# Theory of Fractional Hybrid Problems in the Frame ofψ-Hilfer Fractional Operators

Authors: Saeed M. Ali; Wedad Albalawi; Mohammed S. Abdo; Heba Y. Zahran; Abdel-Haleem Abdel-Aty
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the present manuscript, we develop and extend a qualitative analysis for two classes of boundary value problems for nonlinear hybrid fractional differential equations with hybrid boundary conditions involving aψ-Hilfer fractional order derivative introduced by Sousa and de Oliveira (2018). First, we derive the equivalent fractional integral equations to the proposed problems from some properties of the ψ-fractional calculus. Next, we establish the existence theorems in the weighted spaces via equivalent fractional integral equations with the help of Dhage’s fixed-point theorem (2004). Besides, for an adequate choice of the kernel ψ, we recover most of all the preceding results on fractional hybrid equations. Finally, two examples are constructed to make our main findings effective.


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## Body

## 1. Introduction

Recently, a lot of keen interest in the topic of fractional calculus (FC) has been shown by many researchers and investigators in view of its theoretical development and extensive applications in the applied and natural sciences. Different types of differential and integral operators of arbitrary orders have been introduced by Kilbas et al. [1]. In the same regard, Atangana and Baleanu [2] proposed a new fractional derivative (FD) based on a nonsingular and nonlocal kernel. On the advanced improvement of the FC without a singular kernel of the sinc function, the Yang-Gao-Machado-Baleanu FD was introduced in [3]. Some properties of the FD without a singular kernel were introduced by Lozada and Nieto [4].Hilfer in [5] proposed a generalization of the Riemann-Liouville fractional derivative (RLFD) and Caputo fractional derivative (CFD) when the author deliberated fractional time evolution in physical phenomena. The author named it a generalized FD, whereas more recently, it was named the Hilfer fractional derivative (HFD). This operator carries two parameters (α,β) that may be decreased to the RLFD and CFD definitions if β=0 and β=1, respectively. So, such a derivative incorporates between the RLFD and CFD. Some important laws and applications of this operator are obtained in [6, 7] and the references therein.Initial value problems (IVPs) involving the HFD were investigated by many authors, like Furati et al. [8], Gu and Trujillo [9], and Wang and Zhang [10]. Some existence and uniqueness results of IVPs for ψ-Hilfer-type coupled hybrid equations were obtained in [11]. Boundary value problems (BVPs) with the nonlocal conditions and the HFD were studied in [12]. The authors in [13] investigated the existence and stability of the solution of BVPs for ψ-Hilfer-type fractional integrodifferential equations with boundary conditions (BCs).ψ-Fractional derivatives (ψ-FDs) have been considered in [1] to be a generalization of RLFD. Some properties of these operators have been given by Agrawal in [14]. These operators are different from the other classical operators because the kernel is shown to be linked to another function ψ. For instance, Almeida [15] gave a new generalization of CFD with some interesting properties. In addition, the authors in [16] introduced a generalized type of the Laplace transform of generalized fractional operators in the frame of both RLFD and CFD.The new version of the HFD with respect to another functionψ has been introduced by Sousa and de Oliveira [17]. Recently, the investigation of diverse qualitative properties of solutions to several fractional differential equations (FDEs) involving generalized FDs has become the key theme of applied mathematics research. Many interesting results concerning the existence and stability of solutions by using various kinds of fixed-point techniques were formulated; e.g., Abdo et al. [18, 19] investigated various types of the Ulam-Hyers stability for ψ-Hilfer fractional problems with infinite delay and without infinite delay, respectively. The Ulam-Hyers-Rassias stability for FDEs using the ψ-Hilfer operator was discussed by Sousa and de Oliveira [20].On the other hand, hybrid-type FDEs have attained a considerable saucepan of interest and investigations of several researchers. This category of hybrid FDEs comprises the perturbations of primitive differential equations in various manners. For instance, Dhage and Lakshmikantham [21] considered the following IVPs for hybrid DEs:
(1)ddϑϰϑqϑ,ϰϑ=pϑ,ϰϑ,a.e.ϑ∈℧≔0,b,ϰϑϑ=ϑ0=ϰ0∈ℝ,where p∈C℧×ℝ,ℝ and q∈C℧×ℝ,ℝ\0. Zhao et al. [22] studied the following hybrid FDEs with RLFD:
(2)DRL0+θ1ϰϑqϑ,ϰϑ=pϑ,ϰϑ,a.e.ϑ∈℧,ϰϑϑ=ϑ0=ϰ0∈ℝ,where p∈C℧×ℝ,ℝ and q∈C℧×ℝ,ℝ\0.On the other hand, Benchohra et al. [23] considered the following BVP for the Caputo-type FDE:
(3)DC0+θ1ϰϑ=pϑ,ϰϑ,ϑ∈℧,c1ϰϑϑ=0+c2ϰϑϑ=b=d,where 0<θ1<1,c1,c2,d∈ℝ with c1+c2≠0, and p∈C℧×ℝ,ℝ.Hilal and Kajouni [24] considered the BVP for Caputo-type hybrid FDEs with BC:
(4)DC0+θ1ϰϑqϑ,ϰϑ=pϑ,ϰϑ,ϑ∈℧,c1ϰϑqϑ,ϰϑϑ=0+c2ϰϑqϑ,ϰϑϑ=b=d,where 0<θ1<1,c1,c2,d∈ℝ with c1+c2≠0, p∈C℧×ℝ,ℝ, and q∈C℧×ℝ,ℝ\0.However, as far as we could possibly know, no one considered the existence of solution for the hybrid-type BVPs involving HFD with respect toψ. Here, we discuss the existence theorems of BVPs for Hilfer-type FDE with respect to ψ and BCs:
(5)Da+θ1,θ2;ψϰϑ=pϑ,ϰϑ,ϑ∈a,b,c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=d.Also, we consider the Hilfer-type hybrid FDE with respect toψ and hybrid BCs:
(6)Da+θ1,θ2;ψϰϑ−zϑ,ϰϑqϑ,ϰϑ=pϑ,ϰϑ,ϑ∈a,b,c1Ia+1−θ;ψϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=a+c2ϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=b=d,where 0<θ1<1 and 0≤θ2≤1,θ=θ1+θ21−θ1,c1,c2,d∈ℝ, Da+θ1,θ2;ψ, and Ia+1−θ;ψ are the HFD and RLFD with respect to ψ, respectively, p,z∈CI×ℝ,ℝ, q∈CI×ℝ,ℝ\0, and I=a,b.Observe that the considered problems (5) and (6) are the first investigations of hybrid FDEs with hybrid BCs involving ψ-HFD. Moreover, we have chosen this operator, besides the fact that it is a global operator and it generalizes more than twenty the freedom of choice of the ordinary differential operator; some of its advantages have been explained in Remark 3. So, we are sure that the obtained results will be a beneficial contribution and an extension of the current results in the literature. We will also refer here to some recent results related to the subject of our study (see [25–31]).The rest of the work is displayed as follows. In Section2, we give some advantageous preliminaries related to our work. In Section 3, we derive the equivalent solutions to linear problems corresponding to the proposed problems. Then, we prove the existence of solutions to given problems via Dhage’s fixed-point theorem. Finally, two examples to justify reported results are offered in Section 4.


## 2. Preliminaries

Let us initially present some imperative definitions and primer ideas related to our work.Letθ=θ1+θ21−θ1 where, 0<θ1<1 and 0≤θ2≤1, and let I=a,b,I′=a,b. Consider the functional spaces CI,ℝ and C1−θψI,ℝ as follows:
(7)CI,ℝ=φ:I⟶ℝ:φiscontinuous,C1−θψI,ℝ=φ:I′⟶ℝ:ψϑ−ψa1−θφϑ∈CI,ℝ,equipped with the norms
(8)φC≔maxφϑ:ϑ∈I,φC1−θψ=ψϑ−ψa1−θφϑC.Obviously,C1−θψI,ℝ,·C1−θψ is the Banach space.In the next expressions, we will considerψ to be an increasing function such that ψ′ϑ≠0 for all ϑ∈I.Definition 1 (see [1]).
Letθ1>0θ1∈ℝ,φ∈L1I,ℝ, and ψ∈C1I,ℝ. Then, ψ-RL fractional integral of φ is defined by
(9)Ia+θ1;ψφϑ=1Γθ1∫aϑψ′sψϑ−ψsθ1−1φsds.Definition 2 (see [17]).
Letn−1<θ1<n∈ℕ, 0≤θ2≤1, and ψ,φ∈CnI,ℝ. Then, the ψ-HFD of φ is defined by
(10)Da+θ1,θ2;ψφϑ=Ia+θ2n−θ1;ψ1ψ′ϑddϑnIa+1−θ2n−θ1;ψφϑ.
Specifically, if0<θ1<1, we have
(11)Da+θ1,θ2;ψφϑ=Ia+θ21−θ1;ψ1ψ′ϑddϑIa+1−θ21−θ1;ψφϑ.Remark 3.
The operatorDa+θ1,θ2;ψ is an interpolator of the following FDs:
(i)
Classical HFD (forψϑ⟶ϑ, see [5]), Hilfer-Hadamard FD (for ψϑ⟶logϑ, see [32]), and Hilfer-Katugampola FD (for ψϑ⟶ϑρ,ρ>0, see [33])(ii)
Standard RLFD (forψϑ⟶ϑ,θ2⟶0, see [1]) and standard CFD (for ψϑ⟶ϑ,θ2⟶1, see [1])(iii)
Generalized RLFD (forθ2⟶0, see [1]) and generalized CFD (for θ2⟶1, see [15])(iv)
Generalized Liouville (forθ2⟶0,a=0, see [1]) and generalized Weyl (for θ2⟶0,a=−∞, see [34])Lemma 4 (see [1, 17]).
Letθ1,θ2>0,η>0, and Kψϑ,a≔ψϑ−ψa. Then,
(12)Ia+θ1;ψKψη−1ϑ,a=ΓηΓη+θ1Kψθ1+η−1ϑ,a,Da+θ1,θ2;ψKψθ−1ϑ,a=0,
Ia+θ1;ψIa+θ2;ψφϑ=Ia+θ1+θ2;ψφϑ,forφ∈CI,ℝ.Lemma 5 (see [17]).
Let0<θ1<1 and 0≤θ2≤1, where θ=θ1+θ21−θ1. If Ia+1−θ;ψφϑ∈CI,ℝ, then
(13)Ia+θ1;ψDa+θ1,θ2;ψφϑ=φϑ−Ia+1−θ;ψφaΓθψϑ−ψaθ−1,Da+θ1,θ2;ψIa+θ1;ψφϑ=φϑ.Lemma 6 (see [17]).
Letn−1≤θ<n and φ∈CθI,ℝ. Then, Ia+θ1;ψ is bounded in CθI,ℝ. Moreover, we have
(14)Ia+θ1;ψφa=limϑ⟶a+Ia+θ1;ψφϑ=0,n−1≤θ<θ1.Theorem 7 [35].
LetK be a nonempty, convex, closed subset of the Banach algebra L. Let the operators B1, B2:L⟶L, and B3:K⟶L such that (i) B1 and B2 are Lipschitzian, with Lipschitz constants κ1 and κ2, respectively; (ii) B3 is continuous and compact; (iii) ω=B1ϖ1B3ϖ2+B2ϖ2∈K⇒ϖ2∈K for each ϖ1∈K; and (iv) κ1M+κ2<1, where M=B3K. Then, there exists w∈K such that B1wB3w+B2w=w.


## 3. Main Results

In this section, we pay attention to deriving equivalent solutions to linear problems associated with problems (5) and (6); then, we prove the existence of solution to problems (5) and (6) using Dhage’s fixed-point technique.


### 3.1. Fractional Integral Equations (FIEs)

The forthcoming results give the equivalent of solution formulas for the proposed problems. For brevity, we set the following symbols:(15)Kψθ−1ϑ,a=ψϑ−ψaθ−1,Fψθϑ,s=ψ′sψϑ−ψsθ−1,(16)1Γθ1∫aϑFψθ1ϑ,sds=1Γθ1∫aϑψ′sψϑ−ψsθ1−1ds=ψϑ−ψaθ1Γθ1+1=Kψθ1ϑ,aΓθ1+1,(17)1Γθ1∫aϑFψθ1ϑ,sKψθ−1s,ads=1Γθ1∫aϑψ′sψϑ−ψsθ1−1ψs−ψaθ−1ds=Ia+θ1;ψψϑ−ψaθ−1=ΓθΓθ+θ1ψϑ−ψaθ1+θ−1=ΓθΓθ+θ1Kψθ1+θ−1ϑ,a.Lemma 8.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2, and H:I′⟶ℝ is continuous. Then, the function ϰ∈C1−θ,ψI,ℝ is a solution of the linear fractional BVP:
(18)Da+θ1,θ2;ψϰϑ=Hϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=d,if and only if ϰ satisfies the FIE:
(19)ϰϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ,ϑ∈I′,where Πb≔c1Γθ+c2Kψθ−1b,a≠0.Proof.
Letϰ∈C1−θ,ψI,ℝ be a solution of (18). We need to prove that ϰ is also a solution of (19). By the definition of C1−θ,ψI,ℝ and Lemma 6, we have Ia+1−θ;ψϰϑ∈CI,ℝ.
Now, by applyingIa+θ1;ψ to the first equation in (18) and using Lemma 5, we can write
(20)ϰϑ=Ia+1−θ;ψϰa+Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ.
SetY0≔Ia+1−θ;ψϰa+. Then,
(21)ϰϑ=Y0Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ.
Taking the limitϑ⟶b, we get
(22)ϰb=Y0Γθψb−ψaθ−1+Ia+θ1;ψHsb.
From the mixed boundary conditionsc1Y0+c2ϰb=d, we get
(23)Y0=dc1−c2c1Y0Γθψb−ψaθ−1+Ia+θ1;ψHsb,which implies
(24)Y0=Γθc1Γθ+c2Kψθ−1b,ad−c2Ia+θ1;ψHsb.
Then, (21) becomes
(25)ϰϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ,which shows that formula (19) is satisfied, where
(26)Ia+θ1;ψHsr=1Γθ1∫arψ′sψr−ψsθ1−1Hsds.
Conversely, letϰ∈C1−θ,ψI,ℝ satisfy (19) which can be written as (25). In (19), taking the limit as ϑ⟶a and ϑ⟶b and then using Lemma 6, we get
(27)c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=c1ΓθΠbd−c2Ia+θ1;ψHsb+c2Kψθ−1b,aΠbd−c2Ia+θ1;ψHsb+c2Ia+θ1;ψHsb=c1ΓθΠb+c2Kψθ−1b,aΠbd−c1c2Γθc1Γθ+c2Kψθ−1b,aIa+θ1;ψHsb+c2−c2c2Kψθ−1b,aΠbIa+θ1;ψHsb=d.
In another direction, by applyingDa+θ1,θ2;ψ on (19) and using Lemmas 4 and 5, we obtain
(28)Da+θ1,θ2;ψϰϑ=Da+θ1,θ2;ψKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Da+θ1,θ2;ψIa+θ1;ψHsϑ=Da+θ1,θ2;ψIa+θ1;ψHsϑ=Hϑ.
This finishes the proof.Lemma 9.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2, and let H,Z∈CI′,ℝ and W∈CI′,ℝ\0. Then, the function ϰ∈C1−θ,ψI,ℝ is a solution of the following linear fractional hybird BVP:
(29)Da+θ1,θ2;ψϰϑ−ZϑWϑ=Hϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑ−ZϑWϑϑ=a+c2ϰϑ−ZϑWϑϑ=b=d,if and only if ϰ satisfies the FIE:
(30)ϰϑ=WϑKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ+Zϑ,where Πb is defined as in Lemma 8.Proof.
Letϰ∈C1−θ,ψI,ℝ be a solution of (29). We need to prove that ϰ is also a solution of (30). By the definition of C1−θ,ψI,ℝ and Lemma 6, we have Ia+1−θ;ψϰϑ∈CI,ℝ.
Now, by applyingIa+θ1;ψ on the first equation of (29) and using Lemma 5, we can write
(31)ϰϑ−ZϑWϑ=ψϑ−ψaθ−1ΓθIa+1−θ;ψϰa−ZaWa+Ia+θ1;ψHsϑ.
SetY1=Ia+1−θ;ψϰa−Za/Wa; it follows that
(32)ϰϑ=WϑY1Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ+Zϑ.
Taking the limitϑ⟶b, we get
(33)ϰb=WbY1Γθψb−ψaθ−1+Ia+θ1;ψHsb+Zb,which implies
(34)ϰb−ZbWb=Y1Γθψb−ψaθ−1+Ia+θ1;ψHsb.
To determine the constantY1, we use the BC of (29). It was followed by
(35)Y1=dc1−c2c1Y1Γθψb−ψaθ−1+Ia+θ1;ψHsb,which implies
(36)Y1=Γθc1Γθ+c2Kψθ−1b,ad−c2Ia+θ1;ψHsb.
Then, (32) becomes
(37)ϰϑ=WϑKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ+Zϑ,which shows that formula (30) is satisfied.
Conversely, letϰ∈C1−θ,ψI,ℝ satisfy (29) which can be written as (37). In (30), taking the limit as ϑ⟶a and ϑ⟶b and then using Lemma 6, we get
(38)c1Ia+1−θ;ψϰϑ−ZϑWϑϑ=a+c2ϰϑ−ZϑWϑϑ=b=d.
In the same context, we have from (30) that
(39)ϰϑ−ZϑWϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ.
OperateDa+θ1,θ2;ψ on both sides of (39); then, use Lemmas 4 and 5 to get
(40)Da+θ1,θ2;ψϰϑ−ZϑWϑ=Da+θ1,θ2;ψKψθ−1s,aϑΠbd−c2Ia+θ1;ψHsb+Da+θ1,θ2;ψIa+θ1;ψHsϑ=Da+θ1,θ2;ψIa+θ1;ψHsϑ=Hϑ.
This finishes the proof.


### 3.2. Existence Theorems

In this portion, we prove the existence theorems to problems (5) and (6) in the weighted space by means of Dhage’s fixed-point technique.By a solution of (6), we mean a function ϰ∈C1−θ,ψI,ℝ such that
(1)
the functionϰϑ−zϑ,ϰϑ/qϑ,ϰϑ∈C1−θ,ψI,ℝ if ψϑ−ψa1−θϰϑ−zϑ,ϰϑ/qϑ,ϰϑ∈CI,ℝ(2)
ϰsatisfies the equations in (6)Before pursuing the main findings, we state the following assumptions:(A1) Let q:I′×ℝ⟶ℝ\0,z:I′×ℝ⟶ℝ, and p:I′×ℝ⟶ℝ be continuous functions such that q·,ϰ·,z·,ϰ·,p·,ϰ·∈C1−θψI,ℝ, for each ϰ∈C1−θψI,ℝ.(A2) There exist two positive functions μq,μz∈CI,ℝ such that
(41)qϑ,ϰ−qϑ,ϰ¯≤μqϰ−ϰ¯,zϑ,ϰ−zϑ,ϰ¯≤μzϰ−ϰ¯,for each ϑ∈I′ and ϰ,ϰ¯∈ℝ.(A3) There exist ρ,σ∈CI,ℝ such that
(42)pϑ,ϰ≤ρϑ+σϑϰC1−θψ,foreachϑ,ϰ∈I′×ℝ.(A4) There exists a constant μ>0 such that
(43)μ≥q0R+z01−μqCR+μzC,μqCR+μzC<1,(44)R≔Kψθ−1b,aΠbd+Kψθ−1b,ac2Πb+1Kψθ1b,aΓθ1+1ρC+σCμ,where q0≔maxϑ∈Iψϑ−ψa1−θqϑ,0 and z0≔maxϑ∈Iψϑ−ψa1−θzϑ,0.As a result of Lemma8, we present the next lemma.Lemma 10.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2 and p:I′×ℝ⟶ℝ are continuous. Then, the nonlinear fractional BVP
(45)Da+θ1,θ2;ψϰϑ=pϑ,ϰϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=dis equivalent to
(46)ϰϑ=Kψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.As a result of Lemma9, we present the next lemma.Lemma 11.
Let0<θ1<1, 0≤θ2≤1, where θ=θ1+θ2−θ1θ2 and p:I′×ℝ⟶ℝ are continuous. Then, the nonlinear fractional hybrid BVP
(47)Da+θ1,θ2;ψϰϑ−zϑ,ϰϑqϑ,ϰϑ=pϑ,ϰϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=a+c2ϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=b=dis equivalent to
(48)ϰϑ=zϑ,ϰϑ+qϑ,ϰϑKψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds.Theorem 12.
Suppose that (A1)–(A4) hold. Then, problem (6) has at least one solution in C1−θψI,ℝ.Proof.
Set(49)Bμ=ϰ∈C1−θψI,ℝ:ϰC1−θψ≤μ.
Obviously,Bμ is a convex, closed, bounded subset of C1−θ,ψI,ℝ. By Lemma 11, problem (6) is equivalent to
(50)ϰϑ=zϑ,ϰϑ+qϑ,ϰϑKψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.
Define the operatorsA,B:C1−θψI,ℝ⟶C1−θψI,ℝ and C:Bμ⟶C1−θψI,ℝby
(51)Aϰϑ=qϑ,ϰϑ,ϑ∈I′,Bϰϑ=zϑ,ϰϑ,ϑ∈I′,Cϰϑ=Kψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.
Then, we can express equation (50) as follows:
(52)ϰϑ=Bϰϑ+Cϰϑ·Aϰϑ,ϑ∈I′.
Now, we will show thatA, B, and C meet all the requirements for Theorem 7. This will be accomplished in a series of next steps.
Step 1.A,B:C1−θψI,ℝ⟶C1−θψI,ℝ are Lipschitzian on C1−θψI,ℝ.
Letϰ,ω∈C1−θψI,ℝ and ϑ∈I′. Then, by (A2), we have
(53)Aϰ−AωC1−θψ=maxϑ∈Iψϑ−ψa1−θAϰϑ−Aωϑ=maxϑ∈Iψϑ−ψa1−θqϑ,ϰϑ−qϑ,ωϑ≤maxϑ∈Iψϑ−ψa1−θμqϑϰϑ−ωϑ≤μqCϰ−ωC1−θψ.
Therefore,A is Lipschitzian on C1−θψI,ℝ with the Lipschitz constant μqC.
Similarly, we conclude thatB is Lipschitzian on C1−θψI,ℝ with the Lipschitz constant μzC, i.e.,
(54)Bϰ−BωC1−θψ≤μzCϰ−ωC1−θψ.
Step 2.C:Bμ⟶C1−θψI,ℝ is completely continuous.
First, we show thatC:Bμ⟶C1−θψI,ℝ is continuous. Let ϰn be a sequence such that ϰn⟶ϰ in Bμ. Then,
(55)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤1Πbc2Γθ1∫abFψθ1b,slimn⟶∞ps,ϰns−ps,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,slimn⟶∞ps,ϰns−ps,ϰsds=1Πbc2Γθ1∫abFψθ1b,sKψθ−1s,a×limn⟶∞Kψ1−θs,aps,ϰns−ps,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,sKψθ−1s,a×limn⟶∞Kψ1−θs,aps,ϰns−ps,ϰsds.
Sincep·,ϰ· is a continuous function with p·,ϰ·∈C1−θψI,ℝ, we have
(56)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤1Πbc2Γθ1∫abFψθ1b,sKψθ−1s,alimn⟶∞p·,ϰn·−p·,ϰ·C1−θψds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,sKψθ−1s,alimn⟶∞p·,ϰn·−p·,ϰ·C1−θψds.
Sinceψ is increasing and using (17), we obtain
(57)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤c2Πb+Kψ1−θb,aΓθΓθ+θ1Kψθ1+θ−1b,a×limn⟶∞p·,ϰn·−p·,ϰ·C1−θψ⟶0asn⟶∞.
This shows thatC:Bμ⟶C1−θψI,ℝ is continuous on Bμ.
Next, we show thatCBμ is uniformly bounded in Bμ. Indeed, for any ϰ∈Bμ, we have
(58)CϰC1−θψ=maxϑ∈Iψϑ−ψa1−θCϰϑ≤dΠb+c2Πb1Γθ1∫abFψθ1b,smaxϑ∈Ips,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,smaxϑ∈Ips,ϰsds≤dΠb+c2Πb1Γθ1∫abFψθ1b,smaxϑ∈Iρs+σsϰC1−θψds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,smaxϑ∈Iρs+σsϰC1−θψds≤dΠb+c2ΠbKψθb,aΓθ1+1ρC+σCϰC1−θψ+Kψ1b,aΓθ1+1ρC+σCϰC1−θψ≤dΠb+c2ΠbKψθ1b,a+Kψ1b,aρC+σCμΓθ1+1.
By (44), then CϰC1−θψ≤R for each ϰ∈Bμ.
Now, we prove thatCBμ is an equicontinuous set in C1−θψI,ℝ.
Letϰ∈Bμ and ϑ1,ϑ2∈I′ with ϑ1<ϑ2. Then,
(59)ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1=Kψ1−θϑ2,aΓθ1∫aϑ2Fψθ1ϑ2,sps,ϰsds−Kψ1−θϑ1,aΓθ1∫aϑ1Fψθ1ϑ1,sps,ϰsds=Kψ1−θϑ2,aΓθ1∫aϑ2Fψθ1ϑ2,sKψθ−1s,aψs−ψa1−θps,ϰsds−Kψ1−θϑ1,aΓθ1∫aϑ1Fψθ1ϑ1,sKψθ−1s,aψs−ψa1−θps,ϰsds.
Sincep·,ϰ·∈C1−θψI,ℝ for any ϰ∈C1−θψI,ℝ and ψ·−ψa1−θp·,ϰ·∈CI,ℝ, there exist ξ∈ℝ such that
(60)ψs−ψa1−θps,ϰs≤ξforallϑ∈I′.
Hence,(61)ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1≤Kψ1−θϑ2,aΓθ1ξ∫aϑ2Fψθ1ϑ2,sKψθ−1s,ads−Kψ1−θϑ1,aΓθ1ξ∫aϑ1Fψθ1ϑ1,sKψθ−1s,ads=Kψ1−θϑ2,aξΓθΓθ+θ1ψϑ2−ψaθ1+θ−1−Kψ1−θϑ1,aξΓθΓθ+θ1ψϑ1−ψaθ1+θ−1=ξΓθΓθ+θ1ψϑ2−ψaθ1−ψϑ1−ψaθ1.
The continuity ofψ shows that ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1⟶0 as ϑ2−ϑ1⟶0.
This confirms thatCBμ is an equicontinuous set in C1−θψI,ℝ. As a result of the Arzelà-Ascoli theorem, C is completely continuous.
Step 3.AϰCω+Bϰ∈Bμ for ϰ∈C1−θψI,ℝ and ω∈Bμ.
Letϰ∈C1−θψI,ℝ, and ω∈Bμ such that ϰ=AϰCω+Bϰ. Then,
(62)ϰϑ≤AϰϑCωϑ+Bϰϑ≤qϑ,ϰϑ−qϑ,0+qϑ,0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Γθ1∫abFψθ1b,sps,ωsds+1Γθ1∫aϑFψθ1ϑ,sps,ωsdϰ+zϑ,ϰϑ−zϑ,0+zϑ,0≤μqϑϰϑ+qϑ,0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Ia+θ1;ψρs+σsωC1−θψb+Ia+θ1;ψρs+σsωC1−θψϑ+μzϑϰϑ+zϑ,0≤Kψθ−1ϑ,aμqCϰC1−θψ+q0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Kψθ1b,aΓθ1+1ρC+σCωC1−θψ+ρC+σCωC1−θψKψθ1ϑ,aΓθ1+1+Kψθ−1ϑ,aμzCϰC1−θψ+z0.
Therefore,(63)ϰC1−θψ≤μqCϰC1−θψ+q0R+μzCϰC1−θψ+z0,which implies
(64)ϰC1−θψ≤q0R+z01−μqCR+μzC≤μ.
Step 4. Condition (iv) in Theorem7 holds. That is, κ1M+κ2<1.
From Step 2 and (44), we have
(65)M=CBμ=supϰ∈Bμsupϑ∈ICϰϑ≤Kψθ−1b,aΠbd+Kψθ−1b,ac2Πb+1Kψθ1b,aΓθ1+1ρC+σCμ=R.
Hence,(66)μqCM+μzC≤μqCR+μzC<1,where κ1=μqC and κ2=μzC. Thus, all the assumptions of Theorem 7 are fulfilled, so the equation ϰ=AϰCϰ+Bϰ has a solution in Bμ. As a result, problem (6) has a solution on I.Theorem 13.
Suppose that (A1)–(A4) hold. Then, the ψ-Hilfer problem (5) has at least one solution on I.Proof.
The proof of this theorem is quite similar to that of Theorem12 with consideration that zϑ,ϰϑ≡0 and qϑ,ϰϑ≡1 in problem (6). Thus, we omit the details.Remark 14.
(1)
The conditions (A2) and (A3) are also correct if we replace the functions ρ,σ,μq,μz∈CI,ℝ with constants(2)
Problem (6) reduces to problem (5) when zϑ,ϰϑ≡0 and qϑ,ϰϑ≡1. Consequently, Theorem 12 is applied to problem (5)(3)
Problem (5) reduces to problem (3) when ψϑ=ϑ, a=0, and θ2=1 (see [23])(4)
Problem (6) reduces to problem (4) when ψϑ=ϑ, a=0, θ2=1, and zϑ,ϰϑ≡0 (see [24])(5)
In particular, ifψϑ=ϑ, then the obtained results correspond to Caputo-type FDE and RL-type FDE, for θ2=1 and θ2=0, respectively(6)
The corresponding fractional problems involving the Hilfer-Katugampola type and Hilfer-Hadamard type appear as a special case of our proposed problems forψϑ=ϑρ, ρ>0, and ψϑ=logϑ, respectively


## 3.1. Fractional Integral Equations (FIEs)

The forthcoming results give the equivalent of solution formulas for the proposed problems. For brevity, we set the following symbols:(15)Kψθ−1ϑ,a=ψϑ−ψaθ−1,Fψθϑ,s=ψ′sψϑ−ψsθ−1,(16)1Γθ1∫aϑFψθ1ϑ,sds=1Γθ1∫aϑψ′sψϑ−ψsθ1−1ds=ψϑ−ψaθ1Γθ1+1=Kψθ1ϑ,aΓθ1+1,(17)1Γθ1∫aϑFψθ1ϑ,sKψθ−1s,ads=1Γθ1∫aϑψ′sψϑ−ψsθ1−1ψs−ψaθ−1ds=Ia+θ1;ψψϑ−ψaθ−1=ΓθΓθ+θ1ψϑ−ψaθ1+θ−1=ΓθΓθ+θ1Kψθ1+θ−1ϑ,a.Lemma 8.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2, and H:I′⟶ℝ is continuous. Then, the function ϰ∈C1−θ,ψI,ℝ is a solution of the linear fractional BVP:
(18)Da+θ1,θ2;ψϰϑ=Hϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=d,if and only if ϰ satisfies the FIE:
(19)ϰϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ,ϑ∈I′,where Πb≔c1Γθ+c2Kψθ−1b,a≠0.Proof.
Letϰ∈C1−θ,ψI,ℝ be a solution of (18). We need to prove that ϰ is also a solution of (19). By the definition of C1−θ,ψI,ℝ and Lemma 6, we have Ia+1−θ;ψϰϑ∈CI,ℝ.
Now, by applyingIa+θ1;ψ to the first equation in (18) and using Lemma 5, we can write
(20)ϰϑ=Ia+1−θ;ψϰa+Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ.
SetY0≔Ia+1−θ;ψϰa+. Then,
(21)ϰϑ=Y0Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ.
Taking the limitϑ⟶b, we get
(22)ϰb=Y0Γθψb−ψaθ−1+Ia+θ1;ψHsb.
From the mixed boundary conditionsc1Y0+c2ϰb=d, we get
(23)Y0=dc1−c2c1Y0Γθψb−ψaθ−1+Ia+θ1;ψHsb,which implies
(24)Y0=Γθc1Γθ+c2Kψθ−1b,ad−c2Ia+θ1;ψHsb.
Then, (21) becomes
(25)ϰϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ,which shows that formula (19) is satisfied, where
(26)Ia+θ1;ψHsr=1Γθ1∫arψ′sψr−ψsθ1−1Hsds.
Conversely, letϰ∈C1−θ,ψI,ℝ satisfy (19) which can be written as (25). In (19), taking the limit as ϑ⟶a and ϑ⟶b and then using Lemma 6, we get
(27)c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=c1ΓθΠbd−c2Ia+θ1;ψHsb+c2Kψθ−1b,aΠbd−c2Ia+θ1;ψHsb+c2Ia+θ1;ψHsb=c1ΓθΠb+c2Kψθ−1b,aΠbd−c1c2Γθc1Γθ+c2Kψθ−1b,aIa+θ1;ψHsb+c2−c2c2Kψθ−1b,aΠbIa+θ1;ψHsb=d.
In another direction, by applyingDa+θ1,θ2;ψ on (19) and using Lemmas 4 and 5, we obtain
(28)Da+θ1,θ2;ψϰϑ=Da+θ1,θ2;ψKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Da+θ1,θ2;ψIa+θ1;ψHsϑ=Da+θ1,θ2;ψIa+θ1;ψHsϑ=Hϑ.
This finishes the proof.Lemma 9.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2, and let H,Z∈CI′,ℝ and W∈CI′,ℝ\0. Then, the function ϰ∈C1−θ,ψI,ℝ is a solution of the following linear fractional hybird BVP:
(29)Da+θ1,θ2;ψϰϑ−ZϑWϑ=Hϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑ−ZϑWϑϑ=a+c2ϰϑ−ZϑWϑϑ=b=d,if and only if ϰ satisfies the FIE:
(30)ϰϑ=WϑKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ+Zϑ,where Πb is defined as in Lemma 8.Proof.
Letϰ∈C1−θ,ψI,ℝ be a solution of (29). We need to prove that ϰ is also a solution of (30). By the definition of C1−θ,ψI,ℝ and Lemma 6, we have Ia+1−θ;ψϰϑ∈CI,ℝ.
Now, by applyingIa+θ1;ψ on the first equation of (29) and using Lemma 5, we can write
(31)ϰϑ−ZϑWϑ=ψϑ−ψaθ−1ΓθIa+1−θ;ψϰa−ZaWa+Ia+θ1;ψHsϑ.
SetY1=Ia+1−θ;ψϰa−Za/Wa; it follows that
(32)ϰϑ=WϑY1Γθψϑ−ψaθ−1+Ia+θ1;ψHsϑ+Zϑ.
Taking the limitϑ⟶b, we get
(33)ϰb=WbY1Γθψb−ψaθ−1+Ia+θ1;ψHsb+Zb,which implies
(34)ϰb−ZbWb=Y1Γθψb−ψaθ−1+Ia+θ1;ψHsb.
To determine the constantY1, we use the BC of (29). It was followed by
(35)Y1=dc1−c2c1Y1Γθψb−ψaθ−1+Ia+θ1;ψHsb,which implies
(36)Y1=Γθc1Γθ+c2Kψθ−1b,ad−c2Ia+θ1;ψHsb.
Then, (32) becomes
(37)ϰϑ=WϑKψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ+Zϑ,which shows that formula (30) is satisfied.
Conversely, letϰ∈C1−θ,ψI,ℝ satisfy (29) which can be written as (37). In (30), taking the limit as ϑ⟶a and ϑ⟶b and then using Lemma 6, we get
(38)c1Ia+1−θ;ψϰϑ−ZϑWϑϑ=a+c2ϰϑ−ZϑWϑϑ=b=d.
In the same context, we have from (30) that
(39)ϰϑ−ZϑWϑ=Kψθ−1ϑ,aΠbd−c2Ia+θ1;ψHsb+Ia+θ1;ψHsϑ.
OperateDa+θ1,θ2;ψ on both sides of (39); then, use Lemmas 4 and 5 to get
(40)Da+θ1,θ2;ψϰϑ−ZϑWϑ=Da+θ1,θ2;ψKψθ−1s,aϑΠbd−c2Ia+θ1;ψHsb+Da+θ1,θ2;ψIa+θ1;ψHsϑ=Da+θ1,θ2;ψIa+θ1;ψHsϑ=Hϑ.
This finishes the proof.


## 3.2. Existence Theorems

In this portion, we prove the existence theorems to problems (5) and (6) in the weighted space by means of Dhage’s fixed-point technique.By a solution of (6), we mean a function ϰ∈C1−θ,ψI,ℝ such that
(1)
the functionϰϑ−zϑ,ϰϑ/qϑ,ϰϑ∈C1−θ,ψI,ℝ if ψϑ−ψa1−θϰϑ−zϑ,ϰϑ/qϑ,ϰϑ∈CI,ℝ(2)
ϰsatisfies the equations in (6)Before pursuing the main findings, we state the following assumptions:(A1) Let q:I′×ℝ⟶ℝ\0,z:I′×ℝ⟶ℝ, and p:I′×ℝ⟶ℝ be continuous functions such that q·,ϰ·,z·,ϰ·,p·,ϰ·∈C1−θψI,ℝ, for each ϰ∈C1−θψI,ℝ.(A2) There exist two positive functions μq,μz∈CI,ℝ such that
(41)qϑ,ϰ−qϑ,ϰ¯≤μqϰ−ϰ¯,zϑ,ϰ−zϑ,ϰ¯≤μzϰ−ϰ¯,for each ϑ∈I′ and ϰ,ϰ¯∈ℝ.(A3) There exist ρ,σ∈CI,ℝ such that
(42)pϑ,ϰ≤ρϑ+σϑϰC1−θψ,foreachϑ,ϰ∈I′×ℝ.(A4) There exists a constant μ>0 such that
(43)μ≥q0R+z01−μqCR+μzC,μqCR+μzC<1,(44)R≔Kψθ−1b,aΠbd+Kψθ−1b,ac2Πb+1Kψθ1b,aΓθ1+1ρC+σCμ,where q0≔maxϑ∈Iψϑ−ψa1−θqϑ,0 and z0≔maxϑ∈Iψϑ−ψa1−θzϑ,0.As a result of Lemma8, we present the next lemma.Lemma 10.
Let0<θ1<1,0≤θ2≤1, where θ=θ1+θ2−θ1θ2 and p:I′×ℝ⟶ℝ are continuous. Then, the nonlinear fractional BVP
(45)Da+θ1,θ2;ψϰϑ=pϑ,ϰϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑϑ=a+c2ϰϑϑ=b=dis equivalent to
(46)ϰϑ=Kψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.As a result of Lemma9, we present the next lemma.Lemma 11.
Let0<θ1<1, 0≤θ2≤1, where θ=θ1+θ2−θ1θ2 and p:I′×ℝ⟶ℝ are continuous. Then, the nonlinear fractional hybrid BVP
(47)Da+θ1,θ2;ψϰϑ−zϑ,ϰϑqϑ,ϰϑ=pϑ,ϰϑ,ϑ∈I′,c1Ia+1−θ;ψϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=a+c2ϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=b=dis equivalent to
(48)ϰϑ=zϑ,ϰϑ+qϑ,ϰϑKψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds.Theorem 12.
Suppose that (A1)–(A4) hold. Then, problem (6) has at least one solution in C1−θψI,ℝ.Proof.
Set(49)Bμ=ϰ∈C1−θψI,ℝ:ϰC1−θψ≤μ.
Obviously,Bμ is a convex, closed, bounded subset of C1−θ,ψI,ℝ. By Lemma 11, problem (6) is equivalent to
(50)ϰϑ=zϑ,ϰϑ+qϑ,ϰϑKψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.
Define the operatorsA,B:C1−θψI,ℝ⟶C1−θψI,ℝ and C:Bμ⟶C1−θψI,ℝby
(51)Aϰϑ=qϑ,ϰϑ,ϑ∈I′,Bϰϑ=zϑ,ϰϑ,ϑ∈I′,Cϰϑ=Kψθ−1ϑ,aΠbd−c2Γθ1∫abFψθ1b,sps,ϰsds+1Γθ1∫aϑFψθ1ϑ,sps,ϰsds,ϑ∈I′.
Then, we can express equation (50) as follows:
(52)ϰϑ=Bϰϑ+Cϰϑ·Aϰϑ,ϑ∈I′.
Now, we will show thatA, B, and C meet all the requirements for Theorem 7. This will be accomplished in a series of next steps.
Step 1.A,B:C1−θψI,ℝ⟶C1−θψI,ℝ are Lipschitzian on C1−θψI,ℝ.
Letϰ,ω∈C1−θψI,ℝ and ϑ∈I′. Then, by (A2), we have
(53)Aϰ−AωC1−θψ=maxϑ∈Iψϑ−ψa1−θAϰϑ−Aωϑ=maxϑ∈Iψϑ−ψa1−θqϑ,ϰϑ−qϑ,ωϑ≤maxϑ∈Iψϑ−ψa1−θμqϑϰϑ−ωϑ≤μqCϰ−ωC1−θψ.
Therefore,A is Lipschitzian on C1−θψI,ℝ with the Lipschitz constant μqC.
Similarly, we conclude thatB is Lipschitzian on C1−θψI,ℝ with the Lipschitz constant μzC, i.e.,
(54)Bϰ−BωC1−θψ≤μzCϰ−ωC1−θψ.
Step 2.C:Bμ⟶C1−θψI,ℝ is completely continuous.
First, we show thatC:Bμ⟶C1−θψI,ℝ is continuous. Let ϰn be a sequence such that ϰn⟶ϰ in Bμ. Then,
(55)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤1Πbc2Γθ1∫abFψθ1b,slimn⟶∞ps,ϰns−ps,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,slimn⟶∞ps,ϰns−ps,ϰsds=1Πbc2Γθ1∫abFψθ1b,sKψθ−1s,a×limn⟶∞Kψ1−θs,aps,ϰns−ps,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,sKψθ−1s,a×limn⟶∞Kψ1−θs,aps,ϰns−ps,ϰsds.
Sincep·,ϰ· is a continuous function with p·,ϰ·∈C1−θψI,ℝ, we have
(56)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤1Πbc2Γθ1∫abFψθ1b,sKψθ−1s,alimn⟶∞p·,ϰn·−p·,ϰ·C1−θψds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,sKψθ−1s,alimn⟶∞p·,ϰn·−p·,ϰ·C1−θψds.
Sinceψ is increasing and using (17), we obtain
(57)limn⟶∞ψϑ−ψa1−θCϰnϑ−Cϰϑ≤c2Πb+Kψ1−θb,aΓθΓθ+θ1Kψθ1+θ−1b,a×limn⟶∞p·,ϰn·−p·,ϰ·C1−θψ⟶0asn⟶∞.
This shows thatC:Bμ⟶C1−θψI,ℝ is continuous on Bμ.
Next, we show thatCBμ is uniformly bounded in Bμ. Indeed, for any ϰ∈Bμ, we have
(58)CϰC1−θψ=maxϑ∈Iψϑ−ψa1−θCϰϑ≤dΠb+c2Πb1Γθ1∫abFψθ1b,smaxϑ∈Ips,ϰsds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,smaxϑ∈Ips,ϰsds≤dΠb+c2Πb1Γθ1∫abFψθ1b,smaxϑ∈Iρs+σsϰC1−θψds+Kψ1−θϑ,aΓθ1∫aϑFψθ1ϑ,smaxϑ∈Iρs+σsϰC1−θψds≤dΠb+c2ΠbKψθb,aΓθ1+1ρC+σCϰC1−θψ+Kψ1b,aΓθ1+1ρC+σCϰC1−θψ≤dΠb+c2ΠbKψθ1b,a+Kψ1b,aρC+σCμΓθ1+1.
By (44), then CϰC1−θψ≤R for each ϰ∈Bμ.
Now, we prove thatCBμ is an equicontinuous set in C1−θψI,ℝ.
Letϰ∈Bμ and ϑ1,ϑ2∈I′ with ϑ1<ϑ2. Then,
(59)ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1=Kψ1−θϑ2,aΓθ1∫aϑ2Fψθ1ϑ2,sps,ϰsds−Kψ1−θϑ1,aΓθ1∫aϑ1Fψθ1ϑ1,sps,ϰsds=Kψ1−θϑ2,aΓθ1∫aϑ2Fψθ1ϑ2,sKψθ−1s,aψs−ψa1−θps,ϰsds−Kψ1−θϑ1,aΓθ1∫aϑ1Fψθ1ϑ1,sKψθ−1s,aψs−ψa1−θps,ϰsds.
Sincep·,ϰ·∈C1−θψI,ℝ for any ϰ∈C1−θψI,ℝ and ψ·−ψa1−θp·,ϰ·∈CI,ℝ, there exist ξ∈ℝ such that
(60)ψs−ψa1−θps,ϰs≤ξforallϑ∈I′.
Hence,(61)ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1≤Kψ1−θϑ2,aΓθ1ξ∫aϑ2Fψθ1ϑ2,sKψθ−1s,ads−Kψ1−θϑ1,aΓθ1ξ∫aϑ1Fψθ1ϑ1,sKψθ−1s,ads=Kψ1−θϑ2,aξΓθΓθ+θ1ψϑ2−ψaθ1+θ−1−Kψ1−θϑ1,aξΓθΓθ+θ1ψϑ1−ψaθ1+θ−1=ξΓθΓθ+θ1ψϑ2−ψaθ1−ψϑ1−ψaθ1.
The continuity ofψ shows that ψϑ2−ψa1−θCϰϑ2−ψϑ1−ψa1−θCϰϑ1⟶0 as ϑ2−ϑ1⟶0.
This confirms thatCBμ is an equicontinuous set in C1−θψI,ℝ. As a result of the Arzelà-Ascoli theorem, C is completely continuous.
Step 3.AϰCω+Bϰ∈Bμ for ϰ∈C1−θψI,ℝ and ω∈Bμ.
Letϰ∈C1−θψI,ℝ, and ω∈Bμ such that ϰ=AϰCω+Bϰ. Then,
(62)ϰϑ≤AϰϑCωϑ+Bϰϑ≤qϑ,ϰϑ−qϑ,0+qϑ,0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Γθ1∫abFψθ1b,sps,ωsds+1Γθ1∫aϑFψθ1ϑ,sps,ωsdϰ+zϑ,ϰϑ−zϑ,0+zϑ,0≤μqϑϰϑ+qϑ,0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Ia+θ1;ψρs+σsωC1−θψb+Ia+θ1;ψρs+σsωC1−θψϑ+μzϑϰϑ+zϑ,0≤Kψθ−1ϑ,aμqCϰC1−θψ+q0×Kψθ−1ϑ,aΠbd+Kψθ−1ϑ,aΠbc2Kψθ1b,aΓθ1+1ρC+σCωC1−θψ+ρC+σCωC1−θψKψθ1ϑ,aΓθ1+1+Kψθ−1ϑ,aμzCϰC1−θψ+z0.
Therefore,(63)ϰC1−θψ≤μqCϰC1−θψ+q0R+μzCϰC1−θψ+z0,which implies
(64)ϰC1−θψ≤q0R+z01−μqCR+μzC≤μ.
Step 4. Condition (iv) in Theorem7 holds. That is, κ1M+κ2<1.
From Step 2 and (44), we have
(65)M=CBμ=supϰ∈Bμsupϑ∈ICϰϑ≤Kψθ−1b,aΠbd+Kψθ−1b,ac2Πb+1Kψθ1b,aΓθ1+1ρC+σCμ=R.
Hence,(66)μqCM+μzC≤μqCR+μzC<1,where κ1=μqC and κ2=μzC. Thus, all the assumptions of Theorem 7 are fulfilled, so the equation ϰ=AϰCϰ+Bϰ has a solution in Bμ. As a result, problem (6) has a solution on I.Theorem 13.
Suppose that (A1)–(A4) hold. Then, the ψ-Hilfer problem (5) has at least one solution on I.Proof.
The proof of this theorem is quite similar to that of Theorem12 with consideration that zϑ,ϰϑ≡0 and qϑ,ϰϑ≡1 in problem (6). Thus, we omit the details.Remark 14.
(1)
The conditions (A2) and (A3) are also correct if we replace the functions ρ,σ,μq,μz∈CI,ℝ with constants(2)
Problem (6) reduces to problem (5) when zϑ,ϰϑ≡0 and qϑ,ϰϑ≡1. Consequently, Theorem 12 is applied to problem (5)(3)
Problem (5) reduces to problem (3) when ψϑ=ϑ, a=0, and θ2=1 (see [23])(4)
Problem (6) reduces to problem (4) when ψϑ=ϑ, a=0, θ2=1, and zϑ,ϰϑ≡0 (see [24])(5)
In particular, ifψϑ=ϑ, then the obtained results correspond to Caputo-type FDE and RL-type FDE, for θ2=1 and θ2=0, respectively(6)
The corresponding fractional problems involving the Hilfer-Katugampola type and Hilfer-Hadamard type appear as a special case of our proposed problems forψϑ=ϑρ, ρ>0, and ψϑ=logϑ, respectively


## 4. Examples

In this part, we construct two examples to explain the main results.Example 15.
Consider theψ-Hilfer hybrid FDE with hybrid BC:
(67)Da+θ1,θ2;ψϰϑ−zϑ,ϰϑqϑ,ϰϑ=pϑ,ϰϑ,ϑ∈0,1,c1Ia+1−θ;ψϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=a+c2ϰϑ−zϑ,ϰϑqϑ,ϰϑϑ=b=d.
Definez:0,1×ℝ⟶ℝ, q:0,1×ℝ⟶ℝ\0, and p:0,1×ℝ⟶ℝ by
(68)zϑ,ϰϑ=12cosϑ3ϰϑ1+ϰϑ+e−ϑ,qϑ,ϰϑ=1+sinϑ12ϰϑ,pϑ,ϰϑ=ψϑ−ψ0100ϰϑ1+ϰϑ+2.
It is easy to show that for anyϰ,ω∈ℝ, we have
(69)zϑ,ϰϑ−zϑ,ωϑ≤12cosϑ3ϰϑ−ωϑ,qϑ,ϰϑ−qϑ,ωϑ≤sinϑ12ϰϑ−ωϑ,and for each ϰ∈C1−θψ0,1,ℝ, we have
(70)pϑ,ϰϑ≤ψϑ−ψ0100ϰϑ+ψϑ−ψ050≤ψϑ−ψ0θ100ϰ1−θ;ψ+ψϑ−ψ050.
Hence, the hypotheses (A1)–(A3) hold with μqϑ=sinϑ/12,μzϑ=1/2cosϑ/3,ρϑ=ψϑ−ψ0θ/100, and σϑ=ψϑ−ψ0/50. Then, μqC=1/12, μzC=1/2, q0=maxϑ∈0,1ψϑ−ψ01−θ, and z0=maxϑ∈0,1ψϑ−ψ01−θcosϑ/3e−ϑ. Taking θ1=1/2, θ2=0,θ=1/2, c1=1/3, c2=1/3, d=1, and ψϑ=ϑ, we get ρC=1/100, σC=1/50, q0=1, and z0=1/ecos1/3. From the condition (A4), μqCR+μzC<1 when R<6, and μ≥1/1−1/12R+1/2R+1/ecos1/3. Also, Πb≠0; it follows from (44) that R=3/1+π+1/1+π+11/50π+1/25πμ<6. Hence, μ<2−149π−300π/−4−2π. Using the MATLAB program, μ satisfies the inequality 64.17<μ<159.65. Hence, all assumptions of Theorem 12 are satisfied, so problem (67) has at least one solution on 0,1.Example 16.
As a special case whenzϑ,ϰϑ=0 and qϑ,ϰϑ=1, we consider the ψ-Hilfer FDE with BC:
(71)D0+1/2,ϑϰϑ=ϰϑ1+ϰϑ+1100,ϑ∈0,1,100Ia+1−θ;ψϰϑϑ=0+ϰϑϑ=1=1.
Comparing problem (71) with problem (5), we obtain
(72)θ1=12,θ2=1,c1=c2=100,d=1,ψϑ=ϑ,pϑ,ϰϑ=ϰϑ1+ϰϑ+1100.
It is obvious thatθ=1, and in this case, the space C1−θψ0,1,ℝ is reduced to the space of continuous functions C0,1,ℝ. Hence, (A1) holds. It is easy to show that for each ϰ∈ℝ, we have
(73)pϑ,ϰϑ≤ϰϑ+1100≤ϰC+1100.
Thus, the hypothesis (A3) holds with ρϑ=1/100, and σϑ=1. Then, μqC=1, μzC=0, q0=1, and z0=0. So, we get ρC=1/100, σC=1. From the condition (A4), we have μ≥R/1−R. Finally, we need to show that R<1. Indeed, from (44), we have R=1/200+3/100π+3/πμ. It follows that there exists μ>0 with μ<1/600199π−6 such that R<1. Therefore, problem (71) can be applied to Theorem 12.


## 5. Concluding Remarks

We have acquired further existence results for the solution of BVPs for theψ-Hilfer problem (6), and the ψ-Hilfer hybrid problem (5) relies on the reduction of proposed problems to FIEs. Dhage’s hybrid fixed-point theorem in the Banach algebra has been applied. The reported results in the current paper are also valid for the hybrid FDEs involving RLFD and CFD, and they are also true for special cases of the function ψ. We confirm that the obtained results of this work are recent and generalize some of the previous results in the literature. More precisely, when taking different values of function ψ and parameter θ2, the studied problems cover several problems involving classical fractional operators such as RLFD, CFD, HFD, Hilfer-Katugampola FD, and Hilfer-Hadamard FD, as mentioned in Remark 14, which have been incorporated into the operators used in our investigation. Using these investigations, other qualitative analyses of the solution such as stability and continuous dependence results can be discussed, and this is what we desire to think about in a future paper.



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