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Nonlinear Differential Problem with p-Laplacian and via Phi-Hilfer Approach: Solvability and

Stability Analysis

Hamid Beddani1,∗, Moustafa Beddani2, Zoubir Dahmani3
1Laboratory of Complex Systems of the Higher School of Electrical and Energy Engineering of Oran,

31000, Algeria
beddanihamid@gmail.com

2Department of Mathematics, University of Sidi Bel-Abbès 22000, Algeria
beddani2004@yahoo.fr

3Laboratory of Pure and Applied Mathematics, Abdelhamid Bni Badis University, 27000, Algeria
zzdahmani@yahoo.fr

∗Correspondence: beddanihamid@gmail.com

Abstract. This paper we consider a study of a general class of nonlinear singular fractional DEs withp-Laplacian for the existence and uniqueness solution and the Hyers-Ulam (HU) stability. result via
ϕ−Hilfer derivative is studied. Then, an existence of one solution is investigated. Some illustrativeexamples are discussed at the end.

1. Introduction
Recently, fractional differential equations with boundary conditions are being studied by manyinterested people. This is because fractional differential equations describe many more real op-erations than classical differential equations. Therefore, partial differential equations appearin many engineering and technological disciplines that include several sciences; See for exam-ple [1, 3–6, 8, 17, 18, 20, 22, 23, 31].
Currently there are several different definitions of fractional integrals and derivatives, from themost famous of which are the Riemann-Liouville and Caputo fractional derivatives to other less wellknown definitions. A generalization of the derivatives of both Riemann-Liouville and Caputo wasgiven by R. Hilfer in [11], known as the fractional Hilfer derivative of order α and type β ∈ [0, 1].Some properties and applications of the Helfer derivative are given in [12, 13] and the referencesmentioned therein. Prime value problems involving fractional Hilfer derivatives have been studiedby several authors, see [9, 10, 26]. However, in the literature there are few papers on the boundary
Received: 8 Oct 2021.
Key words and phrases. ϕ−Hilfer derivative; existence of solution; fixed point; Hyers-Ulam stability.164

https://adac.ee
https://doi.org/10.28924/ada/ma.1.164


Eur. J. Math. Anal. 1 (2021) 165
value problems of the fractional Hilfer derivatives. The authors set out in [2] non-local value prob-lems for derivatives of Helfer’s fractions. For some recent work on boundary value problems withfractional Hilfer derivatives, we refer to the papers in [28–30].Some authors have worked on the EU of solutions for fractional DEs with p−Laplacian operator.We cite, for example; Li., Wang., Khan et al. [15, 19, 27] studieds a nonlinear fractional DE withp-Laplacian operator for the EU of solutions.

H. Khan, T. Abdeljawad, M. Aslam, R. A. Khan and A. Khan [16]. worked on the followingproposal for the existence of a positive solution (EPS) and stability analysis:


Dr1ψp [Dr2 (u(t) −v1(t,u(t)))] = −A(t)v2(t,u(t −τ)),
ψp [Dr2 (u(t) −v1(t,u(t)))]|t=0 = ψp

[
Dr2 (u(t) −v1(t,u(t)))′

]∣∣
t=0

= 0,

u(0) = u(1) = 0,[
I2−r2 (u(t) −v1(t,u(t)))

]∣∣
t=0

= 0,

where 0 < r1 < 1 < r2 < 2, and v1,v2 are continuous but singular at some points. Thefractional derivatives Dr1 and Dr2 are taken in the Caputo sense and in the Riemann–Liouvillesense, respectively, and ψp(z) = |z|p−2 z denotes the p−Laplacian operator and satisfies 1p + 1q =
1, (ψp)

−1
= ψq.A. Devi, A. Kumar, D. Baleanu and A. Khan [7]. worked on the EU and HU stability results, fornonliner FDEs involving Caputo fractional derivatives of distinct orders with ψp Laplacian operator:



cDr1ψp
[
cDr2

(
u(t) −

∑m
i=1 vi (t)

)]
= −w(t,u(t)),t ∈ (0, 1]

ψp
[
cDr2

(
u(t) −

∑m
i=1 vi (t)

)]∣∣
t=0

= 0,

u(0) =
∑m
i=1 vi (0),

u′(1) =
∑m
i=1 v

′
i (1),

uj(0) =
∑m
i=1 v

j
i
(0), for j = 2, 3, ...,n− 1,

where 0 < r1 ≤ 1,n − 1 < r2 ≤ n,n ≥ 4, and vi,w are continuous functions. cDr1 and
cDr2 denotes the derivative of fractional order r1 and r2 in Caputo’s sense, respectively, and
ψp(z) = |z|p−2 z denotes the p−Laplacian operator and satisfies 1p + 1q = 1, (ψp)−1 = ψq.
In the present research work, we study the existence and uniqueness of a solution (EPS) andstability analysis which includes the ϕ−Hilfer fractional-order of the form:



Eur. J. Math. Anal. 1 (2021) 166


HDα1,β1;ϕ
a+

ψp

(
HDα2,β2;ϕ

a+
u
)

(t) = h(t,u(t),RLDµ;ϕ
a+
u(t)), t ∈ J = (a,b]

u(a) = 0,u(b) =
n∑
i=1

λiu (ζi ) ,

ψp

(
HDα2,β2;ϕ

a+
u
)

(a) = 0,and ψp (HDα2,β2;ϕa+ u(b)) = Iρ;ϕa+ u (ζ) , a < ζ,ζi < b,
(1.1)

Here, we take HDα1,β;ϕ
0+

,H Dα2,β;ϕ
0+

, are the ϕ−Hilfer fractional derivative of orders α1,α2, 1 <
α1,α2 < 2 and β1,β2 two parameters 0 ≤ β1,β2 ≤ 1, RLDκ;ϕa+ the ϕ-Riemann-Liouville fractionalderivative of order µ where µ < α2, and Iρ;ϕ0+ the left-sided ϕ−Riemann Liouville fractional integralof order ρ, where ρ > 0, and ψp(z) = |z|p−2 z denotes the p−Laplacian operator and satisfies
1
p

+ 1
q

= 1, (ψp)
−1

= ψq, and ϕ : J → R be an increasing function such that ϕ′(t) 6= 0, for all
t ∈ J, and f : J ×R×R→R, is given function will be "well defined" later.

2. Phi-Hilfer Derivatives Calculus
In this section, we introduce some notations and definitions of Phi-Hilfer Derivatives Calculusand present preliminary results needed in our proofs later, for details, see [17, 24, 25].Let ϕ : [a,b] → R be an increasing function with ϕ′(t) 6= 0, for all t ∈ J, and let C([a,b] ,R) bethe Banach space.
For all υ > −1 and s,t ∈ [0,∞), (t ≥ s), we pose ϕυ(t,s) = (ϕ(t) −ϕ(s))υ.

Definition 1. Let (a,b), (−∞≤ a < b ≤∞) be a finite or infinite interval of the half-axis (0,∞)
and α > 0. In addition, let ϕ(t) be a positive increasing function on (a,b], which has a continuous
derivative ϕ′(t) on (a,b). The ϕ−Riemann–Liouville fractional integral of a function u with respect
to another function ϕ on [a,b] is defined by

Iα;ϕ
a+
u(t) =

1

Γ(α)

t∫
a

ϕ′(s)ϕα−1(t,s)u(s)ds, (2.1)
where Γ (.) is the Gamma function.

Definition 2. Let n ∈ N and let ϕ,u ∈ Cn (J) be two functions such that ϕ is increasing and
ϕ′(t) 6= 0, for all t ∈ (a,b]. The left-sided ϕ−Riemann Liouville fractional derivative of a function
u of order α is defined by

Dα;ϕ
a+
u(t) =

(
1

ϕ′(t)

d

dt

)n
In−α;ϕ
a+

u(t)

=
1

Γ(n−α)

(
1

ϕ′(t)

d

dt

)n t∫
a

ϕ′(s)ϕn−α−1(t,s)u(s)ds,



Eur. J. Math. Anal. 1 (2021) 167
where n = [α] + 1, [α] represents the integer part of the real number α.

Definition 3. Let n− 1 < α < n with n ∈N, [a,b] is the interval such that −∞≤ a < b ≤∞ and
ϕ,u ∈ Cn ([a,b] ,R) two functions such that ϕ is increasing and ϕ′(t) 6= 0, for all t ∈ [a,b]. The
ϕ-Hilfer fractional derivative of a function u of order a and type 0 ≤ β ≤ 1 is defined by

HDα,β;ϕ
a+

u(t) = Iβ(n−α);ϕ
a+

(
1

ϕ′(t)

d

dt

)n
I(1−β)(n−α);ϕ
a+

u(t) = Iγ−α;ϕ
a+

Dγ;ϕ
a+
u(t),

where n = [α] + 1, γ −α = β (n−α) .

2.1. Auxiliary Lemma.
Lemma 1. Let α,ρ > 0. Then, we have the following semigroup property given by

Iα;ϕ
a+
Iρ;ϕ
a+
u(t) = Iα+ρ;ϕ

a+
u(t), t > a.

Next, we present the ϕ-fractional integral and derivatives of a power function.

Proposition 1. Let α ≥ 0,σ > 0 and t > a. Then, ϕ-fractional integral and derivative of a power
function are given by

(1) Iα,ϕ
a+
ϕσ−1(t,a)(t) =

Γ(σ)
Γ(α+σ)

ϕσ+α−1(t,a).(2) HDα,β;ϕ
a+

ϕσ−1(t,a)(t) =
Γ(σ)

Γ(σ−α)ϕσ−α−1(t,a),n− 1 < α < n,σ > n.

Lemma 2. If u ∈ Cn([a,b],R),n− 1 < α < n, 0 ≤ β ≤ 1 and γ = α + β(n−α). Then

Iα,ϕ
a+

(HDα,β;ϕ
a+

u)(t) = u(t) −
k=n∑
k=1

ϕγ−k(t,s)

Γ(γ −k + 1)
∇[n−k]ϕ I

(1−β)(n−α);ϕ
a+

u(a), t ∈ [a,b],

where ∇[n]ϕ u(t) :=
(

1
ψ′(t)

d
dt

)n
u(t).

Lemma 3. Let u ∈ Cn [a,b] and 0 < q < 1, we have∣∣Iq;ϕ
a+
u(t2) −I

q;ϕ
a+
u(t1)

∣∣ ≤ 2‖u‖
Γ (q + 1)

ϕq(t2,t1).

Lemma 4. ( [14]) For the p−Laplacian operator ψp, the following conditions hold true:

(1) If |δ1| , |δ2| ≥ ρ > 0, 1 < p ≤ 2,δ1δ2 > 0, then
|ψp(δ1) −ψp(δ2)| ≤ (p− 1) ρp−2 |δ1 −δ2| .

(2) If p > 2, |δ1| , |δ2| ≤ ρ∗ > 0, then
|ψp(δ1) −ψp(δ2)| ≤ (p− 1) ρp−2∗ |δ1 −δ2| .

Lemma 5. [9] For nonnegative ai, i = 1, ...,k,(
k∑
i=1

ai

)q
≤ kq−1

(
k∑
i=1

a
q
i

)
,q ≥ 1.



Eur. J. Math. Anal. 1 (2021) 168
Lemma 6. Let a ≥ 0, 1 < α1,α2 < 2, 0 ≤ β1,β2 ≤ 1, and 2 −γ1 = (1 −β1) (2 −α1) , 2 −γ2 =
(1 −β2) (2 −α2) . For f ∈ C(J,,R,R), the unique solution of the sequential Hilfer fractional
boundary value problem

HDα1,β1;ϕ
a+

ψp

(
HDα2,β2;ϕ

a+
u
)

(t) = f (t), t ∈ J = [a,b] , (2.2)


u(a) = 0,u(b) =
n∑
i=1

λiu (ζi ) ,

ψp

(
HDα2,β2;ϕ

a+
u
)

(a) = 0,

and ψp
(
HDα2,β2;ϕ

a+
u(b)

)
= Iρ;ϕ

a+
u (ζ) , a < ζ,ζi < b,

(2.3)
is given by

u(t) =
1

Γ(α2)

t∫
a

ϕ′(s)ϕα2−1(t,s)X(s,a)ds

−
ϕγ2−1 (t,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)X(t,a)dt

+
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi ) .

where

X(s,a) = ψq

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)f (z)dz +

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)


Iρ;ϕ

0+
u (ζ) =

1

Γ(ρ)

ζ∫
a

ϕ′(s)ϕρ(ζ,s)u (s) ds,

Iα1;ϕ
0+

f (b) =
1

Γ(α1)

b∫
a

ϕ′(s)ϕα1−1(b,s)f (s)ds.

Proof. Assume that u is a solution of the sequential nonlocal boundary value Problems (3.6) and(2.3). Applying the two operators Iα1;ϕ
a+

, Iα2;ϕ
a+

to both sides of Equation (3.6) and using Lemma 2and Proposition 1, we obtain
ψp

(
HDα2,β2;ϕ

a+
u
)

(t) = Iα1;ϕ
a+

f (t) +
m0

Γ (γ1 − 1)
ϕγ1−2 (t,a) +

m1
Γ (γ1)

ϕγ1−1 (t,a) , (2.4)
where m0,m1 ∈R, and 2 −γ1 = (1 −β1) (2 −α1) . From the boundary condition
ψp

(
HDα2,β2;ϕ

a+
u
)

(a) = 0, and if t → a then ϕγ1−2 (t,a) →∞, we get
m0 = 0.



Eur. J. Math. Anal. 1 (2021) 169
and by ψp (HDα2,β2;ϕa+ u) (b) = Iρ;ϕa+ u (ζ) , we obtain

m1 =
Γ (γ1)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

)
.

So
HDα2,β2;ϕ

a+
u(t) = ψq

(
Iα1;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

))
,

by (2.4)we have
u(t) = Iα2;ϕ

a+

[
ψq

(
Iα1;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

))]
+

m2
Γ (γ2 − 1)

ϕγ2−2 (t,a) +
m3

Γ (γ2)
ϕγ2−1 (t,a) ,

where m2,m3 ∈R, and 2 − (1 −β2) (2 −α2) = γ2.and if t → a then ϕγ2−2 (t,a) →∞, we getBy conditions u(a) = 0, and lim
t→0

tγ2−2 = ∞, we get
m2 = 0.

So
u(t) = Iα2;ϕ

a+

[
ψq

(
Iα1;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

))]
+

m3
Γ (γ2)

ϕγ2−1 (t,a) .

By conditions u(b) = n∑
i=1

λiu (ζi ) , we get
m3 =

Γ (γ2)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

−
Γ (γ2)

ϕγ2−1 (b,a)
Iα2;ϕ
a+

[
ψq

(
Iα1;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
f (b)

))]
t=b

.

Then
u(t) = Iα2 ;ϕ

a+

[
ψq

(
Iα1 ;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1 ;ϕ

0+
f (b)

))]
+
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

−
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)
Iα2 ;ϕ
a+

[
ψq

(
Iα1 ;ϕ
a+

f (t) +
ϕγ1−1 (t,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iα1 ;ϕ

0+
f (b)

))]
t=b

.

This finishes the proof. �



Eur. J. Math. Anal. 1 (2021) 170
Conjecture 1.

RLDµ;ϕ
a+
u(t) =

1

Γ(α2 −µ)

t∫
a

ϕ′(s)ϕα2−µ−1(t,s)X(s,a)ds

+
Γ (γ2)

Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

−
Γ (γ2)

Γ(α2)Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)X(t,a)dt.

3. Main Results
In this section, we present to the reader our main results on the existence and stability for theabove problem. We begin by considering the space

Cµϕ =
{
u : u,RLDµ;ϕ

a+
u ∈ C ([a,b] ,R)

}
,

with the norm
‖u‖Cµϕ = ‖u‖C +

∥∥RLDµ;ϕ
a+
u
∥∥
C
,

such that
‖u‖C = sup

t∈[a,b]
|u(t)| , and ∥∥RLDµ;ϕ

a+
u
∥∥
C

= sup
t∈[a,b]

∣∣RLDµ;ϕ
a+
u(t)

∣∣ .
3.1. Criteria For Uniqueness Solution. Now, wee need to consider the following assumptions:
H1) h is continuous function.
H2) There exists a constant Υ > 0, such that

|h(t,u,v) −h(t,x,y)| ≤ Υ (|u −x| + |v −y|) ,

with t ∈ [a,b] , (u,v,x,y) ∈R4.
H3) There exists two continuous functions π1,π2 : [a,b] →R+, such that

|h(t,u,v)| ≤ π1(t) |u(t)| + π2(t) |v(t)| ,

where
π∗1 = sup

t∈[a,b]
|π1(t)| , and π∗2 = sup

t∈[a,b]
|π2(t)| .

Now, we define the following quantities:



Eur. J. Math. Anal. 1 (2021) 171

ϕq(b,a) = M
q,

Ω = 3q−2

[(
2Mα1

Γ(α1 + 1)

)q−1 (
(π∗1 )

q−1
+ (π∗2 )

q−1
)

+

(
Mρ

Γ (ρ + 1)

)q−1]
Λ1 =

2.Ω.Mα2

Γ(α2 + 1)
,

Λ2 =

(
n∑
i=1

|λi|

)
,

Λ3 =
Ω.Mα2−µ

Γ(α2 −µ + 1)
+

Ω.Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)
,

Λ4 =
Γ (γ2) M

−µ

Γ (γ2 −µ)
Λ2.

Based on the above hypotheses, we present to the reader the following result.
Theorem 1. Under H2 and H3 the equation (1.1) has a solution.

Proof. Firstly: We begin this proof by defining the operator G : Cµϕ → Cµϕ by:
(Gu) (t) =

1

Γ(α2)

t∫
a

ϕ′(s)ϕα2−1(t,s)Xu(s,a)ds −
ϕγ2−1 (t,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

+
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi ) .

where
Xu(s,a) = ψq

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz +

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
hu(b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)

 ,
where

hu(t) = h(t,u(t),
RLDµ;ϕ

a+
u(t)).

We consider the set Ur = {u ∈ Cµϕ : ‖u‖Cµϕ ≤ r} , so that
max

{
(2 (Λ1 + Λ3))

1
2−q , 2 (Λ2 + Λ4)

}
≤ r.



Eur. J. Math. Anal. 1 (2021) 172
We show that GUr ⊂Ur . For any u ∈Ur , and by Lemma 5 we have
|Xu(s,a)|

=

∣∣∣∣∣∣
ψq

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz +

(
Iρ;ϕ

0+
u (ζ) −Iα1 ;ϕ

0+
hu(b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)


∣∣∣∣∣∣

≤ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α1)
s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz + I
ρ;ϕ
0+
u (ζ) + Iα1 ;ϕ

0+
hu(b)

∣∣∣∣∣∣
q−1

≤ 3q−2 sup
t∈[a,b]

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz

q−1 + (Iρ;ϕ
0+
u (ζ)

)q−1
+
(
Iα1 ;ϕ

0+
hu(b)

)q−1
≤ 3q−2

[
2

(
π∗1M

α1

Γ(α1 + 1)
‖u‖C +

π∗2M
α1

Γ(α1 + 1)

∥∥RLDµ;ϕ
a+
u
∥∥
C

)q−1
+

(
Mρ

Γ (ρ + 1)

)q−1
(‖u‖C)

q−1

]

≤ 3q−2
[(

2.π∗1M
α1

Γ(α1 + 1)

)q−1
(‖u‖C)

q−1
+

(
2.π∗2M

α1

Γ(α1 + 1)

)q−1 (∥∥RLDµ;ϕ
a+
u
∥∥
C

)q−1
+

(
Mρ

Γ (ρ + 1)

)q−1
(‖u‖C)

q−1

]

≤ 3q−2
[(

2Mα1

Γ(α1 + 1)

)q−1 (
(π∗1 )

q−1
+ (π∗2 )

q−1
)

+

(
Mρ

Γ (ρ + 1)

)q−1]
rq−1

≤ Ω.rq−1.

Then
sup
t∈[a,b]

|(Gu) (t)| (3.1)
≤ sup

t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2)
t∫
a

ϕ′(s)ϕα2−1(t,s)Xu(s,a)ds +
ϕγ2−1 (t,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

+
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

∣∣∣∣∣
≤

2Mα2

Γ(α2 + 1)
|Xu| +

(
n∑
i=1

|λi|

)
sup
t∈[a,b]

|u (t)|

≤
2.Ω.Mα2

Γ(α2 + 1)
rq +

(
n∑
i=1

|λi|

)
r.

≤ Λ1rq + Λ2r.

Also, we have
sup
t∈[a,b]

∣∣(RLDµ;ϕ
a+
Gu
)

(t)
∣∣ (3.2)

≤ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2 −µ)
t∫
a

ϕ′(s)ϕα2−µ−1(t,s)Xu(s,a)ds +
Γ (γ2)

Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )



Eur. J. Math. Anal. 1 (2021) 173
−

Γ (γ2)

Γ(α2)Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

∣∣∣∣∣∣
≤

[
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

]
|Xu| +

Γ (γ2) M
−µ

Γ (γ2 −µ)

(
n∑
i=1

|λi|

)
sup
t∈[a,b]

|u (t)|

≤
[

Ω.Mα2−µ

Γ(α2 −µ + 1)
+

Ω.Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

]
rq−1 +

Γ (γ2) M
−µ

Γ (γ2 −µ)

(
n∑
i=1

|λi|

)
r

≤ Λ3rq−1 + Λ4r.

By (3.1) and (3.2), we find
‖u‖Cµϕ = sup

t∈[a,b]
|(Gu) (t)|C + sup

t∈[a,b]

∣∣(RLDµ;ϕ
a+
Gu
)

(t)
∣∣
C

(3.3)
≤ (Λ1 + Λ3) rq−1 + (Λ2 + Λ4) r

≤ r.

that is GUr belongs to Ur on [a,b].
Next, we prove that G is completely continuous. For any u ∈ Ur and t1,t2 ∈ [a; b] such that

t1 < t2, by Lemma 3, we have
sup
t∈[a,b]

|(Gu) (t2) − (Gu) (t1)|

≤ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2)
t2∫
a

ϕ′(s)ϕα2−1(t2,s)Xu(s,a)ds −
1

Γ(α2)

t1∫
a

ϕ′(s)ϕα2−1(t1,s)Xu(s,a)ds

+
ϕγ2−1 (t2,a) −ϕγ2−1 (t1,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

+
ϕγ2−1 (t2,a) −ϕγ2−1 (t1,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

∣∣∣∣∣
≤

Ω.rq−1

Γ(α2 + 1)
ϕα2 (t2,t1) +

Ω.Mα2.rq−1 + Γ(α2 + 1)Λ2r

Γ(α2 + 1)ϕγ2−1 (b,a)
ϕγ2−1 (t2,t1) .

Hence,
sup
t∈[a,b]

|(Gu) (t2) − (Gu) (t1)|→ 0, as t2 → t1.



Eur. J. Math. Anal. 1 (2021) 174
Also, we can say that

sup
t∈[a,b]

∣∣(RLDµ;ϕ
a+
Gu
)

(t2) −
(
RLDµ;ϕ

a+
Gu
)

(t1)
∣∣

≤ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2 −µ)
t2∫
a

ϕ′(s)ϕα2−µ−1(t2,s)Xu(s,a)ds −
1

Γ(α2 −µ)

t1∫
a

ϕ′(s)ϕα2−µ−1(t1,s)Xu(s,a)ds

+
Γ (γ2)

Γ (γ2 −µ)
ϕγ2−µ−1 (t2,a) −ϕγ2−µ−1 (t1,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

+
Γ (γ2)

Γ(α2)Γ (γ2 −µ)
ϕγ2−µ−1 (t2,a) −ϕγ2−µ−1 (t1,a)

ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

∣∣∣∣∣∣
≤

Ω.rq−1

Γ(α2 −µ + 1)
ϕα2−µ(t2,t1)

+

(
Γ (γ2) Λ2r

Γ (γ2 −µ) ϕγ2−1 (b,a)
+

Γ (γ2) .Ω.r
q−1.Mα2

Γ(α2 + 1)Γ (γ2 −µ) ϕγ2−1 (b,a)

)
ϕγ2−µ−1 (t2,t1) .

Hence,
sup
t∈[a,b]

∣∣(RLDµ;ϕ
a+
Gu
)

(t2) −
(
RLDµ;ϕ

a+
Gu
)

(t1)
∣∣ → 0, as t2 → t1.

As a consequence of the above three steps and thanks to Arzela–Ascoli theorem, we conclude that
G is completely continuous.The proof of Theorem 1 is thus completely achieved. �
3.2. Criteria For Existence of a Solution.
Theorem 2. Assume that H2 and H3 are satisfied. Suppose that

Υ1 + Υ2 < 1,

where

Υ1 =
2 (q − 1) ∆q−2Mα2

Γ(α2 + 1)

(
4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
+ Λ2,

and

Υ2 = (q − 1) ∆q−2
(

4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
+

Γ (γ2) Λ2M
−µ

Γ (γ2 −µ)
.

Then, (1.1) has a uniqueness solution.



Eur. J. Math. Anal. 1 (2021) 175
Proof. We pass to prove that G is a contraction. For any u,v ∈Ur , we have the following estimate

|Xu(s,a) −Xv (s,a)|

=

∣∣∣∣∣∣ψq
 1

Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz +

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
hu(b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)


− ψq

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hv (z)dz +

(
Iρ;ϕ

0+
v (ζ) −Iα1;ϕ

0+
hv (b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)


∣∣∣∣∣∣

≤ (q − 1) yq−2
∣∣∣∣∣∣ 1Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hu(z)dz −
1

Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hv (z)dz

+
ϕγ1−1 (s,a)

ϕγ1−1 (b,a)

(
Iρ;ϕ

0+
u (ζ) −Iρ;ϕ

0+
v (ζ)

)
+
ϕγ1−1 (s,a)

ϕγ1−1 (b,a)

(
Iα1;ϕ

0+
hu(b) −I

α1;ϕ
0+

hv (b)
)∣∣∣∣

≤ (q − 1) yq−2
(

2Mα1

Γ(α1 + 1)
sup
t∈[a,b]

|hu(t) −hv (t)| +
Mρ

Γ (ρ + 1)
sup
t∈[a,b]

|u(t) −v(t)|

)

≤ (q − 1) yq−2
((

2ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
sup
t∈[a,b]

|u(t) −v(t)|

+
2ΥMα1

Γ(α1 + 1)
sup
t∈[a,b]

∣∣RLDµ;ϕ
a+
u(t) −RL Dµ;ϕ

a+
v(t)

∣∣)
≤ (q − 1) ∆q−2

(
4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
‖u −v‖Cµϕ .

where 
∆ > 2ΥM

α1

Γ(α1+1)
+ M

ρ

Γ(ρ+1)
, if q > 2,ou

0 < ∆ ≤ 2ΥM
α1

Γ(α1+1)
+ M

ρ

Γ(ρ+1)
, if 1 < q ≤ 2.

Then
sup
t∈[a,b]

|(Gu) (t) − (Gv) (t)| (3.4)
≤ sup

t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2)
t∫
a

ϕ′(s)ϕα2−1(t,s) (Xu −Xv ) (s,a)ds

∣∣∣∣∣∣
+ sup
t∈[a,b]

∣∣∣∣∣∣ ϕγ2−1 (t,a)Γ(α2)ϕγ2−1 (b,a)
b∫
a

ϕ′(t)ϕα2−1(b,t) (Xu −Xv ) (t,a)dt

∣∣∣∣∣∣
+ sup
t∈[a,b]

∣∣∣∣∣ϕγ2−1 (t,a)ϕγ2−1 (b,a)
n∑
i=1

λi (u (ζi ) −v(ζi ))

∣∣∣∣∣



Eur. J. Math. Anal. 1 (2021) 176
≤

2Mα2

Γ(α2 + 1)
sup
t∈[a,b]

|Xu(t,a) −Xv (t,a)| + Λ2 sup
t∈[a,b]

(|u (t) −v(t)|)

≤
(

2 (q − 1) ∆q−2Mα2
Γ(α2 + 1)

(
4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
+ Λ2

)
‖u −v‖Cµϕ

≤ Υ1 ‖u −v‖Cµϕ .Also
sup
t∈[a,b]

∣∣(RLDµ;ϕ
a+
Gu
)

(t) −
(
RLDµ;ϕ

a+
Gv
)

(t)
∣∣ (3.5)

≤ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2 −µ)
t∫
a

ϕ′(s)ϕα2−µ−1(t,s) (Xu −Xv ) (s,a)ds

+
Γ (γ2)

Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λi (u (ζi ) −v(ζi ))

+
Γ (γ2)

Γ(α2)Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t) (Xu −Xv ) (t,a)dt

∣∣∣∣∣∣
≤

(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
sup
t∈[a,b]

|Xu −Xv|

+
Γ (γ2) Λ2M

−µ

Γ (γ2 −µ)
sup
t∈[a,b]

(|u (t) −v(t)|)

≤
{

(q − 1) ∆q−2
(

4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
+

Γ (γ2) Λ2M
−µ

Γ (γ2 −µ)

}
‖u −v‖Cµϕ

≤ Υ2 ‖u −v‖Cµϕ .By (3.4) and (3.5), yields the following inequality
‖Gu −Gv‖Cµϕ ≤ (Υ1 + Υ2)‖u −v‖Cµϕ .Where Υ1 + Υ2 < 1. Hence G is a contraction operator and the contraction mapping principleimplies that (1.1) has a unique solution. �

3.3. Ulam Type Stability. We introduce the following two definitions
Definition 4. The problem (1.1) is Ulam–Hyers stable if ∃ λ ∈R∗+, such that for each ε > 0,t ∈ J,
and for each u ∈ Cµϕ solution of the following inequality∥∥∥HDα1,β1;ϕ

a+
ψp

(
HDα2,β2;ϕ

a+
u
)

(t) −h(t,u(t),RLDµ;ϕ
a+
u(t))

∥∥∥
C
µ
ϕ

< ε, (3.6)
∃v ∈ Cµϕ solution of (1.1), i.e.

HDα1,β1;ϕ
a+

ψp

(
HDα2,β2;ϕ

a+
v
)

(t) = h(t,v(t),RLDµ;ϕ
a+
v(t)), (3.7)



Eur. J. Math. Anal. 1 (2021) 177
such that, the inequality

‖u −v‖Cµϕ ≤ λε,

holds.

Definition 5. The equation (1.1) has the Ulam–Hyers stability in the generalized sense if ∃ ϕ ∈
C (J,R+), such that for each ε > 0,t ∈ J, and for each u ∈ Cµϕ solution of:∥∥∥HDα1,β1;ϕ

a+
ψp

(
HDα2,β2;ϕ

a+
u
)

(t) −h(t,u(t),RLDµ;ϕ
a+
u(t))

∥∥∥
C
µ
ϕ

< ε, (3.8)
∃v ∈ Cµϕ solution of (1.1) that satisfies

‖u(t) −v(t)‖Cµϕ ≤ εϕ(t).

In the light of the first definition and using the above existence and uniqueness theorem, wepresent to the reader the following result.
Theorem 3. If the assumptions (H2) are satisfied, then Eq (1.1) is Ulam–Hyers stable under the
condition that N1 + N2 < 1, where

N1 =
2 (q − 1) ∆q−2Mα2

Γ(α2 + 1)

(
4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
,

and

N2 =

(
4Υ (q − 1) ∆q−2Mα1

Γ(α1 + 1)
+

(q − 1) ∆q−2Mρ

Γ (ρ + 1)

)(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
.

Proof. Let u ∈ Cµϕ be a solution of the inequality (3.6), i.e.∥∥∥HDα1,β1;ϕ
a+

ψp

(
HDα2,β2;ϕ

a+
u
)

(t) −h(t,u(t),RLDµ;ϕ
a+
u(t))

∥∥∥
C
µ
ϕ

< ε, ∀t ∈ J. (3.9)
Let v ∈ Cµϕ be a unique solution of:

HDα1,β1;ϕ
a+

ψp

(
HDα2,β2;ϕ

a+
v
)

(t) = h(t,v(t),RLDµ;ϕ
a+
v(t)), ∀t ∈ J,

and 
u(a) = v(a), u(b) = v(b)and

ψp

(
HDα2,β2;ϕ

a+
u
)

(a) = ψp

(
HDα2,β2;ϕ

a+
v
)

(a),

ψp

(
HDα2,β2;ϕ

a+
u
)

(b) = ψp

(
HDα2,β2;ϕ

a+
v
)

(b),



Eur. J. Math. Anal. 1 (2021) 178
By using Proof of Lemma 6

v(t) =
1

Γ(α2)

t∫
a

ϕ′(s)ϕα2−1(t,s)Xv (s,a)ds

−
ϕγ2−1 (t,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xv (t,a)dt

+
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi ) ,

where
Xv (s,a) = ψq

 1
Γ(α1)

s∫
a

ϕ′(s)ϕα1−1(s,z)hv (z)dz +

(
Iρ;ϕ

0+
u (ζ) −Iα1;ϕ

0+
hv (b)

)
ϕγ1−1 (b,a)

ϕγ1−1 (s,a)

 .
By integration of inequality (3.9), for any t ∈ J, we have∥∥∥∥∥∥u(t) − 1Γ(α2)

t∫
a

ϕ′(s)ϕα2−1(t,s)Xu(s,a)ds (3.10)
+

ϕγ2−1 (t,a)

Γ(α2)ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t)Xu(t,a)dt

−
ϕγ2−1 (t,a)

ϕγ2−1 (b,a)

n∑
i=1

λiu (ζi )

∥∥∥∥∥
C

≤ Iα2;ϕ
a+

ψq
(
Iα1;ϕ
a+

ε
)

=
Mq−1ϕα1+α2 (t,a)

Γ (α1 + α2 + 1)
ε.

On the other hand, for any u,v ∈ Cµϕ, we have the following estimate
‖u(t) −v(t)‖C (3.11)

<
Mq−1ϕα1+α2 (t,a)

Γ (α1 + α2 + 1)
ε

+ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2)
t∫
a

ϕ′(s)ϕα2−1(t,s) (Xu −Xv ) (s,a)ds

∣∣∣∣∣∣
+ sup
t∈[a,b]

∣∣∣∣∣∣ ϕγ2−1 (t,a)Γ(α2)ϕγ2−1 (b,a)
b∫
a

ϕ′(t)ϕα2−1(b,t) (Xu −Xv ) (t,a)dt

∣∣∣∣∣∣
<

Mα1+α2+q1

Γ (α1 + α2 + 1)
ε +

2 (q − 1) ∆q−2Mα2
Γ(α2 + 1)

(
4ΥMα1

Γ(α1 + 1)
+

Mρ

Γ (ρ + 1)

)
‖u −v‖Cµϕ

<
Mα1+α2+q−1

Γ (α1 + α2 + 1)
ε + N1 ‖u −v‖Cµϕ .



Eur. J. Math. Anal. 1 (2021) 179
Also, for any t ∈ J, we have∥∥RLDµ;ϕ

a+
(u(t) −v(t))

∥∥
C

(3.12)
≤

Mq−1ϕα1 +α2−µ (t,a)

Γ (α1 + α2 −µ + 1)
ε

+ sup
t∈[a,b]

∣∣∣∣∣∣ 1Γ(α2 −µ)
t∫
a

ϕ′(s)ϕα2−µ−1(t,s) (Xu −Xv ) (s,a)ds

+
Γ (γ2)

Γ(α2)Γ (γ2 −µ)
ϕγ2−µ−1 (t,a)

ϕγ2−1 (b,a)

b∫
a

ϕ′(t)ϕα2−1(b,t) (Xu −Xv ) (t,a)dt

∣∣∣∣∣∣
≤

Mα1 +α2 +q−µ−1

Γ (α1 + α2 −µ + 1)
ε +

(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
sup
t∈[a,b]

|Xu −Xv|

≤
Mα1 +α2 +q−µ−1

Γ (α1 + α2 −µ + 1)
ε

+

(
4Υ (q − 1) ∆q−2Mα1

Γ(α1 + 1)
+

(q − 1) ∆q−2Mρ

Γ (ρ + 1)

)(
Mα2−µ

Γ(α2 −µ + 1)
+

Γ (γ2) M
α2−µ

Γ(α2 + 1)Γ (γ2 −µ)

)
‖u −v‖Cµϕ

≤
Mα1 +α2 +q−µ−1

Γ (α1 + α2 −µ + 1)
ε + N2 ‖u −v‖Cµϕ .

So, by (3.11) and (3.12) we have
‖u −v‖Cµϕ ≤ ε

(
Mα1+α2+q−1

Γ (α1 + α2 + 1)
+

Mα1+α2+q−µ−1

Γ (α1 + α2 −µ + 1)

)
+ (N1 + N2)‖u −v‖Cµϕ .

Therefore, we get
‖u −v‖Cµϕ ≤ λε,such that

λ =
1

1 − (N1 + N2)

(
Mα1+α2+q−1

Γ (α1 + α2 + 1)
+

Mα1+α2+q−µ−1

Γ (α1 + α2 −µ + 1)

)
,

for any t ∈ J. This implies that the Ulam-Hyers stability condition is satisfied. �
3.4. Illustrative exemple. Consider the following problem

HD
13
10
, 6

7
;t2

0+
ψp

((
HD

17
10
, 2

3
;t2

0+
u

))
(t) = h(t,u(t),RLD

1
2

;t2

a+
u(t)),t ∈ J = [0, 2] , (3.13)

u(a) = 0,u(2) =

n∑
i=1

(
3i

11

)
u

(
i

2 + i

)
,

ψp

((
HD

17
10
, 2

3
;t2

0+
u

))
(0) = 0,ψp

((
HD

17
10
, 2

3
;t2

0+
u

))
(2) = I

3
2

;t2

a+
u

(
4

3

)
f (t,u(t),v(t)) = exp

(
1

7 (1 + t2)

)
u(t) +

v(t)

(1 + et)
,



Eur. J. Math. Anal. 1 (2021) 180
then assumptions (H1), (H2) and (H3) are satisfied with

Υ = π∗2 =
1

2
,π∗1 = e

1
7 , and M = 4.

We conclude that (3.13) has an unique solution.
References

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	1. Introduction
	2. Phi-Hilfer Derivatives Calculus
	2.1. Auxiliary Lemma.

	3. Main Results
	3.1. Criteria For Uniqueness Solution.
	3.2. Criteria For Existence of a Solution.
	3.3. Ulam Type Stability.
	3.4. Illustrative exemple.

	References

