Using Krasnoselskii's theorem to investigate the Cauchy and neutral fractional q-integro-differential equation via numerical technique

Xiao-Guang Yue; Mohammad Esmael Samei; Azam Fathipour; Mohammed K. A. Kaabar; Artion Kashuri

doi:10.1515/nleng-2022-0023

Open Access Published by De Gruyter June 11, 2022

Using Krasnoselskii's theorem to investigate the Cauchy and neutral fractional q-integro-differential equation via numerical technique

Xiao-Guang Yue , Mohammad Esmael Samei , Azam Fathipour , Mohammed K. A. Kaabar and Artion Kashuri

From the journal Nonlinear Engineering

https://doi.org/10.1515/nleng-2022-0023

Abstract

This article discusses the stability results for solution of a fractional q-integro-differential problem via integral conditions. Utilizing the Krasnoselskii’s, Banach fixed point theorems, we demonstrate existence and uniqueness results. Based on the results obtained, conditions are provided to ensure the generalized Ulam and Ulam–Hyers–Rassias stabilities of the original system. The results are illustrated by two examples.

Keywords: nonlinear fractional integro-differential equation; neutral differential equation; Cauchy differential equation; existence and stability

MSC 2010: 34A08; 34B16; 39A13

1 Introduction and formulation of the problem

The fractional derivative can be considered as a global operator that has greater degrees of flexibility as compared to integer-order derivative, because a classical derivative with integer-order could be a nearby operator. A few researchers have demonstrated that fractional-order derivatives play a noteworthy part in electrochemical analysis to explain the mechanistic behavior of the concentration of a substrate at the electrode surface to the current [1,2, 3,4,5, 6,7,8, 9,10,11, 12,13]. Some interesting applications of fractional calculus in science and engineering have been discussed [14,15,16, 17,18].

It is interesting to study solution to fractional q-integro-differential problem with integral conditions, which will allow a generalized stability. It is shown in [4] that, in a real k -dimensional Euclidean space, the local and global solutions exist for the following Cauchy problem:

(1.1) D 0 + σ C [ y ] ( t ) = h ( t , y ( t ) ) + ∫ t 0 t Θ ( t , ξ , y ( ξ ) ) d ξ , y ( 0 ) = y 0 ,

where 0 < σ ≤ 1 , h ∈ C ( I ¯ × R k , R k ) , Θ ∈ C ( I ¯ 2 × R k , R k ) , I ¯ ≔ [ 0 , 1 ] , and D 0 + σ C is the Caputo fractional operator. A class of abstract delayed fractional neutral integro-differential equations was introduced in [19], for σ ∈ ( 1 , 2 ) ,

(1.2) D t σ ( y t ) = a 1 N ( y t ) + ∫ 0 t a 2 ( t − ξ ) N ( y ξ ) d ξ + h ( t , y ρ ( t , y t ) ) , y ( 0 ) = η , y ′ ( 0 ) = 0 .

Using the Leray–Schauder alternative fixed point theorem, the existence results were obtained (for more details, see [3]). Recently, much attention has been paid to the study of differential equations with fractional derivatives [20,21,22, 23,24,25]. Note that in [22], the authors introduced and studied a related problem. Shah et al. [8] investigated the following problem under delay differential equations involving Caputo fractional derivative and under nonlocal initial condition with non-monotone term as

(1.3) D 0 + σ RL ( y ( t ) ) = h ( t , y ( t ) , D 0 + σ y ( α t ) ) , t ∈ [ 0 , ϱ ] , α ∈ ( 0 , 1 ) , t 1 − σ y ( 0 ) = y 0 + ψ ( t , y ( t ) ) ,

where D 0 + σ RL ( y ( t ) ) represent Riemann–Liouville fractional derivative of order σ ∈ ( 0 , 1 ] and h ∈ C ( [ 0 , ϱ ] × R 2 , R ) , ψ ∈ C ( [ 0 , ϱ ] × R , R ) . Ruzhansky et al. studied particularly the existence for the following problem:

(1.4) D 0 + σ C [ D 0 + ν C [ y ] ( t ) + h 1 ( t , y ( t ) ) ] = h 2 ( t , y ( t ) ) , t ∈ I ¯ ≔ [ 0 , 1 ] , y ( 0 ) = ∑ j = 1 m η j y ( b j ) c y ( 1 ) = a ∫ 0 1 y ( ξ ) d H ( ξ ) + ∑ j = 1 n α j ∫ τ 0 j τ 1 j y ( ξ ) d ( ξ ) ,

where 0 < τ 0 j < τ 1 j < 1 , σ , ν ∈ I ≔ ( 0 , 1 ) , η j , α j ∈ R and D 0 + σ c is the CFD of order σ , h 1 , h 2 are given continuous functions. By using the classical tools of fixed point theory, the existence and uniqueness results are obtained. On an arbitrary domain, in [21], the authors studied an FDE with non-conjugate Riemann–Stieltjes integro-multipoint boundary conditions by using new tools on function analysis. For some more related works, refer to [26,27]. Shah et al. considered the following system is investigated under Atangana, Baleanu, and Caputo fractional order derivative

(1.5) D 0 + σ A B C ( y 1 ( t ) ) = h 1 ( t , y 1 ( t ) , y 2 ( t ) , y 3 ( t ) ) , y 1 ( 0 ) = η 1 D 0 + σ A B C ( y 2 ( t ) ) = h 2 ( t , y 1 ( t ) , y 2 ( t ) , y 3 ( t ) ) , y 2 ( 0 ) = η 2 D 0 + σ A B C ( y 3 ( t ) ) = h 3 ( t , y 1 ( t ) , y 2 ( t ) , y 3 ( t ) ) , y 3 ( 0 ) = η 3 ,

where t ∈ [ 0 , ϱ ] , σ ∈ ( 0 , 1 ] , h i : [ 0 , ϱ ] × R 3 → R , ( i = 1 , 2 , 3 ) are continuous functions [28]. Derbazi et al. determined the existence criteria of extremal solutions for the following θ -Caputo-type fractional differential equation in a Caputo sense under nonlinear boundary conditions

(1.6) D a + σ ; θ C ( y ( t ) ) = h 1 ( t , y ( t ) ) , t ∈ [ a , b ] , α ∈ ( 0 , 1 ) , h 2 ( y ( a ) , y ( b ) ) = 0 ,

where D a + σ ; θ C is the θ -fractional operator of order 0 < σ ≤ 1 in the Caputo sense and h 1 ∈ C ( [ a , b ] × R , R ) , h 2 ∈ C ( R 2 , R ) [29]. Abbas [7] investigated the following Langevin equation with the generalized proportional fractional derivatives with respect to another function

(1.7) D a σ 1 ; θ , ϑ ( D a σ 2 ; θ , ϑ + λ ) y ( t ) = h 1 ( t , y ( t ) ) , t ∈ [ a , b ] , λ ∈ R , I a 1 − σ 2 ; θ , ϑ y ( a ) = y a , y ( b ) + μ y ( c ) = 0 , a < c < b , y a ∈ R ,

where θ > 0 , D a σ i ; θ , ϑ , and I a 1 − σ 2 ; θ , ϑ are the generalized proportional fractional derivative and integral with respect to another continuous function ϑ of order σ i ∈ ( 0 , 1 ] and 1 − σ 2 , respectively, and h 1 ∈ C ( [ a , b ] × R , R ) is the given function. The authors in [30] studied qualitative aspects of the system of fractional differential equations via Caputo–Hadamard derivative given as

(1.8) D 1 + σ C y ( t ) + h 1 ( t , y ( α t ) , z ( t ) ) = 0 , t ∈ [ 1 , e ] , D 1 + σ C z ( t ) + h 2 ( t , y ( t ) , z ( α t ) ) = 0 ,

under boundary conditions

y ( 1 ) = z ( 1 ) = y ′ ( 1 ) = z ′ ( 1 ) = 0 = y ′ ( e ) = z ′ ( e ) ,

and y ( e ) = ϕ 1 ( y ) , z ( e ) = ϕ 2 ( z ) with σ ∈ ( 3 , 4 ] , α ∈ ( 0 , 1 ) , functions h 1 , h 2 : [ 1 , e ] × R 2 → R and ϕ 1 , ϕ 2 : Y → R are continuous functions, where Y is complete norm space.

The authors in [2], considered the problem for the system (1.9), and we generalized the system in the q FDE which is not explicitly presented, and therefore it makes sense to consider for t ∈ I ¯ , σ , ν ∈ I , the problem for system as follows:

(1.9) D q σ + ν C [ y ] ( t ) = h 1 ( t , y ( t ) ) + I q σ [ h 2 ] ( t , y ( t ) ) + ∫ 0 t Θ ( t , ξ , y ( ξ ) ) d ξ ,

under boundary condition

y ( 0 ) = η ∫ 0 τ ∗ y ( ξ ) d ξ , ( τ ∗ ∈ I ) ,

where η is a real constant, D q σ + ν C is the Caputo fractional q-derivative of order σ + ν , I q σ denotes the left-sided Riemann–Liouville fractional q-integral of order σ , and h i : I ¯ × H → H ( i = 1 , 2 ) , Θ : I ¯ 2 × H → H , are appropriate functions satisfying some conditions which will be stated later. H is a Banach space equipped with the norm ‖ . ‖ .

Here, this study is focused on the question of existence and uniqueness in Section 3. In addition, Section 4 is devoted to show a generalized stability. Note that this representation also allows us to generalize the results obtained recently in the literature. The article is ended by two examples illustrating our results.

2 Notations and preliminaries

We recall some essential preliminaries that are used for the results of the subsequent sections. Let t 0 ∈ R and q ∈ I . The time scale T t 0 is defined by

T t 0 = { 0 } ∪ { t : t = t 0 q n , ∀ n ∈ N } .

If there is no confusion concerning t 0 we shall denote T t 0 by T . Let s ∈ R . Define [ s ] q = ( 1 − q s ) / ( 1 − q ) (see [31]). The q-factorial function ( y − z ) q ( n ) is defined by

(2.1) ( y − z ) q ( n ) = ∏ k = 0 n − 1 ( y − z q k ) , n ∈ N 0 ,

and ( y − z ) q ( 0 ) = 1 , where y , z ∈ R and N 0 ≔ { 0 } ∪ N (see [32]). Also, we have

(2.2) ( y − z ) q ( σ ) = y σ ∏ k = 0 ∞ y − z q k y − z q σ + k , σ ∈ R , s ≠ 0 .

Algorithms 1 and 2 simplify q-factorial functions ( y − z ) q ( n ) and ( y − z ) q ( σ ) , respectively. In [33], the authors proved

( y − z ) q ( σ + ν ) = ( y − z ) q ( σ ) ( y − q σ z ) q ( ν ) , and ( s y − s z ) q ( σ ) = s σ ( y − z ) q ( σ ) .

If z = 0 , then it is clear that y ( σ ) = y σ . The q -gamma function is given by [31].

Γ q ( y ) = ( 1 − q ) 1 − y ( 1 − q ) q ( y − 1 ) , ( y ∈ R \ { ⋯ , − 2 , − 1 , 0 } ) .

In fact, by using (2.2), we have

(2.3) Γ q ( y ) = ( 1 − q ) 1 − y ∏ k = 0 ∞ 1 − q k + 1 1 − q y + k − 1 .

Algorithm 3 shows the MATLAB lines for calculation of Γ q ( y ) which we tend n to infinity in it. Note that, Γ q ( y + 1 ) = [ y ] q Γ q ( y ) [33, Lemma 1]. For any positive numbers σ and ν , the q-beta function is defined by

(2.4) B q ( σ , ν ) = ∫ 0 1 ( 1 − ξ ) q ( σ − 1 ) ξ ν − 1 d q ξ = Γ q ( σ ) Γ q ( ν ) Γ q ( σ + ν ) .

For a function w : T → R , the q -derivative of w , is

(2.5) D q [ y ] ( t ) = d d t q y ( t ) = y ( q t ) − y ( t ) t ( 1 − q ) ,

for all t ∈ T ⧹ { 0 } , and D q [ y ] ( 0 ) = lim t → 0 D q [ y ] ( t ) (see [32]). Also, the higher order q -derivative of the function y is defined by

D q n [ y ] ( t ) = D q [ D q n − 1 [ y ] ] ( t ) , ∀ n ≥ 1 ,

where D q 0 [ y ] ( t ) = y ( t ) (see [32]). In fact,

(2.6) D q n [ y ] ( t ) = 1 t n ( 1 − q ) n ∑ k = 0 n ( 1 − q − n ) q ( k ) ( 1 − q ) q ( k ) q k y ( t q k ) ,

for t ∈ T ⧹ { 0 } (see ref. [27]).

Remark 2.1

[9] By using Eq. (2.1), we can change Eq. (2.6) as follows:

(2.7) D q n [ y ] ( t ) = 1 t n ( 1 − q ) n ∑ k = 0 n ∏ i = 0 k − 1 1 − q i − n 1 − q i + 1 q k y ( t q k ) .

Algorithms 4 and 5 show the MATLAB codes for calculation of Eqs. (2.5) and (2.7), respectively. The q -integral of the function y is defined by

(2.8) I q [ y ] ( t ) = ∫ 0 t y ( ξ ) d q ξ = t ( 1 − q ) ∑ k = 0 ∞ q k y ( t q k ) ,

for 0 ≤ t ≤ b , provided the series absolutely converges (see [32]). By using Algorithm 6, we can obtain the numerical results of I q [ y ] ( t ) when n → ∞ . If s in [ 0 , b ] , then

∫ s b y ( ξ ) d q ξ = I q [ y ] ( b ) − I q [ y ] ( s ) = ( 1 − q ) ∑ k = 0 ∞ q k [ b y ( b q k ) − s y ( s q k ) ] ,

whenever the series exists. The operator I q n is given by I q 0 [ y ] ( t ) = y ( t ) and [32]

I q n [ y ] ( t ) = I q [ I q n − 1 [ y ] ] ( t ) , ∀ n ≥ 1 , y ∈ C ( [ 0 , b ] ) .

It has been proved that

D q [ I q [ y ] ] ( t ) = y ( t ) , I q [ D q [ y ] ] ( t ) = y ( t ) − y ( 0 ) ,

whenever the function y is continuous at t = 0 (see [32]). The fractional Riemann–Liouville-type q -integral of the function y is defined by

(2.9) I q σ [ y ] ( t ) = ∫ 0 t ( t − ξ ) q ( σ − 1 ) y ( ξ ) Γ q ( σ ) d q ξ , I q 0 [ y ] ( t ) = y ( t ) ,

for t ∈ [ 0 , 1 ] and σ > 0 (see refs [27,34]).

Remark 2.2

[9] By using Eqs. (2.2), (2.3), and (2.8), we have

(2.10) I q σ [ y ] ( t ) = t σ ( 1 − q ) σ lim n → ∞ ∑ k = 0 n q k ∏ i = 0 n × ( 1 − q σ + i − 1 ) ( 1 − q k + i ) ( 1 − q i + 1 ) ( 1 − q σ + k + i − 1 ) y ( t q k ) .

Algorithm 7 shows the MATLAB codes of numerical technique. The Caputo fractional q -derivative of the function y is defined by

(2.11) D q σ C [ y ] ( t ) = I q [ σ ] − σ [ D q [ σ ] [ y ] ] ( t ) = ∫ 0 t ( t − ξ ) q ( [ σ ] − σ − 1 ) D q [ σ ] [ y ] ( ξ ) Γ q ( [ σ ] − σ ) d q ξ

for t ∈ [ 0 , 1 ] and σ > 0 (see [34,35]). It has been proved that

I q ν [ I q σ [ y ] ] ( t ) = I q σ + ν [ y ] ( t ) ,

and D q σ C [ I q σ [ y ] ] ( t ) = y ( t ) , where σ , ν ≥ 0 [34]. Also, (see [34])

I q σ [ D q n [ y ] ] ( t ) = D q n [ I q σ [ y ] ] ( t ) − ∑ k = 0 n − 1 t σ + k − n D q k [ y ] ( 0 ) Γ q ( σ + k − n + 1 ) , σ > 0 , n ≥ 1 .

Remark 2.3

From Eq. (2.3), Remark 2.1, and Eq. (2.10) in Remark 2.2, we obtain

(2.12) D q σ c [ y ] ( t ) = 1 t σ ( 1 − q ) σ − [ σ ] lim n → ∞ ∑ k = 0 n × ∏ i = 0 n ( 1 − q [ σ ] − σ + i − 1 ) ( 1 − q k + i ) ( 1 − q i + 1 ) ( 1 − q [ σ ] − σ − 1 + k + i ) × ∑ m = 0 [ σ ] ∏ i = 0 m − 1 ( 1 − q i − [ σ ] ) ( 1 − q i + 1 ) q m y ( t q k + m ) .

Algorithm 8 shows the MATLAB codes of numerical technique. One can find other algorithms in [36]. Now, we introduce some basic definitions, lemmas, and theorems, which are used in the subsequent sections.

Lemma 2.4

[37] Let y ∈ A C n [ t 1 , t 2 ] . Then, one has

For σ , σ ́ > 0 ,
I q σ [ D q σ C [ y ] ] ( t ) = y ( t ) + ∑ i = 0 n − 1 c i ( t − t 1 ) i , ( c 0 , c 1 , … , c n − 1 ∈ R ) ,
for n − 1 < σ ≤ n , n ∈ N ;
I q σ [ I q σ ́ [ y ] ] ( t ) = I q σ ́ [ I q σ [ y ] ] ( t ) = I q σ + σ ́ [ y ] ( t ) ;
D q σ C [ I q σ ́ [ y ] ] ( t ) = y ( t ) .

Lemma 2.5

[37] Let n − 1 < σ ≤ n , n ∈ N , and y ∈ C [ t 1 , t 1 ] . Then for all t ∈ [ t 1 , t 2 ] , we have C D t 1 σ [ I t 1 σ [ y ] ] ( t ) = y ( t ) .

Lemma 2.6

[37] Let σ ∈ ( 0 , 1 ) . Then for each y ∈ A C [ 0 , 1 ] , I σ [ D σ [ y ] ] ( t ) = y ( t ) for a.e. t ∈ [ 0 , 1 ] , where

D σ [ y ] ( t ) = d d t ∫ 0 t ( t − ξ ) − σ y ( ξ ) Γ ( 1 − σ ) d ξ .

Lemma 2.7

([38], Banach fixed point theorem) Let B be a non-empty complete metric space and T : B → B is contraction mapping. Then, there exists a unique point y ∈ B such that T ( y ) = y .

Lemma 2.8

([38], Krasnoselskii fixed point theorem) Let E be bounded, closed, and convex subset in a Banach space B . If T 1 , T 2 : E → E are two applications satisfying the following conditions: (A1) T 1 ( y ) + T 2 ( z ) ∈ E for every y , z ∈ E ; (A2) T 1 is a contraction; (A3) T 2 is compact and continuous. Then there exists v ∗ ∈ B such that T 1 ( v ∗ ) + T 2 ( v ∗ ) = v ∗ .

3 Existence results

Before presenting our main results, we need the following auxiliary lemma.

Lemma 3.1

Let σ + ν ∈ I and η τ ∗ ≠ 1 . Assume that h 1 , h 1 , and Θ are three continuous functions. If y ∈ C ( I ¯ , H ) , then y is solution of (1.9) iff y satisfies the IE

(3.1) y ( t ) = ∫ 0 t ( t − ξ ) q ( σ + ν − 1 ) Γ q ( σ + ν ) h 1 ( ξ , y ( ξ ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − a ) q ( σ − 1 ) Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ + η 1 − η τ ∗ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) h 1 ( s , y ( s ) ) + ∫ 0 s Θ ( s , r , y ( r ) ) d r + ∫ 0 s ( s − r ) q ( σ − 1 ) Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s .

Proof

Let y ∈ C ( I ¯ , H ) be a solution of system (1.9). First, we show that y is a solution of integral Eq. (3.1). By Lemma 2.4 and using boundary condition, we obtain

(3.2) I q σ + ν [ D q σ + ν C [ y ] ( t ) ] = I q σ + ν D q σ + ν C ∫ 0 t ( t − ξ ) q ( σ + ν − 1 ) Γ q ( σ + ν ) [ h 1 ( ξ , y ( ξ ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − a ) q ( σ − 1 ) Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ + η 1 − η τ ∗ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) h 1 ( s , y ( s ) ) + ∫ 0 s Θ ( s , r , y ( r ) ) d r + ∫ 0 s ( s − r ) q ( σ − 1 ) Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s = y ( t ) + ∑ i = 0 n − 1 c i ( t − t 1 ) i = y ( t ) − y ( 0 ) .

From Eq. (1.9) we have

(3.3) I q σ + ν [ D q σ + ν C [ y ] ( t ) ] = I q σ + ν h 1 ( t , y ( t ) ) + I q σ [ h 2 ] ( t , y ( t ) ) ∫ 0 t Θ ( t , ξ , y ( ξ ) ) d ξ = ∫ 0 t ( t − ξ ) q ( σ + ν − 1 ) Γ q ( σ + ν ) h 1 ( t , y ( t ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ .

By substituting (3.3) in (3.2) with nonlocal condition in problem (3.1), we obtain

(3.4) y ( t ) = ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) h 1 ( ξ , y ( ξ ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ + y ( 0 ) .

From integral boundary condition of our problem with using Fubini’s theorem and after some computations, we obtain

y ( 0 ) = η ∫ 0 τ ∗ y ( ξ ) d ξ = η ∫ 0 τ ∗ ∫ 0 ξ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) h 1 ( s , y ( s ) ) d q s d q ξ + η ∫ 0 τ ∗ ∫ 0 ξ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) ∫ 0 s Θ ( s , r , y ( r ) ) d r d q s d q ξ + η ∫ 0 τ ∗ ∫ 0 ξ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s × d q ξ + η τ ∗ y ( 0 )

= η ∫ 0 τ ∗ ∫ s τ ∗ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) d q ξ h 1 ( s , y ( s ) ) d q s + η ∫ 0 τ ∗ ∫ s τ ∗ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) d q ξ ∫ 0 s Θ ( s , r , y ( r ) ) d r d q s + η ∫ 0 τ ∗ ∫ s τ ∗ ( ξ − s ) q σ + ν − 1 Γ q ( σ + ν ) d q ξ × ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s + η τ ∗ y ( 0 ) ,

that is,

(3.5) y ( 0 ) = η 1 − η τ ∗ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν ) × h 1 ( s , y ( s ) ) + ∫ 0 s Θ ( s , r , y ( r ) ) d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s .

Finally, by substituting (3.5) in (3.4), we find (3.1). Conversely, from Lemma 3.1 and by applying the operator D q σ + ν C on both sides of (3.1), we find

(3.6) D q σ + ν C [ y ] ( t ) = D q σ + ν C I q σ + ν h 1 ( t , y ( t ) ) + ∫ 0 t Θ ( t , ξ , y ( ξ ) ) d s + I q σ h 2 ( t , y ( t ) ) + D q σ + ν C y ( 0 ) = h q ( t , y ( t ) ) + I q σ h 2 ( t , y ( t ) ) + ∫ 0 t Θ ( t , ξ , y ( ξ ) ) d ξ .

This means that y satisfies the equation in problem (1.9). Furthermore, by substituting t by 0 in integral Eq. (3.1), we are clear that the integral boundary condition in (1.9) holds. Therefore, y is solution of problem (1.9), which completes the proof.

In order to prove the existence and uniqueness of solution for problem (1.9) in C ( I ¯ , H ) , we use two fixed point theorem. First, we transform the system (1.9) into fixed point problem as y = U y , where U : ( I ¯ , H ) → ( I ¯ , H ) is an operator defined by following

(3.7) U y ( t ) = ∫ 0 t ( t − ξ ) q σ + ν Γ q ( σ + ν ) h 1 ( ξ , y ( ξ ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ + η 1 − η τ ∗ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) h 1 ( s , y ( s ) ) + ∫ 0 s Θ ( s , r , y ( r ) ) d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s .

3.1 Existence result by Krasnoselskii’s fixed point

Theorem 3.2

Consider continuous functions h 1 , h 2 : I ¯ × H → H and Θ : I ¯ 2 × H → H such that satisfying: ( H 1 ) the inequalities

‖ h j ( t , y ( t ) ) − h j ( t , z ( t ) ) ‖ ≤ μ j ‖ y ( t ) − z ( t ) ‖ , j = 1 , 2 ,

and

‖ Θ ( t , s , y ( s ) ) − Θ ( t , s , z ( s ) ) ‖ ≤ μ ∗ ‖ y ( s ) − z ( s ) ‖ ,

where μ ∗ , μ j ≥ 0 , ( j = 1 , 2 ) with μ = max { μ 1 , μ 2 , μ ∗ } . ( H 2 ) there exist three functions ϱ ∗ , ϱ j ∈ L ∞ ( I ¯ , R + ) , ( j = 1 , 2 ) , such that

‖ h j ( t , y ( t ) ) ‖ ≤ ϱ j ( t ) ‖ y ( t ) ‖ , j = 1 , 2 ,

and

‖ Θ ( t , s , y ( s ) ) ‖ ≤ ϱ ∗ ( t ) ‖ y ( s ) ‖ ,

∀ t ∈ I ¯ , y , z ∈ H and ( t , s ) ∈ G ≔ { ( t , s ) : 0 ≤ s ≤ t ≤ 1 } . If λ ≤ 1 and μ λ ∗ ≤ 1 , then problem (1.9) has at least one solution on I ¯ , where

(3.8) λ = ‖ ϱ 1 ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ B q ( σ + 1 , σ + ν ) Γ q ( σ + 1 ) Γ q ( σ + ν ) + ∣ η ∣ ‖ ϱ 1 ‖ L ∞ τ ∗ σ + ν + 1 + ∣ η ∣ ‖ ϱ ∗ ‖ L ∞ τ ∗ σ + ν + 1 ∣ 1 − η τ ∗ ∣ Γ q ( σ + ν + 2 ) + ∣ η ∣ ‖ ϱ 2 ‖ L ∞ τ ∗ 2 σ + ν + 1 B q ( σ + 1 , σ σ + ν + 1 ) ∣ 1 − ν τ ∗ Γ q ( σ + 1 ) Γ q ( σ + ν + 1 )

and

(3.9) λ ∗ = ∣ η ∣ ∣ 1 − η τ ∗ ∣ 2 τ ∗ σ + ν + 1 Γ q ( σ + ν + 2 ) + τ ∗ 2 σ + ν + 1 B q ( σ + 1 , σ + ν + 1 ) Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) .

Proof

For any function y ∈ C ( I ¯ , H ) , we define the norm

‖ y ‖ ∗ ≔ max { e − t ‖ y ( t ) ‖ : t ∈ I ¯ } ,

and consider the closed ball

B ℓ ≔ { y ∈ C ( I ¯ , H ) : ‖ y ‖ ∗ ≤ ℓ } .

Next, let us define the operators U 1 , U 2 on B ℓ as follows:

(3.10) U 1 y ( t ) = ∫ 0 t ( t − ξ ) q σ + ν + 1 Γ q ( σ + ν ) h 1 ( ξ , y ( ξ ) ) + ∫ 0 ξ Θ ( ξ , s , y ( s ) ) d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) h 2 ( s , y ( s ) ) d q s d q ξ

and

(3.11) U 2 y ( t ) = η 1 − η τ ∗ ∫ 0 τ ( τ − s ) q σ + ν Γ q ( σ + ν + 1 ) × h 1 ( s , y ( s ) ) + ∫ 0 s Θ ( s , r , y ( r ) ) d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , y ( r ) ) d q r d q s .

For y , z ∈ B ℓ , t ∈ I ¯ and by the assumption ( H 2 ) , we find

‖ U 1 y ( t ) + U 2 z ( t ) ‖ ≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) ‖ h 1 ( ξ , y ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( ξ , y ( ξ ) ) ‖ d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × ‖ h 1 ( s , z ( s ) ) ‖ + ∫ 0 s ‖ Θ ( s , r , z ( r ) ) ‖ d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) ‖ h 2 ( r , z ( r ) ) ‖ d q r d q s

≤ ∫ 0 t ( t − ξ ) q σ + ν + 1 Γ q ( σ + ν ) ϱ 1 ( ξ ) ‖ y ( ξ ) ‖ + ∫ 0 ξ ϱ ∗ ( ξ ) ‖ y ( s ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ϱ ∗ ( s ) ‖ y ( s ) ‖ d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) σ + ν Γ q ( σ + ν + 1 ) × ϱ 1 ( s ) ‖ z ( s ) ‖ + ∫ 0 s ϱ ∗ ( s ) ‖ z ( r ) ‖ d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) ϱ 2 ( r ) ‖ z ( r ) ‖ d q r d q s

≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) [ ‖ ϱ 1 ‖ μ ∞ ‖ y ‖ ∗ e ξ + ‖ ϱ ∗ ‖ L ∞ ‖ y ‖ ∗ ( e ξ − 1 ) + ‖ ϱ 2 ‖ L ∞ ‖ y ‖ ∗ ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) e s d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) σ + ν Γ q ( σ + ν + 1 ) × [ ‖ ϱ 1 ‖ L ∞ ‖ z ‖ ∗ e s + ‖ ϱ ∗ ‖ L ∞ ‖ z ‖ ∗ ( e s − 1 ) + ‖ ϱ 2 ‖ L ∞ ‖ z ‖ ∗ ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) e r d r d q s .

Therefore,

‖ U 1 y + U 2 z ‖ ∗ ≤ ∫ 0 t ( t − ξ ) σ + ν − 1 Γ q ( σ + ν ) ‖ ϱ 1 ‖ L ∞ ‖ y ‖ ∗ e ξ e t + ‖ ϱ ∗ ‖ L ∞ ‖ y ‖ ∗ ( e ξ − 1 ) e t + ‖ ϱ 2 ‖ L ∞ ‖ y ‖ ∗ ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) e s e t d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) σ + ν Γ q ( σ + ν + 1 ) × ‖ ϱ 1 ‖ L ∞ ‖ z ‖ ∗ e s e t + ‖ ϱ ∗ ‖ L ∞ ‖ z ‖ ∗ ( e s − 1 ) e t + ‖ ϱ 2 ‖ L ∞ ‖ z ‖ ∗ ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) e r e t d q r d q s

(3.12) ≤ ℓ ‖ ϱ 1 ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ Γ q ( σ + 1 ) Γ q ( ν + 1 ) ∫ 0 1 ( 1 − ξ ) q σ + ν + 1 ξ σ d q ξ + ∣ η ∣ ‖ ϱ 1 ‖ L ∞ τ ∗ σ + ν + 1 + ∣ η ∣ ‖ ϱ ∗ ‖ L ∞ τ ∗ σ + ν + 1 ∣ 1 − η τ ∗ ∣ Γ q ( σ + ν + 1 ) + ∣ η ∣ ‖ ϱ 2 ‖ L ∞ ∣ 1 − η τ ∗ ∣ Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) × ∫ 0 τ ∗ ( τ ∗ − s ) σ + ν s σ d q s = ℓ ‖ ϱ 1 ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ ν ( σ + 1 , σ + ν ) Γ q ( σ + 1 ) Γ q ( ν + 1 ) + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ‖ ϱ 1 ‖ L ∞ τ ∗ σ + ν + 1 + ‖ ϱ ∗ ‖ L ∞ τ ∗ σ + ν + 1 Γ q ( σ + ν + 2 ) + ‖ ϱ 2 ‖ L ∞ τ ∗ 2 σ + ν + 1 ν ( σ + 1 , σ + ν + 1 ) Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) = ℓ λ ≤ ℓ .

This implies that ( U 1 y + U 2 z ) ∈ B ℓ . Here we used the computations

∫ 0 1 ( 1 − ξ ) q σ + ν ξ σ d q ξ = β q ( σ + 1 , σ + ν ) , ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν s σ d q ξ = τ ∗ 2 σ + ν + 1 ν ( σ + 1 , σ + ν + 1 ) ,

and the estimations:

e ξ e t ≤ 1 , e s e t ≤ 1 , e r e t ≤ 1 .

In this step, we show that U 2 is a contraction mapping. Let y , z ∈ H , t ∈ I ¯ . We have

‖ U 2 y ( t ) − U 2 z ( t ) ‖ ≤ ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × [ ‖ h 1 ( s , y ( s ) ) − h 1 ( s , ν ( s ) ) ‖ + ∫ 0 s ‖ Θ ( s , r , y ( r ) ) − Θ ( s , r , z ( r ) ) ‖ d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) ‖ h 2 ( r , y ( r ) ) − h 2 ( r , z ( r ) ) ‖ d q r d q s

≤ ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × [ μ 1 ‖ y − z ‖ ∗ e s + ∫ 0 s μ ∗ ‖ y − z ‖ ∗ e r d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) μ 2 ‖ y − z ‖ ∗ e r d r d q s ≤ ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × [ μ ‖ y − z ‖ ∗ e s + μ ‖ y − z ‖ ∗ ( e s − 1 ) + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) μ ‖ y − z ‖ ∗ e r d q r d q s .

Thus,

‖ U 2 y − U 2 z ‖ ∗ ≤ ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × μ ‖ y − z ‖ ∗ e s e t − μ ‖ y − z ‖ ∗ ( e s − 1 ) e t + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) μ ‖ y − z ‖ ∗ e r e t d q r d q s ≤ ∣ η ∣ μ ∣ 1 − η τ ∗ ∣ 2 τ ∗ σ + ν + 1 Γ q ( σ + ν + 2 ) + τ ∗ 2 σ + ν + 1 ν ( σ + 1 , σ + ν + 1 ) Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) ‖ y − z ‖ ∗ .

Then since μ λ ∗ ≤ 1 , U 2 is a contraction mapping. The continuity of the functions h 1 , h 2 , and Θ implies that U 1 is continuous and U 1 B ℓ ⊂ B ℓ , for each y ∈ B ℓ , i.e., U 1 is uniformly bounded on B ℓ as

‖ ( U 1 y ) ( t ) ‖ ≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) × ‖ h 1 ( ξ , y ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , y ( s ) ) ‖ d q s d q ξ ,

which implies that

(3.13) ‖ U 1 y ‖ ∗ ≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) ‖ ϱ 1 ‖ L ∞ ‖ y ‖ ∗ e ξ e t + ϱ ∗ ‖ L ∞ ‖ y ‖ ∗ ( e ξ − 1 ) e t + ϱ 2 ‖ L ∞ ‖ y ‖ ∗ ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) e s e t d q s d q ξ ≤ ℓ ‖ ϱ ∗ ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ ν ( σ + 1 , σ + ν ) Γ q ( σ + 1 ) Γ q ( ν + 1 ) ≤ ℓ λ ≤ ℓ .

Finally, we will show that ( U 1 B ℓ ) is equi-continuous. For this end, we put

h ¯ j = sup ( t , y ( t ) ) ∈ I ¯ × B ℓ ‖ h j ( t , y ( t ) ) ‖ , j = 1 , 2 Θ ¯ = sup ( t , s , y ( s ) ) ∈ G × B ℓ ∫ 0 ξ ‖ Θ ( t , ξ , y ( ξ ) ) ‖ d ξ .

Let for any y ∈ B ℓ and for each t 1 , t 2 ∈ I ¯ with t 1 ≤ t 2 , we have

‖ ( U 1 y ) ( t 2 ) − ( U 1 y ) ( t 1 ) ‖ ≤ ∫ t 1 t 2 ( t 2 − ξ ) σ + ν − 1 Γ q ( σ + ν ) ‖ h 1 ( ξ , y ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) ‖ d q s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , y ( s ) ) ‖ d s d q ξ + 1 Γ q ( σ ) ∫ 0 t 1 [ ( t 1 − ξ ) q σ + ν − 1 − ( t 2 − ξ ) q σ + ν − 1 ] × ‖ h 1 ( ξ , y ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , y ( s ) ) ‖ d s d q ξ

≤ ∫ t 1 t 2 ( t 2 − ξ ) q σ + ν − 1 Γ ( σ + ν ) h ¯ 1 + Θ ¯ + ∫ 0 ξ h ¯ 2 ( ξ − s ) q σ − 1 Γ q ( σ ) d s d q ξ + 1 Γ q ( σ + ν ) ∫ 0 t 1 [ ( t 1 − ξ ) q σ + ν − 1 − ( t 2 − ξ ) q σ + ν − 1 ] × h ¯ 1 + Θ ¯ + ∫ 0 ξ h ¯ 2 ( ξ − s ) q σ − 1 Γ q ( σ ) d q s d q ξ

≤ ∫ t 1 t 2 ( t 2 − ξ ) σ + ν − 1 Γ q ( σ + ν ) h ¯ 1 + Θ ¯ + h ¯ 2 Γ q ( σ + 1 ) d q ξ + 1 Γ q ( σ + ν ) ∫ 0 t 1 [ ( t 1 − ξ ) q σ + ν − 1 − ( t 2 − ξ ) q σ + ν − 1 ] × h ¯ 1 + Θ ¯ + h ¯ 1 Γ q ( σ + 1 ) d q ξ + 1 Γ q ( σ + ν + 1 ) [ h ¯ 1 + Θ ¯ + h ¯ 2 Γ q ( σ + 1 ) [ 2 ( t 2 − t 1 ) q σ + ν + t 1 σ + ν − t 2 σ + ν ] .

The RHS of the last inequality is independent of y and tends to zero when ∣ t 2 − t 1 ∣ → 0 , this means that ∣ U 1 y ( t 2 ) − U 1 y ( t 1 ) ∣ → 0 , which implies that U 1 B ℓ is equi-continuous, then U 1 is relatively compact on B ℓ . Hence, by the Arzelá–Ascoli theorem, U 1 is compact on B ℓ . Now, all hypotheses of Theorem 3.2 hold; therefore, the operator U 1 + U 2 has a fixed point on B ℓ . So problem (1.9) has at least one solution on I ¯ . This proves the theorem.

3.2 Existence and uniqueness result

Theorem 3.3

Assume that ( H 1 ) holds. If μ λ < 1 , then the boundary value problems (1.9) has a unique solution on I ¯ .

Proof

Define m = max { m 1 , m 2 , m ∗ } , where m j and m ∗ are positive numbers such that

m j = sup t ∈ I ¯ ‖ h j ( t , 0 ) ‖ , ( j = 1 , 2 ) , m ∗ = sup ( t , s ) ∈ G ‖ Θ ( t , s , 0 ) ‖ .

We fix ℓ ≥ m ∗ λ 1 − μ λ and we consider

N ℓ = { y ∈ C ( I ¯ , H ) : ‖ y ‖ ∗ ≤ ℓ } .

Then, in view of the assumption ( H 1 ) , we have

‖ h q ( t , y ( t ) ) ‖ = ‖ h 1 ( t , y ( t ) ) − h 1 ( t , 0 ) + h 1 ( t , 0 ) ‖ ≤ ‖ h q ( t , y ( t ) ) − h q ( t , 0 ) ‖ + ‖ h 1 ( t , 0 ) ‖ ≤ μ 1 ‖ y ‖ ∗ + m 1 , ‖ h 2 ( t , y ( t ) ) ‖ ≤ μ 2 ‖ y ‖ ∗ + m 2 ,

and ‖ Θ ( t , s , y ( s ) ) ‖ ≤ μ ∗ ‖ y ‖ ∗ + m ∗ . In the first step, we show that U N ℓ ⊂ N ℓ . For each t ∈ I ¯ and for any y ∈ N ℓ ,

‖ U y ( t ) ‖ ≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) × ‖ h 1 ( ξ , y ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , y ( s ) ) ‖ d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × ‖ h 1 ( s , z ( s ) ) ‖ + ∫ 0 ξ ‖ Θ ( s , r , z ( r ) ) ‖ d r + ∫ 0 ξ ( s − r ) q σ − 1 Γ q ( σ ) ‖ h 2 ( r , z ( r ) ) ‖ d r d q s ≤ ( μ ℓ + m ) λ ≤ ℓ .

Hence, U N ℓ ⊂ N ℓ . Now, in the second step, we shall show that U : N ℓ → N ℓ is a contraction. From the assumption ( H 1 ) we have for any y , z ∈ N ℓ and for each t ∈ I ¯

(3.14) ‖ U y ( t ) − U z ( t ) ‖ ≤ ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) [ ‖ h 1 ( ξ , y ( ξ ) ) − h 1 ( ξ , z ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , y ( s ) ) − Θ ( ξ , s , z ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , y ( s ) ) − h 2 ( s , z ( s ) ) ‖ d q s d q ξ + ∣ η ∣ ∣ 1 − η τ ∗ ∣ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ − q ( σ + ν + 1 ) [ ‖ h 1 ( s , y ( s ) ) − h 1 ( s , z ( s ) ) ‖ + ∫ 0 s ‖ Θ ( s , r , y ( r ) ) − Θ ( s , r , z ( r ) ) ‖ d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) ‖ h 2 ( r , y ( r ) ) − h 2 ( r , z ( r ) ) ‖ d q r d q s ≤ μ λ ‖ y − z ‖ ∗ .

Since μ λ < 1 , it follows that U is a contraction. All assumptions of Lemma (3.1) are satisfied, then there exists y ∈ C ( I ¯ , H ) such that U y = y , which is the unique solution of problem (1.1) in C ( I ¯ , H ) .

4 Generalized Ulam stabilities

The aim is to discuss the Ulam stability for problem (1.9), by using the integration

z ( t ) = ∫ 0 t ( t − ξ ) q σ + ν − 1 Γ q ( σ + ν ) ‖ h 1 ( ξ , z ( ξ ) ) ‖ + ∫ 0 ξ ‖ Θ ( ξ , s , z ( s ) ) ‖ d s + ∫ 0 ξ ( ξ − s ) q σ − 1 Γ q ( σ ) ‖ h 2 ( s , z ( s ) ) ‖ d q s d q ξ + η 1 − η τ ∗ ∫ 0 τ ∗ ( τ ∗ − s ) q σ + ν Γ q ( σ + ν + 1 ) × h 1 ( s , z ( s ) ) + ∫ 0 s Θ ( s , r , z ( r ) ) d r + ∫ 0 s ( s − r ) q σ − 1 Γ q ( σ ) h 2 ( r , z ( r ) ) d q r d q s .

Here z ∈ C ( I ¯ , H ) possess a fractional derivative of order σ + ν , where 0 < σ + ν < 1 and h j : I ¯ × H → H and Θ : I ¯ 2 × H → H , are continuous functions. Then we define the nonlinear continuous operator P : C ( I ¯ , H ) → C ( I ¯ , H ) as follows:

P z ( t ) = D q σ + ν C z ( t ) − h 1 ( t , v ( t ) ) − I q σ h 2 ( t , v ( t ) ) − ∫ 0 t Θ ( t , ξ , z ( ξ ) ) d ξ .

For each ε > 0 and for each solution z of problem (1.9), such that

(4.1) ‖ P z ‖ ∗ ≤ ε ,

problem (1.9) is said to be Ulam–Hyers stable if we can find a solution y ∈ C ( I ¯ , HU ) of problem (1.9) and γ ∈ R ≥ 0 , satisfying the inequality ‖ y − z ‖ ∗ ≤ γ ε ∗ , is a positive real number depending on ε . Consider function ℘ in C ( R + , R + ) such that for each solution z of problem (1.9), we can find a solution u ∈ C ( I ¯ , H ) of problem (1.9) such that

‖ y ( t ) − z ( t ) ‖ ∗ ≤ ℘ ( ε ) , ∀ t ∈ I ¯ .

Then problem (1.9) is said to be generalized Ulam–Hyers stable. For each ε > 0 and for each solution z of problem (1.9), problem (1.9) is called Ulam–Hyers–Rassias stable with respect to ϱ ∈ C ( I ¯ , R + ) if

(4.2) ‖ P z ( t ) ‖ ∗ ≤ ε ϱ ( t ) , t ∈ I ¯ ,

and there exist a real number γ > 0 and a solution z ∈ C ( I ¯ , H ) of problem (1.9) such that

‖ y ( t ) − z ( t ) ‖ ≤ γ ε ∗ ϱ ( t ) , t ∈ I ¯ ,

where ε ∗ is a positive real number depending on ε .

Theorem 4.1

Under assumption ( H 1 ) in Theorem 3.1, with μ λ < 1 , problem (1.1) is both Ulam–Hyers and generalized Ulam–Hyers stable.

Proof

Let y ∈ C ( I ¯ , H ) be a solution of problem (1.9), satisfying (3.1) in the sense of Theorem 3.2. Let z be any solution satisfying (4.1). Lemma 2.4 implies the equivalence between the operators P and T − I d (where I d is the identity operator) for every solution z ∈ C ( I ¯ , H ) of problem (1.9) satisfying μ λ < 1 . Therefore, we deduce by the fixed-point property of the operator T that

‖ z ( t ) y ( t ) ‖ ∗ = ‖ z ( t ) T z ( t ) + T z ( t ) y ( t ) ‖ ∗ = ‖ z ( t ) T z ( t ) + T z ( t ) T y ( t ) ‖ ∗ ≤ ‖ T z ( t ) T y ( t ) ‖ + ‖ T z ( t ) z ( t ) ‖ ∗ ≤ μ λ ‖ y z ‖ ∗ + ε ,

because μ λ < 1 and ε > 0 , we find

‖ u − v ‖ ∗ ≤ ε 1 − μ λ .

Fixing ε ∗ = ε 1 − μ λ and γ = 1 , we obtain the Ulam–Hyers stability condition. In addition, the generalized Ulam–Hyers stability follows by taking ℘ ( ε ) = ε 1 − μ λ .

Theorem 4.2

Assume that ( H 1 ) holds with μ < λ 1 , and there exists a function ϱ ∈ C ( I ¯ , R + ) satisfying the condition 4.2. Then problem (1.9) is Ulam–Hyers–Rassias stable with respect to ϱ .

Proof

We have from the proof of Theorem 4.1,

‖ y ( t ) − z ( t ) ‖ ∗ ≤ ε ∗ ϱ ( t ) , ∀ t ∈ I ¯ ,

where ε ∗ = ε 1 − μ λ , and so the proof is completed.

5 Illustrative of our outcome

First we present Example 5.1, for illustrative our main result.

Example 5.1

Consider the following fractional integro-differential problem:

(5.1) D q 68 77 C [ y ] ( t ) = ( 15 − 2 t ) y ( t ) 25 + I q 5 11 ( 5 − t ) sin ( y ( t ) ) 43 + ∫ 0 t y ( ξ ) exp ( − ( t + ξ ) ) 20 d ξ ,

with boundary condition

y ( 0 ) = − 15 2 ∫ 0 0.6 y ( ξ ) d ξ , ∀ t ∈ I .

Clearly σ + ν = 68 77 , σ = 5 11 , τ ∗ = 0.6 , and η = − 15 2 . To illustrate our results in Theorems 3.2 and 4.1, we take for y , z ∈ H = R + and t ∈ [ 0 , 1 ] the following continuous functions:

h 1 ( t , y ( t ) ) = ( 15 − 2 t ) y ( t ) 25 , h 2 ( t , y ( t ) ) = ( 5 − t ) sin ( y ( t ) ) 43 ,

and

Θ ( t , s , y ( s ) ) = y ( s ) exp ( − ( t + s ) ) 20 .

Now, for y , z ∈ H , we have

‖ h 1 ( t , y ( t ) ) − h 1 ( t , z ( t ) ) ‖ = ( 15 − 2 t ) y ( t ) 25 − ( 15 − 2 t ) z ( t ) 25 = 15 − 2 t 25 ( y ( t ) − z ( t ) ) = 15 − 2 t 25 ∥ y ( t ) − z ( t ) ∥ ≤ 3 5 ∥ y ( t ) − z ( t ) ∥ ,

‖ h 2 ( t , y ( t ) ) − h 2 ( t , z ( t ) ) ‖ = ( 5 − 2 t ) sin ( y ( t ) ) 43 − ( 5 − 2 t ) sin ( z ( t ) ) 43 = 5 − 2 t 43 ( sin ( y ( t ) ) − sin ( z ( t ) ) ) = 5 − 2 t 43 ∥ y ( t ) − z ( t ) ∥ ≤ 5 43 ∥ y ( t ) − z ( t ) ∥ ,

and

‖ Θ ( t , s , y ( s ) ) − Θ ( t , s , z ( s ) ) ‖ = y ( s ) exp ( − ( t + s ) ) 20 − y ( s ) exp ( − ( t + s ) ) 20 = exp ( − ( t + s ) ) 20 ( y ( s ) − z ( s ) ) = exp ( − ( t + s ) ) 20 ∥ y ( s ) − z ( s ) ∥ ≤ 1 20 ∥ y ( s ) − z ( s ) ∥ ,

for each t , s ∈ I and ( t , s ) ∈ G . Hence, μ 1 = 17 25 , μ 2 = 7 43 , μ ∗ = 1 20 , and so

μ = max { μ 1 , μ 2 , μ ∗ } = 17 25 .

Also, we obtain

‖ h 1 ( t , y ( t ) ) ‖ = ( 15 − 2 t ) y ( t ) 25 ≤ 15 − 2 t 25 ‖ y ( t ) ‖ , ‖ h 2 ( t , y ( t ) ) ‖ = ( 5 − 2 t ) sin ( y ( t ) ) 43 ≤ 5 − 2 t 43 ‖ y ( t ) ‖ , ‖ Θ ( t , s , y ( s ) ) ‖ ≤ y ( s ) exp ( − ( t + s ) ) 20 ≤ exp ( − ( t + s ) ) 20 ‖ y ( s ) ‖ ,

for each t , s ∈ I . Hence,

ϱ 1 ( t ) = 15 − 2 t 25 , ϱ 2 ( t ) = 5 − 2 t 43 , ϱ ∗ ( t ) = exp ( − t ) 20 ,

for all t ∈ I ¯ , y , z ∈ H and ( t , s ) ∈ G . By the above, we find that

(5.2) λ = ‖ ϱ 1 ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ B q ( σ + 1 , σ + ν ) Γ q ( σ + 1 ) Γ q ( σ + ν ) + ∣ η ∣ ‖ ϱ 1 ‖ L ∞ τ ∗ σ + ν + 1 + ∣ η ∣ ‖ ϱ ∗ ‖ L ∞ τ ∗ σ + ν + 1 ∣ 1 − η τ ∗ ∣ Γ q ( σ + ν + 2 ) + ∣ η ∣ ‖ ϱ 2 ‖ L ∞ τ ∗ 2 σ + ν + 1 B q ( σ + 1 , σ + ν + 1 ) ∣ 1 − η τ ∗ ∣ Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) = 3 5 + 1 20 Γ q 5 11 + 3 7 + 1 + 5 43 B q 5 11 + 1 , 5 11 + 3 7 Γ q 5 11 + 1 Γ q 5 11 + 3 7 + − 15 2 × 3 5 × 0.6 5 11 + 3 7 + 1 + − 15 2 × 1 20 × 0.6 5 11 + 3 7 + 1 1 − 0.6 − 15 2 Γ q 5 11 + 3 7 + 2 + − 15 2 × 5 43 × 0 . 6 10 11 + 3 7 + 1 B q 5 11 + 1 , 5 11 + 3 7 + 1 1 − 5 11 × 0.6 Γ q 5 11 + 1 Γ q 5 11 + 3 7 + 1 ,

and

(5.3) λ ∗ = ∣ η ∣ ∣ 1 − η τ ∗ ∣ 2 τ ∗ σ + ν + 1 Γ q ( σ + ν + 2 ) + τ ∗ 2 σ + ν + 1 B q ( σ + 1 , σ + ν + 1 ) Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) = − 15 2 1 − 0.6 − 15 2 2 × 0 . 6 5 11 + 3 7 + 1 Γ q 5 11 + 3 7 + 2 + 0 . 6 2 5 11 + 3 7 + 1 B q 5 11 + 1 , 5 11 + 3 7 + 1 Γ q 5 11 + 1 Γ q 5 11 + 3 7 + 1 .

Considering q = 3 8 , 1 2 , 8 9 , we can see the results of λ and λ ∗ in Table 1. These results are plotted in Figure 1. Then, we obtain

λ j ≃ 0.95547 , q j = 3 8 , 0.76172 , q j = 1 2 , 0.16793 , q j = 8 9 , < 1 , λ j ∗ ≃ 1.41986 , q j = 3 8 , 1.13395 , q j = 1 2 , 0.25096 , q j = 8 9 , < 1 , μ λ j ∗ ≃ 0.9655 , q j = 3 8 , 0.7711 , q j = 1 2 , 0.1707 , q j = 8 9 , < 1 .

All assumptions of Theorem 3.2 are satisfied. Hence, there exists at least one solution for problem (5.1) on I ¯ . One can use Algorithm 9 to obtain these results. By taking the same functions, we result the assumption

μ λ j ≃ 0.6497 , q j = 3 8 , 0.5180 , q j = 1 2 , 0.1142 , q j = 8 9 , < 1 ,

then system (5.1) is Ulam–Hyers stable, then it is generalized Ulam–Hyers stable. It is Ulam–Hyers–Rassias stable if there exists a continuous and positive function ϱ j ∈ C ( I ¯ , R + ) such that

‖ y ( t ) − z ( t ) ‖ ≤ ε j ∗ ϱ ( t ) = ε j ϱ ( t ) 1 − μ λ j ,

which it satisfies in assumption of Theorem 4.2.

Table 1

Numerical results of λ and λ ∗ for q = 3 8 , 1 2 , 8 9 in Example 5.1

n	q = 3 8		q = 1 2		q = 8 9
	λ	λ ∗	λ	λ ∗	λ	λ ∗
1	0.93177	1.34571	0.71630	0.99360	0.11402	0.07701
2	0.94654	1.39205	0.73885	1.06376	0.11895	0.09638
3	0.95212	1.40943	0.75025	1.09885	0.12377	0.11354
4	0.95422	1.41595	0.75598	1.11640	0.12828	0.12878
5	0.95500	1.41840	0.75885	1.12518	0.13242	0.14232
6	0.95530	1.41931	0.76029	1.12957	0.13618	0.15436
7	0.95541	1.41966	0.76101	1.13176	0.13957	0.16506
8	0.95545	1.41978	0.76137	1.13286	0.14262	0.17458
9	0.95546	1.41983	0.76155	1.13341	0.14536	0.18304
10	0.95547 ̲	1.41985	0.76164	1.13368	0.14781	0.19057
11	0.95547	1.41986 ̲	0.76168	1.13382	0.15001	0.19727
12	0.95547	1.41986	0.76170	1.13389	0.15197	0.20322
13	0.95547	1.41986	0.76172 ̲	1.13392	0.15372	0.20852
14	0.95547	1.41986	0.76172	1.13394	0.15528	0.21323
15	0.95547	1.41986	0.76172	1.13395 ̲	0.15667	0.21741
16	0.95547	1.41986	0.76173	1.13395	0.15791	0.22114
17	0.95547	1.41986	0.76173	1.13396	0.15901	0.22445
18	0.95547	1.41986	0.76173	1.13396	0.16000	0.22739
⋮	⋮	⋮	⋮	⋮	⋮	⋮
76	0.95547	1.41986	0.76173	1.13396	0.16792	0.25095
77	0.95547	1.41986	0.76173	1.13396	0.16792	0.25096 ̲
78	0.95547	1.41986	0.76173	1.13396	0.16793 ̲	0.25096
79	0.95547	1.41986	0.76173	1.13396	0.16793	0.25096
80	0.95547	1.41986	0.76173	1.13396	0.16793	0.25096

$Figure 1 Graphical representation of λ \lambda , λ ∗ {\lambda }^{\ast } , and μ λ \mu \lambda , μ λ ∗ \mu {\lambda }^{\ast } for q = 3 8 q=\frac{3}{8} , 1 2 \frac{1}{2} , 8 9 \frac{8}{9} in Example 5.1. (a) λ \lambda , Eq. (3.8). (b) μ λ \mu \lambda . (c) λ ∗ {\lambda }^{\ast } , Eq. (3.9). (d) μ λ ∗ \mu {\lambda }^{\ast } .$

Figure 1

Graphical representation of λ , λ ∗ , and μ λ , μ λ ∗ for q = 3 8 , 1 2 , 8 9 in Example 5.1. (a) λ , Eq. (3.8). (b) μ λ . (c) λ ∗ , Eq. (3.9). (d) μ λ ∗ .

In the next example, we review and check Theorem 3.3 numerically.

Example 5.2

Consider the following fractional integro-differential problem:

(5.4) D q 29 45 C [ y ] ( t ) = ( 16 − t ) tan − 1 ( y ( t ) ) 75 + I q 4 9 2 t sin − 1 ( y ( t ) ) 21 + ∫ 0 t y ( ξ ) exp ( − ( 3 t + ξ ) ) 10 d ξ ,

with boundary condition

y ( 0 ) = − 5 2 ∫ 0 0.95 y ( ξ ) d ξ , ∀ t ∈ I .

Clearly σ + ν = 29 45 , σ = 4 9 , τ ∗ = 0.95 , and η = 5 2 . To illustrate our results in Theorem 3.3, we take for y , z ∈ H = R + and t ∈ I ¯ the following continuous functions:

h 1 ( t , y ( t ) ) = ( 16 − t ) tan − 1 ( y ( t ) ) 75 , h 2 ( t , y ( t ) ) = 2 t sin − 1 ( y ( t ) ) 21 ,

and

Θ ( t , s , y ( s ) ) = y ( s ) exp ( − ( 3 t + s ) ) 10 .

Now, for y , z ∈ H , we have

‖ h 1 ( t , y ( t ) ) − h 1 ( t , z ( t ) ) ‖ = ( 16 − t ) tan − 1 ( y ( t ) ) 75 − ( 16 − t ) tan − 1 ( z ( t ) ) 75 = 16 − t 75 ( y ( t ) − z ( t ) ) = 16 − t 75 ∥ y ( t ) − z ( t ) ∥ ≤ 17 75 ∥ y ( t ) − z ( t ) ∥ ,

‖ h 2 ( t , y ( t ) ) − h 2 ( t , z ( t ) ) ‖ = 2 t sin − 1 ( y ( t ) ) 21 − 2 t sin − 1 ( z ( t ) ) 43 = 2 t 21 ( sin ( y ( t ) ) − sin ( z ( t ) ) ) = 2 t 21 ∥ y ( t ) − z ( t ) ∥ ≤ 2 21 ∥ y ( t ) − z ( t ) ∥ ,

and

‖ Θ ( t , s , y ( s ) ) − Θ ( t , s , z ( s ) ) ‖ = y ( s ) exp ( − ( 3 t + s ) ) 10 − y ( s ) exp ( − ( 3 t + s ) ) 10 = exp ( − ( 3 t + s ) ) 10 ( y ( s ) − z ( s ) ) = exp ( − ( 3 t + s ) ) 10 ∥ y ( s ) − z ( s ) ∥ ≤ 1 10 ∥ y ( s ) − z ( s ) ∥ ,

for each t , s ∈ I and ( t , s ) ∈ G . Hence, μ 1 = 17 75 , μ 2 = 2 21 , μ ∗ = 1 10 , and so

μ = max { μ 1 , μ 2 , μ ∗ } = 17 25 .

Also, we obtain

‖ h 1 ( t , y ( t ) ) ‖ = ( 16 − t ) tan − 1 ( y ( t ) ) 75 ≤ 16 − t 75 ‖ y ( t ) ‖ , ‖ h 2 ( t , y ( t ) ) ‖ = 2 t sin − 1 ( y ( t ) ) 21 ≤ 2 t 21 ‖ y ( t ) ‖ , ‖ Θ ( t , s , y ( s ) ) ‖ ≤ y ( s ) exp ( − ( 3 t + s ) ) 10 ≤ exp ( − ( 3 t + s ) ) 10 ‖ y ( s ) ‖ ,

for each t , s ∈ I . Hence,

ϱ 1 ( t ) = 16 − t 75 , ϱ 2 ( t ) = 2 t 21 , ϱ ∗ ( t ) = exp ( − 3 t ) 10 ,

for all t ∈ I ¯ , y , z ∈ H , and ( t , s ) ∈ G . By the above, we find that

(5.5) λ = ‖ ϱ 1 ‖ L ∞ + ‖ ϱ ∗ ‖ L ∞ Γ q ( σ + ν + 1 ) + ‖ ϱ 2 ‖ L ∞ B q ( σ + 1 , σ + ν ) Γ q ( σ + 1 ) Γ q ( σ + ν ) + ∣ η ∣ ‖ ϱ 1 ‖ L ∞ τ ∗ σ + ν + 1 + ∣ η ∣ ‖ ϱ ∗ ‖ L ∞ τ ∗ σ + ν + 1 ∣ 1 − η τ ∗ ∣ Γ q ( σ + ν + 2 ) + ∣ η ∣ ‖ ϱ 2 ‖ L ∞ τ ∗ 2 σ + ν + 1 B q ( σ + 1 , σ + ν + 1 ) ∣ 1 − η τ ∗ ∣ Γ q ( σ + 1 ) Γ q ( σ + ν + 1 ) = 16 75 + 1 10 Γ q 4 9 + 1 5 + 1 + 2 21 B q 4 9 + 1 , 4 9 + 1 5 Γ q 4 9 + 1 Γ q 4 9 + 1 5 + ∣ 2.5 ∣ × 16 75 0.9 5 4 9 + 1 5 + 1 + ∣ 2.5 ∣ 1 10 0.9 5 4 9 + 1 5 + 1 ∣ 1 − 2.5 × 0.95 ∣ Γ q 4 9 + 1 5 + 2 + ∣ 2.5 ∣ 4 9 × 2 21 × 0.9 5 8 9 + 1 5 + 1 B q 4 9 + 1 , 4 9 + 1 5 + 1 ∣ 1 − 2.5 × 0.95 ∣ Γ q 4 9 + 1 Γ q 4 9 + 1 5 + 1 .

Considering q = 2 7 , 1 2 , 9 11 , we can see the results of λ and λ ∗ in Table 2. These results are plotted in Figure 2. Then, we obtain

λ j ≃ 0.81987 , q j = 3 8 , 0.57290 , q j = 1 2 , 0.20831 , q j = 8 9 , μ λ j ≃ 0.55751 , q j = 3 8 , 0.38957 , q j = 1 2 , 0.14165 , q j = 8 9 , < 1 .

All assumptions of Theorem 3.3 are satisfied. Hence, there exists at least one solution for problem (5.4) on I ¯ .

Table 2

Numerical results of λ and μ λ for q = 2 7 , 1 2 , 9 11 in Example 5.2

n	q = 2 7		q = 1 2		q = 9 11
	λ	μ λ	λ	μ λ	λ	μ λ
1	0.81214	0.55225	0.54150	0.36822	0.15811	0.10752
2	0.81764	0.55600	0.55700	0.37876	0.16610	0.11295
3	0.81923	0.55708	0.56491	0.38414	0.17332	0.11785
4	0.81969	0.55739	0.56890	0.38685	0.17947	0.12204
5	0.81982	0.55748	0.57090	0.38821	0.18462	0.12554
6	0.81986	0.55750	0.57190	0.38889	0.18887	0.12843
7	0.81987 ̲	0.55751 ̲	0.57240	0.38923	0.19238	0.13082
8	0.81987	0.55751	0.57265	0.38940	0.19526	0.13278
9	0.81987	0.55751	0.57278	0.38949	0.19763	0.13439
10	0.81987	0.55751	0.57284	0.38953	0.19956	0.13570
11	0.81987	0.55751	0.57287	0.38955	0.20115	0.13678
12	0.81987	0.55751	0.57289	0.38956	0.20245	0.13767
13	0.81987	0.55751	0.57290 ̲	0.38957 ̲	0.20352	0.13839
14	0.81987	0.55751	0.57290	0.38957	0.20439	0.13898
15	0.81987	0.55751	0.57290	0.38957	0.20510	0.13947
⋮	⋮	⋮	⋮	⋮	⋮	⋮
43	0.81987	0.55751	0.57290	0.38957	0.20830	0.14164
44	0.81987	0.55751	0.57290	0.38957	0.20830	0.14165 ̲
45	0.81987	0.55751	0.57290	0.38957	0.20830	0.14165
46	0.81987	0.55751	0.57290	0.38957	0.20831 ̲	0.14165
47	0.81987	0.55751	0.57290	0.38957	0.20831	0.14165

$Figure 2 Graphical representation of λ \lambda and μ λ \mu \lambda for q = 2 7 q=\frac{2}{7} , 1 2 \frac{1}{2} , 9 11 \frac{9}{11} in Example 5.2. (a) λ \lambda , Eq. (3.8). (b) μ λ \mu \lambda .$

Figure 2

Graphical representation of λ and μ λ for q = 2 7 , 1 2 , 9 11 in Example 5.2. (a) λ , Eq. (3.8). (b) μ λ .

6 Conclusion

Determining the answer of differential equations from the order of fractions in the discrete state simplifies many problems. The q-integro-differential boundary equations and their applications have attracted several researchers’ interests in the field of fractional q-calculus and its applications in various phenomena from science and technology. q-Integro-differential boundary value problems occur in the mathematical modeling of a variety of physical operations. Using the Krasnoselskii’s, Banach fixed point theorems, we prove existence and uniqueness results. Based on the results obtained, conditions are provided to ensure the generalized Ulam stability of the original system. The results of Eq. (1.9) investigation on a time scale are illustrated by two numerical examples.

mesamei@gmail.com, artionkashuri@gmail.com

Funding information: There is no funding to declare for this research study.
Author contributions: X-GY: Actualization, methodology, formal analysis, validation, investigation, initial draft. MES: Actualization, methodology, formal analysis, validation, investigation, software, simulation, initial draft, and was a major contributor in writing the manuscript. AF: Actualization, methodology, formal analysis, validation, investigation, and initial draft. MKAK: Actualization, methodology, formal analysis, validation, investigation, initial draft, and supervision of the original draft and editing. AK: Actualization, validation, methodology, formal analysis, investigation, and initial draft. All authors read and approved the final manuscript.
Conflict of interest: The authors declare that they have no competing interests.
Data availability statement: Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

Appendix

Algorithm 1: MATLAB lines for calculation q -factorial function ( x − y ) q ( m )
	Input: y , z , q , m
	Output: H
1	if m=0
2	⌊ totalout=1
3	else
4 5 6 7	totalout=1 ; for n = 0 : m − 1 do ⌊ totalout = totalout ∗ ( y − z ∗ q n ) ; H=totalout;

Algorithm 2: MATLAB lines for calculation q -factorial function ( y − z ) q ( σ ) .
	Input: y , z , q , sigma, m
	Output: H
1	if m=0 then
2	⌊ H=1
3	else
4 5 6 7	totalout=1 ; for n = 0 : m − 1 do ⌊ totalout = totalout ∗ ( y − z ∗ q n ) / ( y − z ∗ q s i g m a + n ) ; H=totalout ∗ y s i g m a ;

Algorithm 3: MATLAB lines for calculation Γ q ( y ) .
	Input: q, y, m
	Output: H
1	totalout=1;
2	for n = 0 : m do
3	⌊ totalout = totalout ∗ ( 1 − q ( n + 1 ) ) / ( 1 − q ( n + k − 1 ) ) ;
4	H = totalout ∗ ( 1 − q ) ( 1 − y ) ;

Algorithm 4: MATLAB lines for calculation D q [ y ] ( t ) .
	Input: q, s, fun
	Output: H
1	if s = 0 then
2	⌊ H=limit((subs(fun,s) - subs(fun,q ∗ s))/((1-q) ∗ s),s,0);
3	else
4	⌊ H=(eval(subs(fun,s)) - eval(subs(fun,q ∗ s)))/((1-q) ∗ s);

Algorithm 5: MATLAB lines for calculation D q m [ y ] ( t ) .
	Input: q, s, m, func
	Output: H
1	totalout=0;
2	for n = 0 : m do
3 4 5 6 7	V = 1 ; for i = 0 : m − 1 do ⌊ V = V ∗ ( 1 − q ( i − m ) ) / ( 1 − q ( i + 1 ) ) ; V = V ∗ q n ∗ eval ( subs ( func , s ∗ q n ) ) ; totalout = totalout + V ;
8	H=totalout/( s m ∗ ( 1 − q ) m );

Algorithm 6: MATLAB lines for calculation I q [ y ] ( t ) .
	Input: q, s, m, func
	Output: H
1	totalout=1;
2	for n = 0 : m do
3	⌊ totalout = totalout + q n ∗ eval(subs(funx, totalout ∗ q n ));
4	H = s ∗ (1-q) ∗ totalout;

Algorithm 7: MATLAB lines for calculation I q σ [ y ] ( t ) .
	Input: q, sigma, s, m, func
	Output: H
1	totalout=0;
2	for n = 0 : m do
3 4 5 6	V = 1 ; for i = 0 : m do ⌊ V = V ∗ ( 1 − q ( s i g m a + i − 1 ) ) ∗ ( 1 − q ( n + i ) ) / ( ( 1 − q ( i + 1 ) ) ∗ ( 1 − q ( s i g m a + n + i − 1 ) ) ) ; totalout = totalout + q k ∗ V ∗ eval ( subs ( func , s ∗ q n ) ) ;
7	H = round(totalout ∗ ( s s i g m a ) ∗ ( 1 − q ) s i g m a , 6 );

Algorithm 8: MATLAB lines for calculation D q σ C [ y ] ( t ) .
	Input: q, sigma, s, m, func
	Output: H
1	Tootalout=0;
2	for n = 0 : m do
3 4 5 6 7 8 9 10 11 12 13 14 15	V = 1 ; for i = 0 : m do V = V * ( 1 − q ( f l o o r ( s i g m a ) − s i g m a + i − 1 ) ) … * ( 1 − q ( k + i ) ) / ( ( 1 − q ( i + 1 ) ) … * ( 1 − q ( f l o o r ( s i g m a ) − s i g m a + k + i − 1 ) ) ) ; Tootalout2=0 ; for k = 0 : f l o o r ( s i g m a ) do V2=1 ; for i = 0 : k − 1 do V 2 = V 2 * ( 1 − q ( i − f l o o r ( s i g m a ) ) ) / ( 1 − q ( i + 1 ) ) ; V 2 = V 2 * q k * eval ( subs ( fun , s * q ( n + k ) ) ) ; Tootalout2 = Tootalout2 + V2 ; Tootalout = Tootalout + V * Tootalout2 ;
16	H = round(totalout ∗ ( s s i g m a ) ∗ ( 1 − q ) s i g m a , 6 );

Algorithm 9: MATLAB lines for calculating values of λ and λ ∗ in Example ?? for q = 3 8 , 1 2 , 8 9 .
	Input: q, sigma, nu, eta, taustar
	Output: H
1	clear;
2	format long;
3	column=1
4	for s = 1 : y q do
5 6 7 8 9 10 11 12 13 14 15	for n = 1 : k paramsmatrix(n, column)=n ; C1=qGamma(q(s),sigma+1,n); C2=qGamma(q(s),sigma+nu,n); C3=qGamma(q(s),sigma+nu+1,n); C4=qGamma(q(s),2sigma+nu+1,n); C5=qGamma(q(s),sigma+nu+2,n); C6=qGamma(q(s),2sigma+nu+2,n); paramsmatrix(n, column+1) = round ( (normvarrho1 + normvarrhostar)/C3 + normvarrho2/C4 + ( abs(eta) ∗ normvarrho1 ∗ t a u s t a r ( s i g m a + n u + 1 ) + abs(eta) ∗ * normvarrhostar ∗ taustar ( sigma + n u + 1 ) ) ∕ ( abs ( 1 − eta ∗ taustar ) * C 5 ) + abs ( eta ) ∗ normvarrho2 ∗ taustar ( 2 ∗ sigma + nu + 1 ) ∕ C 6 , 6 ) ; paramsmatrix ( n , column + 2 ) = round ( abs ( eta ) ∕ abs ( 1 − eta * taustar ) * ( 2 * taustar ( sigma + nu + 1 ) / C 5 + taustar ( 2 * sigma + nu + 1 ) / C 6 ) , 6 ) ; column=column +3 ;

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Received: 2022-03-15

Revised: 2022-05-03

Accepted: 2022-05-18

Published Online: 2022-06-11

This work is licensed under the Creative Commons Attribution 4.0 International License.

Using Krasnoselskii's theorem to investigate the Cauchy and neutral fractional q-integro-differential equation via numerical technique

Abstract

1 Introduction and formulation of the problem

2 Notations and preliminaries

Remark 2.1

Remark 2.2

Remark 2.3

Lemma 2.4

Lemma 2.5

Lemma 2.6

Lemma 2.7

Lemma 2.8

3 Existence results

Lemma 3.1

Proof

3.1 Existence result by Krasnoselskii’s fixed point

Theorem 3.2

Proof

3.2 Existence and uniqueness result

Theorem 3.3

Proof

4 Generalized Ulam stabilities

Theorem 4.1

Proof

Theorem 4.2

Proof

5 Illustrative of our outcome

Example 5.1

Example 5.2

6 Conclusion

Appendix

References

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