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Article

Mild Solutions of the (ρ1,ρ2,k1,k2,φ)-Proportional Hilfer–Cauchy Problem

Department of Mathematics, Qiongtai Normal University, Haikou 571100, China
*
Author to whom correspondence should be addressed.
Symmetry 2024, 16(10), 1349; https://doi.org/10.3390/sym16101349
Submission received: 26 August 2024 / Revised: 8 October 2024 / Accepted: 9 October 2024 / Published: 11 October 2024

Abstract

:
Inspired by prior research on fractional calculus, we introduce new fractional integral and derivative operators: the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional Hilfer fractional derivative. Numerous previous studied fractional integrals and derivatives can be considered as particular instances of the novel operators introduced above. Some properties of the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral are discussed, including mapping properties, the generalized Laplace transform of the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional Hilfer fractional derivative. The results obtained suggest that the most comprehensive formulation of this fractional calculus has been achieved. Under the guidance of the findings from earlier sections, we investigate the existence of mild solutions for the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional Hilfer fractional Cauchy problem. An illustrative example is provided to demonstrate the main results.

1. Introduction

This paper introduces the definitions of the novel ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional Hilfer (H) fractional derivative. Fractional calculus is widely used in various fields such as physics, engineering, and medicine; see [1,2,3,4,5]. At the present time, more general fractional integrals and derivatives have been developed, including the Caputo (C) and Riemann–Liouville (R-L) derivatives. The R-L derivative and the C derivative represent symmetrical concepts. The former involves integration followed by differentiation, while the latter proceeds in the opposite manner. Building upon these, R. Hilfer [4] introduced the H fractional derivative, an advancement that encapsulates the characteristics of both R-L and C derivatives. When v = 1 2 , the H fractional derivative is a symmetric derivative. The H derivative is distinguished by its technical properties, offering broader applicability and versatility compared to its predecessors in the field of fractional calculus. The application of H fractional differential equations is pivotal in characterizing a wide array of processes and materials within the realms of electrical circuits, control theory, and electromagnetism. Owing to its versatility, the H fractional derivative stands out as a more comprehensive tool for examining real-world occurrences and fostering subsequent technological innovations [6]. In 2018, Sousa et al. [7] introduced a new fractional derivative, the φ -Hilfer ( φ -H) fractional derivative, which further extends the generalization of previously established fractional derivatives, such as the H, C, and R-L fractional derivatives. The φ -H derivative’s primary advantage lies in its adaptability, allowing for the selection of the appropriate function φ , which facilitates the consolidation and revival of numerous prior investigations into fractional differential equations. However, the C, R-L, H, and φ -H fractional derivatives all bear a significant relationship to the gamma function. In 2007, Díaz et al. [8] introduced the generalized gamma function Γ k ( · ) and the Beta function B k ( · ) , marking a new era in the evolution and expansion of fractional derivatives. For instance, Kucche et al. [9] proposed the ( k , φ ) -H fractional derivative, which surpasses the φ -H fractional derivative in terms of generality. By strategically selecting the appropriate kernel φ and the parameters v [ 0 , 1 ] and k > 0 , exceptional cases, like the k-Hilfer–Hadamard fractional derivative, ( k , φ ) -C, ( k , φ ) -R-L, and others can be derived.
In [10], Jarad et al. introduced the generalized proportional fractional (GPF) integral and derivative. The primary advantage of this type of fractional derivative is its flexibility, allowing for the selection of the proportion parameter ρ within the range ( 0 , 1 ] . Furthermore, the GPF derivative not only extends the R-L fractional derivative but also generalizes several earlier fractional derivatives from a distinct perspective. Naturally, the question arises: which fractional derivative best represents real-world problems? Finding a definitive answer is challenging. A key motivation for this paper is to present a more general form of fractional integrals and derivatives, although we do not claim to solve the aforementioned question. Inspired by the ( k , φ ) -H and GPF derivatives, we define a new fractional derivative, the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H derivative. To the best of our knowledge, no prior research has been published on the definition and properties of this ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional derivative. We highlight the main contributions of this work as follows:
1.
We introduce a novel ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional derivative. In the context of this fractional derivative, it is possible to retrieve a spectrum of fractional derivatives and integrals.
2.
We derive and are now examining the properties intrinsic to this novel type of fractional calculus. A subset of these properties has been rigorously demonstrated.
3.
We present a characterization of the mild solution for the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional Cauchy problem. Futhermore, we delve into the existence of the mild solutions, thereby addressing and correcting inaccuracies found in several prior studies.
We have endeavored to articulate the fundamental principles of ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional fractional calculus. In Section 2, we focus on presenting the definitions of the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional derivative. Subsequently, we delve into an exploration of their properties, including the mapping properties of the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral with the weighted spaces L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] and C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] . Additionally, we provide the outcomes of generalized ( k 2 , φ ) -Laplace transform for both the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral and the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional derivative. Section 3 is dedicated to the discourse on the existence of mild solutions for a ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional Cauchy problem. Utilizing the properties of probability density [11] and semigroup theory [12], we establish the appropriate definition of a mild solution. Subsequently, we employ the Banach contraction fixed-point theorem to demonstrate the existence of mild solutions for the Cauchy problem. We also present an example of a ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional Cauchy problem, which serves to elucidate the main results developed and to indicate potential avenues for further research in this domain.

2. The ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -Proportional Integrals and H Derivatives

Throughout this paper, we let φ : R R be a function with φ ( t ) > 0 .
Definition 1
([13]). If f and h are two functions, the general convolution of functions f and h is given by
( f φ h ) ( t ) = a t f ( s ) h ¯ ( t , s ) φ ( s ) d s ,
where h ¯ ( t , s ) = h φ 1 ( φ ( t ) φ ( s ) + φ ( a ) ) .
Let ρ 1 , ρ 2 , k 1 , k 2 > 0 , η R for short, denoted as
g ( t , s ) = φ ( t ) φ ( s ) , g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , s ) = g η k 1 1 ( t , s ) ρ 1 η k 1 k 1 Γ k 1 ( η ) e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 , t , s R ,
g a , ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t ) = 0 , t a , g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , a ) , t > a ,
and
g b , ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t ) = 0 , t b , g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( b , t ) , t < b ,
where Γ k ( z ) denotes the k-gamma function
Γ k ( z ) = 0 s z 1 e s k k d s , Re ( z ) > 0 .
Definition 2.
If ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , then the first-order left ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional derivative D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h is defined by
D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = ρ 1 δ k 1 , φ 1 h ( t ) + ( 1 ρ 2 ) k 1 k 2 h ( t ) .
The first-order right ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional derivative D 1 , ρ 1 , ρ 2 ; φ    k 1 , k 2 h is given by
D k 1 , ρ 1 , ρ 2 ; φ    k 1 , k 2 h ( t ) = ρ 1 δ k 1 , φ 1 h ( t ) + ( 1 ρ 2 ) k 1 k 2 h ( t ) ,
where δ k 1 , φ 1 = k 1 φ ( t ) d d t .
Remark 1.
If ρ 1 = ρ 2 = ρ 1 , k 1 = k 2 = 1 with φ ( t ) = t , then the ( ρ , ρ , 1 , 1 , t ) -proportional derivative reduces to D 1 , ρ h ( t ) [10].
Remark 2.
e ρ 2 1 ρ 1 k 2 g ( t , a ) is not a constant for ρ 2 < 1 , but D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 e ρ 2 1 ρ 1 k 2 g ( t , a ) = 0 .
Definition 3.
Let ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , h L 1 [ a , b ] , then
I a + k 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = a t g ^ ρ 1 , ρ 2 , k 1 , k 2 , k 1 ; φ ( t , s ) h ( s ) φ ( s ) d s
is said to be the k 1 -order left ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral of function h.
Definition 4
([14]). Let r , k 2 > 0 and h : [ a , ) R . Then, the ( k 2 , φ ) -Laplace transform of h is defined as
L k 2 , a r ; φ ( h ( t ) ) ( λ ) = a e λ k 2 1 r k 2 g ( t , a ) h ( t ) φ ( t ) d t .
Definition 5.
If ρ 1 , k 1 , k 2 , η > 0 , ρ 2 [ 0 , 1 ) , h is integrable on [ a , b ] , then
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = a t g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , s ) h ( s ) φ ( s ) d s
is the left ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral of h, and η is the order of h. The right ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral of h, where η is the order of h, is given by
I b η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = t b g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( s , t ) h ( s ) φ ( s ) d s .
Remark 3.
We set I a + 0 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = I b 0 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = h ( t ) .
Definition 6.
Let η , ρ 1 , k 1 , k 2 > 0 , and ρ 2 [ 0 , 1 ) , m = η k 1 , v [ 0 , 1 ] ; when the function h is continuously differentiable m times on [ a , b ] , then
D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H h ( t ) = I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t )
is the left ( ρ 1 , ρ 2 , k , φ ) -proportional H fractional derivative of h; moreover, η and v are the order and type of h, respectively. Similarly, we say that
D b η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H h ( t ) = I b v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ    k 1 , k 2 I b ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t )
is the right ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H fractional derivative of h.
Below, we introduce some properties of two operators that are subsequently essential.
Lemma 1.
Let ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , and m = η k 1 , β > 0 , then we have
(i)
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = g ^ ρ 1 , ρ 2 , k 1 , k 2 , η + β ; φ ( x , a ) , for η 0 ;
(ii)
I b η , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( b , t ) ( x ) = g ^ ρ 1 , ρ 2 , k 1 , k 2 , η + β ; φ ( b , x ) , for η 0 ;
(iii)
D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = g ^ ρ 1 , ρ 2 , k 1 , k 2 , β η ; φ ( x , a ) , for { v [ 0 , 1 ] , β > η + v ( m k 1 η ) } and η 0 ;
(iv)
D b η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( b , t ) ( x ) = g ^ ρ 1 , ρ 2 , k 1 , k 2 , β η ; φ ( b , x ) , for { v [ 0 , 1 ] , β > η + v ( m k 1 η ) } and η 0 .
Proof. 
We will prove the relations of (i) and (iii), since the other cases may be established in a similar fashion.
(i)
If η = 0 , this is made obvious by Remark 3. If η > 0 , using Definition 5 and taking u = g ( s , a ) g ( x , a ) , we can obtain
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = a x g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( x , s ) g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( s , a ) φ ( s ) d s = a x g η k 1 1 ( x , s ) ρ 1 η + β k 1 k 1 2 Γ k 1 ( η ) Γ k 1 ( β ) e ( ρ 2 1 ) g ( x , s ) ρ 1 k 2 φ ( s ) e ( ρ 2 1 ) g ( s , a ) ρ 1 k 2 g β k 1 1 ( s , a ) d s = 1 ρ 1 η + β k 1 k 1 2 Γ k 1 ( η ) Γ k 1 ( β ) e ( ρ 2 1 ) g ( x , a ) ρ 1 k 2 a x g η k 1 1 ( x , s ) g β k 1 1 ( s , a ) φ ( s ) d s = 1 ρ 1 η + β k 1 k 1 2 Γ k 1 ( η ) Γ k 1 ( β ) e ( ρ 2 1 ) g ( x , a ) ρ 1 k 2 g η + β k 1 1 ( x , a ) 0 1 ( 1 u ) η k 1 1 u β k 1 1 d u = g ^ ρ 1 , ρ 2 , k 1 , k 2 , η + β ; φ ( x , a ) .
(ii)
If β > η + v ( m k 1 η ) , where v ( 0 , 1 ) . Using Definition 6 and relation (i), we can see that
D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 1 ρ 1 ( 1 v ) ( m k 1 η ) + β k 1 k 1 Γ k 1 ( β + ( 1 v ) ( m k 1 η ) ) × e ( ρ 2 1 ) g ( x , a ) ρ 1 k 2 g ( 1 v ) ( m k 1 η ) + β k 1 1 ( x , a ) = I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 ρ 1 m ρ 1 ( 1 v ) ( m k 1 η ) + β k 1 k 1 Γ k 1 ( β + ( 1 v ) ( m k 1 η ) ) k 1 m e ( ρ 2 1 ) g ( x , a ) ρ 1 k 2 × ( 1 v ) ( m k 1 η ) + β k 1 1 ( 1 v ) ( m k 1 η ) + β k 1 m × g ( 1 v ) ( m k 1 η ) + β k 1 m 1 ( x , a ) = I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , ( 1 v ) ( m k 1 η ) + β m k 1 ; φ ( x , a ) = g ^ ρ 1 , ρ 2 , k 1 , k 2 , β η ; φ ( x , a ) ,
where we used the fact that
D m , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) e ρ 2 1 ρ 1 k 2 g ( t , a ) = ρ 1 m e ρ 2 1 ρ 1 k 2 g ( t , a ) δ k 1 , φ m h ( t ) .
When v = 0 or v = 1 , the proof is similar. □
Remark 4.
If β = η + v ( m k 1 η ) , then D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H g ^ ρ 1 , ρ 2 , k 1 , k 2 , β ; φ ( t , a ) ( x ) = 0 for ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) and v [ 0 , 1 ] .
Let Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) = e ( ρ 2 1 ) g ( t , a ) ρ 1 k 2 and p 1 , and consider the space L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] = { x ( t ) : ( a , b ] R : φ ( t ) | Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) x ( t ) | p } to be integrable. Then, L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] is a Banach space, which norm is given by x L ρ 1 , ρ 2 , k 2 , φ p = a b φ ( t ) | Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) x ( t ) | p d t 1 p . If 0 γ < 1 , let Q ρ 1 , ρ 2 , k 2 , φ , γ ( t , a ) = Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) g γ ( t , a ) and the weighted space C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] = { x ( t ) : ( a , b ] R : Q ρ 1 , ρ 2 , k 2 , φ , γ ( t , a ) x ( t ) C [ a , b ] } , clearly, C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] is a Banach space with the norm x C ρ 1 , ρ 2 , k 2 , φ , γ = max t [ a , b ] | Q ρ 1 , ρ 2 , k 2 , φ , γ ( t , a ) x ( t ) | .
Theorem 1.
Let η , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , p 1 , then I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 : L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] is bounded and for any h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] ,
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h L ρ 1 , ρ 2 , k 2 , φ p 1 ρ 1 η k 1 η Γ k 1 ( η ) g η k 1 ( b , a ) h L ρ 1 , ρ 2 , k 2 , φ p .
Proof. 
The case of p = 1 is obvious; we will omit it here. For the case of p > 1 and h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , first, from the Hölder inequality, the inequality can be easily obtained as follows:
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) a t g η k 1 1 ( t , s ) φ ( s ) Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) | h ( s ) | d s = 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) a t g η k 1 q k 1 ( t , s ) [ φ ( s ) ] 1 q × g η k 1 p k 1 ( t , s ) [ φ ( s ) ] 1 p Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) | h ( s ) | d s 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) a t g η k 1 k 1 ( t , s ) φ ( s ) d s 1 q × a t g η k 1 k 1 ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s 1 p 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) k 1 η g η k 1 ( b , a ) 1 q × a t g η k 1 k 1 ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s 1 p ,
where 1 p + 1 q = 1 . Secondly, by interchanging the order of integration, we obtain
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h L ρ 1 , ρ 2 , k 2 , φ p 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) k 1 η g η k 1 ( b , a ) 1 q × a b φ ( t ) a t g η k 1 k 1 ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s d t 1 p = 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) k 1 η g η k 1 ( b , a ) 1 q × a b φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s s b g η k 1 k 1 ( t , s ) φ ( t ) d t 1 p 1 ρ 1 η k 1 η Γ k 1 ( η ) g η k 1 ( b , a ) h L ρ 1 , ρ 2 , k 2 , φ p .
The proof is completed □
Theorem 2.
If ρ 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , 0 < η < k 1 , 1 < p < k 1 η , then I a + η , ρ ; φ k : L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] is bounded, where 1 p < p k 1 k 1 p η .
Proof. 
Let 0 < ϵ < η k 1 1 p + 1 p , ϵ 1 = p ϵ p 1 and ϵ 2 = p p η p k 1 k 1 p p ϵ + k 1 p k 1 p , then recalling the generalized Hölder inequality, we see that for h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] ,
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) a t g ϵ 1 + 1 p ( t , s ) [ φ ( s ) ] 1 1 p × g η k 1 1 p ϵ ( t , s ) [ φ ( s ) ] 1 p | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p p × [ φ ( s ) ] 1 p 1 p | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p p p d s 1 ρ η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) a t g p ϵ p + 1 p 1 ( t , s ) φ ( s ) d s 1 1 p × a t g p η k 1 k 1 p ϵ k 1 p p ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s 1 p × a t φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s 1 p 1 p 1 ρ η k 1 k 1 Γ k 1 ( η ) 1 Q ρ 1 , ρ 2 , k 2 , φ ( t , a ) h L ρ 1 , ρ 2 , k 2 , φ p 1 p p 1 ϵ 1 g ϵ 1 ( b , a ) 1 1 p
× a t g p η k 1 k 1 p ϵ k 1 p p ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s 1 p .
Hence,
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h L ρ 1 , ρ 2 , k 2 , φ p 1 ρ η k 1 k 1 Γ k 1 ( η ) h L ρ 1 , ρ 2 , k 2 , φ p 1 p p 1 ϵ 1 g ϵ 1 ( b , a ) 1 1 p × a b φ ( t ) a t g p η k 1 k 1 p ϵ k 1 p p ( t , s ) φ ( s ) | Q ρ 1 , ρ 2 , k 2 , φ ( s , a ) h ( s ) | p d s d t 1 p 1 ρ η k 1 k 1 Γ k 1 ( η ) h L ρ 1 , ρ 2 , k 2 , φ p 1 ϵ 1 p 1 p ϵ 2 1 p g ( p 1 ) ϵ 1 p + ϵ 2 p ( b , a ) ,
Consequently, we complete the proof. □
Lemma 2.
If ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , η 1 , η 2 0 , p 1 . Then, for h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , we obtain
I a + η 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 I a + η 2 , ρ 1 , ρ 2 ; φ h ) ( t ) = I a + η 2 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 I a + η 1 , ρ 1 , ρ 2 ; φ h ) ( t ) = ( k 1 , k 2 I a + η 1 + η 2 , ρ 1 , ρ 2 ; φ h ) ( t ) .
Proof. 
Using Definition 5 and then interchanging the order of integration, we derive
I a + η 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 I a + η 2 , ρ 1 , ρ 2 ; φ h ) ( t ) = a t a s g ^ ρ 1 , ρ 2 , k 1 , k 2 , η 1 ; φ ( t , s ) g ^ ρ 1 , ρ 2 , k 1 , k 2 , η 2 ; φ ( s , τ ) φ ( s ) φ ( τ ) h ( τ ) d τ d s = 1 ρ 1 η 1 + η 2 k 1 k 1 2 Γ k 1 ( η 1 ) Γ k 1 ( η 2 ) a t e ( ρ 2 1 ) g ( t , τ ) ρ 1 k 2 φ ( τ ) h ( τ ) d τ τ t g η 1 k 1 1 ( t , s ) g η 2 k 1 1 ( s , τ ) φ ( s ) d s = B ( η 2 k 1 , η 1 k 1 ) ρ 1 η 1 + η 2 k k 1 2 Γ k 1 ( η 1 ) Γ k 1 ( η 2 ) a t e ( ρ 2 1 ) g ( t , τ ) ρ 1 k 2 φ ( τ ) g η 1 + η 2 k 1 1 ( t , τ ) h ( τ ) d τ = 1 ρ 1 η 1 + η 2 k 1 k 1 Γ k 1 ( η 1 + η 2 ) a t e ( ρ 2 1 ) g ( t , τ ) ρ 1 k 2 φ ( τ ) g η 1 + η 2 k 1 1 ( t , τ ) h ( τ ) d τ = ( k 1 , k 2 I a + η 1 + η 2 , ρ 1 , ρ 2 ; φ h ) ( t ) ,
which implies that the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral has the semigroup property. □
Theorem 3.
If ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , γ [ 0 , 1 ) with k 1 γ < η < k 1 , then I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 : C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] C [ a , b ] is bounded, and for any h C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] ,
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h C [ a , b ] h C ρ 1 , ρ 2 , k 2 , φ , γ Γ k 1 ( k 1 ( 1 γ ) ) ρ 1 η k 1 Γ k 1 ( η + k 1 ( 1 γ ) ) sup t [ a , b ] e ( ρ 2 1 ) g ( t , a ) ρ 1 k 2 g η k 1 γ ( t , a ) .
Proof. 
Let h C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] and a t 1 < t 2 b . If t 1 = a < t 2 b , then note that
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t 2 ) I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t 1 ) = | a t 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t 2 , s ) φ ( s ) h ( s ) d s | = | e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 a t 2 g η k 1 1 ( t 2 , s ) ρ 1 η k 1 k 1 Γ k 1 ( η ) φ ( s ) g γ ( s , a ) Q ρ 1 , ρ 2 , k 2 , φ , γ ( s , a ) h ( s ) d s | h C ρ 1 , ρ 2 , k 2 , φ , γ B ( 1 γ , η k 1 ) ρ 1 η k 1 k 1 Γ k 1 ( η ) e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 g η k 1 γ ( t 2 , a ) 0 as t 2 a .
If a < t 1 < t 2 b , we obtain
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t 2 ) I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t 1 ) = | a t 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t 2 , s ) φ ( s ) h ( s ) d s a t 1 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t 1 , s ) φ ( s ) h ( s ) d s | = | e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 a t 2 g η k 1 1 ( t 2 , s ) ρ 1 η k 1 k 1 Γ k 1 ( η ) φ ( s ) g γ ( s , a ) Q ρ 1 , ρ 2 , k 2 , φ , γ ( s , a ) h ( s ) d s e ( ρ 2 1 ) g ( t 1 , a ) ρ 1 k 2 a t 1 g η k 1 1 ( t 1 , s ) ρ 1 η k 1 k 1 Γ k 1 ( η ) φ ( s ) g γ ( s , a ) Q ρ 1 , ρ 2 , k 2 , φ , γ ( s , a ) h ( s ) d s | h C ρ 1 , ρ 2 , k 2 , φ , γ ρ 1 η k 1 k 1 Γ k 1 ( η ) a t 1 e ( ρ 2 1 ) g ( t 1 , a ) ρ 1 k 2 g η k 1 1 ( t 1 , s ) e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 g η k 1 1 ( t 2 , s ) φ ( s ) g γ ( s , a ) d s + h C ρ 1 , ρ 2 , k 2 , φ , γ ρ 1 η k 1 k 1 Γ k 1 ( η ) e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 t 1 t 2 g η k 1 1 ( t 2 , s ) φ ( s ) g γ ( s , a ) d s h C ρ 1 , ρ 2 , k 2 , φ , γ B ( 1 γ , η k 1 ) ρ 1 η k 1 k 1 Γ k 1 ( η ) { e ( ρ 2 1 ) g ( t 1 , a ) ρ 1 k 2 g η k 1 γ ( t 1 , a ) e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 g η k 1 γ ( t 2 , a ) + 2 k 1 η e ( ρ 2 1 ) g ( t 2 , a ) ρ 1 k 2 g γ ( t 1 , a ) g η k 1 ( t 2 , t 1 ) } 0 as t 2 t 1 .
With the continuity of φ as our basis, we can determine that I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h C [ a , b ] . By a similar argument, one may show that
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h C [ a , b ] h C ρ 1 , ρ 2 , k 2 , φ , γ Γ k 1 ( k 1 ( 1 γ ) ) ρ 1 η k 1 Γ k 1 ( η + k 1 ( 1 γ ) ) e ( ρ 2 1 ) g ( t , a ) ρ 1 k 2 g η k 1 γ ( t , a ) ,
and this completes the proof. □
Corollary 1.
Let ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , h C ρ 1 , ρ 2 , k 2 , φ , γ [ a , b ] , where γ [ 0 , 1 ) , k 1 γ < η < k 1 , then I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( a ) = 0 .
By repeating a similar process to that in Theorem 3, we obtain the following mapping property.
Theorem 4.
Let η , ρ 1 , k 1 , k 2 > 0 and ρ 2 [ 0 , 1 ) , then I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 : C [ a , b ] C [ a , b ] .
Lemma 3.
Let ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , p 1 , n Z + , η > n k 1 with h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , then
D n , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ) ( t ) = ( k 1 , k 2 I a + η n k 1 , ρ 1 , ρ 2 ; φ h ) ( t ) .
Proof. 
Observing that
D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ) ( t ) = ( 1 ρ 2 ) k 1 k 2 ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ) ( t ) + ρ 1 δ k 1 , φ 1 ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ) ( t ) = ( k 1 , k 2 I a + η k 1 , ρ 1 , ρ 2 ; φ h ) ( t ) .
Subsequently, employing the method of induction, we derive
D n , ρ 1 , ρ 2 ; φ k 1 , k 2 ( I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) = D n 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) = = ( k 1 , k 2 I a + η n k 1 , ρ 1 , ρ 2 ; φ h ) ( t ) .
and the proof of the Theorem is now complete. □
Lemma 4.
Let η 1 , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , p 1 , v [ 0 , 1 ] , η 2 > m k 1 , m = η 1 k 1 with h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , then
D a + η 1 , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H I a + η 2 , ρ ; φ k 1 , k 2 h ( t ) = I a + η 2 η 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) .
Proof. 
Using Definition 6 and Lemmas 2 and 3, we obtain
D a + η 1 , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H I a + η 2 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = I a + v ( m k 1 η 1 ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η 1 ) , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + η 2 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = I a + v ( m k 1 η 1 ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η 1 ) + η 2 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = I a + v ( m k 1 η 1 ) , ρ ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η 1 ) + η 2 m k 1 , ρ ; φ k 1 , k 2 h ( t ) = I a + η 2 η 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) .
We complete the proof. □
Lemma 5.
Let ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , m Z + , η > ( m 1 ) k 1 , and h A C m [ a , b ] , then
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 ( D m , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) = D m , ρ 1 , ρ 2 ; φ k 1 , k 2 ( I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) ρ 1 k 1 i = 1 m g ^ ρ 1 , ρ 2 , k 1 , k 2 , η k 1 i + k 1 ; φ ( t , a ) D m i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( a ) .
Proof. 
For m = 1 , from Definition 4 and Lemmas 4, (2), and (3), the following three equalities hold true:
L k 2 , a r ; φ I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 ( D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 L k 2 , a r ; φ ( D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ) ( λ ) = k 2 k 1 η k 1 1 ( ρ 1 λ k 2 2 r k 2 ρ 2 + 1 ) L k , a r ; φ ( h ( t ) ) ( λ ) ρ 1 k 2 h ( a ) ( ρ 1 λ k 2 2 r k 2 ρ 2 + 1 ) η k 1 ,
L k 2 , a r ; φ D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) ( λ ) = L k 2 , a r ; φ I a + η k 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 1 L k 2 , a r ; φ ( h ( t ) ) ( λ ) ,
L k 2 , a r ; φ ρ 1 k 1 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , a ) h ( a ) ( λ ) = k 2 k 1 η k 1 ρ 1 k 1 h ( a ) ( ρ 1 λ k 2 2 r k 2 ρ 2 + 1 ) η k 1 .
The relationship (1) holds true when m = 1 . Now, we assume that (1) is satisfied for m = l . Next, we verify that (1) is also satisfied for the case of m = l + 1 ,
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 ( D l + 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) = D l , ρ 1 , ρ 2 ; φ k 1 , k 2 ( I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) ρ 1 k 1 i = 1 l g ^ ρ 1 , ρ 1 , k 1 , k 2 , η k 1 i + k 1 ; φ ( t , a ) D l + 1 i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( a ) = D l + 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 ( I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) ρ 1 k 1 D l , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , a ) h ( a ) ρ 1 k 1 i = 1 l g ^ ρ 1 , ρ 1 , k 1 , k 2 , η k 1 i + k 1 ; φ ( t , a ) D l + 1 i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( a ) ,
observing that
D l , ρ 1 , ρ 2 ; φ k 1 , k 2 g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , a ) = g ^ ρ , k , η k 1 l ; φ ( t , a ) .
The relationship (1) is proved by mathematical induction. □
Theorem 5.
Let η , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , p 1 , v [ 0 , 1 ] , k 1 > ( 1 v ) ( m k 1 η ) , where m = η k 1 . h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , and moreover I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h A C m [ a , b ] , then
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 , H D a + η , ρ 1 , ρ 2 , v ; φ h ) ( t ) = h ( t ) ρ 1 k 1 j = 0 m 1 g ^ ρ 1 , ρ 2 , k 1 , k 2 , k 1 ( j + 1 ) ( 1 v ) ( m k 1 η ) ; φ ( t , a ) × ( D j , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( a ) .
Proof. 
From Lemmas 2, 3, and 5 and Definition 6, we obtain
I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 ( k 1 , k 2 , H D a + η , ρ 1 , ρ 2 , v ; φ h ) ( t ) = I a + η , ρ ; φ k 1 , k 2 ( I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( t ) = I a + η + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) = D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + m k 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ρ 1 k 1 i = 1 m g ^ ρ 1 , ρ 2 , k 1 , k 2 , η + v ( m k 1 η ) ( i 1 ) k 1 ; φ ( t , a ) × ( D m i , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( a ) = h ( t ) ρ 1 k 1 j = 0 m 1 g ^ ρ 1 , ρ 2 , k 1 , k 2 , k 1 ( j + 1 ) ( 1 v ) ( m k 1 η ) ; φ ( t , a ) × ( D j , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ) ( a ) .
The proof is completed □
Lemma 6.
If η , ρ 1 , k 1 , k 2 , r > 0 , ρ 2 [ 0 , 1 ) , p 1 and h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] , then
L k 2 , a r ; φ ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ( t ) ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 L k 2 , a r ; φ ( h ( t ) ) ( λ ) ,
for λ > ρ 2 1 ρ 1 k 2 2 r k 2 .
Proof. 
If we apply L k 2 , a r ; φ to I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) , then we obtain
L k 2 , a r ; φ ( k 1 , k 2 I a + η , ρ 1 , ρ 2 ; φ h ( t ) ) ( λ ) = L k 2 , a r ; φ g a , ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t ) φ h ( t ) ( λ ) .
By selecting u such that u = ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ρ 1 k 2 g ( t , a ) , we ascertain that
L k 2 , a r ; φ g a , ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t ) ( λ ) = a e λ k 2 1 r k 2 g ( t , a ) g ^ ρ 1 , ρ 2 , k 1 , k 2 , η ; φ ( t , a ) φ ( t ) d t = 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) a e ρ 2 1 ρ 1 k 2 λ k 2 1 r k 2 g ( t , a ) g η k 1 1 ( t , a ) φ ( t ) d t = 1 ρ 1 η k 1 k 1 Γ k 1 ( η ) ρ 1 k 2 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 Γ ( η k 1 ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 .
The convolution clearly demonstrates the validity of Equation (2). □
Lemma 7.
Let ρ 1 , k 1 , k 2 , r > 0 , ρ 2 [ 0 , 1 ) , m Z + and h A C m [ a , b ] , then
L k 2 , a r ; φ D m , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 m L k 2 , a r ; φ ( h ( t ) ) ( λ ) ρ 1 k 1 i = 0 m 1 ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 m i 1 D i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a
for λ > 0 .
Proof. 
Using Definition 6, integration by parts leads to
L k 2 , a r ; φ ( δ k 1 , φ 1 h ( t ) ) ( λ ) = a e λ k 2 1 r k 2 g ( t , a ) k 1 d d t h ( t ) d t = k 1 h ( a ) + k 1 k 2 1 r k 2 λ L k r ; φ ( h ( t ) ) ( λ ) .
For m = 1 , from Definitions 2 and (5), we obtain
L k 2 , a r ; φ D 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = L k 2 , a r ; φ ρ 1 δ k 1 , φ 1 h ( t ) + ( 1 ρ 2 ) k 1 k 2 h ( t ) = ρ 1 k 1 h ( a ) + k 1 k 2 1 r k 2 λ L k 2 , a r ; φ ( h ( t ) ) ( λ ) + ( 1 ρ 2 ) k 1 k 2 L k 2 , a r ; φ ( h ( t ) ) ( λ ) = ρ 1 k 1 h ( a ) + ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 L k 2 , a r ; φ ( h ( t ) ) ( λ ) .
Assume that (4) is true for m = l . Next, we verify that the equality is also satisfied for the case of m = l + 1 . In fact, using (5), we obtain
L k 2 , a r ; φ D l + 1 , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = L k 2 , a r ; φ ρ 1 δ k 1 , φ 1 D l , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) + ( 1 ρ 2 ) k 1 k 2 D l , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 L k 2 , a r ; φ D l , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) ρ 1 k 1 D l , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a = ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 l + 1 L k 2 , a r ; φ ( h ( t ) ( λ ) ρ 1 k 1 D l , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a ρ 1 k 1 i = 0 l 1 ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 l i D i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a = ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 l + 1 L k , a r ; φ ( h ( t ) ) ( λ ) ρ 1 k 1 i = 0 l ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 l i D i , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a .
The relationship (4) is proved by mathematical induction. □
Theorem 6.
If η , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , m = η k 1 , v [ 0 , 1 ] , p 1 with h L ρ 1 , ρ 2 , k 2 , φ p [ a , b ] and I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h A C m [ a , b ] , then
L k 2 , a r ; φ ( k 1 , k 2 , H D a + η , ρ 1 , ρ 2 , v ; φ h ( t ) ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 L k 2 , a r ; φ ( h ( t ) ) ( λ ) ρ 1 k 1 i = 0 m 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 m + i + 1 + v ( m k 1 η ) k 1 × D i , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a ,
for λ > 0 .
Proof. 
In view of Definition 6 and Lemma 6, we obtain
L k 2 , a r ; φ D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H h ( t ) ( λ ) = L k 2 , a r ; φ I a + v ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( m k 1 η ) k 1 × L k 2 , a r ; φ D m , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) ( λ ) .
Moreover, with the support of Lemmas 6 and 7, we can confidently assert that
L k 2 , a r ; φ ( k 1 , k 2 , H D a + η , ρ 1 , ρ 2 , v ; φ h ( t ) ) ( λ ) = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( m k 1 η ) k 1 { ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 m × L k 2 , a r ; φ ( k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ h ( t ) ) ( λ ) ρ 1 k 1 i = 0 m 1 ρ 1 k 1 k 2 1 r k 2 λ + ( 1 ρ 2 ) k 1 k 2 m i 1 × D i , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a } = k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 L k 2 , a r ; φ ( h ( t ) ) ( λ ) ρ 1 k 1 i = 0 m 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 m + i + 1 + v ( m k 1 η ) k 1 × D i , ρ 1 , ρ 2 ; φ k 1 , k 2 I a + ( 1 v ) ( m k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) t = a .

3. Definition of Mild Solutions

Let η , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , v [ 0 , 1 ] . X is a Banach space with the norm · and γ = ( 1 v ) ( k 1 η ) k 1 , and we set E = C ρ 1 , ρ 2 , k 2 , φ , γ ( [ 0 , b ] , X ) = { x ( t ) : ( 0 , b ] X : Q ρ 1 , ρ 2 , k 2 , φ , γ ( t , 0 ) x ( t ) C ( [ 0 , b ] , X ) } . Clearly, E is a Banach space with the norm x E = max t [ 0 , b ] Q ρ 1 , ρ 2 , k 2 , φ , γ ( t , 0 ) x ( t ) .
In this section, we first give the definition of a mild solution for the following ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -H fractional differential equations:
D 0 + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H x ( t ) = A x ( t ) + f ( t , x ( t ) ) , t ( 0 , b ] , η ( 0 , k 1 ] , I 0 + ( 1 v ) ( k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 x ( t ) t = 0 = x 0 ,
where A generates a strongly continuous semigroup T ( t ) = e A t , t 0 . We assume that M = sup t 0 T ( t ) . f : [ 0 , b ] × X X is a given function satisfying some assumptions. Next, we show the existence of mild solutions.
The one-sided stable k-probability density is given by
ψ k , q ( θ ) = 1 π i = 1 ( 1 ) i 1 θ q i 1 Γ k ( k i q + k ) i ! k q i sin ( i π q ) , k , θ > 0 ,
and the integration property
0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) θ ψ k , q ( θ ) d θ = e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) q , 0 < q < 1 .
Lemma 8.
The Cauchy problem (6) has a solution which satisfies the following integral equation:
x ( t ) = T 1 ( t ) x 0 + 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 T 2 ( t , s ) f ( s , x ( s ) ) d s ,
where
T 1 ( t ) x = I 0 + v ( k 1 η ) , ρ 1 , ρ 2 ; φ k 1 , k 2 T 3 ( t ) x ,
T 2 ( t , s ) x = η k 1 0 θ h k 1 , η k 1 ( θ ) T g ( t , s ) ρ 1 k 1 η k 1 θ x d θ ,
T 3 ( t ) x = ρ 1 η 0 e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 h k 1 , η k 1 ( θ ) T g ( t , 0 ) ρ 1 k 1 η k 1 θ g η k 1 1 ( t , 0 ) ( ρ 1 k 1 ) η k 1 θ x d θ
and
h k 1 , η k 1 ( θ ) = k 1 η θ k 1 η 1 ψ k 1 , η k 1 ( θ k 1 η ) .
Proof. 
By applying the ( k 2 , φ ) -Laplace transform to the first equality of (6) and using Theorem 6, we obtain
k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 η k 1 L k 2 , 0 r ; φ ( x ( t ) ) ( λ )
= ρ 1 k 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( k 1 η ) k 1 x 0 + A L k 2 , 0 r ; φ ( x ( t ) ) ( λ ) + L k 2 , 0 r ; φ ( f ( t , x ( t ) ) ) ( λ ) .
Consequently, we can deduce that
L k 2 , 0 r ; φ ( x ( t ) ) ( λ ) = ρ 1 k 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( k 1 η ) k 1 k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 I A 1 x 0 + k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 I A 1 L k 2 , 0 r ; φ ( f ( t , x ( t ) ) ) ( λ ) .
Subsequently, analyzing the semigroup’s expression allows us to derive
L k 2 , 0 r ; φ ( x ( t ) ) ( λ ) = ρ 1 k 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( k 1 η ) k 1 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 s T ( s ) d s x 0 + 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 s T ( s ) d s L k 2 , 0 r ; φ ( f ( t , x ( t ) ) ) ( λ ) .
By using relation (7), we obtain
0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 s T ( s ) d s = η k 1 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 t η k 1 T ( t η k 1 ) t η k 1 1 d t = η k 1 0 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) t θ ψ k 1 , η k 1 ( θ ) T ( t η k 1 ) t η k 1 1 d θ d t = η k 1 0 0 e λ k 2 1 r k 2 ρ 1 k 1 t θ e k 1 k 2 ( 1 ρ 2 ) t θ ψ k 1 , η k 1 ( θ ) T ( t η k 1 ) t η k 1 1 d θ d t = η k 1 0 e λ k 2 1 r k 2 g ( τ , 0 ) φ ( τ ) 0 e ρ 2 1 ρ 1 k 2 g ( τ , 0 ) ψ k 1 , η k 1 ( θ ) T g ( τ , 0 ) ρ 1 k 1 θ η k 1 × g ( τ , 0 ) ρ 1 k 1 θ η k 1 1 1 ρ 1 k 1 θ d θ d τ ,
On the other hand,
0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 s T ( s ) d s L k 2 , 0 r ; φ ( f ( t , x ( t ) ) ) ( λ ) = 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 s T ( s ) 0 e λ k 2 1 r k 2 g ( w , 0 ) φ ( w ) f ( w , x ( w ) ) d w d s = η k 1 0 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) η k 1 u η k 1 T ( u η k 1 ) e λ k 2 1 r k 2 g ( w , 0 ) φ ( w ) f ( w , x ( w ) ) d w × u η k 1 1 d u = η k 1 0 0 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) u θ ψ k 1 , η k 1 ( θ ) d θ T ( u η k 1 ) e λ k 2 1 r k 2 g ( w , 0 ) × φ ( w ) f ( w , x ( w ) ) d w u η k 1 1 d u = η k 1 0 0 0 e k 1 k 2 ( ρ 1 k 2 2 r k 2 λ + 1 ρ 2 ) g ( σ , 0 ) θ ψ k 1 , η k 1 ( θ ) d θ T ( g η k 1 ( σ , 0 ) ) × e λ k 2 1 r k 2 g ( w , 0 ) φ ( w ) f ( w , x ( w ) ) d w g η k 1 1 ( σ , 0 ) φ ( σ ) d σ = η k 1 0 0 0 e λ k 2 1 r k 2 g ( σ , 0 ) θ ρ 1 k 1 e k 1 ( ρ 2 1 ) k 2 g ( σ , 0 ) θ ψ k 1 , η k 1 ( θ ) d θ × T ( g η k 1 ( σ , 0 ) ) e λ k 2 1 r k 2 g ( w , 0 ) φ ( w ) f ( w , x ( w ) ) d w g η k 1 1 ( σ , 0 ) φ ( σ ) d σ = η k 1 0 0 0 e λ k 2 1 r k 2 g ( δ , 0 ) e k 1 ( ρ 2 1 ) k 2 g ( δ , 0 ) ρ 1 k 1 ψ k 1 , η k 1 ( θ ) d θ
× T g ( δ , 0 ) ρ 1 k 1 θ η k 1 e λ k 2 1 r k 2 g ( w , 0 ) φ ( w ) f ( w , x ( w ) ) d w g ( δ , 0 ) ρ 1 k 1 θ η k 1 1 φ ( δ ) ρ 1 k 1 θ d δ = η k 1 0 e λ k 2 1 r k 2 [ g ( δ , 0 ) + g ( w , 0 ) ] 0 0 e ( ρ 2 1 ) g ( δ , 0 ) ρ 1 k 2 ψ k 1 , η k 1 ( θ ) × T g ( δ , 0 ) ρ 1 k 1 θ η k 1 φ ( w ) f ( w , x ( w ) ) d w g ( δ , 0 ) ρ 1 k 1 θ η k 1 1 φ ( δ ) ρ 1 k 1 θ d θ d δ = η k 1 0 e λ k 2 1 r k 2 g ( τ , 0 ) φ ( τ ) 0 0 τ e ( ρ 2 1 ) g ( τ , w ) ρ 1 k 2 ψ k 1 , η k 1 ( θ ) × T g ( τ , w ) ρ 1 k 1 θ η k 1 f ( w , x ( w ) ) g ( τ , w ) ρ 1 k 1 θ η k 1 1 φ ( w ) ρ 1 k 1 θ d w d θ d τ .
Combining the equalities of (9), (10), and (11) produces the following representation:
L k 2 , 0 r ; φ ( x ( t ) ) ( λ ) = ρ 1 k 1 k 2 k 1 1 ρ 1 k 2 2 r k 2 λ + 1 ρ 2 v ( k 1 η ) k 1 η k 1 0 e λ k 2 1 r k 2 g ( τ , 0 ) φ ( τ ) × 0 e ρ 2 1 ρ 1 k 2 g ( τ , 0 ) ψ k 1 , η k 1 ( θ ) T g ( τ , 0 ) ρ 1 k 1 θ η k 1 g ( τ , 0 ) ρ 1 k 1 θ η k 1 1 1 ρ 1 k 1 θ d θ d τ x 0 + η k 1 0 e λ k 2 1 r k 2 g ( τ , 0 ) φ ( τ ) 0 0 τ e ( ρ 2 1 ) g ( τ , w ) ρ 1 k 2 ψ k 1 , η k 1 ( θ ) × T g ( τ , w ) ρ 1 k 1 θ η k 1 f ( w , x ( w ) ) g ( τ , w ) ρ 1 k 1 θ η k 1 1 φ ( w ) ρ 1 k 1 θ d w d θ d τ = L k 2 , 0 r ; φ g 0 , ρ 1 , ρ 2 , k 1 , k 2 , v ( k 1 η ) ; φ ( t ) ( λ ) × L k 2 , 0 r ; φ ρ 1 η 0 e ( ρ 2 1 ) g ( τ , 0 ) ρ 1 k 2 ψ k 1 , η k 1 ( θ ) T g ( τ , 0 ) ρ 1 k 1 θ η k 1 g η k 1 1 ( τ , 0 ) ( ρ 1 k 1 θ ) η k 1 d θ ( λ ) x 0 + η k 1 L k 2 , 0 r ; φ ( 0 0 τ e ( ρ 2 1 ) g ( τ , w ) ρ 1 k 2 ψ k 1 , η k 1 ( θ ) T g ( τ , w ) ρ 1 k 1 θ η k 1 f ( w , x ( w ) ) × g ( τ , w ) ρ 1 k 1 θ η k 1 1 φ ( w ) ρ 1 k 1 θ d w d θ ) ( λ ) = L k 2 , 0 r ; φ g 0 , ρ 1 , ρ 2 , k 1 , k 2 , v ( k 1 η ) ; φ ( t ) ( λ ) × L k 2 , 0 r ; φ ρ 1 η 0 e ( ρ 2 1 ) g ( τ , 0 ) ρ 1 k 2 h k 1 , η k 1 ( θ ) T g ( τ , 0 ) ρ 1 k 1 η k 1 θ g η k 1 1 ( τ , 0 ) ( ρ 1 k 1 ) η k 1 θ d θ ( λ ) x 0 + η k 1 L k 2 , 0 r ; φ ( 0 0 τ e ( ρ 2 1 ) g ( τ , w ) ρ 1 k 2 h k 1 , η k 1 ( θ ) T g ( τ , w ) ρ 1 k 1 η k 1 θ f ( w , x ( w ) ) × g ( τ , w ) ρ 1 k 1 η k 1 1 φ ( w ) ρ 1 k 1 θ d w d θ ) .
With the help of the inverse ( k 2 , φ ) -generalized Laplace transform, we can see that x ( t ) satisfies (7). □
Definition 7.
x is said to be a mild solution of (6) if x E and satisfies (8).
Lemma 9.
The operators T 1 and T 2 have the following properties:
(i)
For 0 s t , T 1 and T 2 are linear in X with
T 1 ( t ) x M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) + η k 1 1 Γ k 1 ( v ( k 1 η ) + η ) g ( v 1 ) ( k 1 η ) k 1 ( t , 0 ) x ,
T 2 ( t ) x M k 1 η k 1 1 Γ k 1 ( η ) x : = K x .
(ii)
T 2 is strongly continuous for 0 s t .
Proof. 
(i)
For any x X ,
T 1 ( t ) x = 0 t g v ( k 1 η ) k 1 1 ( t , s ) e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) k 1 k 1 Γ k 1 ( v ( k 1 η ) ) φ ( s ) 0 ρ 1 η h k 1 , η k 1 ( θ ) × T g ( s , 0 ) ρ 1 k 1 η k 1 θ g η k 1 1 ( s , 0 ) ( ρ 1 k 1 ) η k 1 θ x d θ d s η M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) + η k 1 1 k 1 1 + η k 1 Γ k 1 ( v ( k 1 η ) ) 0 t g v ( k 1 η ) k 1 1 ( t , s ) g η k 1 1 ( s , 0 ) φ ( s ) d s × 0 θ h k 1 , η k 1 ( θ ) d θ x = η M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 B ( v ( k 1 η ) k 1 , η k 1 ) g ( v 1 ) ( k 1 η ) k 1 ( t , 0 ) ρ 1 v ( k 1 η ) + η k 1 1 k 1 1 + η k 1 Γ k 1 ( v ( k 1 η ) ) k 1 η k 1 Γ k 1 ( k 1 + η ) x = M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) + η k 1 1 Γ k 1 ( v ( k 1 η ) + η ) g ( v 1 ) ( k 1 η ) k 1 ( t , 0 ) x ,
T 2 ( t , s ) x = η k 1 0 θ h k 1 , η k 1 ( θ ) T g ( t , s ) ρ 1 k 1 η k 1 θ x d θ η k 1 M x 0 θ h k 1 , η k 1 ( θ ) d θ = η k 1 M k 1 η k 1 Γ k 1 ( k 1 + η ) x = K x .
(ii)
For any x X and 0 t 1 t 2 b , we have
T 2 ( t 2 , s ) x T 2 ( t 1 , s ) x = η k 1 0 θ h k 1 , η k 1 ( θ ) T g ( t 2 , s ) ρ 1 k 1 η k 1 θ T g ( t 1 , s ) ρ 1 k 1 η k 1 θ x d θ M η k 1 0 θ h k 1 , η k 1 ( θ ) T g ( t 2 , s ) ρ 1 k 1 η k 1 θ g ( t 1 , s ) ρ 1 k 1 η k 1 θ I x d θ .
According to the strongly continuity of T ( t ) , we can see that T 2 is strongly continuity, which completes the proof. □
Theorem 7.
Assume that η , ρ 1 , k 1 , k 2 > 0 , ρ 2 [ 0 , 1 ) , v [ 0 , 1 ] , and γ = ( 1 v ) ( k 1 η ) k 1 .
(H1)
For almost all t ( 0 , b ] , f ( t , · ) C ( X , X ) , and for each x X , the function f ( · , x ) : ( 0 , b ] X is strongly measurable;
(H2)
There exists l > 0 such that for any x , y X , f ( t , x ) f ( t , y ) l x ( t ) y ( t ) for all t ( 0 , b ] . Moreover, f ( t , 0 ) is bounded in X.
Then, problem (6) has a unique mild solution if
Λ = K l ( ρ 1 k 1 ) η k 1 B η k 1 , 1 γ g η k 1 ( b , 0 ) < 1 .
Proof. 
Set
Λ 1 = M ρ 1 v ( k 1 η ) + η k 1 1 Γ k 1 ( v ( k 1 η ) + η ) ,
Λ 2 = e ( ρ 2 1 ) g ( b , 0 ) ρ 1 k 2 K sup t [ 0 , b ] f ( t , 0 ) ( ρ 1 k 1 ) η k 1 k 1 η g η k 1 + γ ( b , 0 ) .
Choosing r such that r > Λ 1 + Λ 2 1 Λ , let us consider the operator F : E r E by
( F x ) ( t ) = T 1 ( t ) x 0 + 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 T 2 ( t , s ) f ( s , x ( s ) ) d s ,
where E r = { x E : x E r } .
Firstly,
( F x ) ( t ) M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) + η k 1 1 Γ k 1 ( v ( k 1 η ) + η ) g ( v 1 ) ( k 1 η ) k 1 ( t , 0 ) x 0 + K 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 f ( s , x ( s ) ) f ( s , 0 ) d s + K 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 f ( s , 0 ) d s M e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 ρ 1 v ( k 1 η ) + η k 1 1 Γ k 1 ( v ( k 1 η ) + η ) g ( v 1 ) ( k 1 η ) k 1 ( t , 0 ) x 0 + K l ( ρ 1 k 1 ) η k 1 e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 B ( η k 1 , 1 γ ) g η k 1 γ ( t , 0 ) x E + K ( ρ 1 k 1 ) η k 1 k 1 η g η k 1 ( t , 0 ) sup t [ 0 , b ] f ( t , 0 ) ,
which implies that
F x E Λ 1 x 0 + Λ x E + Λ 2 r .
Hence, F is a mapping from E r to itself.
On the other hand, note that for any x , y E r and t [ 0 , b ] , we obtain
( F x ) ( t ) ( F y ) ( t ) 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 T 2 ( t , s ) [ f ( s , x ( s ) ) f ( s , y ( s ) ) ] d s K 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 f ( s , x ( s ) ) f ( s , y ( s ) ) d s K l 0 t e ( ρ 2 1 ) g ( t , s ) ρ 1 k 2 g ( t , s ) ρ 1 k 1 η k 1 1 φ ( s ) ρ 1 k 1 x ( s ) y ( s ) d s = K l ( ρ 1 k 1 ) η k 1 e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 0 t g η k 1 1 ( t , s ) g γ ( s , 0 ) φ ( s ) d s x y E = K l ( ρ 1 k 1 ) η k 1 e ( ρ 2 1 ) g ( t , 0 ) ρ 1 k 2 B ( η k 1 , 1 γ ) g η k 1 γ ( t , 0 ) x y E .
Thus,
F x F y E K l ( ρ 1 k 1 ) η k 1 B ( η k 1 , 1 γ ) g η k 1 ( b , 0 ) x y E .
From the above, it is clear that F is a contraction. Thus, F has a unique fixed point in E r from the Banach contraction principle. The proof is complete. □
Example 1.
Leting X = L 2 [ 0 , π ] , we study the ( 2 , 1 2 , 2 , 2 , t 3 ) -proportional H fractional initial problem:
D 0 + 3 2 , 2 , 1 2 , 1 4 ; t 3 2 , 2 , H x ( t , z ) = 2 z 2 x ( t , z ) x ( t , z ) + e t x ( t , z ) 1 + t , ( t , z ) ( 0 , 1 ] × [ 0 , π ] , I 0 + 3 8 , 2 , 1 2 ; t 3 2 , 2 x ( t ) t = 0 = x 0 ( z ) , x ( t , 0 ) = x ( t , π ) = 0 ,
where A : D ( A ) X X and A x = 2 x z 2 x , with the domain D ( A ) = { x X : x H 2 [ 0 , π ] , x ( t , 0 ) = x ( t , π ) = 0 } . Then, A can be written as A x = i = 1 ( 1 + i 2 ) x , x i x i , and x i ( z ) = 2 π sin i z are the eigenfunctions corresponding to the eigenvalues ( 1 + i 2 ) . We have T ( t ) x = i = 1 e ( 1 + i 2 ) t x , x i x i for x X . Obviously, A generates a strongly continuous semigroup T ( t ) with T ( t ) 1 .
Moreover, T 1 and T 2 are defined by
T 1 ( t ) x = I 0 + 1 8 , 2 , 1 2 ; t 3 2 , 2 0 e t 3 8 h 2 , 3 4 ( θ ) T t 3 4 3 4 θ x d θ ,
T 2 ( t , s ) x = 3 4 0 θ h 2 , 3 4 ( θ ) T t 3 s 3 4 3 4 θ x d θ .
Clearly, T 1 ( t ) exp ( 1 8 t 3 ) t 3 8 2 3 16 Γ 2 ( 13 8 ) , T 2 ( t , s ) 2 1 4 Γ 2 ( 3 2 ) 0.8160 . Let γ = 3 16 , then Λ 0.4571 . So, using Theorem 7, problem (12) has a unique mild solution.

4. Discussion

Novel definitions of the fractional integral and derivative have been introduced, which we term the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral, denoted as I a + η , ρ 1 , ρ 2 ; φ k 1 , k 2 h ( t ) , and the ( ρ 1 , ρ 2 , k 1 , k 2 ; φ ) -proportional H derivative, denoted as D a + η , ρ 1 , ρ 2 , v ; φ k 1 , k 2 , H h ( t ) , respectively. At present, we can not definitively assert that the newly proposed fractional integral and derivative are the most universally applicable. A notable limitation is that our new fractional derivative remains primarily within the theoretical realm; we have yet to identify suitable physical models that can effectively connect with these concepts.
The second main result of this paper delves into the properties of this novel fractional calculus. To elucidate the mapping properties of the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional integral, we introduce a set of weighted spaces, namely L ρ 1 , ρ 2 , k 2 , φ p and C ρ 1 , ρ 2 , k 2 , φ , γ . Furthermore, we explore the general ( k 2 , φ ) -Laplace transform for the ( ρ 1 , ρ 2 , k 1 , k 2 ; φ ) -proportional integrals and H derivative. In this paper, we address a common misconception regarding the concept of mild solutions as used in various studies. Our third principal finding introduces a corrected formulation of mild solutions for the Cauchy problem by utilizing the ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional H derivative, its associated properties, the C 0 -semigroup, and the probability density function. This approach not only provides a more accurate definition of mild solutions but also establishes their existence. The results presented herein can be extended to investigate the stability and controllability [15] of ( ρ 1 , ρ 2 , k 1 , k 2 , φ ) -proportional fractional differential equations and stochastic differential equations.

Author Contributions

Conceptualization, H.W.; methodology, H.W.; validation, H.W.; formal analysis, J.Z.; investigation, J.Z.; writing—original draft preparation, H.W.; writing—review and editing, H.W. and J.Z.; supervision, J.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by the Hainan Provincial Natural Science Foundation of China (122MS088) and Qiongtai Normal University (QTjg2022-4, QTjg2022-49, QTNB2022010).

Data Availability Statement

No new data were created or analyzed in this study.

Acknowledgments

We are very thankful to the anonymous reviewers for their constructive comments and suggestions which helped us to improve the manuscript.

Conflicts of Interest

The authors declare no conflicts of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript; or in the decision to publish the results.

Abbreviations

The following abbreviations are used in this manuscript:
R-LRiemann–Liouville
CCaputo
HHilfer
GPFGeneralized proportional fractional

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Wang, H.; Zhao, J. Mild Solutions of the (ρ1,ρ2,k1,k2,φ)-Proportional Hilfer–Cauchy Problem. Symmetry 2024, 16, 1349. https://doi.org/10.3390/sym16101349

AMA Style

Wang H, Zhao J. Mild Solutions of the (ρ1,ρ2,k1,k2,φ)-Proportional Hilfer–Cauchy Problem. Symmetry. 2024; 16(10):1349. https://doi.org/10.3390/sym16101349

Chicago/Turabian Style

Wang, Haihua, and Jie Zhao. 2024. "Mild Solutions of the (ρ1,ρ2,k1,k2,φ)-Proportional Hilfer–Cauchy Problem" Symmetry 16, no. 10: 1349. https://doi.org/10.3390/sym16101349

APA Style

Wang, H., & Zhao, J. (2024). Mild Solutions of the (ρ1,ρ2,k1,k2,φ)-Proportional Hilfer–Cauchy Problem. Symmetry, 16(10), 1349. https://doi.org/10.3390/sym16101349

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