A singular ABC-fractional differential equation with p-Laplacian operator

Chaos, Solitons and Fractals 129 (2019) 56–61

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Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos

A singular ABC-fractional differential equation with p-Laplacian operator Hasib Khan a,b, Fahd Jarad c, Thabet Abdeljawad d,∗, Aziz Khan d a

College of Engineering Mechanics and Materials, Hohai University, Nanjing 211100, PR China Department of Mathematics, Shaheed Benazir Bhutto University, Khybar Pakhtunkhwa, Dir Upper 18000, Pakistan c Department of Mathematics, Çankaya University, Etimesgut, Ankara 06790, Turkey d Department of Mathematics and General Sciences, Prince Sultan University, P. O. Box 66833, Riyadh 11586, Saudi Arabia b

a r t i c l e

i n f o

Article history: Received 19 May 2019 Revised 3 August 2019 Accepted 14 August 2019

a b s t r a c t In this article, we have focused on the existence and uniqueness of solutions and Hyers–Ulam stability for ABC-fractional DEs with p-Laplacian operator involving spatial singularity. The existence and uniqueness of solutions are derived with the help of the well-known Guo-Krasnoselskii theorem. Our work is a continuation of the study carried out in the recently published article ” Chaos Solitons & Fractals. 2018;117:1620.” To manifest the results, we include an example with specific parameters and assumptions.

MSC: 34B82 26A33 45N05

© 2019 Elsevier Ltd. All rights reserved.

Keywords: ABC-fractional differential equation Existence of positive solution HU-stability Applications of the results

1. Introduction Among all areas of research in mathematics, one of the most rapidly growing area is the fractional calculus. One of the most important reasons of this growth are the results recorded by many researched when they benefited from the fractional operators for the sake of modelling real world problems [1–5]. Another interesting fact is that there are varieties of fractional operators with different kernels. This fact gives the researchers the opportunity to choose the operator which is the most appropriate for the model they investigate. Recently, ABC-fractional differential equations were studied in both the theoretical and applied aspects. In the theoretical aspect, existence and uniqueness of solutions to the ABC-fractional DEs are under development. To the best of our studies, no one considered the existence of positive solutions for the ABC-fractional DEs involving singularities. Till 2015, all of the fractional operators had singularities in the kernel they embody their kernels. These singularities are believed to be troublesome especially when these operators are applied to

∗

Corresponding author. E-mail addresses: [email protected] (F. Jarad), [email protected] (T. Abdeljawad). https://doi.org/10.1016/j.chaos.2019.08.017 0960-0779/© 2019 Elsevier Ltd. All rights reserved.

model some physical phenomena. To overcome this deficiency, Caputo et al. [6] suggested a new fractional derivative with nonsingular kernel involving an exponential function. Just one year later, Atangana et al. [7] suggested a fractional differential derivative including the Mittag-Leffler functions which generalize the exponential function. Then researchers have started to discover the theoretical properties of these operators and their discrete analogue. For more details about both continuous and discrete case we suggest the reader inspect the work in [8–18]. The existence of solutions and stability analysis have been considered as hot research topics in fractional calculus and a lot of new ideas and notions have been introduced [19–22]. In this paper, we are presenting an analytical study of fractional DEs including the existence of positive solutions and Hyer–Ulam stability. Here, we highlight some recent developments related to our work. Very recently, the authors in [12,13] introduced and studied an extension by using generalized Mittag-Leffler kernel. In fact, the author in [13] analyzed the use of this extension in gaining a semigroup property for the correspondent integral operator and in releasing the vanishing condition of the right hand side in the ABC− fractional dynamical system. Existence and uniqueness of solutions to boundary value problems in the frame work of the singular Riemann–Liouville

H. Khan, F. Jarad and T. Abdeljawad et al. / Chaos, Solitons and Fractals 129 (2019) 56–61

fractional derivative were analyzed in [23]. Using Schauder’s fixed point theorem and upper-lower solution techniques, existenceuniqueness analysis of fractional differential equations with nonlocal boundary values was executed in [24]. Existence and uniqueness of solutions to fractional differential equations with singular kernel and in the presence of p-Laplacian were considered in [25]. Meanwhile many researchers studied existence and uniqueness of positive solution to fractional differential equations with p-Laplacian operator. We refer the reader to [26] and the references therein. Some more interesting results for the existence and uniqueness of delay fractional differential equations were considered in [27,28]. Recently, Jarad et al. [29] have recently studied the following equation

ABC a

D ϑ0 x(t ) = (t , x(t )),

(1.1)

x ( a ) = x0 ,

ϑ0 is the ABC-fractional order differential operator where ABC a D ϑ0 and ϑ0 ∈ (0, 1), ABC a D x (t ), (t, x (t )) ∈ C[a, b]. To the best of our knowledge, no one has investigated existence and uniqueness of solution to ABC-fractional DEs with  p -operator with singularity. In order to overcome the deficiency in the study, we follow [29] for the EPS and HU-stability of a nonlinear ABC-fractional DE given by:

ABC 

0

D 0





C ϑ0 p [0 D x

C ϑ0 p [0 D x

(t )] = −(t, x(t )),

(t )]|t=0 = 0, x(1 ) = x(n) (0 ), x(k) (0 ) = 0 = x(n) (1 ),

where k = 1, 2, . . . , n − 1, 0 , ∈ (0, 1], ϑ0 ∈ (n, n + 1] and (t, x(t)) ∈ C[0,1] are continuous functions. The ABC D 0 is the ABC0 fractional differential operator of order  0 ∈ (0,1] and C0 D ϑ0 is the Caputo’s fractional order differential operators of order ϑ0 ∈ (n, n + 1]. The  p (r ) = |r| p−2 r is the p-Laplacian-operator such that 1/p + 1/q = 1 and  p−1 = q . By a positive solution x(t) of (1.2), we mean x(t) > 0 for t ∈ (0,1] satisfying (1.2). Our problem ABC-fractional DE with the operator  p (1.2), is more complicated and more general than (1.1). To investigate the existence uniqueness of solution to (1.2) and its stability in the sense of Hyers and Ulam stability, we will transform the problem into an equivalent AB-fractional integral form by using the classical results and AB I 0 , 0 I ϑ0 and a Green function. 0 The Green’s function will be tested for its nature. For an application, we present and analyze an important example. The readers may reconsider (1.2) for the multiplicity results. Definition 1.1 [7,30]. The fractional Atangane–Baleanu derivative in the Caputo setting of the function f ∈ H∗ (a, b), where b > a, η∗ ∈ [0, 1] is defined by ABC

η∗

a Dτ

where

f (τ ) =

B (η ∗ )

B (η 1−μ

a





f (ϑ0 )Eη∗

− s )μ

−η (τ 1 − η∗ ∗

ds,

(1.3)

B (η ∗ ) d η∗ a Dτ f ( τ ) = ∗

1 − η dτ

η∗

a Iτ

η∗

a Dτ



f ( τ ) = f ( τ ) − f ( a ).

(1.6)

The discrete version of (1.6) was announced in [10] as well, and the Caputo–Fabrizio analogous of both continuous and discrete cases were proved in [11]. Definition 1.5. The Riemann–Liouville fractional integral of a function ψ of order η∗ > 0, ψ : (0, +∞ ) → R, is given by 0I

η∗ ψ (t ) =



1

(η ) ∗

t 0

(t − s )η −1 ψ (s ) ds, ∗

where for Re(η∗ ) > 0 we have



(η∗ ) =

+∞

e−s sη

∗

−1

ds.

0

Definition 1.6. The Caputo fractional derivative of a continuous function ψ (t ) : (0, +∞ ) → R reads ϑ0 0 D ψ (t ) =

1 (k − ϑ0 )



t 0

(t − s )k−ϑ0 −1 ψ (k) (s )ds,

for k = [ϑ0 ] + 1, where [ϑ0 ] is the greatest integer less than ϑ0 .

η∗

0D

η∗ ψ (t ) =

ψ (t ) + m0 + m1t + m2t 2 + . . . + mn−1t n−1 ,

where mi , i = 0, 1, . . . , n − 1 are constant. To prove the existence of solution, we benefit from the following Guo–Krasnoselskii theorem. Theorem 1.2 [33,34]. Let Y a Banach space and P ∗ ∈ Y be a cone. Suppose that B1∗ , B2∗ are two bounded subsets of Y such that 0 ∈ B1∗ , B1∗ ⊂ B2∗ and let the operator J ∗ : P ∗ ∩ (B2∗ \B1∗ ) → P ∗ be continuous such that (A1 ) J ∗ z ≤ z if z ∈ P∗ ∩ ∂ B1∗ and J ∗ z ≥ z if z ∈ P∗ ∩ ∂ B2∗ , or (A2 ) J ∗ z ≥ z if z ∈ P∗ ∩ ∂ B1∗ and J ∗ z ≤ z if z ∈ P∗ ∩ ∂ B2∗ . Then J ∗ has a fixed point in P ∗ ∩ (B2∗ \B1∗ ). Lemma 1.3 [25]. Let  p be a nonlinear operator. Then (1) for 1 < p ≤ 2, η1∗ η2 > 0 and |η1∗ |, |η2∗ | ≥ ρ > 0, then

| p (η1∗ ) −  p (η2∗ )| ≤ ( p − 1 )ρ p−2 |η1∗ − η2∗ |. (2) If p > 2, and |η1∗ |, |η2∗ | ≤ ρ ∗ , then

| p (η1∗ ) −  p (η2∗ )| ≤ ( p − 1 )ρ ∗ p−2 |η1∗ − η2∗ |. 2. Green’s function and its properties

is satisfied the property B(0 ) = B(1 ) = 1.

 τ a



−η∗ (τ − s )η f ( ϑ0 )Eη 1 − η∗ ∗

∗



dϑ .

(1.4)

Definition 1.3 [31,32]. The fractional Atangane–Baleanu integral of the function f ∈ H∗ (a, b), b > a, 0 < η∗ < 1 is given by AB

η∗ ABC

a Iτ



Definition 1.2 [7]. The fractional Atangana–Baleanu derivative in Riemann–Liouville settingss of the function f ∈ H∗ (a, b), where b > a, η∗ ∈ [0,1] is described as follows: ABR

AB

0I

(1.2)

 ) τ

Lemma 1.4. [10] The ABC fractional derivative and AB fractional integral of the function f satisfies the Newton–Leibniz formula

Lemma 1.1. [4] For a fractional order η∗ ∈ (n − 1, n], ψ ∈ C n−1 , the following is satisfied



∗

57

f (τ ) =

1 − η∗ η∗ f (τ ) + B (η ∗ ) B(η∗ )(η∗ )

 τ a

f (ϑ0 )(τ − s )η

∗

−1

dϑ .

(1.5)

Theorem 2.1. For k = 1, 2, . . . , n − 1, 0 , ∈ (0, 1], ϑ0 ∈ (n, n + 1] and (t, x(t)) ∈ C[0, 1] such that (0, x(0 )) = 0 , x(t) is a solution of (1.2) if and only if

x(t ) =

1 0

Hϑ0 (t , s )q

AB 0





I 0 (t , x(t )) dt ,

(2.1)

where

⎧ (1 − s )ϑ0 −1 t n (1 − s )ϑ0 −n−1 (t − s )ϑ0 −1 ⎪ + − , s ≤ t, ⎨ (ϑ0 ) n! (ϑ0 − n ) (ϑ0 ) Hϑ0 (t, s ) = ϑ −1 n ϑ −n−1 ⎪ ⎩ (1 − s ) 0 + t (1 − s ) 0 , s ≥ t. (ϑ0 ) n! (ϑ0 − n ) (2.2)

58

H. Khan, F. Jarad and T. Abdeljawad et al. / Chaos, Solitons and Fractals 129 (2019) 56–61

Proof. If we apply the AB-fractional integral operator AB I 0 on 0 (1.2) and make use of Lemma 1.4, then problem (1.2) becomes as below

C



0  p 0 Dϑ0 x(t ) = −AB 0 I





(t, x(t )) + d0 .

(2.3)

Proof. In order to prove (A1 ), we consider two cases. Case 1: For s ≤ t, consider

(1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 (t − s )ϑ0 −1 + − (ϑ0 ) n! (ϑ0 − n ) (ϑ0 ) (1 − ts )ϑ0 −1 (1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 = + − t ϑ0 −1 (ϑ0 ) n! (ϑ0 − n ) (ϑ0 ) ϑ −1 n ϑ −1 0 0 (1 − s ) t (1 − s ) (1 − s )ϑ0 −1 ≥ t ϑ0 −1 + −t ϑ0 −1 (ϑ0 ) n! (ϑ0 − n ) (ϑ0 )

Hϑ0 (t, s ) =

The conditions  p [C 0 D ϑ0 x(t )]|t=0 = 0 and (0, x(0 )) = 0, force d0 = 0. This implies

    0 (t, x (t )) .  p C0 Dϑ0 x(t ) = −AB 0 I From (2.4), we further have C ϑ0 0D x

(t ) = −q



AB 0 0 I



(2.4)

≥ 0.



(t, x(t )) .

(2.5)

Performing 0 I ϑ0 on both sides (2.5) and using Lemma 1.1 give

x(t ) = − 0 I ϑ0 q



AB 0 0 I [

(t, x(t ))] + c0 + c1 t

+ . . . + cn−1 t n−1 + cn t n .

(2.6)

Using the condition x(k ) (0 ) = 0 for k = 1, 2, . . . , n − 1 implies that c1 = . . . = cn−1 = 0. Putting the values in (2.6), we have



x(t ) = − 0 I ϑ0 q



AB 0 0 I [

(t , x(t ))] + c0 + cn t n .

(2.7)

Case 2: For t ≤ s, we have

Hϑ0 (t, s ) =

 1 ϑ0 −n  AB I 0 (t, x (t )) | [ ] t=1 . q 0 0I n!

c0 =

0

I ϑ0 

AB

q 0

I 0

[(t, x(t ))] |t=1 .

(2.9)

(t, x(t ))] |t=1   tn ϑ0 −n  AB I 0 (t, x (t )) + [ ] |t=1 q 0 0I n!    0 − 0 I ϑ0 q AB 0 I [(t, x (t ))]  1  (1 − s )ϑ0 −1 t n 1 (1 − s )ϑ0 −n−1 = + (ϑ0 ) n! 0 (ϑ0 − n ) 0    t ϑ −n −1 (t − s ) 0 1 − ϑ0 − q [(s, x(s ))] (ϑ0 ) B ( ϑ0 ) 0 0I

ϑ0 



AB 0 q 0 I [

  τ ϑ0 + (τ − ξ )( , x( ))d ds B(ϑ0 )(ϑ0 − 1 ) 0   1 1 − ϑ0 = Hϑ0 (t, s )q [(s, x(s ))] B ( ϑ0 ) 0   τ ϑ0 + (τ − ξ )( , x( ))d ds. (2.10) B(ϑ0 )(ϑ0 − 1 ) 0 This accomplishes the required proof.

=

t n−1 (1 − s )ϑ0 −1 (t − s )ϑ0 −2 − (n − 1 )! (ϑ0 − n ) (ϑ0 − 1 )

≥

(1 − s )ϑ0 −2 t n−1 (1 − s )ϑ0 −1 − t ϑ0 −2 > 0. (2.13) (n − 1 )! (ϑ0 − n ) (ϑ0 − 1 )

Case 2: For t ≤ s, we get

(2.7) and (2.8), we have

x(t ) =

(2.12)

  ∂ ϑ0 ∂ (1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 (t − s )ϑ0 −1 H (t, s ) = + − ∂t ∂t (ϑ0 ) n! (ϑ0 − n ) (ϑ0 )

(2.8)

Finally, using the condition x(1 ) = x(n ) (0 ), we have

(1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 + > 0. (ϑ0 ) n! (ϑ0 − n )

With (2.11) and (2.12), it is proved that Hϑ0 (t, s ) > 0 for all 0 < s, t < 1. For (A2 ), we consider that: Case 1 when s ≤ t, we proceed

With the help of x(n ) (1 ) = 0, we have

cn =

(2.11)



Remark 2.1. From the proof of Theorem 2.1 above, to guarantee that d0 = 0, the conditions  p [C 0 D ϑ0 x(t )]|t=0 = 0 and (0, x(0 )) = 0 can be alternatively replaced by that  p [C 0 Dϑ0 x(t )]|t=0 = − B1(−0) (0, x(0 )). Also, the results through0

out this article can be easily carried over and interval [a, b] in place of [0,1]. Lemma 2.2. The function Hϑ0 (t, s ) defined by the Eq. (2.2), satisfies:

(A1 ) 0 < Hϑ0 (t, s ) for all s, t ∈ (0, 1); (A2 ) the function Hϑ0 (t, s ) is increasing and Hϑ0 (1, s ) = maxt∈[0,1] Hϑ0 (t, s ) and (A3 ) Hϑ0 (t, s ) ≥ t ϑ0 −1 maxt∈[0,1] Hϑ0 (t, s ) for s, t ∈ (0, 1).

∂ ϑ0 t n−1 (1 − s )ϑ0 −1 H (t, s ) = > 0. ∂t (n − 1 )! (ϑ0 − n )

(2.14)

Benefiting from (2.13) and (2.14), we conclude that ∂∂t Hϑ0 (t, s ) > 0

for s, t ∈ (0, 1). Therefore, Hϑ0 (t, s ) is increasing with respect to t. Thus, for t ≥ s, we get

max Hϑ0 (t, s ) =

t∈[0,1]

1 (1 − s )ϑ0 −1 = H ϑ0 ( 1, s ). n! (ϑ0 − n )

(2.15)

Similarly, for s ≥ t, we get

max Hϑ0 (t, s ) =

t∈[0,1]

(1 − s )ϑ0 −1 1 (1 − s )ϑ0 −1 + = Hϑ0 (1, s ). (2.16) (ϑ0 ) n! (ϑ0 − n )

For (A3 ), we presume two cases. Case 1: For t ≥ s, then

(1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 (t − s )ϑ0 −1 + − (ϑ0 ) n! (ϑ0 − n ) (ϑ0 ) ϑ −1 n ϑ −1 0 0 (1 − ts )ϑ0 −1 (1 − s ) t (1 − s ) = + − t ϑ0 −1 (ϑ0 ) n! (ϑ0 − n ) (ϑ0 ) ϑ0 −1 ϑ0 −1 (1 − s )ϑ0 −1 ( 1 − s ) t ≥ t ϑ0 −1 + (ϑ0 ) n! (ϑ0 − n ) s ϑ0 −1 (1 − t ) − t ϑ0 −1 (ϑ0 ) = t ϑ0 −1 max Hϑ0 (t, s ) = t ϑ0 −1 Hϑ0 (1, s ). (2.17)

Hϑ0 (t, s ) =

t∈[0,1]

Case 2: For s ≥ t, then

(1 − s )ϑ0 −1 t n (1 − s )ϑ0 −1 + (ϑ0 ) n! (ϑ0 − n ) ϑ −1 0 (1 − s ) t ϑ0 −1 (1 − s )ϑ0 −1 ≥ t ϑ0 −1 + (ϑ0 ) n! (ϑ0 − n ) = t ϑ0 −1 max Hϑ0 (t, s ) = t ϑ0 −1 Hϑ0 (1, s ).

Hϑ0 (t, s ) =

t∈[0,1]

By (2.17) and (2.18), proof of (A3 ) is completed.



(2.18)

H. Khan, F. Jarad and T. Abdeljawad et al. / Chaos, Solitons and Fractals 129 (2019) 56–61



3. Existence results

≤

Let Y be the Banach space C[0, 1] endowed by the norm x = maxt∈[0,1] {|x(t )| : x ∈ Y } and P be the cone containing non-negative functions in the space Y defined by P = {x ∈ Y : x(t ) ≥ t ϑ0 x , t ∈ [0, 1]}. Let W (r ) = {x ∈ P : x < r}, ∂ W (r ) = {x ∈ P : x = r}. With the help of Theorem 2.1, an alternate form of (1.2) is

x(t ) =

 0

1

Hϑ0 (t, s )q

AB 0



I 0 [(ξ , x(ξ ))] ds.

≤



1

0

Hϑ0 (t, s )q

(3.1)

≤

AB 0



I 0 [(ξ , x(ξ ))] ds.

(3.2)

Hence, the solution of (1.2), x(t), is nothing but the fixed point of J , i.e.,

x(t ) = J ∗ x(t ).

(3.3)

Assume the following conditions: (P1 ) y∗ : ((0, 1 ) × (0, +∞ )) → [0, +∞ ) is continuous and y1 ( ∗ t , x(t ) ≤ < +∞; (P2 ) A : (0, 1 ) → [0, +∞ ) is discontinuous on (0,1) and non vanishing and A = maxt∈[0,1] |A ∗ (t )| < +∞; (P3 ) For a constant value λ1 > 0 and u1 , u2 ∈ Y, the functions , y∗2 , fulfill

|(t, u1 (t )) − (t, u2 (t ))| ≤ λ1 |u1 (t ) − u2 (t )|.

J ∗ x(t ) =



0

 ≤ and

J ∗ x(t ) =

1

1

0

 0

1

Hϑ0 (t, s )q

AB 0

Hϑ0 (1, s )q

Hϑ0 (t, s )q

≥ t ϑ0 −1



1 0

2.2

AB 0

Eq.

(3.2),

for





Hϑ0 (1, s )q

AB 0



I 0 [(ξ , x(ξ ))] ds.

(3.5)

Making use of equations (3.4) and (3.5), we obtain that

J ∗ x(t ) ≥ t ϑ0 −1 J ∗ u , t ∈ [0, 1].

(3.6)

This implies J ∗ : W (r2 )\W (r1 ) → P. Now, for the sake of proving that J ∗ is continuous, we have to show that J ∗ (un ) − J ∗ (u ) → 0 as n → ∞.

1 1 n! (ϑ0 − n + 1 )



+



1

0



0 Hϑ0 (t2 , s )q (AB 0 I (ζ , x (ζ )) ds

   ϑ0  H (t1 , s ) − Hϑ0 (t2 , s ) 0 1 −  0 q (t, x(t )) B(0 )  1  0 + (τ − ζ )0 −1 (t, x(ζ )) dζ ds B(0 )(0 ) 0  t n (1 − s )ϑ0 −n t n (1 − s )ϑ0 −n ≤ 1 − 2 n! (ϑ0 − n + 1 ) n! (ϑ0 − n + 1 ) (t1 − s )ϑ0 (t2 − s )ϑ0  + − (ϑ0 + 1 ) (ϑ0 + 1 ) 1 −  

0 q

+ ( 1 − 0 ) 0 . B(0 ) B(0 )(0 ) 

≤

(3.4)

I 0 [(ξ , x(ξ ))] ds



(ϑ0 + 1 ) n! (ϑ0 − n + 1 ) 1 −  

0 q

+ ( 1 − 0 ) 0 < ∞ . B(0 ) B(0 )(0 )

−



I 0 [(ξ , x(ξ ))] ds,

+



I 0 (ξ , x(ξ )) ds

0

x∈

any

2

0

(3.8)

|J ∗ x(t1 ) − J ∗ x(t2 )|  1  0 = Hϑ0 (t1 , s )q (AB 0 I (ζ , x (ζ )) ds

I 0 [(ξ , x(ξ ))] ds

AB 0

and

0



AB

By (3.8), J ∗ : B ∗ (r2 )\B ∗ (r1 ) is uniformly bounded. To show the equicontinuity J ∗ , we make use of (P3 ), Theorems (2.1) and (3.2). In fact, ∀t1 , t2 ∈ [0, 1], we have

Theorem 3.1. If conditions (P1 ) − (P3 ) hold, J ∗ is a completely continuous operator. Proof. From both Lemma W (r2 )\W (r1 ), we have

|Hϑ0 (1, s )|q

(ϑ0 + 1 ) 1 −  0 × q (t, x(ζ )) B(0 )  τ  0 + (τ − ζ )0 −1 (t, x(ζ )) dζ ds B(0 )(0 ) 0   2 1 1

Define J ∗ : P \{0} → Y by

J ∗ x(t ) =

1

59

(3.9)

1

(3.10)

With the tendency of t1 → t2 , we the (3.9) reaches to zero. This implies J ∗ (B ∗ (r2 )\B ∗ (r1 )) is an equicontinuous operator and with the help of well-known Arzela–Ascoli theorem J ∗ (B ∗ (r2 )\B ∗ (r1 )) is compact. This further implies that J ∗ is compact in B ∗ (r2 )\B ∗ (r1 ). Consequently; J ∗ : B ∗ (r2 )\B ∗ (r1 ) → P is completely continuous. 



Here, we define height for y∗ (t, x(t)) for r > 0, and

max (t, r ) = maxt∈(0,1) {(t , x(t )) : t ϑ0 −1 r ≤ x ≤ r}, min (t, r ) = mint∈(0,1) {(t , x(t )) : t ϑ0 −1 r ≤ x ≤ r}.

(3.11)

    ϑ0  0 ≤ H (t, s )q AB 0 I [(ξ , xn (ξ ))] 0   0 − q AB (3.7) 0 I [(ξ , x (ξ ))] ds.

Theorem 3.2. Let (P1 ) − (P3 ) hold true and there exist E1 , E2 ∈ R+ such that

1 (C1∗ ) E1 ≤ 0 Hϑ0 (1, s )q (AB I 0 min (ζ , E1 ) ds < +∞ and 0

1 ϑ 0 (1, s )q (AB I 0 max (ζ , E ) ds ≤ E or H 2 0 0 2

0  (C2∗ ) ≤ 01 Hϑ0 (1, s )q (AB I ( ζ , E ) ds < E1 and E2 ≤ max 1 0

1 ϑ 0 (1, s )q (AB I 0  H ( ζ , E ) ds ≤ + ∞, is satisfied. Then 2 min 0 0 the ABC-fractional DE with operator  p (1.2) has a positive solution x ∈ P and E1 ≤ x ≤ E2 .

Because of (3.7) and the continuity of ξ ∗ , we have |J ∗ un (t ) − J ∗ x(t )| → 0 as n → +∞, which shows that J ∗ is continuous. Now, we move forward to prove the uniformly boundedness of J ∗ . Exploiting (3.2) and (1 ), we have

Proof. With no loss of generalization, consider the case (C1∗ ). If x ∈ ∂ B∗ (E1 ), then we have x = E1 and t ϑ0 −1 E1 ≤ x(t ) ≤ E1 , t ∈ [0, 1]. With the help of (3.11), for t ∈ (0, 1), we have  min (t, u) ≤ (t, x(t)), which implies

 1    0 |J ∗ xn (t ) − J ∗ x(t )| =  Hϑ0 (t, s )q AB 0 I [(ξ , xn (ξ ))] ds 0



− 

1 0

Hϑ0 (t, s )q

AB 0

1

 ∗   1     0 J x(t ) =  Hϑ0 (t, s )q AB (ξ , x(ξ )) ds 0 I 0

 

I 0 [(ξ , x(ξ ))] ds

J ∗ x(t ) = max



t∈[0,1] 0

1

Hϑ0 (t, s )q

1 − 

0

B(0 )

(t, x(t ))

60

H. Khan, F. Jarad and T. Abdeljawad et al. / Chaos, Solitons and Fractals 129 (2019) 56–61

 (τ − ζ )0 −1 (t, x(ζ )) dζ ds a  1 1 −  0 ϑ −1 ϑ ≥t 0 H 0 (1, s )q (t, x(t )) B(0 ) 0  τ  0 + (τ − ζ )0 −1 (t, x(ζ )) dζ ds B(0 )(0 ) 0  1 1 −  0 ≥ Hϑ0 (1, s )q min (t, E1 ) B(0 ) 0  τ  0 + (τ − ζ )0 −1 min (t, E1 )dζ ds B(0 )(0 ) 0 ≥ E1 = x . (3.12) ∗ ϑ −1 0 If x(t ) ∈ ∂ B (E2 ), then x = E2 and t E2 ≤ u ≤ E2 , for 0 ≤ t ≤ 1. +

0 B(0 )(0 )

 τ

 −



1

t∈[0,1] 0

Hϑ0 (t, s )q

0 + B(0 )(0 ) 

 τ 0

1 − 

0

B(0 )

a

 −

1



By Lemma 1.2, J ∗ has a fixed point say x ∈ B ∗ (1 )\B ∗ (0 ). Thus, we have 0 ≤ x∗ ≤ 1, which by Lemma 2.2 and Theorem 2.1 implies x∗ (t ) ≥ t ϑ0 −1 x∗ ≥ 0t ϑ0 −1 = 0, for t ∈ (0, 1). Thus x∗ is a positive solution.  4. Stability analysis We have reserved this section for the Hyers–Ulam stability of our suggested nonlinear singular ABC-fractional DE with  p operator (1.2). For this purpose, we follow our recently published articles [19,26] and some more related notions in the literature. Definition 4.1. We say that the fractional integral equations (3.1) is Hyers–Ulam stable if for every λ > 0, there is a constant D ∗ > 0, such that: if

 1   0 x(t ) −  Hϑ0 (t, s )q (AB 0 I (ζ , x (ζ )) ds ≤ λ,

(4.1)

0

b

 τ 0

|Hϑ0 (t, s )

q

a

0



1 0



0 Hϑ0 (t, s )q (AB 0 I (ζ , u (ζ )) ds,

(4.2)

implies

|x(t ) − u(t )| ≤ L∗ λ∗ .





(4.4) 0

1−



5. Illustrative example In this section, an application of the results which have proved in the Sections 3 and 4, is provided. Example 5.1. For t ∈ [0, 1], p = 3, q = 1.5, n = 4, 4.5 = ϑ0 ∈ √ (4, 5], b = 1, 0 = 0.5, a = 0.1, y∗1 (t, x(t )) = x + 13 , clearly x 35

 ∈ C ([0, 1] × (0, +∞ ), [0, +∞ ). Presume a singular ABC-fractional DE with  p -operator:



ABC

0D

0



 √  p [ 0 Dϑ0 x(t )] + t x +

1



3 x 35

= 0,

 p [C0 Dϑ0 x(t )]|t=0 = 0, x(1 ) = x(n) (0 ), x(k) (0 ) = 0 = x(n) (1 ), with t



x(t ) +



1 3

x(t ) 35

(5.1)

|t=0 = 0, k = 1, 2, 3. We consider:

√

max (t, η ) = max{t x + √

min (t, η ) = min{t x +

t

: t 3 η ≤ x ≤ η} ≤ 5

3

5x 35 t

η+

1 1 , 5 t 17 η 353

: t 3 η ≤ x ≤ η} ≥ t 6 η0.5 + 5

3

5x 35

√

5

1 5η 35 3

,

as height functions. Then for t ∈ (0, 1), we have



1

(4.3)

Hϑ0 (1, s )q

Theorem 4.1. The singular ABC-fractional DE with delay and  p operator, the problem (1.2) is Hyers–Ulam stable provided that (P1 ) − (P3 ) are satisfied.

AB

1



Hϑ0 (1, s )q

≤ 0.0562164 < 1, and



Ia0 max (ζ , η ) dζ ds

 1 −   1 1 0 η5 + 1 3 B(0 ) 5 t 7 η 35 0     τ 0 1 1 + (τ − ζ )0 −1 η5 + dζ ds B(0 )(0 ) a 5 t 17 η 353  1 1 −    1 1 0 ≤ Hϑ0 (1, s )q 1+ B(0 ) 5 s 17 0  s  0 1 1 + (s − ζ )0 −1 (1 + ) d ζ ds B(0 )(0 ) a 5 ζ 17 

=

In the following Theorem 4.1, we state and then prove the stability of our assumed problem (1.2).

0

(t, x(t ))

) 1 0 where L∗ = ( p − 1 )ρ p−2 (ϑ2 +1 ) + (ϑ −n + B((1−0 B (0 ) +1 ) 0 )(0 ) 0 0 ∗ λ . Therefore, the fractional integral operator (3.2) is Hyers–Ulam stable. Consequently, the singular ABC-fractional DE with nonlinear  p (1.2) is Hyers–Ulam stable. 

0

Proof. By Theorem 3.2, Definition 4.1, Let x(t) be a solution of the ABC-fractional DE with delay (3.1) and y(t) be an approximate solution and satisfying (4.2). Then, we have    1 ϑ 0 x(t ) − u(t ) =  H 0 (t, s )q (AB 0 I (ζ , x (ζ )) ds

B(0 )

  0  (τ − ζ )0 −1 (t, x(ζ )) dζ ds B(0 )(0 ) 0  b 1 − 0 ≤ ( p − 1 )ρ p−2 Hϑ0 (t, s ) (t, x(t )) − (t, u(t )) B(0 ) a  τ   0 + (τ − ζ )0 −1 (t, x(ζ )) − (t, u(t )) dζ ds B(0 )(0 ) 0   1 −  2 1 0 ≤ ( p − 1 )ρ p−2 + (ϑ0 + 1 ) (ϑ0 − n + 1 ) B(0 )  (1 − 0 )0 + λ∗ x − u , B(0 )(0 )

there is a u(t) satisfying

u(t ) =

0

 (τ − ζ )0 −1 (t, x(ζ )) dζ ds

1 −

 τ

+



1 − 0 Hϑ0 (1, s )q (t, x(t )) B(0 ) 0  τ  0 + (τ − ζ )0 −1 (t, x(ζ )) dζ ds B(0 )(0 ) 0  1 1 −  0 ≤ Hϑ0 (1, s )q max (t, E2 ) B(0 ) 0  τ  0 + (τ − ζ )0 −1 max (t, E2 )dζ ds B(0 )(0 ) 0 ≤ E2 = x . (3.13) ≤

0 B(0 )(0 )

+

(t, x(t ))

(τ − ζ )0 −1 (t, x(ζ )) dζ ds

0



0 Hϑ0 (t, s )q (AB 0 I (ζ , u (ζ )) ds

 b 1 −  0 ≤  Hϑ0 (t, s )q (t, x(t )) B ( )

By (3.11), we get  max (t, u) ≥ y∗ (t, x(t)), which further proceed

J ∗ x(t ) = max

1

(5.2)

H. Khan, F. Jarad and T. Abdeljawad et al. / Chaos, Solitons and Fractals 129 (2019) 56–61



1 0

 (s − ζ )0 −1 ψmin (ζ , η )dζ ds (0 ) a   1  s 1 = Hϑ0 (1, s )q (s − ζ )0 −1 (0 ) a 0   1  ) dζ ds ψmin (ζ , 

1

H ϑ 0 ( 1 , s ) q

s

10 0 0

 ≥



1 0

Hϑ0 (1, s )q

1 −   0

1

ζ5

1 10 0 0 3 + 5 10 0 03



B(0 )  s 0 + (s − ζ )0 −1 B(0 )(0 ) a    1 1 10 0 0 3 + dζ ds ζ5 3 5

10 0 0

= 0.0 0671612 >

1 . 10 0 0

By the help of Theorem 3.2, (5.1) has a solution 1 ∗ 10 0 0 ≤ x ≤ 1.

(5.3) u∗

and satisfying

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