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Article

Nonlocal Impulsive Fractional Integral Boundary Value Problem for (ρk,ϕk)-Hilfer Fractional Integro-Differential Equations

by
Marisa Kaewsuwan
1,†,
Rachanee Phuwapathanapun
1,†,
Weerawat Sudsutad
1,†,
Jehad Alzabut
2,3,†,
Chatthai Thaiprayoon
4,*,† and
Jutarat Kongson
4,†
1
Theoretical and Applied Data Integration Innovations Group, Department of Statistics, Faculty of Science, Ramkhamhaeng University, Bangkok 10240, Thailand
2
Department of Mathematics and Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia
3
Department of Industrial Engineering, OSTİM Technical University, 06374 Ankara, Türkiye
4
Research Group of Theoretical and Computation in Applied Science, Department of Mathematics, Faculty of Science, Burapha University, Chonburi 20131, Thailand
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2022, 10(20), 3874; https://doi.org/10.3390/math10203874
Submission received: 6 September 2022 / Revised: 9 October 2022 / Accepted: 13 October 2022 / Published: 18 October 2022
(This article belongs to the Special Issue Fractional Differential Equations: Theory and Application)

Abstract

:
In this paper, we establish the existence and stability results for the ( ρ k , ϕ k ) -Hilfer fractional integro-differential equations under instantaneous impulse with non-local multi-point fractional integral boundary conditions. We achieve the formulation of the solution to the ( ρ k , ϕ k ) -Hilfer fractional differential equation with constant coefficients in term of the Mittag–Leffler kernel. The uniqueness result is proved by applying Banach’s fixed point theory with the Mittag–Leffler properties, and the existence result is derived by using a fixed point theorem due to O’Regan. Furthermore, Ulam–Hyers stability and Ulam–Hyers–Rassias stability results are demonstrated via the non-linear functional analysis method. In addition, numerical examples are designed to demonstrate the application of the main results.

1. Introduction

Fractional calculus ( FC ) is discussed as the fractional integral operator ( FIO ) and fractional derivative operator ( FDO ), which have a long and illustrious history. FC is popularly used to analyze phenomena in the branch of mathematical analysis, which is noticed to be of outstanding assistance in modifying complex real-world problems in many fields, such as physical sciences [1], financial economics [2], dynamics of particles, fields and media [3], bio-engineering [4], Zika [5], HIV [6], COVID-19 [7], ecology [8], continuum mechanics [9], Navier–Stokes problem [10], social media addiction [11], and references cited therein. For more theoretical details on this topic, see: [12,13,14,15,16]. A variety of types of FDO s are regularly settled in the sense of FIO s. Various types of FDO s with different kernel terms, such as Riemann–Liouville ( RL ), Caputo, Hadamard, Katugampola, Hilfer, and others, are shown in the literature survey on FC .
Recently, in 2018, the concept of FDO with respect to another function was developed by Sousa and Oliveira [17], which is known as the ϕ -Hilfer FDO . Some existence and stability results of the solutions for fractional differential equations ( FDE s) were created in the context of ϕ -Hilfer FDO [18,19,20,21,22] and the references therein. After that, in 2021, Kucche and Mali [23] introduced and demonstrated some properties of the ( ρ , ϕ )-Hilfer FDO . They applied Banach’s type to analyzed the uniqueness result for the non-linear FDE s under ( ρ , ϕ )-Hilfer FDO :
ρ H D a + α , β ; ϕ u ( t ) = f ( t , u ( t ) ) , t ( a , b ] , α ( 0 , ρ ) , β [ 0 , 1 ] , ρ I a + ρ γ r ; ϕ u ( a ) = u a R , γ r = α + β ( ρ α ) , ρ > 0 ,
where ρ H D a + α , β ; ϕ is the ( ρ , ϕ )-Hilfer FDO of order α and type β with α ( 0 , 1 ) , and f C ( [ a , b ] × R , R ) , 0 a < b < . It is worth noting the ( ρ , ϕ )-Hilfer FDO , which can be generalized as various known FDO s (see more details in Remark 2).
The physical and social sciences are explained by applying impulsive differential equations with integer order and fractional order. They are also applicable to dynamical systems, such as evolutionary processes, which show instantaneous state changes at some points. The qualitative theory of impulsive FDE s, such as existence theory and stability results, has been widely employed in engineering and applied sciences throughout the last several decades (see [24,25,26]). Many researchers will attempt to operate in the area of impulsive FDE s with impulses and have presented essential and interesting results through the years that have contributed greatly to the mathematical analysis of FDE s with impulses effect. In 2009, Benchohra and Slimani [27] studied a variety of conditions for the existence of the solutions for the impulsive Caputo-type FDE with initial condition by using Banach’s, Leray–Schauder’s, and Schaefer’s fixed point theorems. Later, the impulsive FDE s in [27] have been extended and studied for their existence results in Banach spaces by Benchohra and Seba [28]. In 2012, Wang et al. [29] investigated the piecewise continuous solutions to the problem in [27,28]. The existence, uniqueness, and Ulam’s stability results of solutions for the impulsive boundary value problems ( BVP s) are obtained by using a fixed point theorem via generalized Gronwall inequalities. In 2014, Wang and Lin [30] investigated the existence of solutions to impulsive Caputo FDE s under anti-periodic boundary conditions via constant coefficients. The formula of solutions to the problem in [30] was constructed in the sense of Mittag–Leffler kernels. At the same time, the Lipschitz and non-linear growth conditions were used to establish the existence results of solutions to the problem in [30]. In 2017, Zuo et al. [31] established the existence and uniqueness results for impulsive anti-periodic BVP s through fractional integro-differential equation ( FIDE ) with constant coefficient based on Banach’s and Krasnoselskii’s types. In 2020, Kucche et al. [32] developed the existence results of solutions for the non-linear ϕ -Hilfer impulsive FDE with initial condition:
H D a + α , β ; ϕ u ( t ) = f ( t , u ( t ) ) , t J k ( a , T ] , t t k , T > a , Δ I a + 1 γ ; ϕ u ( t k ) = ξ k R , k = 1 , 2 , , m , I a + 1 γ ; ϕ u ( a ) = δ R , γ = α + β α β ,
where H D a + α , β ; ϕ is the ϕ -Hilfer FDO of order α ( 0 , 1 ) and type β [ 0 , 1 ] , I a + 1 γ ; ϕ is the ϕ - RL - FIO of order 1 γ > 0 . In addition, they extended the problem (2) to the non-local ϕ -Hilfer FDE . In 2022, based on Banach’s and Schauder’s types, Salim et al. [33] proved the existence and uniqueness of solutions for the non-linear implicit ρ -generalized ϕ -Hilfer FDE - BVP s via retardation and anticipation:
ρ H D a + α , β ; ϕ u ( t ) = f ( t , u t ( t ) , ρ H D a + α , β ; ϕ u ( t ) ) , t ( a , b ] , a 1 ρ I a + ρ ( 1 ξ ) ; ϕ u ( a + ) + a 2 ρ I a + ρ ( 1 γ ) ; ϕ u ( b ) = a 3 , γ = 1 ρ ( β ( ρ α ) + α ) , u ( t ) = ω ( t ) , t [ a λ , a ] , λ > 0 , a 1 , a 2 , a 3 R , u ( t ) = ω ¯ ( t ) , t [ b , b + λ ¯ ] , λ ¯ > 0 , a 1 + a 2 0 ,
where ρ H D a + α , β ; ϕ and ρ I a + ρ ( 1 γ ) ; ϕ are the ρ -generalized ϕ -Hilfer FDO of order α ( 0 , ρ ) , ρ > 0 and type β [ 0 , 1 ] , and the ρ -generalized ϕ -Hilfer FIO of order ρ ( 1 γ ) ( 0 , ρ ) , respectively, and u t ( τ ) = u ( t + τ ) for τ [ λ , λ ¯ ] . Note that several works have been published using concentrated and important tools in mathematical analysis. We suggest modern works on impulsive FDE s on existence, uniqueness, and Ulam’s stability and the reference given therein [34,35,36,37,38,39,40,41,42,43,44,45,46].
To motivate the enrichment of novel literature for interested researchers, in this paper, we establish qualitative results of the solutions for the following non-linear impulsive ( ρ k , ϕ k ) -Hilfer FIDE s with non-local multi-point fractional integral boundary conditions ( NMP - FIBC s) as:
ρ k H D t k + α k , β k ; ϕ k u ( t ) = λ k u ( t ) + f ( t , u ( t ) , ρ k I t k δ k ; ϕ k u ( t ) , ρ k I t k θ k ; ϕ k u ( t ) ) , t J k , t t k , Δ ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k ) = φ k ( u ( t k ) ) , k = 1 , 2 , , m , i = 0 m κ i u ( η i ) = j = 0 n ω j ρ j I t j + μ j ; ϕ j u ( ξ j ) + A , η i ( t i , t i + 1 ] , ξ j ( t j , t j + 1 ] .
where ρ k H D t k + α k , β k ; ϕ k denotes the ( ρ k , ϕ k ) -Hilfer FDO of order α k and type β k on J k , 0 < α k < 1 , 0 β k 1 , ρ k > 0 , J k = ( t k , t k + 1 ] ( 0 , T ] , k = 0 , 1 , , m , J 0 = [ a , t 1 ] , J = [ a , b ] , 0 a = t 0 < t 1 < < t m < t m + 1 = b , ρ k I t k q ; ϕ k is the ( ρ k , ϕ k ) - RL   FIO of order q { δ k , θ k , μ k } > 0 , k = 0 , 1 , , m , φ k C ( R , R ) , Δ ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k ) = ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k + ) ρ k 1 I t k 1 + ρ k 1 ( 1 γ k 1 ) ; ϕ k 1 u ( t k ) where ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k + ) = lim h 0 + ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k + h ) , k = 1 , 2 , , m , f C ( J , R 3 , R ) , κ i , ω j R , η i ( t i , t i + 1 ] , ξ j ( t j , t j + 1 ] , i = 0 , 1 , , m , j = 0 , 1 , , n , λ < 0 and A R . For the sake of use, the problem (4) can be called the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s.
The remaining sections of this work are structured as follows: in Section 2, some concepts of the ( ρ , ϕ )-Hilfer fractional operators related to our discussion are defined along with some essential lemmas are proved. Additionally, the solution of the linear variant of the ( ρ , ϕ )-Hilfer fractional Cauchy problem (11) is derived in the form of the generalized Mittag–Leffler kernel. After that, an equivalent integral equation to the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4). The essential lemma is very important to transform the proposed problem (4) into a fixed-point problem. In Section 3, presenting the first main results of the problem (4), the uniqueness result is proved by Banach’s type and the existence result is studied by a fixed point theorem due to O’Regan. In addition, a variety of Ulam’s stability results for problem (4) are investigated in Section 4. Finally, Section 5 shows illustrative examples to verify the main results.

2. Preliminaries

This section introduces fundamental concepts and constructs several properties of the ( ρ , ϕ ) -Hilfer fractional calculus relevant to our results.

2.1. The ( ρ , ϕ ) -Hilfer Fractional Calculus and Its Properties

Definition 1
([47]). Let f L 1 ( [ a , b ] , b ) and an increasing function ϕ ( t ) : [ a , b ] R via ϕ ( t ) 0 for t [ a , b ] . The ( ρ , ϕ ) - RL - FIO of a function f of order α > 0 is defined by
ρ I a + α ; ϕ f ( t ) = 1 ρ Γ ρ ( α ) a t ϕ ( t ) ϕ ( s ) α ρ 1 ϕ ( s ) f ( s ) d s , ρ > 0 ,
where Γ ρ ( · ) is the ρ-Gamma function which is introduced by Diaz and Pariguan [48],
Γ ρ ( z ) = 0 t z 1 e t ρ ρ d t , z C , R e ( z ) > 0 , ρ > 0 .
Some other useful properties of (5) are well known as follows:
Γ ρ ( z + ρ ) = z Γ ρ ( z ) , Γ ρ ( ρ ) = 1 , Γ ρ ( z ) = ( ρ ) z ρ 1 Γ z ρ , Γ ( z ) = lim ρ 1 Γ ρ ( z ) .
Definition 2
([23]). Let f C n ( [ a , b ] , R ) , ϕ C n ( [ a , b ] , R ) , ϕ ( t ) 0 , for t [ a , b ] , α, ρ R + , and β [ 0 , 1 ] . The ( ρ , ϕ ) -Hilfer FDO of a function f of order α and type β is given by
ρ H D a + α , β ; ϕ f ( t ) = ρ I a + β ( ρ n α ) ; ϕ δ ϕ n ρ I a + ( 1 β ) ( ρ n α ) ; ϕ f ( t ) ,
where δ ϕ n = ρ ϕ ( t ) d d t n and n = α ρ .
Remark 1.
The ( ρ , ϕ )-Hilfer FDO can be rewritten in the sense of the ( ρ , ϕ )- RL - FDO as follows:
ρ H D a + α , β ; ϕ f ( t ) = ρ I a + γ α ; ϕ ρ ϕ ( t ) d d t n ρ I a + n ρ γ ; ϕ f ( t ) = ρ I a + γ α ; ϕ ρ RL D a + γ ; ϕ f ( t ) ,
where
ρ RL D a + γ ; ϕ f ( t ) = ρ ϕ ( t ) d d t n ρ I a + n ρ γ ; ϕ f ( t ) ,
and n ρ γ = ( 1 β ) ( n ρ α ) with γ ρ ( n 1 , n ] .
Remark 2.
It is noticed that:
( A 1 )
If we take β = 0 in (7), then we have the ( ρ , ϕ )- RL - FDO defined in [23], while if ϕ ( t ) = t with β = 0 , then we obtain the ρ- RL - FDO defined in [49].
( A 2 )
If we take β = 1 in (7), then we have the ( ρ , ϕ )-Caputo FDO defined in [23], while if we take ϕ ( t ) = t with β = 0 , then we obtain the ρ-Caputo FDO defined in [23].
( A 3 )
If we take ϕ ( t ) = t σ in (7), then we have the ρ-Hilfer–Katugampola FDO , that is:
(i) 
If we take ϕ ( t ) = t σ with β = 0 in (7), then we have the ρ-Katugampola FDO defined in [50].
( i i )
If we take ϕ ( t ) = t σ with β = 1 in (7), then we have the ρ-Caputo–Katugampola FDO defined in [50].
( A 4 )
If we take ϕ ( t ) = log t in (7), then we have the ρ-Hilfer–Hadamard FDO , that is:
(i) 
If we take ϕ ( t ) = log t with β = 0 in (7), then we have the ρ-Hadamard FDO defined in [23].
( i i )
If we take ϕ ( t ) = log t with β = 1 in (7), then we have the ρ-Caputo–Hadamard FDO defined in [23].
Some important basic properties, which are used throughout this paper, are as follows:
Lemma 1
([23]). Let α, ρ > 0 and β R , such that β ρ > 1 . Then, we have
(i) 
ρ I a + α ; ϕ ϕ ( t ) ϕ ( a ) β ρ = Γ ρ ( β + ρ ) Γ ρ ( β + ρ + α ) ϕ ( t ) ϕ ( a ) β + α ρ .
(ii) 
ρ R L D a + α ; ϕ ϕ ( t ) ϕ ( a ) β ρ = Γ ρ ( β + ρ ) Γ ρ ( β + ρ α ) ϕ ( t ) ϕ ( a ) β α ρ .
(iii) 
ρ I a + α ; ϕ ρ I a + β ; ϕ f ( t ) = ρ I a + α + β ; ϕ f ( t ) = ρ I a + β ; ϕ ρ I a + α ; ϕ f ( t ) .
Lemma 2
([33]). If f C n ( [ a , b ] , R , α ρ ( n 1 , n ) , β [ 0 , 1 ] , where n N and ρ > 0 , then
ρ I a + α ; ϕ ρ H D a + α , β ; ϕ f ( t ) = f ( t ) i = 1 n ϕ ( t ) ϕ ( a ) γ i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) δ ϕ n i ρ I a + ρ ( n γ ) ; ϕ f ( a ) ,
where γ = 1 ρ β ρ n α + α and n = α ρ .
Next, we provide the Mittag–Leffler functions E α ( · ) and E α , β ( · ) that will be employed throughout in this paper.
Lemma 3
([51,52]). Take α ( 0 , 1 ) , c > 0 . Hence, E α ( · ) and E α , c ( · ) are non-negative functions, and for each u < 0 , E α ( u ) 1 , E α , c ( u ) 1 / Γ ( c ) , with
E α ( u ) = n = 0 u n Γ ( α n + 1 ) a n d E α , c ( u ) = n = 0 u n Γ ( α n + c ) , u R .
For the sake of easy for calculation in this paper, we define the symbols:
Φ ϕ c 1 ( t , a ) = ϕ ( t ) ϕ ( a ) c 1 .
Lemma 4.
Assume that α > 0 , c > 0 , ρ > 0 , q > 0 , and λ R . Then, we obtain
ρ I a + α , ϕ Φ ϕ q 1 ( t , a ) E c , q λ ρ 1 Φ ϕ ( t , a ) c = ρ α ρ Φ ϕ q + α ρ 1 ( t , a ) E c , q + α ρ λ ρ 1 Φ ϕ ( t , a ) c ,
where Φ ϕ ( · ) ( t , a ) and E u , v ( · ) are given as in (8) and Lemma 3, respectively.
Proof. 
By applying Definition 1 and Lemma 3 we have
ρ I a + α , ϕ Φ ϕ q 1 ( t , a ) E c , q λ ρ 1 Φ ϕ ( t , a ) c = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( s , t ) Φ ϕ q 1 ( s , a ) E c , q λ ρ 1 Φ ϕ ( s , a ) c ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) Φ ϕ q 1 ( s , a ) n = 0 λ ρ 1 Φ ϕ ( s , a ) c n Γ c n + q ϕ ( s ) d s = 1 ρ Γ ρ ( α ) n = 0 λ ρ c n Γ c n + q a t ϕ ( t ) ϕ ( s ) α ρ 1 ϕ ( s ) ϕ ( a ) c n + q 1 ϕ ( s ) d s = n = 0 λ ρ c n Γ c n + q ρ I a + α , ϕ ϕ ( t ) ϕ ( a ) ρ ( c n + q ) ρ 1 .
By using (ii) of Lemma 1, we have
ρ I a + α , ϕ Φ ϕ q 1 ( t , a ) E c , q λ ρ 1 Φ ϕ ( t , a ) c = n = 0 λ ρ c n Γ c n + q Γ ρ ( ρ ( c n + q ) ) Γ ρ ( ρ ( c n + q ) + α ) ϕ ( t ) ϕ ( a ) ρ ( c n + q ) + α ρ 1 = n = 0 λ ρ c n Γ c n + q ρ c n + q 1 Γ ( c n + q ) ρ c n + q + α ρ 1 Γ ( c n + q + α ρ ) ϕ ( t ) ϕ ( a ) c n + q + α ρ 1 ,
which provides the desired (9). □
Lemma 5.
Suppose that α > 0 , c > 0 , ρ > 0 , λ R , and h C ( [ a , b ] ) . Then, we have
ρ I a + α , ϕ a t Φ ϕ c 1 ( t , s ) E c , c λ ρ 1 Φ ϕ ( t , s ) c h ( s ) ϕ ( s ) d s = ρ α ρ a t Φ ϕ c + α ρ 1 ( s , r ) E c , c + α ρ λ ρ 1 Φ ϕ ( s , r ) c h ( r ) ϕ ( r ) d r ,
where Φ ϕ ( · ) ( t , a ) and E u , v ( · ) are given as in (8) and Lemma 3, respectively.
Proof. 
By applying Definition 1 and Lemma 3 we have
ρ I a + α , ϕ a t Φ ϕ c 1 ( t , s ) E c , c λ ρ 1 Φ ϕ ( t , s ) c h ( s ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) a s Φ ϕ c 1 ( s , r ) E c , c λ ρ 1 Φ ϕ ( s , r ) c h ( r ) ϕ ( r ) d r ϕ ( s ) d s = a t 1 ρ Γ ρ ( α ) r t Φ ϕ α ρ 1 ( t , s ) Φ ϕ c 1 ( s , r ) E c , c λ ρ 1 Φ ϕ ( s , r ) c ϕ ( s ) d s h ( r ) ϕ ( r ) d r = a t ρ I r + α , ϕ Φ ϕ c 1 ( s , r ) E c , c λ ρ 1 Φ ϕ ( s , r ) c h ( r ) ϕ ( r ) d r .
By using Lemma 4, the equality (10) is obtained. □

2.2. The Linear ( ρ , ϕ ) -Hilfer Fractional Cauchy Problem

Consider the linear variant of the ( ρ , ϕ ) -Hilfer fractional Cauchy problem with constant coefficient as follows:
ρ H D a + α , β ; ϕ u ( t ) = λ u ( t ) + h ( t ) , α ( n 1 , n ) , β [ 0 , 1 ] , t ( a , b ] , δ ϕ n i ρ I a + ρ ( n γ ) ; ϕ u ( a ) = c i , i = 1 , 2 , , n , α γ = 1 ρ β ρ n α + α ,
where ρ H D a + α , β ; ϕ denotes the ( ρ , ϕ ) -Hilfer FDO of order α and type β , ρ I a + ρ ( n γ ) , ϕ denotes the ( ρ , ϕ ) - RL - FIO of order ρ ( n γ ) > 0 , c j R , j = 1 , 2 , , n , and λ < 0 . By applying the Picard’s successive approximation technique, we derive to construct an explicit solution to the problem (11) in form of the Mittag–Leffler kernel.
Lemma 6.
Let h C ( [ a , b ] , R ) , λ R , α ( n 1 , n ) , β [ 0 , 1 ] , and ρ > 0 . Then, the explicit solution of the problem (11) is provided by
u ( t ) = 1 ρ α ρ a t Φ ϕ α ρ 1 ( t , s ) E α ρ , α ρ λ ρ 1 Φ ϕ ( t , s ) α ρ h ( s ) ϕ ( s ) d s + i = 1 n c i ρ γ n Φ ϕ γ i ( t , a ) E α ρ , γ i + 1 λ ρ 1 Φ ϕ ( t , a ) α ρ .
Proof. 
Assume u is a solution of the problem (11). By applying Lemma 2, The corresponding an integral equation of the problem (11) can be represented as
u ( t ) = i = 1 n Φ ϕ γ i ( t , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) u ( s ) ϕ ( s ) d s + 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s .
The method of successive approximation is applied to develop an explicit form for the solution in our results. Define
u 0 ( t ) = i = 1 n Φ ϕ γ i ( t , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) , u k ( t ) = u 0 ( t ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) u k 1 ( s ) ϕ ( s ) d s + 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s , k = 1 , 2 , 3 , .
For k = 1 , by using Definition 1, we obtain
u 1 ( t ) = u 0 ( t ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) u 0 ( s ) ϕ ( s ) d s + 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n Φ ϕ γ i ( t , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) i = 1 n Φ ϕ γ i ( s , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n Φ ϕ γ i ( t , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + i = 1 n λ c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) Φ ϕ ρ ( γ i + 1 ) ρ 1 ( s , a ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n Φ ϕ γ i ( t , a ) c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + i = 1 n λ c i ρ i n Γ ρ ( ρ ( γ i + 1 ) ) ρ I a + α ; ϕ Φ ϕ ρ ( γ i + 1 ) ρ 1 ( t , a ) .
From (i) of Lemma 1, we have
u 1 ( t ) = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + i = 1 n λ c i Φ ϕ ρ ( γ i + 1 ) + α ρ 1 ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) + α ) = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n 1 Γ ρ ( ρ ( γ i + 1 ) ) + λ Φ ϕ α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + α ) = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n j = 1 2 λ j 1 Φ ϕ ( j 1 ) α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) .
By the same process, for k = 2 , one has
u 2 ( t ) = u 0 ( t ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) u 1 ( s ) ϕ ( s ) d s + 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + λ ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) { 1 ρ Γ ρ ( α ) a s Φ ϕ α ρ 1 ( s , r ) h ( r ) ϕ ( r ) d r + i = 1 n c i Φ ϕ γ i ( s , a ) ρ i n j = 1 2 λ j 1 Φ ϕ ( j 1 ) α ρ ( s , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) } ϕ ( s ) d s
= 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + λ ρ Γ ρ ( 2 α ) a t Φ ϕ 2 α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i ρ i n j = 1 2 λ j Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) × 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) Φ ϕ ρ ( γ i + 1 ) + ( j 1 ) α ρ 1 ( s , a ) ϕ ( s ) d s = 1 ρ Γ ρ ( α ) a t Φ ϕ α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + λ ρ Γ ρ ( 2 α ) a t Φ ϕ 2 α ρ 1 ( t , s ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + i = 1 n c i ρ i n j = 1 2 λ j ρ I a + α ; ϕ Φ ϕ ρ ( γ i + 1 ) + ( j 1 ) α ρ 1 ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) = 1 ρ a t Φ ϕ α ρ 1 ( t , s ) Γ ρ ( α ) + λ Φ ϕ 2 α ρ 1 ( t , s ) Γ ρ ( 2 α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n Γ ρ ( ρ ( γ i + 1 ) ) + i = 1 n c i ρ i n j = 1 2 λ j Φ ϕ ρ ( γ i + 1 ) + j α ρ 1 ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + j α ) = 1 ρ a t Φ ϕ α ρ 1 ( t , s ) Γ ρ ( α ) + λ Φ ϕ 2 α ρ 1 ( t , s ) Γ ρ ( 2 α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n 1 Γ ρ ( ρ ( γ i + 1 ) ) + j = 1 2 λ j Φ ϕ j α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + j α ) = 1 ρ a t j = 1 2 λ j 1 Φ ϕ j α ρ 1 ( t , s ) Γ ρ ( j α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n j = 1 3 λ j 1 Φ ϕ ( j 1 ) α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) .
For k = 1 , 2 , , we obtain that
u k ( t ) = 1 ρ a t j = 1 k λ j 1 Φ ϕ j α ρ 1 ( t , s ) Γ ρ ( j α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n j = 1 k + 1 λ j 1 Φ ϕ ( j 1 ) α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) .
Taking k with changing the summation index in the last expression, j j + 1 , we obtain
u ( t ) = 1 ρ a t j = 1 λ j 1 Φ ϕ j α ρ 1 ( t , s ) Γ ρ ( j α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n j = 1 λ j 1 Φ ϕ ( j 1 ) α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + ( j 1 ) α ) = 1 ρ a t Φ ϕ α ρ 1 ( t , s ) j = 0 λ j Φ ϕ j α ρ ( t , s ) Γ ρ ( ( j + 1 ) α ) h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ i n j = 0 λ j Φ ϕ j α ρ ( t , a ) Γ ρ ( ρ ( γ i + 1 ) + j α ) .
By using the property (6), we obtain
u ( t ) = 1 ρ α ρ a t Φ ϕ α ρ 1 ( t , s ) j = 0 λ j Φ ϕ j α ρ ( t , s ) ρ j α ρ Γ α ρ j + α ρ h ( s ) ϕ ( s ) d s + i = 1 n c i Φ ϕ γ i ( t , a ) ρ γ n j = 0 λ j Φ ϕ j α ρ ( t , a ) ρ j α ρ Γ α ρ j + γ i + 1 .
Applying Lemma 3, we find that the explicit solution (12). □

2.3. An Auxiliary Lemma

Let us denote the weighted space
C ϕ k 1 γ k ( J , R ) = u : ( a , b ] R | u ( a + ) exists and Φ ϕ 1 γ ( t , a ) u ( t ) C ( J , R ) , γ ( 0 , 1 ] ,
where C ϕ k 1 γ k = C ϕ k 1 γ k ( J , R ) . Next, we provide the weighted space of piecewise continuous functions as follows:
PC ϕ k 1 γ k ( J , R ) = { u : ( a , b ] R | u C ϕ k 1 γ k , k = 0 , 1 , 2 , , m , ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k + ) , ρ k 1 I t k 1 + ρ k 1 ( 1 γ k 1 ) ; ϕ k 1 u ( t k ) exist and ρ k 1 I t k 1 + ρ k 1 ( 1 γ k 1 ) ; ϕ k 1 u ( t k ) = ρ k 1 I t k 1 + ρ k 1 ( 1 γ k 1 ) ; ϕ k 1 u ( t k ) , k = 1 , , m } .
Observe that PC ϕ k 1 γ k = PC ϕ k 1 γ k ( J , R ) is a Banach space equipped with
u PC ϕ k 1 γ k = sup t J Φ ϕ k 1 γ k ( t , t k ) u ( t ) .
For the easy to prove, we set the symbol that will be used throughout this paper.
P k ( j ) = l = j k 1 E α l ρ l λ l ρ l 1 Φ ϕ l ( t l + 1 , t l ) α l ρ l , P k ( j ) 1 ,
Ξ = i = 0 m κ i P i ( 0 ) Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i j = 0 n ω j P j ( 0 ) Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j .
Lemma 7.
Assume that α k ( 0 , 1 ) , β k [ 0 , 1 ] , ρ k > 0 , γ k = ( β k ( ρ k α k ) + α k ) / ρ k , λ k < 0 , ϕ k C ( J , R ) with ϕ k > 0 , k = 0 , 1 , 2 , m , κ i , ω r R , i = 0 , 1 , 2 , m , j = 0 , 1 , 2 , n , h C ϕ k 1 γ k , and Ξ 0 . Then, the following impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s:
ρ k H D t k + α k , β k ; ϕ k u ( t ) = λ k u ( t ) + h ( t ) , t J k J , t t k , Δ ρ k I t k + ρ k ( 1 γ k ) ; ϕ k u ( t k ) = φ k ( u ( t k ) ) , k = 1 , 2 , , m , i = 0 m κ i u ( η i ) = j = 0 n ω j ρ j I t j + μ j ; ϕ j u ( ξ j ) + A , η i ( t i , t i + 1 ] , ξ j ( t j , t j + 1 ] ,
is corresponding to the following integral equation, u PC ϕ k 1 γ k ,
u ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k h ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r h ( s ) ϕ r ( s ) d s + r = 1 k φ r ( u ( t r ) ) P k ( r ) ] Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j h ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r h ( s ) ϕ r ( s ) d s + r = 1 j φ r ( u ( t r ) ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + A i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i h ( s ) ϕ r ( s ) d s i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r h ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) × P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
Proof. 
Let u PC ϕ k 1 γ k be a solution of the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4). We consider the following several cases.
For t 1 [ t 0 , t 1 ] , we obtain
u ( t ) = 1 ρ 0 α 0 ρ 0 t 0 t Φ ϕ 0 α 0 ρ 0 1 ( t , s ) E α 0 ρ 0 , α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 Φ ϕ 0 γ 0 1 ( t , t 0 ) ρ 0 γ 0 1 E α 0 ρ 0 , γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t , t 0 ) α 0 ρ 0 .
Taking the operator ρ 0 I t 0 + ρ 0 1 γ 0 ; ϕ 0 into the above equation with Lemmas 4 and 5, which implies that
ρ 0 I t 0 + ρ 0 1 γ 0 ; ϕ 0 u ( t ) = 1 ρ 0 α 0 ρ 0 + 1 γ 0 t 0 t Φ ϕ 0 α 0 ρ 0 γ 0 ( t , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t , t 0 ) α 0 ρ 0
In particular, for t = t 1 , it follows that
ρ 0 I t 0 + ρ 0 1 γ 0 ; ϕ 0 u ( t 1 ) = 1 ρ 0 α 0 ρ 0 + 1 γ 0 t 0 t 1 Φ ϕ 0 α 0 ρ 0 γ 0 ( t 1 , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , t 0 ) α 0 ρ 0
For t ( t 1 , t 2 ] , we obtain
u ( t ) = 1 ρ 1 α 1 ρ 1 t 1 t Φ ϕ 1 α 1 ρ 1 1 ( t , s ) E α 1 ρ 1 , α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + ρ 1 I t 1 + ρ ( 1 γ 1 ) ; ϕ 1 u ( t 1 + ) Φ ϕ 1 γ 1 1 ( t , t 1 ) ρ 1 γ 1 1 E α 1 ρ 1 , γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , t 1 ) α 1 ρ 1 .
By using the impulsive condition, that is ρ 1 I t 1 + ρ 1 ( 1 γ 1 ) ; ϕ 1 u ( t 1 + ) = ρ 0 I t 0 + ρ 0 ( 1 γ 0 ) ; ϕ 0 u ( t 1 ) + φ 1 ( u ( t 1 ) ) , we obtain
u ( t ) = 1 ρ 1 α 1 ρ 1 t 1 t Φ ϕ 1 α 1 ρ 1 1 ( t , s ) E α 1 ρ 1 , α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + ρ 0 I t 0 + ρ 0 ( 1 γ 0 ) ; ϕ 0 u ( t 1 ) + φ 1 ( u ( t 1 ) ) Φ ϕ 1 γ 1 1 ( t , t 1 ) ρ 1 γ 1 1 E α 1 ρ 1 , γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , t 1 ) α 1 ρ 1 = 1 ρ 1 α 1 ρ 1 t 1 t Φ ϕ 1 α 1 ρ 1 1 ( t , s ) E α 1 ρ 1 , α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + [ 1 ρ 0 α 0 ρ 0 + 1 γ 0 t 0 t 1 Φ ϕ 0 α 0 ρ 0 γ 0 ( t 1 , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , t 0 ) α 0 ρ 0 + φ 1 u ( t 1 ) ] Φ ϕ 1 γ 1 1 ( t , t 1 ) ρ 1 γ 1 1 E α 1 ρ 1 , γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , t 1 ) α 1 ρ 1 .
Taking ρ 1 I t 1 + ρ 1 1 γ 1 ; ϕ 1 into the above equation with Lemmas 4 and 5, one has
ρ 1 I t 1 + ρ 1 1 γ 1 ; ϕ 1 u ( t ) = 1 ρ 1 α 1 ρ 1 + 1 γ 1 t 1 t Φ ϕ 1 α 1 ρ 1 γ 1 ( t , s ) E α 1 ρ 1 , α 1 ρ 1 + 1 γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + [ 1 ρ 0 α 0 ρ 0 + 1 γ 0 t 0 t 1 Φ ϕ 0 α 0 ρ 0 γ 0 ( t 1 , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , t 0 ) α 0 ρ 0 + φ 1 u ( t 1 ) ] E α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t , t 1 ) α 1 ρ 1 .
In particular, for t = t 2 , we have
ρ 1 I t 1 + ρ 1 1 γ 1 ; ϕ 1 u ( t 2 ) = 1 ρ 1 α 1 ρ 1 + 1 γ 1 t 1 t 2 Φ ϕ 1 α 1 ρ 1 γ 1 ( t 2 , s ) E α 1 ρ 1 , α 1 ρ 1 + 1 γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + [ 1 ρ 0 α 0 ρ 0 + 1 γ 0 t 0 t 1 Φ ϕ 0 α 0 ρ 0 γ 0 ( t 1 , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , t 0 ) α 0 ρ 0 + φ 1 u ( t 1 ) ] E α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , t 1 ) α 1 ρ 1 .
For t ( t 2 , t 3 ] with ρ 2 I t 2 + ρ 2 ( 1 γ 2 ) ; ϕ 2 u ( t 2 + ) = ρ 1 I t 1 + ρ 1 ( 1 γ 1 ) ; ϕ 1 u ( t 2 ) + φ 2 ( u ( t 2 ) ) , we obtain
u ( t ) = 1 ρ 2 α 2 ρ 2 t 2 t Φ ϕ 2 α 2 ρ 2 1 ( t , s ) E α 2 ρ 2 , α 2 ρ 2 λ 2 ρ 2 1 Φ ϕ 2 ( t , s ) α 2 ρ 2 h ( s ) ϕ 2 ( s ) d s + { 1 ρ 1 α 1 ρ 1 + 1 γ 1 t 1 t 2 Φ ϕ 1 α 1 ρ 1 γ 1 ( t 2 , s ) E α 1 ρ 1 , α 1 ρ 1 + 1 γ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , s ) α 1 ρ 1 h ( s ) ϕ 1 ( s ) d s + t 0 t 1 Φ ϕ 0 α 0 ρ 0 γ 0 ( t 1 , s ) E α 0 ρ 0 , α 0 ρ 0 + 1 γ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , s ) α 0 ρ 0 h ( s ) ϕ 0 ( s ) d s × E α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , t 1 ) α 1 ρ 1 ρ 0 α 0 ρ 0 + 1 γ 0 + u 0 E α 0 ρ 0 λ 0 ρ 0 1 Φ ϕ 0 ( t 1 , t 0 ) α 0 ρ 0 × E α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , t 1 ) α 1 ρ 1 + φ 1 ( u ( t 1 ) ) E α 1 ρ 1 λ 1 ρ 1 1 Φ ϕ 1 ( t 2 , t 1 ) α 1 ρ 1 + φ 2 u ( t 2 ) } Φ ϕ 2 γ 2 1 ( t , t 2 ) ρ 2 γ 2 1 E α 2 ρ 2 , γ 2 λ 2 ρ 2 1 Φ ϕ 2 ( t , t 2 ) α 2 ρ 2 .
Repeating the previous procedure, for t J k , k = 0 , 1 , , m , it follows form
u ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k h ( s ) ϕ k ( s ) d s + [ r = 0 k 1 1 ρ r α r ρ r + 1 γ r i = r + 1 k 1 E α i ρ i , 1 λ i ρ i 1 Φ ϕ i ( t i + 1 , t i ) α i ρ i × t r t r + 1 Φ ϕ r α j ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r h ( s ) ϕ r ( s ) d s + r = 1 k φ r u ( t r ) i = r k 1 E α i ρ i , 1 λ i ρ i 1 Φ ϕ i ( t i + 1 , t i ) α i ρ i ] × Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + u 0 i = 0 k 1 E α i ρ i , 1 λ i ρ i 1 Φ ϕ i ( t i + 1 , t i ) α i ρ i × Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
By using the symbol (13), the form (17) can be rewritten
u ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k h ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × h ( s ) ϕ r ( s ) d s + r = 1 k φ r ( u ( t r ) ) P k ( r ) ] Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + u 0 P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
From the non-local condition, i = 0 m κ i u ( η i ) = j = 0 n ω j ρ j I t j + μ j ; ϕ j u ( ξ j ) + A , we obtain
i = 0 m κ i u ( η i ) = i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i h ( s ) ϕ r ( s ) d s + i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × h ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } + u 0 i = 0 m κ i P i ( 0 ) Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i .
and
j = 0 n ω j ρ j I t j + μ j ; ϕ j u ( ξ j ) = j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j h ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × h ( s ) ϕ r ( s ) d s + r = 1 j φ r u ( t r ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + u 0 j = 0 n ω j P j ( 0 ) Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j .
Solving the above system (19) and (20), we obtain
u 0 = 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j h ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × h ( s ) ϕ r ( s ) d s + r = 1 j φ r u ( t r ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + A i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i h ( s ) ϕ r ( s ) d s i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × h ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) ,
where Ξ is given by (14). Inserting u 0 into (18), we obtain the solution (16).
Conversely, it is easy to show by direct calculation that the solution u ( t ) is defined by (16) satisfies the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (15). □

3. Existence Results

By Lemma 7, we define an operator Q : PC ϕ k 1 γ k PC ϕ k 1 γ k as
( Q u ) ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k F u ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s + r = 1 k φ r ( u ( t r ) ) P k ( r ) ] Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j F u ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s + r = 1 j φ r u ( t r ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + A i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i F u ( s ) ϕ r ( s ) d s i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) × P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
where F u ( t ) = f ( t , u ( t ) , ρ k I t k δ k ; ϕ k u ( t ) , ρ k I t k θ k ; ϕ k u ( t ) ) . It should be noted that Q has fixed points if and only if the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) has solutions.
We define the notations of constants that will be used throughout this paper.
Δ 1 = Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] ,
Δ 2 = Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i ,
Δ 3 = 1 + 1 Ξ Γ ρ m ρ m γ m j = 0 n ω j + i = 0 m κ i .

3.1. Uniqueness Result via Banach’s Fixed Point Theorem

Lemma 8
(Banach’s fixed point theorem [53]). Assume that B is a non-empty closed subset of X where X is a Banach. Then, any contraction mapping Q from B into itself has a unique fixed point.
Theorem 1.
Let f C ( J × R 3 , R ) and φ k C ( R , R ) for k = 1 , 2 , m . Suppose that
( H 1 )
There exist constants L 1 , L 2 , L 3 > 0 so that
| f ( t , u 1 , v 1 , w 1 ) f ( t , u 2 , v 2 , w 2 ) | Φ ρ k γ k 1 ( t , t k ) L 1 u 1 u 2 PC ϕ k 1 γ k + L 2 v 1 v 2 PC ϕ k 1 γ k + L 3 w 1 w 2 PC ϕ k 1 γ k ,
for all t J and u i , v i , w i R , i = 1 , 2 .
( H 2 )
There exists a constant N 1 > 0 so that
| φ k ( u ) φ k ( v ) | N 1 Φ ρ k γ k 1 ( t , t k ) u v PC ϕ k 1 γ k , u , v R , k = 1 , 2 , , m .
Then, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) has a unique solution if
Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 < 1 ,
where Φ 1 and Φ 2 are given by
Φ 1 : = Γ ρ m ( ρ m γ m ) Φ ρ m δ m ρ m + γ m 1 ( T , t m ) Γ ρ m ( ρ m γ m + δ m ) , Φ 2 : = Γ ρ m ( ρ m γ m ) Φ ρ m θ m ρ m + γ m 1 ( T , t m ) Γ ρ m ( ρ m γ m + θ m ) .
Proof. 
Transformation the problem (4) into a fixed point problem, u = Q u , where Q is define by (21). We know that the fixed points of Q are solutions to the problem (4). We separate the procedure into two steps.
Step 1: We show that Q B Υ 1 B Υ 1 .
Let sup t J | f ( t , 0 , 0 , 0 ) | : = F 1 < and I 1 : = max { | φ k ( 0 ) | : k = 1 , 2 , , m } . Define B Υ 1 : = { u PC ϕ k 1 γ k : u PC ϕ k 1 γ k Υ 1 } with the radius
Υ 1 Δ 1 F 1 + Δ 2 I 1 + A Ξ Γ ρ m ρ m γ m 1 Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 .
Obviously, the set B Υ 1 is a bounded, closed, and convex subset of PC ϕ k 1 γ k .
For every u B Υ 1 , we obtain
Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) Φ ϕ k 1 γ k ( t , t k ) ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k F u ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × F u ( s ) ϕ r ( s ) d s + r = 1 k φ r ( u ( t r ) ) P k ( r ) ] 1 ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k
+ 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j F u ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × F u ( s ) ϕ r ( s ) d s + r = 1 j φ r u ( t r ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 × E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + A + i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) × E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i F u ( s ) ϕ r ( s ) d s + i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) × P k ( 0 ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
Since,
ρ k I t k δ k ; ϕ k u ( t ) = 1 ρ k Γ ρ k ( δ k ) t k t Φ ϕ k δ k ρ k 1 ( t , s ) ϕ k ( s ) u ( s ) d s Γ ρ k ( ρ k γ k ) Γ ρ k ( ρ k γ k + δ k ) Φ ρ k δ k ρ k + γ k 1 ( t , t k ) u PC ϕ k 1 γ k , ρ k I t k θ k ; ϕ k u ( t ) Γ ρ k ( ρ k γ k ) Γ ρ k ( ρ k γ k + θ k ) Φ ρ k θ k ρ k + γ k 1 ( t , t k ) u PC ϕ k 1 γ k .
By applying ( H 1 )–( H 2 ), we have the following inequalities
F u ( t ) f ( t , u ( t ) , ρ k I t k δ k ; ϕ k u ( t ) , ρ k I t k θ k ; ϕ k u ( t ) ) f ( t , 0 , 0 , 0 ) + f ( t , 0 , 0 , 0 ) L 1 u ( t ) + L 2 ρ k I t k δ k ; ϕ k u ( t ) + L 3 ρ k I t k θ k ; ϕ k u ( t ) + f ( t , 0 , 0 , 0 ) L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ,
φ k ( u ( t k ) ) φ k ( u ( t k ) ) φ k ( 0 ) + φ k ( 0 ) N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 .
By using the properties E α ( z ) 1 and E α , β ( z ) 1 / Γ ( β ) with (27) and (28), we obtain that
Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ k α k ρ k Γ α k ρ k Φ ϕ k 1 γ k ( t , t k ) t k t Φ ϕ k α k ρ k 1 ( t , s ) ϕ k ( s ) d s
+ 1 ρ k γ k 1 Γ γ k [ r = 0 k 1 L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) ϕ r ( s ) d s + k N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] + 1 Ξ ( j = 0 n ω j L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ j α j + μ j ρ j Γ α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) ϕ r ( s ) d s + j N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j } + A + i = 0 m κ i L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ i α i ρ i Γ α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) ϕ r ( s ) d s + i = 0 m { [ r = 0 i 1 L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) ϕ r ( s ) d s + i N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i } ) 1 ρ k γ k 1 Γ γ k .
By using (6) and the property (i) in Lemma 1, we have
Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) 1 ρ k Γ ρ k α k t k t Φ ϕ k α k ρ k 1 ( t , s ) ϕ k ( s ) d s Φ ϕ k 1 γ k ( t , t k ) L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + 1 Γ ρ k β k ( ρ k α k ) + α k [ r = 0 k 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s × L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + k N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] + 1 Ξ ( j = 0 n ω j ρ j Γ ρ j α j + μ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) ϕ j ( s ) d s [ L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 ] + j = 0 n { ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j [ r = 0 j 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + j N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] } + A + 1 Ξ i = 0 m κ i ρ i Γ ρ i α i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) ϕ r ( s ) d s L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1
+ i = 0 m { κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i [ r = 0 i 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + i N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 ] } ) 1 Γ ρ k β k ( ρ k α k ) + α k { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m β m ( ρ m α m ) + α m [ r = 0 m 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j r = 0 j 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i r = 0 i 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] } × L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + 1 Γ ρ m β m ( ρ m α m ) + α m { m + 1 Ξ j = 0 n j ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j + i = 0 m i κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i } × N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 + A Ξ Γ ρ m β m ( ρ m α m ) + α m = { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] } × L 1 + L 2 Φ 1 + L 3 Φ 2 u PC ϕ k 1 γ k + F 1 + 1 Γ ρ m ρ m γ m [ m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i ] × N 1 Φ ϕ m γ m 1 ( T , t m ) u PC ϕ k 1 γ k + I 1 + A Ξ Γ ρ m ρ m γ m = Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 u PC ϕ k 1 γ k + Δ 1 F 1 + Δ 2 I 1 + A Ξ Γ ρ m ρ m γ m Υ 1 .
Hence, Q u PC ϕ k 1 γ k Υ 1 which implies that Q B Υ 1 B Υ 1 .
Step 2: We show that Q is a contraction.
For each u, v B Υ 1 and for any t J , we obtain that
Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) ( Q v ) ( t ) Φ ϕ k 1 γ k ( t , t k ) ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k F u ( s ) F v ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × F u ( s ) F v ( s ) ϕ r ( s ) d s + r = 1 k φ r ( u ( t r ) ) φ r ( v ( t r ) ) P k ( r ) ] × 1 ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) × E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j F u ( s ) F v ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) F v ( s ) ϕ r ( s ) d s + r = 1 j φ r u ( t r ) φ r v ( t r ) P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 × E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) × E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i F u ( s ) F v ( s ) ϕ r ( s ) d s + i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r × F u ( s ) F v ( s ) ϕ r ( s ) d s + r = 1 i φ r ( u ( t r ) ) φ r ( v ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 × E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) P k ( 0 ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
By using ( H 1 )–( H 2 ), we have
F u ( t ) F v ( t ) | f ( t , u ( t ) , ρ k I t k δ k ; ϕ k u ( t ) , ρ k I t k θ k ; ϕ k u ( t ) ) f ( t , v ( t ) , ρ k I t k δ k ; ϕ k v ( t ) , ρ k I t k θ k ; ϕ k v ( t ) ) | L 1 + L 2 Φ 1 + L 3 Φ 2 u v PC ϕ k 1 γ k ,
φ k ( u ( t k ) ) φ k ( v ( t k ) ) N 1 Φ ϕ m γ m 1 ( T , t m ) u v PC ϕ k 1 γ k .
By using Lemma 1, Lemma 3 and (29) and (30), implies that
Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) ( Q v ) ( t ) { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m β m ( ρ m α m ) + α m [ r = 0 m 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + i = 0 m κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i r = 0 i 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ] } × L 1 + L 2 Φ 1 + L 3 Φ 2 u v PC ϕ k 1 γ k + N 1 Φ ϕ m γ m 1 ( T , t m ) u v PC ϕ k 1 γ k Γ ρ m β m ( ρ m α m ) + α m { m + 1 Ξ j = 0 n j ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j + i = 0 m i κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i } = { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] } × L 1 + L 2 Φ 1 + L 3 Φ 2 u v PC ϕ k 1 γ k + Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m [ m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i ] N 1 u v PC ϕ k 1 γ k = Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 u v PC ϕ k 1 γ k ,
this yields that Q u Q v PC ϕ k 1 γ k [ Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 ] u v PC ϕ k 1 γ k . Since condition (25) holds, then Q is a contraction map. Hence, by Lemma 8, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) has a unique solution. □

3.2. Existence Result via O’Regan’s Fixed Point Theorem

Lemma 9
(O’Regan’s fixed point theorem [54]). Let K be an open set in a closed, convex set B of a Banach space X , with K ¯ and K representing the closure and boundary of K , respectively. Moreover, it is assumed that 0 K and Q : K ¯ B is such that Q ( K ¯ ) is bounded and that Q = Q 1 + Q 2 , where Q 1 : K B is continuous and completely continuous and Q 2 : K B is non-linear contraction, that is, there exists a non-negative non-decreasing function ψ : [ 0 , ) [ 0 , ) , such that ψ ( z ) < z for z > 0 , and Q 2 u Q 2 v ψ ( u v ) for all u, v K . Then, either ( C 1 ) Q has a fixed point u K ¯ ; or ( C 2 ) there exist a point u K and σ ( 0 , 1 ) , such that u = σ Q ( u ) .
Theorem 2.
Let f C ( J × R 3 , R ) and φ k C ( R , R ) , k = 1 , 2 , m . Suppose that
( H 3 )
There exists M 1 > 0 such that | φ k ( u ) | M 1 for all u R , k = 1 , 2 , , m .
( H 4 )
There exist a continuous non-decreasing function ψ : [ 0 , ) ( 0 , ) and p 1 , p 2 , p 3 C ( [ 0 , T ] , R + ) , such that
| f ( t , u , v , w ) | p 1 ( t ) ψ Φ ρ k γ k 1 ( t , t k ) | u | + Φ ρ k γ k 1 ( t , t k ) p 2 ( t ) | v | + p 3 ( t ) | w | ,
for any ( t , u , v , w ) J × R 3 .
( H 5 )
There exist a continuous non-decreasing function q 1 : [ 0 , ) [ 0 , ) and b 1 > 0 , such that
| φ k ( u ) φ k ( v ) | q 1 Φ ρ k γ k 1 ( t , t k ) | u v |
q 1 Φ ρ k γ k 1 ( t , t k ) | u | b 1 Φ ρ k γ k 1 ( t , t k ) | u | ,
for any u, v R , k = 1 , 2 , , m satisfying Δ 2 Φ ϕ m γ m 1 ( T , t m ) b 1 < 1 where Δ 2 is defined by (23).
( H 6 )
The following condition holds
sup Υ 2 ( 0 , ) Υ 2 Δ 1 p 1 ψ ( Υ 2 ) + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m > 1 1 Δ 1 ( p 2 Φ 1 + p 3 Φ 2 ) ,
with Δ 1 ( p 2 Φ 1 + p 3 Φ 2 ) < 1 and Δ i , ( i = 1 , 2 ) , are given by (22) and (23), respectively, p 1 = sup t J | p 1 ( t ) | , p 2 = sup t J | p 2 ( t ) | , p 3 = sup t J | p 3 ( t ) | .
Then, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) has at least one solution.
Proof. 
We separate Q : PC ϕ k 1 γ k PC ϕ k 1 γ k by (21) into two operators Q 1 and Q 2 as ( Q u ) ( t ) = ( Q 1 u ) ( t ) + ( Q 2 u ) ( t ) for t J , where
( Q 1 u ) ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k F u ( s ) ϕ k ( s ) d s + r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 × E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j F u ( s ) ϕ j ( s ) d s + j = 0 n { r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s × ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i × F u ( s ) ϕ r ( s ) d s i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i { r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) × E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F u ( s ) ϕ r ( s ) d s } ) P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) Ξ ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k ,
( Q 2 u ) ( t ) = Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k r = 1 k φ r ( u ( t r ) ) P k ( r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j r = 1 j φ r u ( t r ) P j ( r ) + A i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i r = 1 i φ r ( u ( t r ) ) P i ( r ) ) × P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
Let B Υ 2 = { u PC ϕ k 1 γ k : u PC ϕ k 1 γ k Υ 2 } satisfying
Υ 2 Δ 1 p 1 ψ ( Υ 2 ) + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m > 1 1 Δ 1 ( p 2 Φ 1 + p 3 Φ 2 ) .
From Theorem 1, we can prove that Q 1 is continuous. By using ( H 4 ), we show that Q 1 ( B Υ 2 ) is bounded. For any u B Υ 2 , we obtain
Φ ϕ k 1 γ k ( t , t k ) ( Q 1 u ) ( t ) p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 { Φ ϕ k 1 γ k ( t , t k ) ρ k Γ ρ k α k t k t Φ ϕ k α k ρ k 1 ( t , s ) ϕ k ( s ) d s + 1 Γ ρ k β k ( ρ k α k ) + α k [ r = 0 k 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j Γ α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) F u ( s ) ϕ j ( s ) d s + j = 0 n ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j r = 0 j 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s + i = 0 m κ i ρ i Γ ρ i α i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) ϕ r ( s ) d s + i = 0 m κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i × r = 0 i 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) ϕ r ( s ) d s ) ] } Δ 1 p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 ,
which implies that Q 1 ( B Υ 2 ) Δ 1 [ p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 ] .
Next, we will show that Q 1 maps bounded set B Υ 2 into equicontinuous set of PC ϕ k 1 γ k . Let τ 1 , τ 2 J k , for k = 0 , 1 , , m with τ 1 < τ 2 and u B Υ 2 . Then, we obtain
( Q 1 u ) ( τ 2 ) ( Q 1 u ) ( τ 1 ) p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 { Φ ϕ m α m ρ m ( τ 2 , t m ) Φ ϕ m α m ρ m ( τ 1 , t m ) Γ ρ m α m + ρ m + Φ ϕ m β m ( ρ m α m ) + α m ρ m 1 ( τ 2 , t m ) Φ ϕ m β m ( ρ m α m ) + α m ρ m 1 ( τ 1 , t m ) Γ ρ m β m ( ρ m α m ) + α m [ r = 0 k 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j r = 0 j 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i r = 0 i 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r ( t r + 1 , t r ) ρ r Γ ρ r α r + ρ r ( 1 γ r ) ) ] } .
It is easy to see that the above result is independent of variable u B Υ 2 , which implies that ( Q 1 u ) ( τ 2 ) ( Q 1 u ) ( τ 1 ) 0 as τ 2 τ 1 . Since, Q 1 maps bounded set B Υ 2 into equicontinuous set of PC ϕ k 1 γ k . Thus, by the Arzelá-Ascoli theorem, we get that Q 1 is completely continuous.
Next, we will show that Q 2 is a non-linear contraction. Define a continuous non-decreasing function ψ : R + R + by ψ ( ϵ ) = Δ 2 b 1 ϵ , for all ϵ 0 . Clearly, ψ ( t ) satisfies ψ ( 0 ) = 0 and by applying Δ 2 b 1 < 1 , we obtain ψ ( ϵ ) < ϵ for all ϵ > 0 . For every u, v B Υ 2 , we have
Φ ϕ k 1 γ k ( t , t k ) ( Q 2 u ) ( t ) ( Q 2 v ) ( t ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k r = 1 k Φ ϕ k 1 γ k ( t , t k ) φ r ( u ( t r ) ) φ r ( v ( t r ) ) × P k ( r ) + Φ ϕ k γ k 1 ( t , t k ) Ξ ( j = 0 n [ ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j × r = 1 j Φ ϕ k 1 γ k ( t , t k ) φ r u ( t r ) φ r v ( t r ) P j ( r ) ] + i = 0 m [ κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 × E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i r = 1 i Φ ϕ k 1 γ k ( t , t k ) φ r ( u ( t r ) ) φ r ( v ( t r ) ) P i ( r ) ] ) × P k ( 0 ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k Φ ϕ m γ m 1 ( T , t m ) ρ m γ m 1 Γ γ m r = 1 m q 1 ( u v PC ϕ k 1 γ k ) + 1 Ξ ( j = 0 n [ ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j × r = 1 j q 1 ( u v PC ϕ k 1 γ k ) ] + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i r = 1 i q 1 ( u v PC ϕ k 1 γ k ) ) Φ ϕ m γ m 1 ( T , t m ) ρ m γ m 1 Γ γ m Φ ϕ m γ m 1 ( T , t m ) ρ m γ m 1 Γ γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i b 1 u v PC ϕ k 1 γ k = Δ 2 b 1 u v PC ϕ k 1 γ k .
By setting ψ ( ϵ ) = Δ 2 b 1 ϵ , note that ψ ( 0 ) = 0 and ψ ( ϵ ) < ϵ for all ϵ > 0 . So,
Q 2 u Q 2 v PC ϕ k 1 γ k ψ ( u v PC ϕ k 1 γ k ) ,
which yields that Q 2 is non-linear contraction.
Next, we will show that Q ( B Υ 2 ) is bounded. By using ( H 3 ), for any u B Υ 2 , we obtain
Φ ϕ k 1 γ k ( t , t k ) ( Q 2 u ) ( t ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k r = 1 k Φ ϕ k 1 γ k ( t , t k ) φ r ( u ( t r ) ) P k ( r ) + 1 Ξ ( j = 0 n [ ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j Φ ϕ k γ k 1 ( t , t k ) × r = 1 j Φ ϕ k 1 γ k ( t , t k ) φ r u ( t r ) P j ( r ) ] + A + i = 0 m [ κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 × E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i Φ ϕ k γ k 1 ( t , t k ) r = 1 i Φ ϕ k 1 γ k ( t , t k ) φ r ( u ( t r ) ) P i ( r ) ] ) × P k ( 0 ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k { Φ ϕ k γ m 1 ( T , t m ) Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i Γ ρ i ρ i γ i + A } M 1 = Δ 2 + A Γ ρ m ρ m γ m M 1 .
This yields that Q ( B Υ 2 ) is bounded with the boundedness of the set Q 1 ( B Υ 2 ) .
Finally, we will show that ( C 2 ) of Theorem 2 does not true. On the contrary, assume that ( C 2 ) holds. Then, there exists σ ( 0 , 1 ) and for every u B Υ 2 so that u = σ Q u . Hence, we have u PC ϕ k 1 γ k Υ 2 and
Φ ϕ k 1 γ k ( t , t k ) u ( t ) = σ Φ ϕ k 1 γ k ( t , t k ) ( Q u ) ( t ) Φ ϕ k 1 γ k ( t , t k ) ( Q 1 u ) ( t ) + ( Q 2 u ) ( t ) p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 ρ m Γ ρ m α m t m T Φ ϕ m α m ρ m 1 ( T , s ) ϕ m ( s ) d s Φ ϕ m 1 γ m ( T , t m ) + p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 Γ ρ m β m ( ρ m α m ) + α m r = 0 m 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s + ( j = 0 n ω j ρ j Γ ρ j α j + μ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) ϕ j ( s ) d s
+ j = 0 n ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j r = 0 j 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s + i = 0 m κ i ρ i Γ ρ i α i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) ϕ r ( s ) d s + i = 0 m κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i r = 0 i 1 1 ρ r Γ ρ r α r + ρ r ( 1 γ r ) × t r t r + 1 Φ ϕ r α r + ρ r ( 1 γ r ) ρ r 1 ( t r + 1 , s ) ϕ r ( s ) d s ) p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 Ξ Γ ρ m β m ( ρ m α m ) + α m + Φ ϕ m γ m 1 ( T , t m ) Γ ρ m β m ( ρ m α m ) + α m [ m + 1 Ξ ( j = 0 n j ω j Φ ϕ j β j ( ρ j α j ) + α j + μ j ρ j 1 ( ξ j , t j ) Γ ρ j β j ( ρ j α j ) + α j + μ j + i = 0 m i κ i Φ ϕ i β i ( ρ i α i ) + α i ρ i 1 ( η i , t i ) Γ ρ i β i ( ρ i α i ) + α i ) ] M 1 + A Ξ Γ ρ m β m ( ρ m α m ) + α m p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] } + Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m [ m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i ] M 1 + A Ξ Γ ρ m ρ m γ m = Δ 1 p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m ,
which yields that
Υ 2 Δ 1 p 1 ψ ( Υ 2 ) + ( p 2 Φ 1 + p 3 Φ 2 ) Υ 2 + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m .
Then, we have
Υ 2 Δ 1 p 1 ψ ( Υ 2 ) + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m 1 1 Δ 1 ( p 2 Φ 1 + p 3 Φ 2 ) ,
which contradicts the condition ( H 6 ). Hence, Q 1 and Q 2 fulfill all the conditions of Lemma 9. Therefore, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) has at least one solution. □

4. Stability Results

First of all, we give the following inequalities for analyzing Ulam’s stability of the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4). Let T C ( J , R + ) be a non-decreasing function, ϵ > 0 , τ 0 , z B , such that for t J k , k = 1 , 2 , , m , the following inequalities are fulfilled:
ρ k H D t k + α k , β k ; ϕ k z ( t ) λ k z ( t ) f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) ϵ , ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) φ k ( z ( t k ) ) ϵ ,
ρ k H D t k + α k , β k ; ϕ k z ( t ) λ k z ( t ) f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) T ( t ) , ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) φ k ( z ( t k ) ) τ ,
ρ k H D t k + α k , β k ; ϕ k z ( t ) λ k z ( t ) f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) ϵ T ( t ) , ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) φ k ( z ( t k ) ) ϵ τ .
Definition 3.
The impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is called Ulam–Hyers ( UH ) stable, if there exists C F > 0 , such that for any ϵ > 0 and for each z B of (37) there is u B of (4) that satisfies
z ( t ) u ( t ) C F ϵ , t J .
Definition 4.
The impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is called generalized Ulam-Hyers ( GUH ) stable, if there exists T C ( R + , R + ) with T ( 0 ) = 0 , such that for any ϵ > 0 and for each z B of (38) there is u B of (4) that satisfies
z ( t ) u ( t ) T ( ϵ ) , t J .
Definition 5.
The impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is called Ulam–Hyers–Rassias UHR stable with respect to ( τ , T ) , if there exists C F , E F > 0 such that for any ϵ > 0 and for each z B of (39) there is u B of (4) that satisfies
z ( t ) u ( t ) C F , E F ϵ ( τ + T ( t ) ) , t J .
Definition 6.
The impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is called generalized Ulam–Hyers–Rassias GUHR stable with respect to ( τ , T ) , if there exists C F , E F > 0 such that for any z B of (38) there is u B of (4) that satisfies
z ( t ) u ( t ) C F , E F ( τ + T ( t ) ) , t J .
Remark 3.
It is easy to see that: (i) Definition 3⇒ Definition 4, ( i i ) Definition 5⇒ Definition 6, and ( i i i ) Definition 5 with τ + T ( t ) = 1 ⇒ Definition 3.
Remark 4.
z B is a solution of (37) if there is x B and x k , k = 1 , 2 , , m , (which depends on z), such that
(i) 
| x ( t ) | ϵ , | x k | ϵ , t J ,
(ii) 
ρ k H D t k + α k , β k ; ϕ k z ( t ) = λ k z ( t ) + f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) + x ( t ) , t J ,
(iii) 
ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) = φ k ( z ( t k ) ) + x k .
Remark 5.
z B is a solution of (38) if there is x B and x k , k = 1 , 2 , , m , (which depends on z), such that
(i) 
| x ( t ) | T ( t ) , | x k | τ , t J ,
(ii) 
ρ k H D t k + α k , β k ; ϕ k z ( t ) = λ k z ( t ) + f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) + x ( t ) , t J ,
(iii) 
ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) = φ k ( z ( t k ) ) + x k .
Remark 6.
z B is a solution of (39) if there is x B and x k , k = 1 , 2 , , m , (which depends on z), such that
(i) 
| x ( t ) | ϵ T ( t ) , | x k | ϵ τ , t J ,
(ii) 
ρ k H D t k + α k , β k ; ϕ k z ( t ) = λ k z ( t ) + f ( t , z ( t ) , ρ k I t k δ k ; ϕ k z ( t ) , ρ k I t k θ k ; ϕ k z ( t ) ) + x ( t ) , t J ,
(iii) 
ρ k I t k + ρ ( 1 γ k ) ; ϕ k z ( t k + ) ρ k 1 I t k 1 + ρ ( 1 γ k 1 ) ; ϕ k 1 z ( t k ) = φ k ( z ( t k ) ) + x k .

4.1. Ulam–Hyers Stability Results

We construct the proof of the following lemma, which gives a base for obtaining a solution to the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4).
Theorem 3.
Let α k ( 0 , 1 ) , β k [ 0 , 1 ] , ρ k > 0 , γ k = ( β k ( ρ k α k ) + α k ) / ρ k , λ k R , ϕ k C ( J , R ) with ϕ k > 0 for k = 1 , 2 , , m . Assume that f C ( J × R 3 , R ) and φ k C ( R , R ) ( k = 1 , 2 , m ). Suppose that ( H 1 )–( H 2 ) hold. Then, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is UH stable if
Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 < 1 .
Proof. 
Suppose that x PC ϕ k 1 γ k is a solution of the problem (37). From Lemma (15) with Remark 4 ( i i )–( i i i ), we obtain
ρ k H D t k + α k , β k ; ϕ k z ( t ) = λ k z ( t ) + F z ( t ) + x ( t ) , t J k J , t t k , Δ ρ k I t k + ρ k ( 1 γ k ) ; ϕ k z ( t k ) = φ k ( z ( t k ) ) + x k , k = 1 , 2 , , m , i = 0 m κ i z ( η i ) = j = 0 n ω j ρ j I t j + μ j ; ϕ j z ( ξ j ) + A , η i ( t i , t i + 1 ] , ξ j ( t j , t j + 1 ] ,
then the solution of (44) can be rewritten as
z ( t ) = 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k F z ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) × E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F z ( s ) ϕ r ( s ) d s + r = 1 k φ r ( z ( t r ) ) P k ( r ) ] Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j F z ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F z ( s ) ϕ r ( s ) d s + r = 1 j φ r z ( t r ) P j ( r ) ] × ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } + A i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i × F z ( s ) ϕ r ( s ) d s i = 0 m { [ r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r F z ( s ) ϕ r ( s ) d s + r = 1 i φ r ( z ( t r ) ) P i ( r ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 ρ k α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) E α k ρ k , α k ρ k λ k ρ k 1 Φ ϕ k ( t , s ) α k ρ k x ( s ) ϕ k ( s ) d s + [ r = 0 k 1 P k ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) × E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r x ( s ) ϕ r ( s ) d s + r = 1 k x r P k ( r ) ] Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) E α j ρ j , α j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , s ) α j ρ j x ( s ) ϕ j ( s ) d s + j = 0 n { [ r = 0 j 1 P j ( r + 1 ) ρ r α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r x ( s ) ϕ r ( s ) d s + r = 1 j x r P j ( r ) ] ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 × E α j ρ j , γ j + μ j ρ j λ j ρ j 1 Φ ϕ j ( ξ j , t j ) α j ρ j } i = 0 m κ i ρ i α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) E α i ρ i , α i ρ i λ i ρ i 1 Φ ϕ i ( η i , s ) α i ρ i x ( s ) ϕ r ( s ) d s i = 0 m { r = 0 i 1 P i ( r + 1 ) ρ r α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) E α r ρ r , α r ρ r + 1 γ r λ r ρ r 1 Φ ϕ r ( t r + 1 , s ) α r ρ r x ( s ) ϕ r ( s ) d s + r = 1 i x r P i ( r ) × κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 E α i ρ i , γ i λ i ρ i 1 Φ ϕ i ( η i , t i ) α i ρ i } ) P k ( 0 ) Φ ϕ k γ k 1 ( t , t k ) ρ k γ k 1 E α k ρ k , γ k λ k ρ k 1 Φ ϕ k ( t , t k ) α k ρ k .
By applying Lemma 3 with P a ( b ) 1 , for any t J , we obtain that
Φ ϕ k 1 γ k ( t , t k ) z ( t ) u ( t ) Φ ϕ k 1 γ k ( t , t k ) ρ k α k ρ k Γ α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) F z ( s ) F u ( s ) ϕ k ( s ) d s + 1 ρ k γ k 1 Γ γ k × [ r = 0 k 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 k φ r ( z ( t r ) ) φ r ( z ( t r ) ) ] + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j Γ α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) × F z ( s ) F u ( s ) ϕ j ( s ) d s + j = 0 n ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j [ r = 0 j 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 j φ r ( z ( t r ) ) φ r ( u ( t r ) ) ] + i = 0 m κ i ρ i α i ρ i Γ α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + i = 0 m { [ r = 0 i 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 i φ r ( z ( t r ) ) φ r ( u ( t r ) ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i } ) 1 ρ k γ k 1 Γ γ k + 1 ρ k α k ρ k Γ α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) x ( s ) ϕ k ( s ) d s + 1 ρ k γ k 1 Γ γ k [ r = 0 k 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r
× t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) x ( s ) ϕ r ( s ) d s + r = 1 k x r ] + 1 Ξ ρ k γ k 1 Γ γ k ( j = 0 n ω j ρ j α j + μ j ρ j Γ α j + μ j ρ j × t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) x ( s ) ϕ j ( s ) d s + j = 0 n ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j × r = 0 j 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) x ( s ) ϕ r ( s ) d s + r = 1 j x r + i = 0 m κ i ρ i α i ρ i Γ α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) x ( s ) ϕ r ( s ) d s + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i × r = 0 i 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) x ( s ) ϕ r ( s ) d s + r = 1 i x r ) .
Thanks to (i) of Remark 4 with ( H 1 )–( H 2 ), we obtain the following result
Φ ϕ k 1 γ k ( t , t k ) z ( t ) u ( t ) { L 1 + L 2 Φ 1 + L 3 Φ 2 ( Φ ϕ m α m ρ m γ m 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] ) + 1 Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i N 1 } × z u PC ϕ k 1 γ k + { Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] + Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i } ϵ ,
which implies that
z u PC ϕ k 1 γ k Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 z u PC ϕ k 1 γ k + Δ 1 + Δ 2 ϵ .
Then z u PC ϕ k 1 γ k C F ϵ , where
C F : = Δ 1 + Δ 2 1 Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 .
Therefore, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is UH stable in B . □
Corollary 1.
Under conditions in Theorem 3, if T ( ϵ ) = C F ϵ so that T ( 0 ) = 0 , then we have the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) becomes GUH stable.

4.2. Ulam–Hyers–Rassias Stability Results

To analyze UHR stability results, we will need the following condition as follows:
( H 7 )
There exists a non-decreasing function T C ( J , R ) and there is E T > 0 , for each ϵ > 0 , such that the following inequality
ρ k I t k + α k ; ϕ k T ( t ) E T T ( t ) .
Theorem 4.
Let f ( J × R 3 , R ) and φ k C ( R , R ) , ( k = 1 , 2 , m ). If ( H 1 ), ( H 2 ), ( H 7 ), and
Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 < 1
are fulfilled. Then the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is UHR stable with respect to ( τ , T ).
Proof. 
Let z B be any solution of (39) and u B be the solution of the problem (4). By the same process in Theorem 3, we have
Φ ϕ k 1 γ k ( t , t k ) z ( t ) u ( t ) Φ ϕ k 1 γ k ( t , t k ) ρ k α k ρ k Γ α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) F z ( s ) F u ( s ) ϕ k ( s ) d s + 1 ρ k γ k 1 Γ γ k × [ r = 0 k 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 k φ r ( z ( t r ) ) φ r ( z ( t r ) ) ] + 1 Ξ ( j = 0 n ω j ρ j α j + μ j ρ j Γ α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) × F z ( s ) F u ( s ) ϕ j ( s ) d s + j = 0 n ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j [ r = 0 j 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r × t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 j φ r ( z ( t r ) ) φ r ( u ( t r ) ) ] + i = 0 m κ i ρ i α i ρ i Γ α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s
+ i = 0 m { [ r = 0 i 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) F z ( s ) F u ( s ) ϕ r ( s ) d s + r = 1 i φ r ( z ( t r ) ) φ r ( u ( t r ) ) ] κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i } ) 1 ρ k γ k 1 Γ γ k + Φ ϕ k 1 γ k ( t , t k ) ρ k α k ρ k Γ α k ρ k t k t Φ ϕ k α k ρ k 1 ( t , s ) x ( s ) ϕ k ( s ) d s + 1 ρ k γ k 1 Γ γ k × r = 0 k 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) x ( s ) ϕ r ( s ) d s + r = 1 k x r + 1 Ξ ρ k γ k 1 Γ γ k ( j = 0 n ω j ρ j α j + μ j ρ j Γ α j + μ j ρ j t j ξ j Φ ϕ j α j + μ j ρ j 1 ( ξ j , s ) x ( s ) ϕ j ( s ) d s + j = 0 n ω j Φ ϕ j γ j + μ j ρ j 1 ( ξ j , t j ) ρ j μ j ρ j + γ j 1 Γ γ j + μ j ρ j [ r = 0 j 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) × x ( s ) ϕ r ( s ) d s + r = 1 j x r ] + i = 0 m κ i ρ i α i ρ i Γ α i ρ i t i η i Φ ϕ i α i ρ i 1 ( η i , s ) x ( s ) ϕ r ( s ) d s + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i [ r = 0 i 1 1 ρ r α r ρ r + 1 γ r Γ α r ρ r + 1 γ r t r t r + 1 Φ ϕ r α r ρ r γ r ( t r + 1 , s ) x ( s ) ϕ r ( s ) d s + r = 1 i x r ] ) .
Thanks to (i) of Remark 6 with ( H 1 ), ( H 2 ), and ( H 7 ), we have the following result
Φ ϕ k 1 γ k ( t , t k ) z ( t ) u ( t ) { L 1 + L 2 Φ 1 + L 3 Φ 2 ( Φ ϕ m α m ρ m γ m + 1 ( T , t m ) Γ ρ m α m + ρ m + 1 Γ ρ m ρ m γ m [ r = 0 m 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + 1 Ξ ( j = 0 n ω j Φ ϕ j α j + μ j ρ j ( ξ j , t j ) Γ ρ j α j + μ j + ρ j + j = 0 n ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j r = 0 j 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) + i = 0 m κ i Φ ϕ i α i ρ i ( η i , t i ) Γ ρ i α i + ρ i + i = 0 m κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i r = 0 i 1 Φ ϕ r α r ρ r γ r + 1 ( t r + 1 , t r ) Γ ρ r α r + ρ r ( 2 γ r ) ) ] ) + Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) Γ ρ i ρ i γ i N 1 } × z u PC ϕ k 1 γ k + { E T T ( t ) 1 + 1 Ξ Γ ρ m ρ m γ m j = 0 n ω j + i = 0 m κ i
+ Φ ϕ m γ m 1 ( T , t m ) Γ ρ m ρ m γ m m + 1 Ξ j = 0 n j ω j Φ ϕ j μ j ρ j + γ j 1 ( ξ j , t j ) Γ ρ j ρ j γ j + μ j + i = 0 m i κ i Φ ϕ i γ i 1 ( η i , t i ) ρ i γ i 1 Γ γ i × E T T ( t ) + τ } ϵ ,
which implies that
z u PC ϕ k 1 γ k Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 z u PC ϕ k 1 γ k + [ Δ 2 + Δ 3 ] E T T ( t ) + Δ 2 τ ϵ Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 z u PC ϕ k 1 γ k + [ Δ 2 + Δ 3 ] E T + Δ 2 ϵ ( τ + T ( t ) ) .
Then z u PC ϕ k 1 γ k C F , E F ϵ ( τ + T ( t ) ) , with
C F , E F : = [ Δ 2 + Δ 3 ] E T + Δ 2 1 Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 .
Hence, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) is UHR stable with respect to ( τ , T ) in B . □
Corollary 2.
Under conditions in Theorem 4, if ϵ = 1 so that T ( 0 ) = 0 , then we have the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (4) becomes GUHR stable.

5. Numerical Examples

This section provides some illustrative examples of the exactness and applicability of our main results.
Example 1.
Consider the following impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s of the form:
k + 2 2 H D t k + 7 3 k 9 3 k , 3 k + 2 10 ; ϕ k u ( t ) = k 2 + 1 k + 1 u ( t ) + f ( t , u ( t ) , k + 2 2 I t k e k 3 ; ϕ k u ( t ) , k + 2 2 I t k sin 2 π k + 2 ; ϕ k u ( t ) ) , Δ k + 2 2 I t k + k + 2 2 ( 1 γ k ) ; ϕ k u ( t k ) = φ k ( u ( t k ) ) , k = 1 , 2 i = 0 2 i + 1 3 u 2 i + 1 5 = j = 0 1 j + 2 2 j + 2 2 I t j + 8 j + 8 10 ; ϕ j u 2 j + 2 9 + 2 .
Form the problem (49), we obtain that α k = ( 7 3 k ) / ( 9 3 k ) , β k = ( 3 k + 2 ) / 10 , ρ k = ( k + 2 ) / 2 , λ k = k 2 + 1 / ( k + 1 ) , δ k = exp ( k ) / 3 , θ k = sin 2 ( π / ( k + 2 ) ) , ϕ k ( t ) = ln ( t + k + 2 ) / ( 2 k + 2 ) , t k = 2 k / 5 , k = 0 , 1 , 2 , T = 6 / 5 , k = 0 , 1 , 2 , κ i = ( i + 2 ) / 3 , η i = ( 2 i + 1 ) / 5 , ω j = ( j + 2 ) / 2 , μ j = ( 8 j + 8 ) / 10 , ξ j = ( 2 j + 2 ) / 9 , i = 0 , 1 , 2 , j = 0 , 1 , and A = 2 . From the given all data, we can find that Ξ 3.350618628 , Δ 1 0.7268674525 , Δ 2 1.676600970 and Δ 3 0.8516099910 . The following functions will be considered for theoretical confirmation:
( i ) . Consider the functions
f ( t , u ( t ) , v ( t ) , w ( t ) ) = 5 t 2 3 2 t 2 + 4 + ( 3 t ) Φ ϕ k 1 γ k ( t , t k ) 7 2 sin 2 π t · | u ( t ) | 5 + | u ( t ) | + 3 e 2 t 1 Φ ϕ k 1 γ k ( t , t k ) 6 cos 2 π t · | v ( t ) | 4 + | v ( t ) | + 4 sin 2 π t Φ ϕ k 1 γ k ( t , t k ) 8 3 e t · | w ( t ) | 4 + | w ( t ) | , φ k ( u ( t k ) ) = 4 Φ ϕ k 1 γ k ( t k + 1 , t k ) u ( t k ) 27 t k + 3 t k .
For u i , v i , w i R , i = 1 , 2 , and t [ 0 , 1 ] , we can find that
f ( t , u 1 , v 1 , w 1 ) f ( t , u 2 , v 2 , w 2 ) 3 25 | u 1 u 2 | + 3 20 | v 1 v 2 | + 1 5 | v 1 v 2 | , φ k ( u 1 ) φ k ( u 2 ) 4 25 | u 1 u 2 | .
The assumption ( H 1 ) ( H 2 ) are satisfied with L 1 = 3 / 5 , L 2 = 3 / 20 , L 3 = 1 / 5 and N 1 = 4 / 25 . Hence,
Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 0.6598729041 < 1 .
Since, all the conditions of Theorem 1 are fulfilled. Then, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (49) has a unique solution on [ 0 , 1 ] . Moreover, by Theorem 3, we also find that
C F : = Δ 1 + Δ 2 1 Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 7.066383275 > 0
Then, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (49) is UH stable on [ 0 , 1 ] . Taking T ( ϵ ) = C F ϵ via T ( 0 ) = 0 , then, by Corollary 1, the following impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (49) is GUH stable on [ 0 , 1 ] . Taking T ( t ) = Φ ϕ k ( t , t k ) with τ = 1 , we obtain
ρ k I t k + α k ; ϕ k T ( t ) = Φ ϕ k α k ρ k ( t , t k ) ρ k Γ ρ k ( α k ) Φ ϕ k ( t , t k ) Φ ϕ m α m ρ m ( T , t m ) ρ m Γ ρ m ( α m ) T ( t )
From ( H 7 ), we obtain
E T = Φ ϕ m α m ρ m ( T , t m ) ρ m Γ ρ m ( α m ) 0.07794511398 > 0 .
Then,
C F , E F : = [ Δ 2 + Δ 3 ] E T + Δ 2 1 Δ 1 L 1 + L 2 Φ 1 + L 3 Φ 2 + Δ 2 N 1 5.508713313 > 0 .
Hence, by all conditions in Theorem 4, the following impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (49) is UHR stable on [ 0 , 1 ] . Moreover, if T ( ϵ ) = C F , E F ϵ with T ( 0 ) = 0 , then, by Corollary 2, the impulsive ( ρ k , ϕ k ) -Hilfer FIDE - NMP - FIBC s (49) is GUHR stable with respect to ( τ , T ) .
( i i ) . Consider the functions
f ( t , u ( t ) , v ( t ) , w ( t ) ) = ( 2 t 1 ) Φ ϕ k 1 γ k ( t , t k ) 12 2 cos π t sin u ( t ) + 2 | u ( t ) | 3 + | u ( t ) | + Φ ϕ k 1 γ k ( t , t k ) 3 cos π t 4 + e t · | v ( t ) | 5 + | v ( t ) | + 6 e t 10 ln ( t + e ) · | w ( t ) | 4 + | w ( t ) | , φ k ( u ( t k ) ) = ( 6 t 2 ) Φ ϕ k 1 γ k ( t , t k ) 5 e t · | u ( t k ) | 5 + 4 | u ( t k ) | .
For u, v, w R , i = 1 , 2 , and t [ 0 , 1 ] , we can find that
| f ( t , u , v , w ) | 2 t 1 12 2 cos π t | u | + 2 + 3 cos π t 20 + 5 e t | v | + 6 e t 40 ln ( t + e ) | w | , | φ k ( u ) φ k ( v ) | 6 25 | u v | , | φ k ( u ) | 6 20 .
The assumption ( H 3 ) ( H 6 ) are satisfied with ψ ( | u | ) = | u | + 2 , p 1 = 0.2 , p 2 = 0.12 , p 3 = 0.15 , M 1 = 0.3 and b 1 = 0.24 . Hence, Δ 2 Φ ϕ m γ m 1 ( T , t m ) b 1 0.8262520217 < 1 , ( 1 [ Δ 1 ( p 2 Φ 1 + p 3 Φ 2 ) ] ) 1 1.31318002 , and
sup Υ 2 ( 0 , ) Υ 2 Δ 1 p 1 ψ ( Υ 2 ) + Δ 2 M 1 + A Ξ Γ ρ m ρ m γ m 6.878833250 .
Since, all the problem (49) has at least one solution on [ 0 , 1 ] .
( i i i ) . Consider the functions f ( t , u ( t ) , v ( t ) , w ( t ) ) = 0 and φ k ( u ( t k ) ) = 7 4 k . By using (14), (22), (23), (24), the numerical values of Ξ, Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } , as shown in Table 1. From Lemma 7, we obtain the implicit solutions of the problem (49), as shown in Figure 1, via fixed values of ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 with vary α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } for k = 0 , 1 , 2 . By using (14), (22), (23), (24), the numerical values of Ξ, Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } , as shown in Table 2. From Lemma 7, we obtain the implicit solutions of the problem (49) as shown in Figure 2 via fixed values of ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 with vary α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } for k = 0 , 1 , 2 . Later, by using (14), (22), (23), (24), the numerical values of Ξ, Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } , as shown in Table 3. From Lemma 7, we obtain the implicit solutions of the problem (49) as shown in Figure 3 via fixed values of ϕ k ( t ) = sin ( π t + k + 1 ( 2 + k ) t + 3 k + 3 ) , β k = cos ( π 5 k ) , and ρ k = ( k + 2 ) tan ( π 6 k ) with vary α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } for k = 0 , 1 , 2 . In addition, we will show the implicit solutions of the problem (49) as shown in Figure 4 for each values of ϕ k ( t ) , α k , β k , ρ k are given as in Table 4.

6. Conclusions

In this paper, we studied the impulsive ( ρ k , ϕ k ) -Hilfer FIDE s with a constant coefficient involving NMP - FIBC s. Firstly, we created some essential properties to apply to our main results. The formula of the solution to the linear ( ρ , ϕ )-Hilfer fractional Cauchy problem was constructed in the form of the Mittag–Leffler kernel. The non-linear impulsive ( ρ k , ϕ k ) -Hilfer fractional Cauchy BVP was converted into a fixed-point problem via an auxiliary lemma regarding a linear variant of the problem. The uniqueness result was investigated by Banach’s fixed point theorem, while the existence result was proved by a fixed point theorem due to O’Regan. In addition, by applying non-linear functional analysis methods and qualitative theory, a variety of UH stability, UHR stability, and their generalization are also examined. To confirm all the achieved theoretical results, numerical examples were given to present the application of our main results in the recent past. Apart from that, our main results are not only novel in the context of the impulsive problem at hand, but they also show some new special situations by adjusting the parameters involved. They have enriched the qualitative theory literature on non-linear impulsive ( ρ k , ϕ k )-Hilfer FIDE s of order in ( 0 , 1 ] equipped with NMP - FIBC s. In future work areas, we recommend working on the qualitative theory literature on non-linear fractional integro-differential equations/inclusions involving a special function, such as the linear Cauchy-type problem with variable coefficients, stability, or the algorithms to solve the ( ρ , ϕ )-Hilfer fractional differential equations in mathematical software.

Author Contributions

Conceptualization, M.K., R.P., W.S., J.A., C.T. and J.K.; methodology, M.K., R.P. and W.S.; software, M.K., W.S. and C.T.; validation, M.K. and W.S.; formal analysis, M.K., W.S., J.A., C.T. and J.K.; investigation, M.K. and W.S.; resources, M.K. and W.S.; data curation, M.K. and W.S.; writing—original draft preparation, M.K., R.P., W.S., J.A., C.T. and J.K.; writing—review and editing, M.K., R.P., W.S., J.A., C.T. and J.K.; visualization, M.K. and W.S.; supervision, W.S. and J.A.; project administration, W.S.; funding acquisition, M.K. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Acknowledgments

M. Kaewsuwan, R. Phuwapathanapun, and W. Sudsutad would like to thank you for financially supporting this paper through Ramkhamhaeng University. J. Alzabut is thankful to Prince Sultan University and OSTİM Technical University for their endless support. J. Kongson and C. Thaiprayoon would like to extend their appreciation to Burapha University.

Conflicts of Interest

The authors declare no conflict of interest.

Abbreviations

The following abbreviations are used in this manuscript:
FC Fractional calculus
RL Riemann–Liouville
NMP Non-local multi-point
BVP Boundary value problem
FIO Fractional integral operator
FDO Fractional derivative operator
FDE Fractional differential equation
FIBC Fractional integral boundary condition
FIDE Fractional integro-differential equation

References

  1. Hilfer, R. Applications of Fractional Calculus in Physics; World Scientific: Singapore, 2000. [Google Scholar]
  2. Fallahgoul, H.A.; Focardi, S.M.; Fabozzi, F.J. Fractional Calculus and Fractional Processes with Applications to Financial Economics; Theory and Application; Elsevier/Academic Press: London, UK, 2017. [Google Scholar]
  3. Tarasov, V.E. Fractional Dynamics: Application of Fractional Calculus to Dynamics of Particles, Fields and Media; Springer & HEP: Berlin, Germany, 2011. [Google Scholar]
  4. Magin, R.L. Fractional Calculus in Bioengineering; Begell House Publishers: Danbury, CT, USA, 2006. [Google Scholar]
  5. Thaiprayoon, C.; Kongson, J.; Sudsutad, W.; Alzabut, J.; Etemad, S.; Rezapour, S. Analysis of a nonlinear fractional system for Zika virus dynamics with sexual transmission route under generalized Caputo-type derivative. J. Appl. Math. Comput. 2022. [Google Scholar] [CrossRef]
  6. Kongson, J.; Thaiprayoon, C.; Neamvonk, A.; Alzabut, J.; Sudsutad, W. Investigation of fractal-fractional HIV infection by evaluating the drug therapy effect in the Atangana-Baleanu sense. Math. Biosci. Eng. 2022, 19, 10762–10808. [Google Scholar] [CrossRef] [PubMed]
  7. Chatterjee, A.N.; Ahmad, B. A fractional-order differential equation model of COVID-19 infection of epithelial cells. Chaos Solitons Fractals 2021, 147, 110952. [Google Scholar] [CrossRef] [PubMed]
  8. Pleumpreedaporn, S.; Pleumpreedaporn, C.; Kongson, J.; Thaiprayoon, C.; Alzabut, J.; Sudsutad, W. Dynamical Analysis of Nutrient-Phytoplankton-Zooplankton Model with Viral Disease in Phytoplankton Species under Atangana-Baleanu-Caputo Derivative. Mathematics 2022, 10, 1578. [Google Scholar] [CrossRef]
  9. Mainardi, F. Some basic problems in continuum and statistical mechanics. In Fractals and Fractional Calculus in Continuum Mechanics; Carpinteri, A., Mainardi, F., Eds.; Springer: Berlin, Germany, 1997; pp. 291–348. [Google Scholar]
  10. Mukhtar, S.; Shah, R.; Noor, S. The numerical investigation of a fractional-order multi-dimensional Model of Navier-Stokes equation via novel techniques. Symmetry 2022, 14, 1102. [Google Scholar] [CrossRef]
  11. Kongson, J.; Sudsutad, W.; Thaiprayoon, C.; Alzabut, J.; Tearnbucha, C. On analysis of a nonlinear fractional system for social media addiction involving Atangana-Baleanu-Caputo derivative. Adv. Differ. Equ. 2021, 2021, 356. [Google Scholar] [CrossRef]
  12. Podlubny, I. Fractional Differential Equations; Academic Press: New York, NY, USA, 1999. [Google Scholar]
  13. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of the Fractional Differential Equations; North-Holland Mathematics Studies; Elsevier: Amsterdam, The Netherlands, 2006; Volume 204. [Google Scholar]
  14. Diethelm, K. The Analysis of Fractional Differential Equations. In Lecture Notes in Mathematics; Springer: New York, NY, USA, 2010. [Google Scholar]
  15. Lakshmikantham, V.; Leela, S.; Devi, J.V. Theory of Fractional Dynamic Systems; Cambridge Scientific Publishers: Cambridge, UK, 2009. [Google Scholar]
  16. Zhou, Y. Basic Theory of Fractional Differential Equations; World Scientific: Singapore, 2014. [Google Scholar]
  17. Sousa, J.V.C.; de Oliveira, E.C. On the ψ-Hilfer fractional derivative. Commun. Nonlinear Sci. Numer. Simul. 2018, 60, 72–91. [Google Scholar] [CrossRef]
  18. Sousa, J.V.C.; de Oliveira, E.C. A Gronwall Inequality and the Cauchy-Type Problem by Means of ψ-Hilfer Operator. Differ. Equ. Appl. 2019, 11, 87–106. [Google Scholar]
  19. Almalahi, A.; Panchal, K. Existence Results of ψ-Hilfer Integro-Differential Equations with Fractional Order in Banach Space. Ann. Univ. Paedagog. Crac. Stud. Math. 2020, 19, 171–192. [Google Scholar] [CrossRef]
  20. Liu, K.; Wang, J.; O’Regan, D. Ulam-Hyers-Mittag-Leffer Stability for ψ-Hilfer Fractional-Order Delay Differential Equations. Adv. Differ. Equ. 2019, 2019, 50. [Google Scholar] [CrossRef]
  21. Alzabut, J.; Adjabi, Y.; Sudsutad, W.; Rehman, M.-U. New Generalizations for Gronwall Type Inequalities Involving a ψ-Fractional Operator and Their Applications. AIMS Math. 2021, 6, 5053–5077. [Google Scholar] [CrossRef]
  22. Thaiprayoon, C.; Sudsutad, W.; Ntouyas, S.K. Mixed Nonlocal Boundary Value Problem for Implicit Fractional Integro-Differential Equations via ψ-Hilfer Fractional Derivative. Adv. Differ. Equ. 2021, 2021, 50. [Google Scholar] [CrossRef]
  23. Kucche, K.D.; Mali, A.D. On the nonlinear (k,ψ)-Hilfer fractional differential equations. Chaos Solitons Fractals 2021, 152, 111335. [Google Scholar] [CrossRef]
  24. Bainov, D.D.; Simeonov, P.S. Impulsive Differential Equations: Periodic Solutions and Applications; Longman Scientific and Technical Group Limited: New York, NY, USA, 1993. [Google Scholar]
  25. Benchohra, M.; Henderson, J.; Ntouyas, S.K. Impulsive Differential Equations and Inclusions; Hindawi Publishing Corporation: New York, NY, USA, 2006; Volume 2. [Google Scholar]
  26. Samoilenko, A.M.; Perestyuk, N.A. Impulsive Differential Equations; World Scientific: Singapore, 1995. [Google Scholar]
  27. Benchohra, M.; Slimani, B.A. Existence and Uniqueness of Solutions to Impulsive Fractional Differential Equations. Elect. J. Diff. Equ. 2009, 2009, 111. [Google Scholar]
  28. Benchohra, M.; Seba, D. Impulsive Fractional Differential Equations in Banach Spaces. Elect. J. Qual. Theory Differ. Equ. 2009, 8, 14. [Google Scholar] [CrossRef]
  29. Wang, J.; Zhou, W.; Fečkan, M. Nonlinear Impulsive Problems for Fractional Differential Equations and Ulam Stability. Comput. Math. Appl. 2012, 64, 3389–3405. [Google Scholar] [CrossRef] [Green Version]
  30. Wang, J.R.; Lin, Z. On the Impulsive Fractional Anti-Periodic BVP Modelling with Constant Coefficients. J. Appl. Math. Comput. 2014, 46, 107–121. [Google Scholar] [CrossRef]
  31. Zuo, M.; Hao, X.; Liu, L.; Cui, Y. Existence Results for Impulsive Fractional Integro-Differential Equation of Mixed Type with Constant Coefficient and Antiperiodic Boundary Conditions. Bound. Value Probl. 2017, 2017, 161. [Google Scholar] [CrossRef] [Green Version]
  32. Kucche, K.D.; Kharade, J.P.; Sousa, J.V.C. On the Nonlinear Impulsive ψ-Hilfer Fractional Differential Equations. Math. Model. Anal. 2020, 25, 642–660. [Google Scholar] [CrossRef]
  33. Salim, A.; Benchohra, M.; Lazreg, J.E.; Henderson, J. On k-Generalized ψ-Hilfer Boundary Value Problems with Retardation and Anticipation. Adv. Theory Nonlinear Anal. Appl. 2022, 6, 173–190. [Google Scholar]
  34. Fečkan, M.; Zhou, Y.; Wang, J. On the Concept and Existence of Solution for Impulsive Fractional Differential Equations. Commun. Nonlinear Sci. Numer Simulat. 2012, 17, 3050–3060. [Google Scholar] [CrossRef]
  35. Guo, T.L.; Jiang, W. Impulsive Fractional Functional Differential Equations. Comput. Math. Appl. 2012, 64, 3414–3424. [Google Scholar] [CrossRef] [Green Version]
  36. Wang, J.; Zhou, Y.; Fečkan, M. On Recent Developments in the Theory of Boundary Value Problems for Impulsive Fractional Differential Equations. Comput. Math. Appl. 2012, 64, 3008–3020. [Google Scholar] [CrossRef]
  37. Shah, K.; Ali, A.; Bushnaq, S. Hyers-Ulam Stability Analysis to Implicit Cauchy Problem of Fractional Differential Equations with Impulsive Conditions. Math. Meth. Appl. Sci. 2018, 41, 8329–8343. [Google Scholar] [CrossRef]
  38. Malti, A.I.N.; Benchohra, M.; Graef, J.R.; Lazreg, J.E. Impulsive Boundary Value Problems for Nonlinear Implicit Caputo- Exponential Type Fractional Differential Equations. Electron. J. Qual. Theory Differ. Equ. 2020, 78, 1–17. [Google Scholar] [CrossRef]
  39. Abbas, M.I. On the Initial Value Problems for the Caputo-Fabrizio Impulsive Fractional Differential Equations. Asian-Eur. J. Math. 2020, 14, 2150073. [Google Scholar] [CrossRef]
  40. Salim, A.; Benchohra, M.; Karapinar, E.; Lazreg, J.E. Existence and Ulam Stability for Impulsive Generalized Hilfer-Type Fractional Differential Equations. Adv. Differ. Equ. 2020, 2020, 601. [Google Scholar] [CrossRef]
  41. Salim, A.; Benchohra, M.; Lazreg, J.E.; Henderson, J. Nonlinear Implicit Generalized Hilfer-Type Fractional Differential Equations with Non-Instantaneous Impulses in Banach Spaces. Adv. Theory Nonlinear Anal. Appl. 2020, 4, 332–348. [Google Scholar] [CrossRef]
  42. Khaminsou, B.; Sudsutad, W.; Thaiprayoon, C.; Alzabut, J.; Pleumpreedaporn, S. Analysis of Impulsive Boundary Value Pantograph Problems via Caputo Proportional Fractional Derivative under Mittag-Leffler Functions. Fractal Fract. 2021, 5, 251. [Google Scholar] [CrossRef]
  43. Kharade, J.P.; Kucche, K.D. On the Impulsive Implicit ψ-Hilfer Fractional Differential Equations with Delay. Math. Meth. Appl. Sci. 2020, 43, 1938–1952. [Google Scholar] [CrossRef] [Green Version]
  44. Savrankumar, S.; Raja, R.; Alzabut, J. Delay-Dependent Passivity Analysis of Non-Deterministic Genetic Regulatory Networks with Leakage and Distributed Delays Against Impulsive Perturbations. Adv. Differ. Equ. 2021, 2021, 353. [Google Scholar] [CrossRef]
  45. Pratap, A.; Raja, R.; Alzabut, J.; Cao, J.; Rajachakit, G.; Hunag, C. Mittag-Leffler Stability and Adaptive Impulsive Synchronization of Fractional Order Neural Networks in Quaternion Field. Math. Meth. Appl. Sci. 2020, 43, 6223–6253. [Google Scholar] [CrossRef]
  46. Afshari, H.; Marasi, H.R.; Alzabut, J. Applications of New Contraction Mappings on Existence and Uniqueness Results for Implicit Φ-Hilfer Fractional Pantograph Differential Equations. J. Inequa. Appl. 2021, 2021, 185. [Google Scholar] [CrossRef]
  47. Kwun, Y.C.; Farid, G.; Nazeer, W.; Ullah, S.; Kang, S.M. Generalized Riemann-Liouville k-fractional integrals associated with Ostrowski type inequalities and error bounds of Hadamard inequalities. IEEE Access 2018, 6, 64946–64953. [Google Scholar] [CrossRef]
  48. Diaz, R.; Pariguan, E. On hypergeometric functions and Pochhammer k-symbol. Divulg. Mat. 2007, 2, 179–192. [Google Scholar]
  49. Dorrego, G.A. An Alternative Definition for the k-Riemann-Liouville Fractional Derivative. Appl. Math. Sci. 2015, 9, 481–491. [Google Scholar] [CrossRef] [Green Version]
  50. Naz, S.; Naeem, M.N. On the Generalization of k-Fractional Hilfer-Katugampola Derivative with Cauchy Problem. Turk. J. Math. 2021, 45, 110–124. [Google Scholar] [CrossRef]
  51. Wang, J.R.; Fečkan, M.; Zhou, Y. Presentation of Solutions of Impulsive Fractional Langevin Equations and Existence Results. Eur. Phys. J. Spec. Top. 2013, 222, 1857–1874. [Google Scholar] [CrossRef]
  52. Almalahi, M.A.; Panchal, S.K. Some existence and stability results for ϕ-Hilfer fractional implicit diferential equation with periodic conditions. J. Math. Anal. Model. 2020, 1, 1–19. [Google Scholar] [CrossRef]
  53. Granas, A.; Dugundji, J. Fixed Point Theory; Springer: New York, NY, USA, 2003. [Google Scholar]
  54. O’Regan, D. Fixed-point theory for the sum of two operators. Appl. Math. Lett. 1996, 9, 1–8. [Google Scholar] [CrossRef]
Figure 1. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 for k = 0 , 1 , 2 .
Figure 1. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 for k = 0 , 1 , 2 .
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Figure 2. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = t ( k + 1 ) 2 + 3 , β k = 1 k + 2 , and ρ k = 3 e k + 1 for k = 0 , 1 , 2 .
Figure 2. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = t ( k + 1 ) 2 + 3 , β k = 1 k + 2 , and ρ k = 3 e k + 1 for k = 0 , 1 , 2 .
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Figure 3. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = sin ( π t + k + 1 ( 2 + k ) t + 3 k + 3 ) , β k = cos ( π 5 k ) , and ρ k = ( k + 2 ) tan ( π 6 k ) for k = 0 , 1 , 2 .
Figure 3. The implicit solutions of Example (49) via α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = sin ( π t + k + 1 ( 2 + k ) t + 3 k + 3 ) , β k = cos ( π 5 k ) , and ρ k = ( k + 2 ) tan ( π 6 k ) for k = 0 , 1 , 2 .
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Figure 4. The implicit solutions of Example (49) via α k , ϕ k ( t ) , β k , and ρ k for k = 0 , 1 , 2 as in Table 4.
Figure 4. The implicit solutions of Example (49) via α k , ϕ k ( t ) , β k , and ρ k for k = 0 , 1 , 2 as in Table 4.
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Table 1. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 for k = 0 , 1 , 2 .
Table 1. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = ln ( t + k + 2 ) 2 k + 2 , β k = 3 k + 2 10 , and ρ k = k + 2 2 for k = 0 , 1 , 2 .
α k Ξ Δ 1 Δ 2 Δ 3
0.22.8926624251.34379811402.0582397910.8286028011
0.43.2132673920.88438199241.9460838740.8450705102
0.63.3280080500.61542639421.9176113700.8499260482
0.83.1991087900.45127235741.9620216470.8435009608
1.02.9036736510.34487659742.0599441340.8272976029
Table 2. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = t ( k + 1 ) 2 + 3 , β k = 1 k + 2 , and ρ k = 3 e k + 1 for k = 0 , 1 , 2 .
Table 2. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = t ( k + 1 ) 2 + 3 , β k = 1 k + 2 , and ρ k = 3 e k + 1 for k = 0 , 1 , 2 .
α k Ξ Δ 1 Δ 2 Δ 3
0.20.9799911124.3770725003.4670015540.6667371782
0.41.0861731114.2869403433.6449591410.6463248146
0.61.1921459464.0396840223.7134339210.6338233539
0.81.2883302093.7022970473.7046774420.6253964583
1.01.3666300963.3201884033.6414345550.6183465441
Table 3. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = sin ( π t + k + 1 ( 2 + k ) t + 3 k + 3 ) , β k = cos ( π 5 k ) , and ρ k = ( k + 2 ) tan ( π 6 k ) for k = 0 , 1 , 2 .
Table 3. Numerical values of Ξ and Δ i , i = 1 , 2 , 3 , for α k { 0.2 , 0.4 , 0.6 , 0.8 , 1.0 } when ϕ k ( t ) = sin ( π t + k + 1 ( 2 + k ) t + 3 k + 3 ) , β k = cos ( π 5 k ) , and ρ k = ( k + 2 ) tan ( π 6 k ) for k = 0 , 1 , 2 .
α k Ξ Δ 1 Δ 2 Δ 3
0.22.7069728845.6932747321.3225846480.7888666540
0.43.0753028703.9507320861.2713888140.8122697082
0.63.4481046012.7480167371.2397771250.8313127993
0.83.8001107631.9243625761.2276189460.8461678966
1.04.0874137051.3634080141.2362120630.8565872889
Table 4. The values of ϕ k ( t ) , α k , β k , ρ k , and Ξ .
Table 4. The values of ϕ k ( t ) , α k , β k , ρ k , and Ξ .
Case ϕ k ( t ) α k β k ρ k Ξ
I ln ( t + k + 2 ) 2 k + 2 7 3 k 9 3 k 3 k + 2 10 k + 2 2 3.350618629
II t ( k + 1 ) 2 + 3 k + 1 2 1 k + 2 3 e k + 1 1.221677115
III sin π t + k + 1 ( 2 + k ) t + 3 k + 3 sin π 4 k cos π 5 k ( k + 2 ) tan π 6 k 3.593850012
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Kaewsuwan, M.; Phuwapathanapun, R.; Sudsutad, W.; Alzabut, J.; Thaiprayoon, C.; Kongson, J. Nonlocal Impulsive Fractional Integral Boundary Value Problem for (ρk,ϕk)-Hilfer Fractional Integro-Differential Equations. Mathematics 2022, 10, 3874. https://doi.org/10.3390/math10203874

AMA Style

Kaewsuwan M, Phuwapathanapun R, Sudsutad W, Alzabut J, Thaiprayoon C, Kongson J. Nonlocal Impulsive Fractional Integral Boundary Value Problem for (ρk,ϕk)-Hilfer Fractional Integro-Differential Equations. Mathematics. 2022; 10(20):3874. https://doi.org/10.3390/math10203874

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Kaewsuwan, Marisa, Rachanee Phuwapathanapun, Weerawat Sudsutad, Jehad Alzabut, Chatthai Thaiprayoon, and Jutarat Kongson. 2022. "Nonlocal Impulsive Fractional Integral Boundary Value Problem for (ρk,ϕk)-Hilfer Fractional Integro-Differential Equations" Mathematics 10, no. 20: 3874. https://doi.org/10.3390/math10203874

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