numbers. In section 3, we prove certain results on a fuzzy fractional integral
and a fuzzy conformable derivative. The extension principle of Zadeh and the concept of
fuzzy conformable derivatives are applied in section 4 using two different ways. The fuzzy
fractional differential equation is demonstrated. The applications are in section 5.

2. Preliminaries

We place a bar over a letter to denote a fuzzy number of R. So, ū, all represent
fuzzy numbers of R. We write µū(t), a number in [0, 1], for the membership function of ū
evaluated at t ∈ R.
Let us denote by RF = {µū : R → [0, 1]} the class of fuzzy subsets of the real axis satisfying
the following properties :

(i) ū is normal i.e, there exists an x0 ∈ R such that µū (x0) = 1,

(ii) ū is fuzzy convex i.e for x, y ∈ R and 0 < λ ≤ 1,

µū(λx+ (1− λ)y) ≥ min [µū(x), µū(y)]

(iii) ū is upper semicontinuous,

(iv) [ū]0 = cl {x ∈ R | µū(x) > 0} is compact.

Then RF is called the space of fuzzy numbers. Obviously, R ⊂ RF . For 0 < α ≤ 1 denote
[ū]α = {x ∈ R | µū(x) ≥ α} , then from (i) to (iv) it follows that the α-level sets [ū]α ∈
PK(R) for all 0 ≤ α ≤ 1 is a closed bounded interval which is denoted by [ū]α = [uα1 , u

α
2 ] .

By PK(R) we denote the family of all nonempty compact convex subsets of R, and define
the addition and scalar multiplication in PK(R) as usual.

Theorem 1. see [1] If ū ∈ RF , then

(i) [ū]α ∈ PK(R) for all 0 ≤ α ≤ 1

(ii) [ū]α2 ⊂ [ū]α1 for all 0 ≤ α1 ≤ α2 ≤ 1

(iii) {αk} ⊂ [0, 1] is a nondecreasing sequence which converges to α then

[ū]α =
⋂
k≥1

[ū]αk



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 559

Conversely, if Aα = {[uα1 , uα2 ] ;α ∈ (0, 1]} is a family of closed real intervals verifying (i)
and (ii), then {Aα} defined a fuzzy number ū ∈ RF such that [ū]α = Aα for 0 < α ≤ 1
and [ū]0 = ∪0<α≤1Aα ⊂ A0.

Definition 1. [7, 9, 11] We represent an arbitrary fuzzy number by an ordered pair of
functions [ū]α = [uα1 , u

α
2 ] , α ∈ [0, 1], which satisfy the following requirements :

1. uα1 is an increasing function over [0, 1];

2. uα2 is a decreasing function on [0, 1];

3. uα1 and uα2 are bounded left continuous on (0, 1], and right continuous at α = 0;

4. uα1 ≤ uα2 , for 0 ≤ α ≤ 1.

Definition 2. For a fuzzy set ū = (u1, u2, u3) , (u1 < u2 < u3) , ū is called triangular fuzzy
number with peak u2, left width u2−u1 > 0 and right width u3−u2 > 0, if its membership
function has the following form:

µū(t) =


1− (u2−t)

u2−u1
, u1 ≤ t ≤ u2,

1− (t−u2)
u3−u2

, u2 ≤ t ≤ u3,

0, otherwise.

Lemma 1. see [3] Let ū, v̄ : R → [0, 1] be the fuzzy sets. Then ū = v̄ if and only if
[ū]α = [v̄]α for all α ∈ [0, 1].

The following arithmetic operations on fuzzy numbers are well known and frequently
used below. If ū, v̄ ∈ RF then

[ū+ v̄]α = [uα1 + vα1 , u
α
2 + vα2 ]

[λū]α = λ[ū]α =

{
[λuα1 , λu

α
2 ] if λ ≥ 0

[λuα2 , λu
α
1 ] if λ < 0,

Definition 3. Let ū, v̄ ∈ RF . If there exists w̄ ∈ RF such as ū = v̄ + w̄ then w̄ is called
the H-difference of ū, v̄ and it is denoted ū	 v̄

3. Fuzzy conformable differentiability and fuzzy fractional integral

Now, we present our new definition, which is the simplest and most natural and efficient
definition of conformable derivative of order q ∈ (0, 1].

Definition 4. [8] Let F̄ : [0, a) → RF be a fuzzy function. qth order fuzzy conformable
derivative of F̄ is defined by

Tq(F̄ )(t) = lim
ε→0+

F̄
(
t+ εt1−q

)
	 F̄ (t)

ε
= lim

ε→0+

F̄ (t)	 F̄
(
t− εt1−q

)
ε



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 560

for all t > 0, q ∈ (0, 1). Let F̄ (q)(t) stands for Tq(F̄ )(t). Hence

F̄ (q)(t) = lim
ε→0+

F̄
(
t+ εt1−q

)
	 F̄ (t)

ε
= lim

ε→0+

F̄ (t)	 F̄
(
t− εt1−q

)
ε

If F̄ is q-differentiable in some (0, a), and limt→0+ F̄ (q)(t) exists, then

F̄ (q)(0) = lim
t→0+

F̄ (q)(t)

and the limits (in the metric d. )

Remark 1. From the definition, it directly follows that if F̄ is q-differentiable then the
multi valued mapping F̄α is q-differentiable for all α ∈ [0, 1] and

TqF̄α =
[
F̄ (q)(t)

]α
(2)

Here TqF̄α is denoted the conformable fractional derivative of F̄α of order q.

Theorem 2. [8] Let F̄ : [0, a) → RF be q-differentiable. Denote F̄α(t) = [fα
1 (t), f

α
2 (t)],

α ∈ [0, 1]. Then fα
1 (t) and fα

2 (t) are q-differentiable and[
F̄ (q)(t)

]α
=

[
(fα

1 )
(q) (t), (fα

2 )
(q) (t)

]
.

Theorem 3. If a function F̄ : [0, a) → RF is q-differentiable at t0 > 0, q ∈ (0, 1] Denote
F̄α(t) = [fα

1 (t), f
α
2 (t)] , α ∈ [0, 1]. Then fα

1 (t) and fα
2 (t) are continuous at t0 so F̄ is

continuous at t0.

Proof. If ε > 0 and α ∈ [0, 1], we have :[
F̄
(
t0 + εt1−q

0

)
	 F̄ (t0)

]α
=

[
fα
1

(
t0 + εt1−q

0

)
− fα

1 (t0) , f
α
2

(
t0 + εt1−q

0

)
− fα

2 (t0)
]

Dividing and multiplying by ε, we have :

[
F̄
(
t0 + εt1−q

0

)
	 F̄ (t0)

]α
=

fα
1

(
t0 + εt1−q

0

)
− fα

1 (t0)

ε
· ε,

fα
2

(
t0 + εt1−q

0

)
− fα

2 (t0)

ε
· ε


Similarly, we obtain:

[
F̄ (t0)	 F̄

(
t0 − εt1−q

0

)]α
=

fα
1 (t0)− fα

1

(
t0 − εt1−q

0

)
ε

· ε,
fα
2 (t0)− fα

2

(
t0 − εt1−q

0

)
ε

· ε


Then

lim
ε→0+

[
F̄
(
t0 + εt1−q

0

)
	 F̄ (t0)

]α
=

 lim
ε→0+

fα
1

(
t0 + εt1−q

0

)
− fα

1 (t0)

ε
· lim
ε→0+

ε

, lim
ε→0+

fα
2

(
t0 + εt1−q

0

)
− fα

2 (t0)

ε
· lim
ε→0+

ε





A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 561

Similarly, we obtain :

lim
ε→0+

[
F̄ (t0)	 F̄

(
t0 − εt1−q

0

)]α
=

 lim
ε→0+

fα
1 (t0)− fα

1

(
t0 − εt1−q

0

)
ε

· lim
ε→0+

ε

, lim
ε→0+

fα
2 (t0)− fα

2

(
t0 − εt1−q

0

)
ε

· lim
ε→0+

ε


Let h = εt1−q

0 . Then

lim
h→0+

[
F̄ (t0 + h)	 F̄ (t0)

]α
=

[
(fα

1 )
(q) (t0) .0, (f

α
2 )

(q) (t0) .0
]

Similarly, we obtain :

lim
h→0+

[
F̄ (t0)	 F̄ (t0 − h)

]α
=

[
(fα

1 )
(q) (t0) .0, (f

α
2 )

(q) (t0) .0
]

which implies that
lim

h→0+

[
F̄ (t0 + h)

]α
=

[
F̄ (t0)

]α
Similarly, we obtain :

lim
h→0+

[
F̄ (t0 − h)

]α
=

[
F̄ (t0)

]α
Hence, F̄ is continuous at t0.

Theorem 4. Let q ∈ (0, 1]

• If F̄ is differentiable and F̄ is q-differentiable then TqF̄ (t) = t1−qF̄ ′(t)

Proof. Let h = εt1−q in Definition 4 , and then ε = tq−1h. Therefore, if ε > 0 and
α ∈ [0, 1], we have[

F̄
(
t+ εt1−q

)
	 F̄ (t)

]α
=

[
fα
1

(
t+ εt1−q

)
− fα

1 (t), f
α
2

(
t+ εt1−q

)
− fα

2 (t)
]
.

Dividing by ε, we have[
F̄
(
t+ εt1−q

)
	 F̄ (t)

]α
ε

=

[
fα
1

(
t+ εt1−q

)
− fα

1 (t)

ε
,
fα
2

(
t+ εt1−q

)
− fα

2 (t)

ε

]
,

and passing to the limit,

lim
ε−→0+

[
F̄
(
t+ εt1−q

)
	 F̄ (t)

]α
ε

= lim
ε−→0+

[
fα
1

(
t+ εt1−q

)
− fα

1 (t)

ε
,
fα
2

(
t+ εt1−q

)
− fα

2 (t)

ε

]

= lim
h−→0+

[
fα
1 (t+ h)− fα

1 (t)

tq−1h
,
fα
2 (t+ h)− fα

2 (t)

tq−1h

]
= t1−q lim

h−→0+

[
fα
1 (t+ h)− fα

1 (t)

h
,
fα
2 (t+ h)− fα

2 (t)

h

]
= t1−q

[
(fα

1 )
′ (t), (fα

2 )
′ (t)

]
.



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 562

Similarly, we obtain[
F̄ (t)	 F̄

(
t− εt1−q

)]α
ε

=

[
fα
1 (t)− fα

1

(
t− εt1−q

)
ε

,
fα
2 (t)− fα

2

(
t− εt1−q

)
ε

]
,

and passing to the limit and ε = tq−1h gives TqF̄ (t) = t1−q
[
(fα

1 )
′ (t), (fα

2 )
′ (t)

]
.

Let q ∈ (0, 1] and F̄ : [0, a) → RF be such that [F̄ (t)]α = [fα
1 (t), f

α
2 (t)] for all α ∈ [0, 1].

Suppose that fα
1 , f

α
2 ∈ C ([0, a),R) ∩ L1 ([0, a),R) for all α ∈ [0, 1] and let

Aα =:

[∫ t

0

fα
1 (x)

x1−q
dx,

∫ t

0

fα
2 (x)

x1−q
dx

]
, t ∈ (0, a) (3)

Lemma 2. [5] The family {Aα;α ∈ [0, 1]} , given by Eq(3), defined a fuzzy number F̄ ∈ RF

such that
[
F̄
]α

= Aα.

Proof. For α < β we have that fα
1 (x) ≤ fβ

1 (x) and fα
2 (x) ≥ fβ

2 (x). It follows that

Aα ⊇ Aβ. Since f0
1 (x) ≤ fαn

1 (x) ≤ f1
1 (x) we have∣∣xq−1fαn

i (x)
∣∣ ≤ max

{
aq−1

∣∣f0
i (x)

∣∣ , aq−1
∣∣f1

i (x)
∣∣} =: gi(x)

for αn ∈ [0, 1] and i = 1, 2. Obviously, gi is integrable on [0, a). Therefore, if αn ↑ α then
by the Lebesque’s Dominated convergence Theorem, we have

lim
n→∞

∫ t

0

fαn
i

x1−q
(x)dx =

∫ t

0

fα
i

x1−q
(x)dx, i = 1, 2

From Theorem (1), the proof is complete.

Remark 2. [12] By using the Weierstrass theorem, it is enough to define the fractional
integral on polynomials. This suggests the following. Let q ∈ (0, 1]. Define Iq (t

p) = tp+q

p+q
for any p ∈ R, and q 6= −p

• If f(t) =
∑n

k=0 bkt
k, then we define Iq(f) =

∑n
k=0 bkIq

(
tk
)
=

∑n
k=0 bk

tk+q

k+q

• If f(t) =
∑∞

k=0 bkt
k, where the series is uniformly convergent, then we define Iq(f) =∑∞

k=0 bk
tk+q

k+q

Clearly, Iq is linear on its domain. Further, if q = 1, then Iq is the usual integral.

Definition 5. Let F̄ ∈ C ([0, a),RF ) ∩ L1 ([0, a),RF ) , Define the fuzzy fractional integral
for a ≥ 0 and q ∈ (0, 1)

Iq(F̄ )(t) = I1
(
tq−1F̄

)
(t) =

∫ t

0

F̄ (x)

x1−q
dx



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 563

by [
Iq(F̄ )(t)

]α
=

[
I1

(
tq−1F̄

)
(t)

]α
=

[∫ t

0

F̄ (x)

x1−q
dx

]α
=

[∫ t

0

fα
1 (x)

x1−q
dx,

∫ t

0

fα
2 (x)

x1−q
dx

]
where the integral

∫ t
0

fα
i

x1−q (x)dx, for i = 1, 2 is the usual Riemann improper integral.
Also, the following properties are obvious.

(i) IqλF̄ (t) = λIqF̄ (t) for each λ ∈ R

(ii) Iq(F̄ + Ḡ)(t) = IqF̄ (t) + IqḠ(t)

Theorem 5. TqIq(F̄ )(t) = F̄ (t), for t ≥ 0, where F̄ is any continuous function in the
domain of Iq.

Proof. Since F̄ is continuous, then Iq(F̄ )(t) is clearly differentiable. Hence,

[
TqIq(F̄ )(t)

]α
=

[
t1−q d

dt
Iq(F̄ )(t)

]α
=

[
t1−q d

dt

∫ t

0

fα
1 (x)

x1−q
dx, t1−q d

dt

∫ t

0

fα
2 (x)

x1−q
dx

]
=

[
t1−q f

α
1 (t)

t1−q
, t1−q f

α
2 (t)

t1−q

]
= [F̄ (t)]α

4. Fuzzy fractional differential equation

In this section, we consider Eq (1) has an unique solution y(t) = G(t, k, c), for t ∈
(0, a), k ∈ Rn, c ∈ R i.e are given

• (0, c) is in (0, a)× I where I be an interval for the y-values.

• F is continuous in (0, a)× I(k is fixed ) and

• ∂F
∂y is continuous in (0, a)× I

Let K̄ =
(
K̄1, . . . , K̄n

)
be a vector of triangular fuzzy numbers and let C̄ be another

triangular fuzzy number. Substitute K̄ for k and C̄ for c in Eq (1) and we get the fuzzy
fractional differential equation

Y q(t) = F̄ (t, Ȳ , K̄), q ∈ (0, 1] (4)
Ȳ (0) = C̄



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 564

4.1. Buckley-Feuring solution

The Buckley-Feuring solution, written BF-solution, to the fuzzy fractional dif- ferential
equation [? ], we first fuzzify the crisp solution y(t) = g(t, k, c) to obtain Ȳ (t) = Ḡ(t, K̄, C̄)
using the extension principle. Alternatively, we get α-cuts as follows :

[Ȳ (t)]α = [yα1 (t), y
α
2 (t)] , (5)

[F̄ (t, Ȳ , K̄)]α = [fα
1 (t), f

α
2 (t)] , (6)

Let W = [K̄]α × [C̄]α. By definition

yα1 (t) = min{G(t, k, c) : (k, c) ∈ W}, (7)

yα2 (t) = max{G(t, k, c) : (k, c) ∈ W}, (8)

fα
1 (t) = min

{
F (t, y, k) : y ∈ [Ȳ (t)]α, k ∈ [K̄]α

}
, (9)

fα
2 (t) = max

{
F (t, y, k) : y ∈ [Ȳ (t)]α, k ∈ [K̄]α

}
, (10)

for t ∈ (0, a) and α ∈ [0, 1].
Let for Ȳ (t) to be a solution to the fuzzy fractional differential equation we need that
Ȳ (q)(t) exist but also (4) must hold. Assume that (y)αi (t) for all i = {1, 2}, is q-differentiable
(conformable derivative) with respect to t ∈ (0, a) for each α ∈ [0, 1] and q ∈ (0, 1].(

y(q)
)α

i
(t) = fα

i (t) (11)

Or (
y(q)

)α

1
(t) = fα

1 (t) (12)(
y(q)

)α

2
(t) = fα

2 (t) (13)

yα1 (0) = cα1 (14)

yα2 (0) = cα2 (15)

where [C̄]α = [cα1 , c
α
2 ] .

We write the partial of yαi (t), i = 1, 2 with respect to t as
(
y(q)

)α
i
, i = 1, 2 and q ∈ (0, 1].

Let
Γ(t, α) =

[(
y(q)

)α

1
(t),

(
y(q)

)α

2
(t)

]
(16)

for t ∈ (0, a), α ∈ [0, 1] and for q ∈ (0, 1].
If Γ(t, α) defines the α-cuts of a fuzzy number for each t ∈ (0, a) we will say that Ȳ , is
q-differentiable and write[

Ȳ (q)(t)
]α

= Γ(t, α) =
[(

y(q)
)α

1
(t),

(
y(q)

)α

2
(t)

]
(17)

for all t ∈ (0, a), α ∈ [0, 1] and for q ∈ (0, 1]. Notice, that Eq (17) is just the conformable
derivative (with respect to t) of Eq (6) . So, Eq (17) could be written

[
Ȳ (q)(t)

]α
.

Sufficient conditions for Γ(t, α) to define the α-cuts of a fuzzy number are: (see [6, 9, 11])



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 565

(i)
(
y(q)

)α
1

and
(
y(q)

)α
2

are continuous on (0, a)× [0, 1] and for q ∈ (0, 1]

(ii)
(
y(q)

)α
1

is an increasing function of α for each t ∈ (0, a) and for q ∈ (0, 1]

(iii)
(
y(q)

)α
2

is a decreasing function of α for each t ∈ (0, a) and for q ∈ (0, 1]

(iv)
(
y(q)

)α
1
≤

(
y(q)

)α
2

all t ∈ (0, a) and q ∈ (0, 1]

Hence, if conditions (i)− (iv) above hold, Ȳ (t) is q-differentiable.
Ȳ (q)(t), ∀q ∈ (0, 1] will be a solution to Eq (4) if,

(a) Ȳ (q)(t), ∀q ∈ (0, 1] is q-differentiable;

(b) (4) holds for Ȳ (t) = Ḡ(t, K̄, C̄);

(c) Ȳ (q)(t), ∀q ∈ (0, 1] satisfies the initial and boundary conditions.

Since there is no specified particular initial and boundary conditions, then only is checked
if (4) holds.

We will say that Ȳ (t) is a solution (without the initial and boundary conditions) if
Ȳ (q)(t) exists and

Ȳ (q)(t) = F̄ (t), ∀q ∈ (0, 1]

or the following equations must hold(
y(q)

)α

1
(t) = fα

1 (t), (18)(
y(q)

)α

2
(t) = fα

2 (t), (19)

for all t ∈ (0, a), q ∈ (0, 1] and all α ∈ [0, 1]. We have the following results regarding
BF-solution = Ȳ (t).

Theorem 6. Assume Ȳ (q)(t), for all q ∈ (0, 1] is q-differentiable for t ∈ (0, a). Then if

(a)
∂F

∂y
> 0,

∂G

∂c
> 0 (20)

and (
∂F

∂kj

)(
∂G

∂kj

)
> 0, j = 1, . . . , n (21)

Then Ȳ (t) is a BF-solution.

(b) If Eq (20) does not hold or Eq (21) does not hold for some j, then Ȳ (t) does not a
BF-solution.



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 566

Proof. Let us assume there is only one ki = k and that ∂G
∂k > 0, ∂F

∂k > 0, the proof
for ∂G

∂k < 0, ∂F
∂k < 0 is similar and omitted. Since ∂G

∂k > 0 and ∂G
∂c > 0 we have

yα1 (t) = G (t, kα1 , c
α
1 ) , (22)

yα2 (t) = G (t, kα2 , c
α
2 ) . (23)

Also, because ∂G
∂y > 0 and ∂F

∂k > 0 we see that

fα
1 (t) = G (t, yα1 (t), k

α
1 ) , (24)

fα
2 (t) = G (t, yα2 (t), k

α
2 ) . (25)

Now, y(t) = G(t, k, c) is unique solution to

y(q)(t) = F (t, y, k), for all q ∈ (0, 1]

y(0) = c

which implies that

G(q)(t) = F (t, G(t, k, c), k), for all q ∈ (0, 1] (26)

and
g(0, k, c) = c (27)

Assuming is q-differentiable we see that

(
y(q)

)α

1
(t) = G(q) (t, kα1 , c

α
1 ) , q ∈ (0, 1]

= F (t, G (t, kα1 , c
α
1 ) , k

α
1 )

= F (t, yα1 (t), k
α
1 )

= fα
1 (t)

and

yα1 (0) = G (0, kα1 , c
α
1 )

= cα1

and also (
y(q)

)α

2
(t) = G(q) (t, kα2 , c

α
2 ) , q ∈ (0, 1]

= F (t, G (t, kα2 , c
α
2 ) , k

α
1 )

= F (t, yα2 (t), k
α
2 )

= fα
2 (t)



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 567

and

yα2 (0) = G (0, kα2 , c
α
2 )

= cα2

for all α ∈ [0, 1] and t ∈ (0, a). Hence Eqs (12)-(15) hold.
Now consider the situation where Eq (20) or (21) does not hold. Let us only look at one
case where ∂G

∂c < 0 ( assume ∂G
∂k < 0, ∂F

∂y > 0 and ∂F
∂k < 0 ). Then we have

fα
1 (t) = F (t, yα1 (t), k

α
2 ) ,

fα
2 (t) = F (t, yα2 (t), k

α
1 ) ,

yα1 (t) = G (t, kα2 , c
α
2 )

and
yα2 (t) = G (t, kα1 , c

α
1 )

Eqs (12)-(15) becames (
y(q)

)α

1
(t) = G(q) (t, kα2 , c

α
2 ) , q ∈ (0, 1]

= F (t, G (t, kα2 , c
α
2 ) , k

α
2 )

= F (t, yα1 (t), k
α
2 )

= fα
1 (t)

and
yα1 (0) = G (0, kα2 , c

α
2 )

= cα2

which is not true.

4.2. Seikkala solution

In this section, we present situations where the BF-solution can, and cannot exist,
with these fuzzy fractional differential equations the main problem is determining where
the Seikkala solution exists ( when the BF-solution does not exist ).
The Seikkala solution, written SS, to the fuzzy fractional differential equations

Xq(t) = F̄ (t, X̄, K̄), q ∈ (0, 1] (28)
X̄(0) = C̄

Eq (28) is equivalent to Eq (4) substituting xi for yi, for X̄ ∈ RF with α-cut [X̄]α(t) =
[xα1 (t), x

α
2 (t)] , α ∈ [0, 1]. Since the fuzzy conformable fractional derivative X̄(q), q ∈ (0, 1]

of fuzzy process X̄ : (0, a) → RF is defined by[
X̄(q)(t)

]α
=

[(
x(q)

)α

1
(t),

(
x(q)

)α

2
(t)

]
, α ∈ [0, 1] and q ∈ (0, 1] (29)



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 568

We call X̄ : (0, a) → RF a fuzzy solution of (28) on (0, a), if(
x(q)

)α

1
(t) = min

{
F (t, y, k) : y ∈ [X̄(t)]α, k ∈ [K̄]α

}
(30)

xα1 (0) = cα1 (31)

(
x(q)

)α

2
(t) = max

{
F (t, y, k) : y ∈ [X̄(t)]α, k ∈ [K̄]α

}
(32)

xα2 (0) = cα2 (33)

for t ∈ (0, a), for all q ∈ (0, 1] and α ∈ [0, 1]. Thus for fixed α, we have an initial value
problem in R2.

If we can solve it (uniquely), we have only to verify that the intervals [xα1 (t), xα2 (t)] , α ∈
[0, 1], define a fuzzy number X̄ ∈ RF

Theorem 7. Let F satisfy

|F (t, x)− F (t, x̃)| ≤ h(t, |x− x̃|), t ≥ 0, x, x̃ ∈ R (34)

where h : R+ ×R+ → R+ is a continuous mapping such that r → h(t, r) is nondecreasing,
the initial value problem

y(q)(t) = h(t, y(t)), y(0) = y0 (35)

has a solution on R+ for y0 > 0 and that y(t) = 0 is the only solution of (35) for y0 = 0.
Then the initial value problem (28) has a unique fuzzy solution.

Proof. Denote F̃ = (F1, F2) , F1(t, x) = min {F (t, y) : y ∈ [x1, x2]} and F2(t, x) =
max {F (t, y) : y ∈ [x1, x2]} where x = (x1, x2) ∈ R2. It can be shown that (26) implies

‖F̃ (t, x)− F̃ (t, x̃)‖ ≤ h(t, ‖x− x̃‖), t ≥ 0, x, x̃ ∈ R2 (36)

where the norm ‖.‖ is defined by ‖x‖ = max {|x1| , |x2|} . It is well known that (36) and
the assumptions on h gurantee tha existence, uniqueness and continuous dependence on
initial value of a solution to

x(q)(t) = F̃ (t, x(t)), x(0) = x0 ∈ R2 (37)

and that for any continuous function x1 : R+ −→ R2 the successive approximations t ≥
0, n = 1, 2, . . .

xn+1(t) = x0 + Iq(F̃ ) (t, xn(t))

= x0 + I
(
t1−qF̃

)
(t, xn(t))

= x0 +

∫ t

0

F̃

x1−q
(s, xn(s)) ds (38)



A. Harir, S. Melliani, L. S. Chadli / Eur. J. Pure Appl. Math, 15 (2) (2022), 557-571 569

converge uniformly on closed subintervals of R+ to the solution of (37) . By choosing
x0 = (cα1 , c

α
2 ) in (37) we get a unique solution xα(t) = (xα1 (t), x

α
2 (t)) to (30),(32) for each

α ∈ [0, 1].
Next we will show that the intervals [xα1 (t), xα2 (t)] , α ∈ [0, 1], define a fuzzy number X̄ ∈ RF
for eatch t ≥ 0, i.e that X̄ is a fuzzy solution to (28).
The successive approximations x1(t) = x0 ∈ RF

xn+1(t) = x0 +

∫ t

0

F

x1−q
(s, xn(s)) ds, t ≥ 0, n = 1, 2, . . .

where the integral is the fuzzy integral, define a sequence of fuzzy numbers xn(t) ∈ RF , ∀t ≥
0 Hence

[xn(t)]
α ⊃ [xn(t)]

β if 0 < α ≤ β ≤ 1

which implies that

[xα1 (t), x
α
2 (t)] ⊃

[
xβ1 (t), x

β
2 (t)

]
, 0 < α ≤ β ≤ 1

since, by the convergence of sequence (38), the end points of [xn(t)]α converge to xα1 (t)
and xα2 (t) respectively. Thus the inclusion (ii) of Theorem (1) holds for the intervals
[xα1 (t), x

α
2 (t)] , α ∈ [0, 1]. For the proof of the continuity Theorem (1) by (iii), let (αk) be

a increasing sequence in [0, 1] converging to α. Then cαk
1 → cα1 and cαk

2 → cα2 because
X̄(0) ∈ RF . But then, by the continous dependence on the initial value of the solution of
(37), xαk

1 → xα1 and xαk
2 → xα2 i.e (iii) holds for the intervals [xα1 (t), x

α
2 (t)] , α ∈ [0, 1]. By

Theorem (1), X̄ ∈ RF , so X̄ is a fuzzy solution of (28) The uniqueness follows from the
uniqueness of the solution of (37).

5. Applications

Now we will solve fuzzy fractional differential equations according to our theorems and
definitions.
Let

y(q)(t) = ky(t), q ∈ (0, 1]

y(0) = c

so that the solution is given by y(t) = cexp
(
k
q t

q
)

. Now we fuzzify F (t, y, k) = ky(t) and

G(t, k, c) = cexp
(
k
q t

q
)

. Clearly let

F̄ (t, Ȳ , K̄) = KY (t), q ∈ (0, 1]

Ȳ (0) = C̄

so that fα
1 (t) = kα1 y

α
1 (t), fα

2 (t) = kα2 y
α
2 (t).

Also Ḡ(t, Ȳ , K̄) = Cexp
(
K̄
q t

q
)
, therefore

yαi (t) = cαi exp

(
kαi
q
tq
)



REFERENCES 570

for i = 1, 2 and q ∈ (0, 1], [K̄]α = [kα1 , k
α
2 ] and [C̄]α = [cα1 , c

α
2 ] , Ȳ is q-differentiable because

(yαi )
(q) (t),∀q ∈ (0, 1], for i = 1, 2 are α-cuts of KY (t) i.e α-cuts of a fuzzy number. Due

to ∂G
∂k > 0, ∂G∂c > 0, ∂F∂k = y > 0 for all t, ∂F∂y = k > 0. So Theorem (6) implies the result

that Ȳ (t) is a BF-solution. we easily see that yαi (0) = cαi for i = 1, 2, so Ȳ (t) also satisfies
the initial condition. The BF-solution may be written as

Ȳ (t) = Cexp

(
K̄

q
tq
)

for all t ∈ (0, a).
So if ∂F

∂y = k < 0, we look for a SS. The function F (t, y, k) = ky and k < 0 satisfies the
assumptions of Theorem (7) with h(t, y) = y and hance the problem

X̄(q)(t) = kX̄(t), q ∈ (0, 1]

X̄(0) = C̄

i.e (
x(q)

)α

1
(t) = kα1 x

α
2 (t)(

x(q)
)α

2
(t) = kα2 x

α
1 (t)

xα1 (0) = cα1

xα2 (0) = cα2

has a unique fuzzy solution x on R+. It is given by the α-cuts [xα1 (t), xα2 (t)] , α ∈ [0, 1], t ∈
(0, a), where

xα1 (t) =
1
2

(
cα1 +

√
kα1
kα2

cα2

)
exp

(√
kα1 k

α
2
tq

q

)
+ 1

2

(
cα1 −

√
kα1
kα2

cα2

)
exp

(
−
√

kα1 k
α
2
tq

q

)
xα2 (t) =

1
2

(
cα1

√
kα2
kα1

+ cα2

)
exp

(√
kα1 k

α
2
tq

q

)
− 1

2

(
cα1

√
kα2
kα1

− cα2

)
exp

(
−
√

kα1 k
α
2
tq

q

)
Then SS exist for t ∈ (0, a).

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