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Article

(ω,ρ)-BVP Solution of Impulsive Hadamard Fractional Differential Equations

by
Ahmad Al-Omari
1,† and
Hanan Al-Saadi
2,*,†
1
Department of Mathematics, Faculty of Sciences, Al al-Bayt University, P.O. Box 130095, Mafraq 25113, Jordan
2
Department of Mathematics, Faculty of Sciences, Umm Al-Qura University, Makkah 24225, Saudi Arabia
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2023, 11(20), 4370; https://doi.org/10.3390/math11204370
Submission received: 12 September 2023 / Revised: 12 October 2023 / Accepted: 18 October 2023 / Published: 20 October 2023
(This article belongs to the Special Issue Fractional Calculus and Mathematical Applications, 2nd Edition)

Abstract

:
The purpose of this research is to examine the uniqueness and existence of the ( ω , ρ ) -BVP solution for a particular solution to a class of Hadamard fractional differential equations with impulsive boundary value requirements on Banach spaces. The notion of Banach contraction and Schaefer’s theorem are used to prove the study’s key findings. In addition, we offer the prerequisites for the set of solutions to the investigated boundary value with impulsive fractional differential issue to be convex. To enhance the comprehension and practical application of our findings, we offer two illustrative examples at the end of the paper to show how the results can be applied.

1. Introduction

There is rising attention to systems with discontinuous trajectories, like impulsive automatic control systems and impulsive systems, as a result of modern technology’s innovation and ongoing development. With applications in a variety of technical issues, threshold biological phenomena.
As examples of explosive rhythms in health and biology, as well as optimum control models in economics, frequency modulated systems and pharmacokinetics, these types of systems have grown significantly in importance and are currently undergoing rapid expansion [1]. When contrasted to the length of the process as a whole, these processes undergo brief disturbances with negligible durations (see [2,3,4,5,6,7]). Considering the tremendous interest in comprehending the characteristics and conduct of these impulsive systems, there is a strong case for investigating the qualitative elements of the solutions. E. Alvarez et al. proposed and investigated the idea of ( ω , c ) -periodic functions in their work [8,9], which naturally grew up in analysis of equations of Mathieu’s y + a y = 2 q cos ( 2 t ) y . These functions are defined as y ( . + ω ) = c y ( . ) , where c C .
M. Fe c ˘ kan, K. Liu, and J.R. Wang [10] developed this idea further by extending it to ( ω , T ) -periodic solutions for the previously specified family of semilinear evolution equations, where T indicates a Banach space X symbolized by ( BS ), which is a linear isomorphism. We broaden our focus to include ( ω , ρ ) -BVP solutions of boundary value with conditions and impulsive fractional differential equations, building on preliminary research into ( ω , c ) and ( ω , T ) -periodic solutions for ordinary and fractional order derivatives in linear and semilinear problems. This study generalizes findings from earlier research.
Regarding the boundary value conditions in the setting of a BS  X, ref. [11] discusses the existence and uniqueness of the ( ω , ρ ) -BVP solution for particular classes of impulsive Caputo fractional differential equations.
Recent research on ( ω , ρ ) -BVP, ( ω , c ) -BVP, and ( ω , T ) -BVP solutions on fractional differential equations for the existence and uniqueness may be found in [12,13,14] and the references therein. It is worth noting that the majority of the studies discussed above are based on Riemann Liouville or Caputo fractional derivatives. Hadamard [15] introduced another class of fractional operators in 1892, which differs from the ones mentioned above (Riemann–Liouville, Caputo), because Hadamard operators involve logarithmic functions of any exponent and are known as Hadamard derivative/Hadamard integral (for more information, see [16,17,18]). Therefore, we continue recent research under specific conditions, and we offer various results on the uniqueness and existence of ( ω , ρ ) –BVP solutions of impulsive Hadamard fractional integral equations with boundary conditions in the main portion of this study. We finish the research with two exemplary cases that show how the obtained conclusions might be applied.
Because the Hadamard operators include logarithmic functions of any exponent, there are several applications can be obtained via Hadamard derivative/Hadamard integral; for example, by (only) focusing on applications between some type statistical distributions, such as Conway–Maxwell–Poisson (COM–Poisson) and integro-differential equations with variable coefficients using Hadamard-type operators and special functions. Also see [19,20] for a link between fractional hyper-Bessel-type equations involving Hadamard operators and Le Roy functions.
This paper’s structure can be summarized concisely as follows. We begin by recalling some definitions from fractional derivatives and preliminary results on impulsive fractional differential equations, and we define ( ω , c ) -BVP functions. Under specific conditions, we show numerous results on the existence and uniqueness of the ( ω , ρ ) -BVP solution of impulsive equations with boundary conditions and the Hadamard fractional differential in the main portion of this study. To improve understanding and practical implementations of our findings, we finish the paper with two illustrated cases demonstrating the applicability of the acquired results.

2. Preliminary Results

A complex BS is denoted here by ( X , . ) . The space of continuous functions K X are denoted by C ( K : X ) , where K is a compact non-empty subset of R . This space is a BS , endowed with the sup-norm. The space of X-valed piecewise continuous functions on [ 1 , w ] given by
P C ( [ 1 , w ] : X ) { z : [ 1 , ω ] X : z C ( ( t i , t i + 1 ] : X ) , z ( t i ) = z ( t i ) and z ( t i + ) i { 0 , 1 , 2 , , m 1 } } ,
such that t 0 = 1 < t 1 < t 2 < < t m = ω and the notion z ( t i ) and z ( t i + ) indicate the function’s left and right limits z ( t ) at the t = t i , i { 0 , 1 , 2 , , m 1 } , respectively. We use that P C ( [ 1 , ω ] : X ) is a BS endowed with the sup-norm. The function Gamma Γ ( q ) is defined as Γ ( q ) = 0 t q 1 e q d t , R ( q ) > 0 . We known that Γ ( q ) = ( q 1 ) ! , and q > 0 , if z is a specified map defined on a interval [ t 0 , t 1 ] .
Definition 1 
([21]). For a function Φ : [ 1 , ) R , the Hadamard fractional integral equations denoted by (HFI) of order q is obtained by
I q Φ ( t ) = 1 Γ ( q ) 1 t log t ζ q 1 Φ ( ζ ) ζ d ζ
for q > 0 , assuming the integral exists.
Definition 2 
([21]). For a function Φ : [ 1 , ) R , the Hadamard fractional derivative denoted by (HFD) of order q is obtained by
D q Φ ( t ) = 1 Γ ( n q ) t d d t n 1 t log t ζ n q 1 Φ ( ζ ) ζ d ζ
q ( n 1 , n ) , n = [ q ] + 1 , where [ q ] signifies the integer component of a real number q and log ( . ) = log e ( . ) .
Lemma 1 
(Contraction mapping principle [22]). Let D X be closed of a BS  X and F : D D a strict contraction, where F x F y k x y for k ( 0 , 1 ) and x , y D . Then, F has a fixed point that is unique.
Let the set ( ω , ρ ) -BVP function, which is piecewise continuous, be symbolized by Φ ω , ρ where ρ : X X is a linear isomorphism, and Φ ω , ρ = { z : z P C ( [ 1 , ω ] : X ) and z ( ω ) = ρ z ( 1 ) } .
We extend the results in [10,11,23] by examining the ( ω , ρ ) -BVP solutions of the impulsive HFI equations:
D q z ( t ) = Φ [ t , z ( t ) , 1 t λ ( t , ζ ) h ( ζ , z ( ζ ) ) d ζ ] , q ( 0 , 1 ) , t t n , t [ 1 , ω ] , Δ z ( t n ) = I n ( z ( t n ) ) , n = 1 , 2 , , m , z ( ω ) = ρ z ( 1 ) ,
where D q is the HFD of order q ( 0 , 1 ) , and Φ : [ 1 , w ] × X × X X and I k : X X and h : [ 1 , w ] × X X are continuous linear mappings and 1 < t 1 < t 2 < < t m 1 < ω . And, we assume λ : [ 1 , w ] × [ 1 , ω ] R is continuous function, and since [ 1 , w ] × [ 1 , ω ] is compact, there exists a constant L > 0 λ ( t , ζ ) < L .
The following conditions are taken into account in this paper:
C(1)
Let E be the identity and ρ be a isomorphism linear on X, where ( ρ E ) is also injective;
C(2)
Let M > 0 and ( ρ E ) 1 < M ;
C(3)
Let I n : X X be a continuous operator and ∃ C 1 [ 0 , 1 m ) with I n ( z ) I n ( v ) C 1 z v , ∀ z , v X and n = 1 , 2 , , m ;
C(4)
There is a finite C 2 > 0 I n ( z ) C 2 z , n = 1 , 2 , , m and ∀ z X ;
C(5)
There are continuous functions μ 1 ( t ) , μ 2 ( t ) : [ 1 , ω ] R + Φ ( t , z , ψ 1 ) Φ ( t , v , ψ 2 ) μ 1 ( t ) z v + μ 2 ( t ) ψ 1 ψ 2 t [ 1 , ω ] and ∀ z , v X and ψ 1 , ψ 2 Ψ = P C ( [ 1 , ω ] , X ) ;
C(6)
There are continuous functions μ 3 ( t ) : [ 1 , ω ] R + and P > 0 Φ ( t , z , ψ ) μ 3 ( t ) z + ψ + P for all z X and ψ Ψ = P C ( [ 1 , ω ] , X ) ;
C(7)
There are constants K > 0 and H > 0 h ( t , z ) K z and h ( t , z ) h ( t , v ) H z v .
Lemma 2. 
Let q ( 0 , 1 ) and Φ : X × I × X X be continuous. A function z Ψ = P C ( [ 1 , ω ] , X ) is called a solution of the HFI equations
z ( t ) = z ( 1 ) 1 Γ ( q ) 1 a log a ζ q 1 Φ [ ζ , z ( ζ ) , f ( ζ ) ] d ζ ζ + 1 Γ ( q ) 1 t log t ζ q 1 Φ [ ζ , z ( ζ ) , f ( ζ ) ] d ζ ζ
if z is a solution of the HFI equations.
D q z ( t ) = Φ [ t , z ( t ) , f ( t ) ] , t [ 1 , ω ] , z ( a ) = z ( 1 ) , a [ 1 , ω ]
Lemma 3. 
The HFI equations
z ( t ) = z ( 1 ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) .
have a solution z Ψ = P C ( [ 1 , ω ] , X ) for t ( t n , t n + 1 ] , n = 1 , 2 , , m if z is a solution of the HFI Equation (1).
Proof. 
First, assume z Ψ = P C ( [ 1 , ω ] , X ) satisfies Problem (1); we have to show that the HFI equations includes at least one solution z Ψ = P C ( [ 1 , ω ] , X ) . Consider F : Ψ Ψ be the operator specified by
( F z ) ( t ) = z ( t ) = z ( 1 ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) n = 1 , 2 , , m .
It is clear that F is defined due to C(3) and C(5). Let { z k } be a sequence and z k z in Ψ . Then, for all t [ 1 , ω ] , next consider:
( F z k ) ( t ) ( F z r ) ( t ) = 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z k ( t n ) ) I n ( z r ( t n ) ) .
1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] + Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] + Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z k ( t n ) ) I n ( z ( t n ) ) + I n ( z ( t n ) ) I n ( z r ( t n ) ) .
1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z k ( t n ) ) I n ( z ( t n ) ) + 1 < t n < t I n ( z r ( t n ) ) I n ( z ( t n ) ) .
Since I n is continuous and Φ is also jointly continuous, then we have
Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] 0
as r , also 1 < t n < t I n ( z k ( t n ) ) I n ( z ( t n ) ) 0 as r , therefore ( F z k ) ( t ) ( F z r ) ( t ) 0 as r , k . Consequently, by assumption C(3) and C(5), it is not difficult to obtain that ( F z r ) ( t ) ( F z ) ( t ) 0 as r , as follows:
( F z r ) ( t ) ( F z ) ( t ) 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z r ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z r ( τ ) ) d τ ] Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z r ( t n ) ) I n ( z ( t n ) ) .
1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 μ 1 ( t ) z r z + L H ω μ 2 ( t ) z r z d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 μ 1 ( t ) z r z + L H ω μ 2 ( t ) z r z d ζ ζ + 1 < t n < t C 1 z r z .
As a result, F is both continuous and entirely continuous. As a result of Schaefer’s theorem, one may conclude that F includes at least one fixed point on Ψ that is a fractional problem solution (1).
Assume, on the other hand, that z satisfies the HFI Equation (2). If t ( 1 , t 1 ] then z ( 1 ) = z 0 , and by utilizing the fact D q is the left inverse of I q , and by Lemma 2, one can obtain D q z ( t ) = Φ ( t , z ( t ) ) . If t ( t n , t n + 1 ] , n = 1 , , m also by Lemma 2 and using that fact the Hadamard derivative of a constant is equal to zero. It can deduced that D q z ( t ) = Φ ( t , z ( t ) ) for t ( t n , t n + 1 ] and z ( t n + ) = z ( t n ) + I n ( z ( t n ) ) , which completes the proof. □

3. Main Results

We investigate the ( ω , ρ ) -BVP solution of Equation (1). Let t 0 = 1 , t m = ω .
If a function z Ψ in Problem (1) is satisfied almost everywhere on [ 1 , ω ] , and the condition of (1) holds, the function z is described a solution of (1).
Proposition 1. 
If ρ : X X is a isomorphism linear operator and ( ρ E ) is injective. Then, a solution z Ψ = P C ( [ 1 , ω ] : X ) of Problem (1) is provided from
z ( t ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t k 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) , t ( t n , t n + 1 ] , n = 0 , 1 , , m 1 .
Proof. 
So, using Lemma 3, the solution z Ψ of Problem (1) holds
z ( t ) = z ( 1 ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) , t ( t n , t n + 1 ] , n = 0 , 1 , , m 1 ,
and we have
z ( ω ) = z ( 1 ) + 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) .
If the requirement for boundary values is used, in our case
z ( 1 ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) .
Hence,
z ( t ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) , t ( t n , t n + 1 ] , n = 0 , 1 , , m 1 .
 □
Theorem 1. 
If C ( 1 ) C ( 5 ) and C ( 7 ) are satisfied, and if 0 < δ < 1 , where
δ = ω ( M + 1 ) ( log ω ) q Γ ( q + 1 ) ( m + 1 ) μ 1 ( t ) + L H μ 2 ( t ) + ( m ) Γ ( q + 1 ) C 1 ω ( log ω ) q ,
then the impulsive HFI Equation (1) has ( ω , ρ ) -BVP solution which is a unique z Φ ω , ρ . Moreover,
z ( M + 1 ) ( m + 1 ) [ ( log ω ) q ] Φ 0 [ μ 1 ( t ) + ω L K μ 2 ( t ) ] Γ ( q + 1 ) ( M + 1 ) [ ( m + 1 ) [ ( log ω ) q ] [ μ 1 ( t ) + ω L K μ 2 ( t ) ] + m C 2 Γ ( q + 1 ) ] ,
where Φ 0 = s u p ζ [ 1 , ω ] Φ ( ζ , 0 ) .
Proof. 
Let the operator Θ : Ψ Ψ by
( Θ z ) ( t ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) , t ( t n , t n + 1 ] , n = 0 , 1 , , m 1 .
The unambiguous dictate of the fixed points of Θ for the ( ω , ρ ) -BVP solution of (1). Furthermore, It is clear that Θ ( Ψ ) Ψ . For all z 1 , z 2 Ψ , we currently have
( Θ z 1 ) ( t ) ( Θ z 2 ) ( t ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z 1 ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z 1 ( t n ) ) , ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z 2 ( t n ) ) 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ 1 < t n < t I n ( z 2 ( t n ) ) ,
( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 × Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] Φ [ ζ , z 2 ( ζ ) , 0 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 0 < t n < ω I n ( z 1 ( t n ) ) I n ( z 2 ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z 1 ( t n ) ) I n ( z 2 ( t n ) ) .
( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 × Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , z 2 ( ζ ) , 0 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 0 < t n < ω I n ( z 1 ( t n ) ) I n ( z 2 ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z 1 ( t n ) ) I n ( z 2 ( t n ) ) .
( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 μ 1 ( t ) z 1 z 2 + μ 2 ( t ) 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 × μ 1 ( t ) z 1 z 2 + μ 2 ( t ) 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + m C 1 z 1 z 2 + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 μ 1 ( ζ ) z 1 z 2 + μ 2 ( t ) × 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 μ 1 ( t ) z 1 z 2 + μ 2 ( t ) 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + C 1 m z 1 z 2 .
M 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 μ 1 ( t ) z 1 z 2 + L H μ 2 ( t ) 1 ζ z 1 z 2 d τ d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 μ 1 ( t ) z 1 z 2 + L H μ 2 ( ζ ) 1 ζ z 1 z 2 d τ d ζ ζ + m C I z 1 z 2 + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 μ 1 ( t ) z 1 z 2 + L H μ 2 ( t ) 1 ζ z 1 z 2 d τ d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 μ 1 ( t ) z 1 z 2 + L H μ 2 ( t ) 1 ζ z 1 z 2 d τ d ζ ζ + C 1 m z 1 z 2 .
M μ 1 ( t ) + L H μ 2 ( t ) ω Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 z 1 z 2 d τ d ζ ζ + μ 1 ( t ) + L H μ 2 ( t ) ω Γ ( q ) t n ω log ω ζ q 1 z 1 z 2 d ζ ζ + m C 1 z 1 z 2 + μ 1 ( t ) + L H μ 2 ( t ) ω Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 z 1 z 2 d ζ ζ + μ 1 ( t ) + L H μ 2 ( t ) ω Γ ( q ) t n t log t ζ q 1 z 1 z 2 d ζ ζ + C 1 m z 1 z 2 .
M ( μ 1 ( t ) + ω L H μ 2 ( t ) ) ( m ( log ω ) q ) Γ ( q + 1 ) + ( μ 1 ( t ) + ω L H μ 2 ( t ) ) ( log ω ) q ) Γ ( q + 1 ) + m C I + ( μ 1 ( t ) + ω L H μ 2 ( t ) ) ( m ( log ω ) q ) Γ ( q + 1 ) + ( μ 1 ( t ) + ω L H μ 2 ( t ) ) ( log ω ) q Γ ( q + 1 ) + C 1 m z 1 z 2 .
ω ( M + 1 ) ( log ω ) q Γ ( q + 1 ) ( m + 1 ) μ 1 ( t ) + L H μ 2 ( t ) + ( m ) Γ ( q + 1 ) C 1 ω ( log ω ) q z 1 z 2 . δ z 1 z 2 .
Since 0 < δ < 1 , we can conclude that it is a contraction of Θ . Hence, a unique fixed points of the operator Θ in such a way z ( 1 ) = ρ z ( ω ) , z Φ ω , ρ exists. Hence, Equation (1) has an ( ω , ρ ) -BVP solution, which is a unique z Φ ω , ρ . And,
z ( t ) ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 × Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ) ] Φ [ ζ , 0 , 0 ] d ζ ζ + ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , 0 , 0 ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , 0 , 0 ] d ζ ζ + ( ρ E ) 1 1 < t n < ω I n ( z 1 ( t n ) ) + 1 < t n < t I n ( z 1 ( t n ) ) .
z ( t ) ( ρ E ) 1 μ 1 ( t ) + ω L K μ 2 ( t ) Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 z d ζ ζ + μ 1 ( t ) + ω L K μ 2 ( t ) Γ ( q ) t n ω log ω ζ q 1 z d ζ ζ + μ 1 ( t ) + ω L K μ 2 ( t ) Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 z d ζ ζ + t n t log t ζ q 1 z d ζ ζ + ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ 0 d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ 0 d ζ ζ + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ 0 d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ 0 d ζ ζ + ( ρ E ) 1 1 < t n < ω I n ( z 1 ( t n ) ) + 1 < t n < t I n ( z 1 ( t n ) ) z [ μ 1 ( t ) + ω L K μ 2 ( t ) ] ) Γ ( q + 1 ) ( m + 1 ) [ ( log ω ) q ] ( M + 1 ) + z m C 2 ( M + 1 ) + Φ 0 [ μ 1 ( t ) + ω L K μ 2 ( t ) ] Γ ( q + 1 ) ( m + 1 ) [ ( log ω ) q ] ( M + 1 ) .
Hence,
z ( M + 1 ) ( m + 1 ) [ ( log ω ) q ] Φ 0 [ μ 1 ( t ) + ω L K μ 2 ( t ) ] Γ ( q + 1 ) ( M + 1 ) [ ( m + 1 ) [ ( log ω ) q ] [ μ 1 ( t ) + ω L K μ 2 ( t ) ] + m C 2 Γ ( q + 1 ) ] .
 □
Theorem 2. 
Consider that C ( 1 ) C ( 4 ) and C ( 6 ) C ( 7 ) hold. Then, the impulsive HFI Equation (1) includes at least one ( ω , ρ ) -BVP solution z Φ ω , ρ .
Proof. 
Assume that B r = { z Ψ : z r } , such that
P ( M + 1 ) ( m + 1 ) ( log ω ) q Γ ( q + 1 ) ( M + 1 ) ( m + 1 ) [ μ 3 ( t ) ( 1 + ω L K ) ] ( log ω ) q + m C 2 Γ ( q + 1 ) r .
We reconsider the operator
( Θ z ) ( t ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z ( t n ) ) , n = 1 , 2 , , m .
Step (1): Let us prove that Θ : Ψ Ψ is bounded operator. It is enough to demonstrate that for all r > 0 , ∃ δ 1 > 0 , in such a way that for each z B r , there are Θ z δ 1 . Let ( z k ) be a bounded a sequence subset B B r . So, using C(4) and C(6), we achieve
( Θ z k ) ( t ) ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 × Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z k ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z k ( t n ) )
( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 μ 3 ( t ) ( 1 + ω L K ) z k + P d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 μ 3 ( t ) ( 1 + ω L K ) z k + P d ζ ζ + C 2 m z k + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 μ 3 ( t ) ( 1 + ω L K ) z k + P d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 μ 3 ( t ) ( 1 + ω L K ) z k + P d ζ ζ + C 2 m z k
M m ( log ω ) q μ 3 ( t ) ( 1 + ω L K ) r + P Γ ( q + 1 ) + ( log ω ) q μ 3 ( t ) ( 1 + ω L K ) r + P Γ ( q + 1 ) + C 2 m r + m ( log ω ) q μ 3 ( t ) ( 1 + ω L K ) r + P Γ ( q + 1 ) + ( log ω ) q μ 3 ( t ) ( 1 + ω L K ) r + P Γ ( q + 1 ) + C 2 m r ( M + 1 ) ( m + 1 ) μ 3 ( t ) ( 1 + ω L K ) r + P ( log ω ) q + C 2 m r Γ ( q + 1 ) Γ ( q + 1 ) = δ 1 .
Hence, ( Θ z k ) is uniformly bounded on B r , which suggests Θ ( B ) is bounded in B r .
Step (2): We demonstrate the Θ : Ψ Ψ is an equicontinuous operator. We let ( z k ) be a bounded a sequence subset B B r . Let t 1 , t 2 [ 1 , ω ] and t 1 < t 2 . So,
( Θ z k ) ( t 2 ) ( Θ z k ) ( t 1 ) = ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z k ( t n ) ) + 1 Γ ( q ) 1 < t n < t 2 t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t 2 log t 2 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < t 2 I n ( z k ( t n ) )
( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z k ( t n ) ) 1 Γ ( q ) 1 < t n < t 1 t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ 1 Γ ( q ) t n t 1 log t 1 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ 1 < t n < t 1 I n ( z k ( t n ) )
= 1 Γ ( q ) 1 < t n < t 2 t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ 1 Γ ( q ) 1 < t n < t 1 t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t 2 log t 2 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ 1 Γ ( q ) t n t 1 log t 1 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < t 2 I n ( z k ( t n ) ) 1 < t n < t 1 I n ( z k ( t n ) )
1 Γ ( q ) 1 t 1 log t 2 ζ q 1 log t 1 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t 1 t 2 log t 2 ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + t 1 < t n < t 2 I n ( z k ( t n ) ) .
Putting t 2 t 1 , we discover that the right-hand side of the previous inequality approaches to zero. So, ( Θ z k ) is equicontinuous.
Step (3): The compact operator Θ : Ψ Ψ is used. In fact, we permitted B Ψ . Since Θ is limited and equicontinuous according to the Arzelà–Ascoli theorem, we may infer that Θ ( B ) is a reasonably compact subset of Ψ . Consequently, Θ : Ψ Ψ is a compact operator.
Step (4): Set F ( Θ ) = { z Ψ : z = δ 1 · Θ ( z ) , for some δ 1 [ 0 , 1 ] } , is bounded. It is now obvious that the fixed points of Θ are solutions of Equation (1). Because Θ is continuous, we must demonstrate that set.
F ( Θ ) = { z Ψ : z = δ 1 · Θ ( z ) , for some δ 1 ( 0 , 1 ) } is bounded.
We let z F ( Θ ) . Then, z = δ 1 · Θ ( z ) for some δ 1 ( 0 , 1 ) . Now,
z ( t ) = δ 1 ( Θ z k ) ( t ) ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z k ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z k ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z k ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z k ( t n ) ) ( M + 1 ) ( m + 1 ) μ 3 ( t ) ( 1 + ω L K ) r + P ( log ω ) q + C 2 m r Γ ( q + 1 ) Γ ( q + 1 ) ,
so F ( Θ ) is bounded in Ψ . By Schaefer’s theorem, we determine that Θ having a fixed point and, according to the preceding, this point is a solution of Problem (1). □
Theorem 3. 
If Assumptions C ( 1 ) , C ( 2 ) , C ( 4 ) , C ( 6 ) and C ( 7 ) hold, the collection of the ( ω , ρ ) -BVP solution to the impulsive HFI equations in Problem (1) is then convex.
Proof. 
We use Theorem 2 to show that the differential Equation (1) with an impulsive fractional has a solution in Ψ . We set δ 1 = 1 . So, the collection of solutions is provided for by F ( Θ ) = { z Ψ : z = Θ ( z ) } . For all z 1 , z 2 F ( Θ ) , 0 λ l , t [ 1 , ω ] we have
λ z 1 ( t ) + ( 1 λ ) z 2 ( t ) = λ ( Θ z 1 ) ( t ) + ( 1 λ ) ( Θ z 2 ) ( t ) = λ ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z 1 ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z 1 ( t n ) ) + ( 1 λ ) ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 < t n < ω I n ( z 2 ( t n ) ) + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ [ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ ] d ζ ζ + 1 < t n < t I n ( z 2 ( t n ) )
= ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 λ Φ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ + ( 1 λ ) Φ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 λ Φ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ + ( 1 λ ) Φ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ 1 < t n < ω [ λ I n ( z 1 ( t n ) ) + ( 1 λ ) I n ( z 2 ( t n ) ) ] + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 λ Φ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ + ( 1 λ ) Φ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 λ Φ ζ , z 1 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 1 ( τ ) ) d τ + ( 1 λ ) Φ ζ , z 2 ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , z 2 ( τ ) ) d τ d ζ ζ + 1 < t n < t λ I n ( z 1 ( t n ) ) + ( 1 λ ) I n ( z 2 ( t n ) )
= ( ρ E ) 1 1 Γ ( q ) 1 < t n < ω t n 1 t n log t n ζ q 1 Φ ζ , ( λ z 1 + ( 1 λ ) z 2 ) ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , ( λ z 1 + ( 1 λ ) z 2 ) ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n ω log ω ζ q 1 Φ ζ , ( λ z 1 + ( 1 λ ) z 2 ) ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , ( λ z 1 + ( 1 λ ) z 2 ) ( τ ) ) d τ d ζ ζ + 1 < t n < ω I n [ ( λ z 1 + ( 1 λ ) z 2 ) ( t n ) ] + 1 Γ ( q ) 1 < t n < t t n 1 t n log t n ζ q 1 Φ ζ , ( λ z 1 + ( 1 λ ) z 2 ) ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , ( λ z 1 + ( 1 λ ) z 2 ) ( τ ) ) d τ d ζ ζ + 1 Γ ( q ) t n t log t ζ q 1 Φ ζ , ( λ z 1 + ( 1 λ ) z 2 ) ( ζ ) , 1 ζ λ ( ζ , τ ) h ( τ , ( λ z 1 + ( 1 λ ) z 2 ) ( τ ) ) d τ d ζ ζ + 1 < t n < t I n [ ( λ z 1 + ( 1 λ ) z 2 ) ( t n ) ] .
Hence,
λ z 1 + ( 1 λ ) z 2 ( t ) = δ 1 Θ λ z 1 + ( 1 λ ) z 2 ( t ) ,
so λ z 1 + ( 1 λ ) z 2 F ( Θ ) , implies that F ( Θ ) is a convex set. This implies that the collection of ( ω , ρ ) -BVP solution of Problem (1) is convex. □
Two excellent examples round up this section:
Example 1. 
The following differential problem with an impulsive fractional is considered:
D 1 3 z ( t ) = a sin 2 t cos z ( t ) + 1 t e ζ t 20 + z ( ζ ) d ζ , t t n , t [ 1 , ) , Δ z ( t n ) = 1 8 m ( z ( t n ) ) , n = 1 , 2 , , m , z ( π + 1 ) = ρ z ( 1 ) ,
where  a R , t n = n π + 2 2 , n = 1 , 2 , , m .  So,
Φ ( t , z ( t ) , ψ ) = a sin 2 t cos z ( t ) + 1 t e ζ t 20 + z ( ζ ) d ζ
and I n ( z ( t n ) ) = 1 8 m ( z ( t n ) ) . We set ω = π + 1 and ρ ( t ) = c t , such that c C { 0 } also λ ( t , ζ ) = e ζ t , h ( ζ , z ( ζ ) ) = 1 20 + z ( ζ ) . Then, there is
  • λ ( t , ζ ) = e ζ t e = L ;
  • Since  I n ( z ( t n ) ) 1 8 m z , C ( 4 ) is fulfilled for C 2 = 1 8 m and since I n ( z 1 ( t n ) ) I n ( z 2 ( t n ) ) 1 8 m z 1 z 2 , C ( 3 ) holds for C 1 = 1 8 m ;
  • h ( ζ , z 1 ( ζ ) ) h ( ζ , z 2 ( ζ ) ) = 1 20 + z 1 ( ζ ) 1 20 + z 2 ( ζ ) = z 1 ( ζ ) z 2 ( ζ ) ( 20 + z 1 ( ζ ) ) ( 20 + z 2 ( ζ ) ) 1 400 z 1 z 2 = H z 1 z 2
    and h ( ζ , z ( ζ ) ) = 1 20 + z ( ζ ) 1 20 z = K z ;
  • For any t [ 1 , ω ] , z 1 , z 2 R we have
    Φ ( t , z 1 , ψ 1 ) Φ ( t , z 2 , ψ 2 ) = a sin 2 t cos z 1 ( t ) + 1 t e ζ t 20 + z 1 ( ζ ) d ζ a sin 2 t cos z 2 ( t ) + 1 t e ζ t 20 + z 2 ( ζ ) d ζ a sin 2 t cos z 1 ( t ) a sin 2 t cos z 2 ( t ) + 1 t e ζ t 20 + z 1 ( ζ ) e ζ t 20 + z 2 ( ζ ) d ζ | a | z 1 z 2 + e 400 z 1 z 2 = μ 1 ( t ) z 1 z 2 + μ 2 ( t ) z 1 z 2 .
    So C ( 5 ) is satisfied for μ 1 ( t ) = | a | and μ 2 ( t ) = e 400 ;
  • Assumptions C ( 1 ) and C ( 2 ) are trivial holds for M = 1 .
Also, if we take | a | and m be so that
6 8 Γ ( 4 3 ) 2 ( m + 1 ) log ( π + 1 ) 3 | a | + ( π + 1 ) e 2 8000 > 0 . Then, by Theorem 1, the in this example has a unique ( ω , ρ ) -BVP solution z Φ ω , ρ .
Moreover, we see that
z 2 ( m + 1 ) log ( π + 1 ) 3 | a | + ( π + 1 ) e 2 8000 Γ ( 4 3 ) 2 ( m + 1 ) log ( π + 1 ) 3 | a | + ( π + 1 ) e 2 8000 2 8 Γ ( 4 3 ) ,
Example 2. 
The following differential problem with an impulsive fractional is considered:
D 2 3 z ( t ) = a sin z ( t ) + 1 5 1 t e z ( ζ ) 10 d ζ , t t n , t [ 1 , ) , Δ z ( t n ) = cos n π , n = 1 , 2 , , m , z ( 2 π + 1 ) = ρ z ( 1 ) ,
where  a R , t n = n π + 2 2 , n = 1 , 2 , , m . Therefore,
Φ ( t , z ( t ) , ψ ) = a sin z ( t ) + 1 5 1 t e z ( ζ ) 10 d ζ
and I n ( z ( t n ) ) = cos n π . We set ω = 2 π + 1 and ρ ( t ) = e t , also λ ( t , ζ ) = 1 , h ( ζ , z ( ζ ) ) = e z ( ζ ) 10 . Then, we have
  • λ ( t , ζ ) 1 = L ;
  • Since I n ( z ( t n ) ) cos n π 1 , C ( 4 ) is fulfilled for C 2 = 1 ;
  • h ( ζ , z 1 ( ζ ) ) h ( ζ , z 2 ( ζ ) ) = e z 1 ( ζ ) 10 e z 2 ( ζ ) 10 1 10 z 1 z 2 = H z 1 z 2
    and h ( ζ , z ( ζ ) ) = e z 2 ( ζ ) 10 1 10 z = K z ;
  • For any t [ 1 , ω ] , z R , so
    Φ ( t , z , ψ ) = a sin z ( t ) + 1 5 1 t e z ( ζ ) 10 d ζ = a sin z ( t ) + 1 5 1 t e z ( ζ ) 10 d ζ μ ( t ) 3 ( z + ψ + P
    It is clear that C ( 6 ) is hold for μ 3 ( t ) = m a x { | a | , 1 50 } and P > 0 ;
  • Assumptions C ( 1 ) and C ( 2 ) are trivial holds for M = 1 .
  • As a result, assuming the assumptions of Theorem 2 are met, the problem with is impulsive fractional in this example includes at least one ( ω , ρ ) -BVP solution z Φ ω , ρ . Furthermore, according to Theorem 3, the collection of solutions in this situation is convex.

4. Conclusions

This project displays well-established results for the existence of ( ω , ρ ) -BVP solutions for a particular class of impulsive Hadamard fractional differential equations with boundary value constraints on BS . The investigation’s primary goal is to provide what is necessary for the existence and uniqueness of the ( ω , ρ ) -periodic solution The Banach contraction mapping idea is used in Equation (1). Furthermore, the paper provides necessary circumstances for the existence of ( ω , ρ ) -BVP solutions to Problem (1), Schaefer’s fixed point theorem. We end the paper by offering two examples that show how the generated results can be used. The authors intend to conduct additional studies into the existence and significance of equations, including some different types of fractional derivatives. For other forms of abstract fractional differentials, the ( ω , ρ ) -BVP solution is unique.

Author Contributions

Writing—original draft, A.A.-O. and H.A.-S.; writing—review & editing, A.A.-O. and H.A.-S. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

No data were used to support this study.

Conflicts of Interest

The authors declare no conflict of interest.

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Al-Omari, A.; Al-Saadi, H. (ω,ρ)-BVP Solution of Impulsive Hadamard Fractional Differential Equations. Mathematics 2023, 11, 4370. https://doi.org/10.3390/math11204370

AMA Style

Al-Omari A, Al-Saadi H. (ω,ρ)-BVP Solution of Impulsive Hadamard Fractional Differential Equations. Mathematics. 2023; 11(20):4370. https://doi.org/10.3390/math11204370

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Al-Omari, Ahmad, and Hanan Al-Saadi. 2023. "(ω,ρ)-BVP Solution of Impulsive Hadamard Fractional Differential Equations" Mathematics 11, no. 20: 4370. https://doi.org/10.3390/math11204370

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