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Ulam stability for nonlinear implicit differential equations with Hilfer-Katugampola fractional derivative and impulses

  • In this paper, we investigate the existence, uniqueness and stability results for a class of nonlinear impulsive Hilfer-Katugampola problems. Our reasoning is founded on the Banach contraction principle and Krasnoselskii's fixed point theorem. In addition, an example is provided to demonstrate the effectiveness of the main results.

    Citation: Soufyane Bouriah, Mouffak Benchohra, Juan J. Nieto, Yong Zhou. Ulam stability for nonlinear implicit differential equations with Hilfer-Katugampola fractional derivative and impulses[J]. AIMS Mathematics, 2022, 7(7): 12859-12884. doi: 10.3934/math.2022712

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  • In this paper, we investigate the existence, uniqueness and stability results for a class of nonlinear impulsive Hilfer-Katugampola problems. Our reasoning is founded on the Banach contraction principle and Krasnoselskii's fixed point theorem. In addition, an example is provided to demonstrate the effectiveness of the main results.



    Numerous authors have been interested in fractional differential equations throughout the years [1,4,5,10,12,13,14,16,18,19,20,28]. Several natural events are known to be modeled using fractional differential equations, which provides for a better description of the true state of the problem as compared to the problem modeled using differential equations of integer order [8,20,24,27].

    Fractional calculus has played an essential role in several domains during the last two decades, including mechanics, chemistry, economics, biology, control theory, and signal and image processing. Furthermore, it has been discovered that fractional order models may accurately capture the dynamical behavior of many complex systems. Such models are appealing not just to engineers and physicists, but also to mathematicians. For further details and applications [2,3,17,20,21,23,29,31,32,33,34,35].

    Several researchers have recently explored impulsive differential equations given their considerable applicability in diverse domains of science and technology. For a detailed study, see for instance [9,11,25].

    Motivated by the aforementioned works and the paper [26], in this paper, we consider the following impulsive problem:

    (ρDζ1,ζ2α+p)(ϑ)=ψ(ϑ,p(ϑ),(ρDζ1,ζ2α+p)(ϑ)), ϑ(ϑν,ϑν+1];ν=0,,ς, (1.1)
    ΔρI1ξα+p|ϑ=ϑν=χνR, ν=1,,ς, (1.2)
    ρI1ξα+p(α)=˜αR, (1.3)

    where ρDζ1,ζ2α+,ρI1ξα+ are the Hilfer-Katugampola fractional derivative of order ζ1(0,1) and type ζ2[0,1] and Katugampola fractional integral of order 1ξ,(ξ=ζ1+ζ2ζ1ζ2) respectively, ψ:Θ×R×RR, and 0<α=ϑ0<ϑ1<<ϑς<ϑς+1=μ, ΔρI1ξα+p|ϑ=ϑν= ρI1ξα+p(ϑ+ν)ρI1ξα+p(ϑν),where ρI1ξα+p(ϑ+ν)=limκ0+ρI1ξα+p(ϑν+κ) and ρI1ξα+p(ϑν)=limκ0ρI1ξα+p(ϑν+κ) represent the right and left limits of ρI1ξα+p(ϑ) at ϑ=ϑν.

    The following is the structure of paper. Section 2 presents certain notations and revisits several notions and auxiliary results. Section 3 presents two results for the problems (1.1)–(1.3) by employing suitable fixed point theorems. In Section 4, the Ulam-Hyers stability for the problems (1.1)–(1.3) is given. Finally, we give an example to illustrate the applicability of our theoretical results.

    Let 0<α<μ,Θ=[α,μ] and C(Θ,R) denotes a Banach space composed of all continuous functions from Θ into R with the norm

    p=sup{|p(ϑ)|:ϑΘ}.

    Consider the weighted spaces

    Cξ,ρ(Θ)={p:(α,μ]R:(ϑραρρ)ξp(ϑ)C(Θ,R)},0ξ<1,
    Cβξ,ρ(Θ)={pCβ1(Θ):p(β)Cξ,ρ(Θ)},βN,C0ξ,ρ(Θ)=Cξ,ρ(Θ),

    with the norms

    pCξ,ρ=supϑΘ|(ϑραρρ)ξp(ϑ)|

    and

    pCβξ,ρ=β1ν=0p(ν)+p(β)Cξ,ρ.

    Consider the weighted Banach space of piecewise continuous functions defined by

    PC1ξ,ρ(Θ,R)={p:ΘR:pC1ξ,ρ((ϑν,ϑν+1],R), ν=0,,ς and thereexist  ρI1ξα+p(ϑν)  and   ρI1ξα+p(ϑ+ν), ν=1,,ς with ρI1ξα+p(ϑν)=ρI1ξα+p(ϑν)},

    with the norm

    pPC1ξ,ρ=supϑΘ|(ϑρϑρνρ)1ξp(ϑ)|.

    For ξ=1, we obtain the space

    PC0,ρ(Θ,R)=PC(Θ,R)={p:ΘR:pC((ϑν,ϑν+1],R), ν=0,,ς and there exist p(ϑν)  and  p(ϑ+ν), ν=1,,ς with p(ϑν)=p(ϑν)},

    with the norm

    pPC=p.

    Now, we consider the weighted spaces:

    PCβ1ξ,ρ(Θ)={pPCβ1(Θ):p(β)PC1ξ,ρ(Θ)},βN,PC01ξ,ρ(Θ)=PC1ξ,ρ(Θ),

    with the norms

    pPCβ1ξ,ρ=β1ν=0p(ν)+p(β)PC1ξ,ρ.

    By Xp˜α(α,ˉα), (˜αR, 1p), we denote the space of those complex-valued Lebesgue measurable functions ψ on [α,ˉα] for which ψXp˜α<, with the norm

    ψXp˜α=(ˉαα|ϑ˜αψ(ϑ)|pdϑϑ)1p,  (1p<,˜αR).

    Definition 2.1. [26] Let ζ1R+,˜αR and ϰXp˜α(α,ˉα). The Katugampola fractional integral of order ζ1 is given by

    (ρIζ1α+ϰ)(ϑ)=ϑαϱρ1(ϑρϱρρ)ζ11ϰ(ϱ)Γ(ζ1)dϱ, ϑ>α,ρ>0.

    Definition 2.2. [26] Let ζ1R+N and ρ>0. The Katugampola fractional derivative ρDζ1α+ of order ζ1 is given by

    (ρDζ1α+ϰ)(ϑ)=δβρ(ρIβζ1α+ϰ)(ϑ)=(ϑ1ρddϑ)βϑαϱρ1(ϑρϱρρ)βζ11ϰ(ϱ)Γ(βζ1)dϱ, ϑ>α,ρ>0,

    where β=[ζ1]+1 and δβρ=(ϑ1ρddϑ)β.

    Theorem 2.3. [26] Let ζ1>0,ζ2>0,1p,0<α<ˉα< and ρ,˜αR,ρ˜α. Then, for ϰXp˜α(α,ˉα) the semigroup property is valid, i.e.

    (ρIζ1α+ ρIζ2α+ϰ)(ϑ)=(ρIζ1+ζ2α+ϰ)(ϑ).

    Lemma 2.4. [22] Let ζ1>0, and 0ξ<1. Then, ρIζ1α+ is bounded from Cξ,ρ(Θ) into Cξ,ρ(Θ).

    Lemma 2.5. [7] Let ϑ>α. Then, for ζ10 and ζ2>0, we have

    [ρIζ1α+(ϱραρρ)ζ21](ϑ)=Γ(ζ2)Γ(ζ1+ζ2)(ϑραρρ)ζ1+ζ21,[ρDζ1α+(ϱραρρ)ζ11](ϑ)=0,  0<ζ1<1.

    Lemma 2.6. [22] Let ζ1>0,0ξ<1 and ϰCξ[α,ˉα]. Then,

    (ρDζ1α+ ρIζ1α+ϰ)(ϑ)=ϰ(ϑ),   forall  ϑ(α,ˉα].

    Lemma 2.7. [22] Let 0<ζ1<1,0ξ<1. If ϰCξ,ρ[α,ˉα] and ρI1ζ1α+ϰC1ξ,ρ[α,ˉα], then

    (ρIζ1α+ ρDζ1α+ϰ)(ϑ)=ϰ(ϑ)(ρI1ζ1α+ϰ)(α)Γ(ζ1)(ϑραρρ)ζ11, forall  ϑ(α,ˉα].

    Definition 2.8. [22] Let the order ζ1 and the type ζ2 satisfy β1<ζ1<β and 0ζ21, with βN. The Hilfer-Katugampola fractional derivative of a function ϰC1ξ,ρ[α,ˉα], is defined by

    (ρDζ1,ζ2α+ϰ)(ϑ)=(ρIζ2(βζ1)α+(ϑρ1ddϑ)β ρI(1ζ2)(βζ1)α+ϰ)(ϑ)=(ρIζ2(βζ1)α+δβρ ρI(1ζ2)(βζ1)α+ϰ)(ϑ).

    In this paper we consider the case β=1 since 0<ζ1<1.

    Property 2.9. [22] The operator ρDζ1,ζ2α+ can be written as

    ρDζ1,ζ2α+= ρIζ2(1ζ1)α+δρ ρI1ξα+= ρIζ2(1ζ1)α+ ρDξα+,  ξ=ζ1+ζ2ζ1ζ2.

    Definition 2.10. We assume that the parameters ζ1,ζ2,ξ satisfy

    ξ=ζ1+ζ2ζ1ζ2,   0<ζ1,ζ2,ξ<1.

    Then, we can define the spaces

    Cζ1,ζ21ξ,ρ(Θ)={pC1ξ,ρ(Θ), ρDζ1,ζ2α+pC1ξ,ρ(Θ)},
    Cξ1ξ,ρ(Θ)={pC1ξ,ρ(Θ), ρDξα+pC1ξ,ρ(Θ)},
    PCζ1,ζ21ξ,ρ(Θ)={pPC1ξ,ρ(Θ), ρDζ1,ζ2α+pPC1ξ,ρ(Θ)},

    and

    PCξ1ξ,ρ(Θ)={pPC1ξ,ρ(Θ), ρDξα+pPC1ξ,ρ(Θ)}.

    Since ρDζ1,ζ2α+p= ρIξ(1ζ1)α+ ρDξα+p, by Lemma 2.4, we get

    Cξ1ξ,ρ(Θ)Cζ1,ζ21ξ,ρ(Θ)C1ξ,ρ(Θ),

    and

    PCξ1ξ,ρ(Θ)PCζ1,ζ21ξ,ρ(Θ)PC1ξ,ρ(Θ).

    Lemma 2.11. [22] Let 0<ζ1<1,0ζ21 and ξ=ζ1+ζ2ζ1ζ2. If pCξ1ξ,ρ(Θ), then

    ρIξα+ ρDξα+p=ρIζ1α+ ρDζ1,ζ2α+p

    and

    ρDξα+ ρIζ1α+p= ρDζ2(1ζ1)α+p.

    Lemma 2.12. [22] Let 0<ζ1<1,0ζ21 and ξ=ζ1+ζ2ζ1ζ2. If φC1ξ[α,ˉα] and ρI1ζ2(1ζ1)α+φC11ξ,ρ[α,ˉα], then ρDζ1,ζ2α+ ρIζ1α+φ exist on (α,ˉα] and

    ρDζ1,ζ2α+ ρIζ1α+φ=φ.

    Definition 2.13. [20] A two-parameter Mittag-Leffler function Eζ1,ζ2(p),ζ1,ζ2,pR with ζ1>0 and ζ2>0, is defined by

    Eζ1,ζ2(p)=ν=0pνΓ(ζ1ν+ζ2).

    If ζ2=1, we obtain:

    Eζ1(p)=ν=0pνΓ(ζ1ν+1).

    Lemma 2.14. [6] Let ζ1>0,p(ϑ),ϖ1(ϑ) be nonnegative functions and ϖ2(ϑ) be nonnegative and nondecreasing function for ϑ[ϑ0,μ),μ>0,ϖ2(ϑ)θ where θ is a constant. If

    p(ϑ)ϖ1(ϑ)+ϖ2(ϑ)ϑϑ0(ϑρϱρρ)ζ11ϱρ1p(ϱ)dϱ,   ϑ[ϑ0,μ).

    Then

    p(ϑ)ϖ1(ϑ)+ϑϑ0[β=1(ϖ2(ϑ)Γ(ζ1))βΓ(βζ1)(ϑρϱρρ)βζ11ϱρ1]ϖ1(ϱ)dϱ, ϑ[ϑ0,μ).

    Corollary 2.15. [6] Assume that the requirements of Lemma 2.14 are met, and that ϖ1(ϑ) is a nondecreasing function for ϑ[ϑ0,μ). Then

    p(ϑ)ϖ1(ϑ)Eζ1(ϖ2(ϑ)Γ(ζ1)(ϑρϑρ0ρ)ζ1),  ϑ[ϑ0,μ).

    Definition 2.16. The Eq (1.1) is Ulam-Hyers stable if there exists a real number cψ,ς>0 such that for each κ>0 and for each solution wPCξ1ξ,ρ(Θ) of the inequality

    {|ρDζ1,ζ2α+w(ϑ)ψ(ϑ,w(ϑ),ρDζ1,ζ2α+w(ϑ))|κ,ϑ(ϑν,ϑν+1],ν=1,,ς|ΔρI1ξα+w|ϑ=ϑνχν|κ,ν=1,,ς; (2.1)

    there exists a solution pPCξ1ξ,ρ(Θ) of Eq (1.1) with

    |w(ϑ)p(ϑ)|cψ,ςκ,ϑΘ.

    Definition 2.17. The Eq (1.1) is generalized Ulam-Hyers stable if there exists ϖψ,ςPC1ξ(R+,R+), ϖψ,ς(0)=0, such that for each solution wPCξ1ξ,ρ(Θ) of the inequality (2.1), there exists a solution pPCξ1ξ,ρ(Θ) of the Eq (1.1) with

    |w(ϑ)p(ϑ)|ϖψ,ς(κ),ϑΘ.

    Definition 2.18. The Eq (1.1) is Ulam-Hyers-Rassias stable with respect to (ω,ϖ) if there exists a real number cψ,ς,ω>0 such that for each κ>0 and for each solution wPCξ1ξ,ρ(Θ) of the inequality

    {|ρDζ1,ζ2α+w(ϑ)ψ(ϑ,w(ϑ),ρDζ1,ζ2α+w(ϑ))|κω(ϑ),ϑ(ϑν,ϑν+1],ν=1,,ς|ΔρI1ξα+w|ϑ=ϑνχν|κϖ,ν=1,...,ς; (2.2)

    there exists a solution pPCξ1ξ,ρ(Θ) of Eq (1.1) with

    |w(ϑ)p(ϑ)|cψ,ς,ωκ(ω(ϑ)+ϖ),ϑΘ.

    Definition 2.19. The Eq (1.1) is generalized Ulam-Hyers-Rassias stable with respect to (ω,ϖ) if there exists a real number cψ,ς,ω>0 such that for each solution wPCξ1ξ,ρ(Θ) of the inequality

    {|ρDζ1,ζ2α+w(ϑ)ψ(ϑ,w(ϑ),ρDζ1,ζ2α+w(ϑ))|ω(ϑ),ϑ(ϑν,ϑν+1],ν=1,,ς|ΔρI1ξα+w|ϑ=ϑνχν|ϖ,ν=1,,ς; (2.3)

    there exists a solution pPCξ1ξ,ρ(Θ) of Eq (1.1) with

    |w(ϑ)p(ϑ)|cψ,ς,ω(ω(ϑ)+ϖ(ϑ)),ϑΘ.

    Remark 2.20. A function wPCξ1ξ,ρ(Θ) is a solution of the inequality (2.2) if and only if there is ηPCξ1ξ,ρ(Θ) and a sequence ην,ν=1,,ς, where

    i) |η(ϑ)|κω(ϑ),ϑ(ϑν,ϑν+1],ν=1,,ς and |ην|κϖ,ν=1,,ς;

    ii) ρDζ1,ζ2α+w(ϑ)=ψ(ϑ,w(ϑ),ρDζ1,ζ2α+w(ϑ))+η(ϑ),ϑ(ϑν,ϑν+1],ν=1,,ς;

    iii) ΔρI1ξα+w|ϑ=ϑν=χν+ην,ν=1,,ς.

    Theorem 2.21. [30] Let DPC(Θ,R). D is relatively compact (i.e ¯D is compact) if:

    (1) D is uniformly bounded i.e, there exists ˉθ>0 where

    |ψ(p)|<ˉθforeveryψDandp(ϑν,ϑν+1],ν=1,,ς.

    (2) D is equicontinuous on (ϑν,ϑν+1] i.e, for every κ>0, there exists δ>0 such that for each p,¯p(ϑν,ϑν+1], |p¯p|δ implies |ψ(p)ψ(¯p)|κ, for every ψD.

    With appropriate modifications, the preceding theorem may be extended to the weighted space PC1ξ(Θ,R).

    Theorem 2.22. [26] (PC1ξ type Arzela-Ascoli theorem). Let DPC1ξ(Θ,R). D is relatively compact (i.e ¯D is compact) if:

    (1) D is uniformly bounded i.e, there exists ˉθ>0 such that

    |ψ(p)|<ˉθforeveryψDandp(ϑν,ϑν+1],ν=1,,ς.

    (2) D is equicontinuous on (ϑν,ϑν+1] i.e, for every κ>0, there exists δ>0 such that for each p,¯p(ϑν,ϑν+1], |p¯p|δ implies |ψ(p)ψ(¯p)|κ, for every ψD.

    In this section, we study the existence of solution for the problems (1.1)–(1.3). We employ the following lemma to establish our main results.

    Lemma 3.1. Let 0<ζ1<1, and 0ζ21,ξ=ζ1+ζ2ζ1ζ2 and γ:ΘR be continuous function. For any ˉαΘ, a function pCξ1ξ,ρ(Θ,R) is a solution of the equation:

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ˉα)(ρI1ξ+ζ1α+)γ(ϑ)|ϑ=ˉα}+(ρIζ1α+γ)(ϑ), (3.1)

    if and only if, p is solution of the Hilfer-Katugampola fractional differential equation

    ρDζ1,ζ2α+p(ϑ)=γ(ϑ);  ϑΘ. (3.2)

    Proof. () Let pCξ1ξ,ρ(Θ) satisfying (3.1). We demonstrate that p also verifies the fractional differential equation (3.2). Applying ρDζ1,ζ2α+ on both sides of the Eq (3.1), we get

    ρDζ1,ζ2α+p(ϑ)=1Γ(ξ){(ρI1ξα+)p(ˉα)(ρI1ξ+ζ1α+)γ(ϑ)|ϑ=ˉα} ρDζ1,ζ2α+(ϱραρρ)ξ1(ϑ)+ρDζ1,ζ2α+ ρIζ1α+γ(ϑ);  ϑΘ.

    From the Lemma 2.4 and by the definition of the space C11ξ,ρ(Θ), we get

    ρI1ζ2(1ζ1)α+γC11ξ,ρ(Θ).

    By Lemma 2.12 and Lemma 2.5 we obtain

    ρDζ1,ζ2α+p(ϑ)=γ(ϑ);  ϑΘ.

    () Let pCξ1ξ,ρ(Θ) by a solution of the fractional differential equation (3.2). We prove that p is also a solution of (3.1). Then

    1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ˉα)(ρI1ξ+ζ1α+)γ(ϑ)|ϑ=ˉα}+(ρIζ1α+γ)(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ˉα)(ρI1ξ+ζ1α+)ρDζ1,ζ2α+p(ϑ)|ϑ=ˉα}+ρIζ1α+ ρDζ1,ζ2α+p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ˉα)ρI1ξα+(ρIζ1α+ ρDζ1,ζ2α+)p(ϑ)|ϑ=ˉα}+ρIζ1α+ ρDζ1,ζ2α+p(ϑ). (3.3)

    By Lemma 2.7, with ζ1=ξ, we obtain

    (ρIξα+ ρDξα+p)(ϑ)=p(ϑ)1Γ(ξ)(ϑραρρ)ξ1 ρI1ξα+p(α),

    where ϑ(α,μ]. By hypothesis, pCξ1ξ,ρ(Θ), using Lemma 2.11, we have

    (ρIξα+ ρDξα+p)(ϑ)=(ρIζ1α+ ρDζ1,ζ2α+p)(ϑ)=p(ϑ)1Γ(ξ)(ϑραρρ)ξ1 ρI1ξα+p(α), (3.4)

    which implies that

    ρI1ξα+(ρIζ1α+ ρDζ1,ζ2α+p)(ϑ)=(ρI1ξα+p)(ϑ)1Γ(ξ) ρI1ξα+p(α)ρI1ξα+(ϱραρρ)ξ1(ϑ).

    By applying Lemma 2.5 we get

    ρI1ξα+(ρIζ1α+ ρDζ1,ζ2α+p)(ϑ)=(ρI1ξα+p)(ϑ)1Γ(ξ) ρI1ξα+p(α)Γ(ξ)=(ρI1ξα+p)(ϑ)ρI1ξα+p(α). (3.5)

    By replacing (3.4) and (3.5) in (3.3) we have

    1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+p)(ˉα)(ρI1ξ+ζ1α+γ)(ϑ)|ϑ=ˉα}+ρIζ1α+γ(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+p)(ˉα)ρI1ξα+p(ϑ)+ρI1ξα+p(α)|ϑ=ˉα}+p(ϑ)1Γ(ξ)(ϑραρρ)ξ1 ρI1ξα+p(α)=1Γ(ξ)(ϑραρρ)ξ1 ρI1ξα+p(α)+p(ϑ)1Γ(ξ)(ϑραρρ)ξ1 ρI1ξα+p(α)=p(ϑ),

    with ϑ(α,ˉα], that is p() satisfies (3.1). This completes the proof of the Lemma.

    Lemma 3.2. Let ξ=ζ1+ζ2ζ1ζ2, where 0<ζ1<1 and 0ζ21. Let Ψ:(α,μ]R is a continuous function. A function pPCξ1ξ,ρ(Θ) is a solution of the fractional integral equation

    p(ϑ)={˜αΓ(ξ)(ϑραρρ)ξ1+(ρIζ1α+Ψ)(ϑ),    ϑ(α,ϑ1]1Γ(ξ)(ϑραρρ)ξ1(˜α+νi=1χi)+(ρIζ1α+Ψ)(ϑ),   ϑ(ϑν,ϑν+1], (3.6)

    where ν=1,,ς, if and only if p is a solution of the problem:

    (ρDζ1,ζ2α+p)(ϑ)=Ψ(ϑ), foreach , ϑ(ϑν,ϑν+1], ν=0,,ς, (3.7)
    ΔρI1ξα+p|ϑ=ϑν=χνR, ν=1,,ς, (3.8)
    ρI1ξα+p(α)=˜αR. (3.9)

    Proof. Assume that pPCξ1ξ,ρ(Θ,R) satisfies the problems (3.7)(3.9).

    If ϑ(α,ϑ1] then

    {(ρDζ1,ζ2α+p)(ϑ)=Ψ(ϑ)ρI1ξα+p(α)=˜α. (3.10)

    Then the problem (3.10) is equivalent to the following fractional integral [22].

    p(ϑ)=˜αΓ(ξ)(ϑραρρ)ξ1+(ρIζ1α+Ψ)(ϑ),   ϑ(α,ϑ1]. (3.11)

    Now, if ϑ(ϑ1,ϑ2] then

    (ρDζ1,ζ2α+p)(ϑ)=Ψ(ϑ);  ϑ(ϑ1,ϑ2] with  ρI1ξα+p(ϑ+1)ρI1ξα+p(ϑ1)=χ1.

    By Lemma 3.1, we have

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ϑ+1)(ρI1ξ+ζ1α+Ψ)(ϑ)|ϑ=ϑ1}+(ρIζ1α+Ψ)(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{(ρI1ξα+)p(ϑ1)+χ1(ρI1ξ+ζ1α+Ψ)(ϑ)|ϑ=ϑ1}+(ρIζ1α+Ψ)(ϑ);  ϑ(ϑ1,ϑ2]. (3.12)

    Now, from (3.11), we have

    ρI1ξα+p(ϑ)=˜α+ρI1ξ+ζ1α+Ψ(ϑ),  ϑ(α,ϑ1].

    This gives

    ρI1ξα+p(ϑ1) ρI1ξ+ζ1α+Ψ(ϑ)|ϑ=ϑ1=˜α. (3.13)

    Using (3.13) in (3.12), we obtain

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+χ1)+ ρIζ1α+Ψ(ϑ),  ϑ(ϑ1,ϑ2]. (3.14)

    Next, if ϑ(ϑ2,ϑ3] then

    ρDζ1,ζ2α+p(ϑ)=Ψ(ϑ);  ϑ(ϑ2,ϑ3]  with ρI1ξα+p(ϑ+2)ρI1ξα+p(ϑ2)=χ2.

    Again, by Lemma 3.1, we have

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{ρI1ξα+p(ϑ+2)ρI1ξ+ζ1α+Ψ(ϑ)|ϑ=ϑ2}+ρIζ1α+Ψ(ϑ)=1Γ(ξ)(ϑραρρ)ξ1{ρI1ξα+p(ϑ2)+χ2ρI1ξ+ζ1α+Ψ(ϑ)|ϑ=ϑ2}+ρIζ1α+Ψ(ϑ); ϑ(ϑ2,ϑ3]. (3.15)

    From (3.14), we have

    ρI1ξα+p(ϑ)=(˜α+χ1)+ρI1ξ+ζ1α+Ψ(ϑ);  ϑ(ϑ1,ϑ2],

    which gives

    ρI1ξα+p(ϑ2)ρI1ξ+ζ1α+Ψ(ϑ)|ϑ=ϑ2=˜α+χ1. (3.16)

    Using (3.16) in (3.15), we get

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+χ1+χ2)+ ρIζ1α+Ψ(ϑ),  ϑ(ϑ2,ϑ3].

    Continuing the above process, we obtain

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+νi=1χi)+(ρIζ1α+Ψ)(ϑ),   ϑ(ϑν,ϑν+1],ν=1ς.

    Conversely, let pPCξ1ξ,ρ(Θ,R) satisfies the fractional integral equation (3.11). Then, for ϑ(α,ϑ1], we have

    p(ϑ)=˜αΓ(ξ)(ϑραρρ)ξ1+ρIζ1α+Ψ(ϑ).

    Applying ρDζ1,ζ2α+ on both sides, we get

    ρDζ1,ζ2α+p(ϑ)=˜αΓ(ξ)ρDζ1,ζ2α+(ϱραρρ)ξ1(ϑ)+ρDζ1,ζ2α+ ρIζ1α+Ψ(ϑ).

    From Lemma 2.4 and by the definition of the space C11λ,ρ(Θ), we can get

    ρI1ζ2(1ζ1)α+ΨC11λ,ρ(Θ). (3.17)

    Using Lemma 2.12 and Lemma 2.5, we obtain

    ρDζ1,ζ2α+p(ϑ)=Ψ(ϑ),ϑ(α,ϑ1].

    Now, for ϑ(ϑν;ϑν+1], we have

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+νi=1χi)+ ρIζ1α+Ψ(ϑ).

    Applying the operator ρDζ1,ζ2α+ on both sides, we get

    ρDζ1,ζ2α+p(ϑ)=1Γ(ξ)(˜α+νi=1χi) ρDζ1,ζ2α+(ϱραρρ)ξ1(ϑ)+ ρDζ1,ζ2α+ ρIζ1α+Ψ(ϑ).

    From (3.17) and using Lemma 2.12 and Lemma 2.5, we obtain

    ρDζ1,ζ2α+p(ϑ)=Ψ(ϑ).

    Thus, p satisfies (3.7). Next, we demonstrate that p also verify (3.8) and (3.9). Applying the operator ρI1ξα+ on both sides of Eq (3.11), we get

    ρI1ξα+p(ϑ)=˜αΓ(ξ)ρI1ξα+(ϱραρρ)ξ1(ϑ)+ ρI1ξ+ζ1α+Ψ(ϑ).

    By Lemma 2.5 we get

    ρI1ξα+p(ϑ)=˜α+ ρI1ξ+ζ1α+Ψ(ϑ),

    which implies that

    ρI1ξα+p(ϑ) ρI1ξ+ζ1α+Ψ(ϑ)=˜α,ρI1ξα+(p(ϑ) ρIζ1α+Ψ(ϑ))=˜α.

    Since p satisfies (3.7) we have

    ρI1ξα+(p(ϑ) ρIζ1α+ ρDζ1,ζ2α+p(ϑ))=˜α.

    Using Lemma 2.11 we have

    ρI1ξα+(p(ϑ) ρIξα+ ρDξα+p(ϑ))=˜α.

    From the definition of Cξ1ξ,ρ(Θ), Lemma 2.4 and using Definition 2.2, we have

    ρI1ξα+pC(Θ)  and  ρDξα+p=δρ ρI1ξα+pC1ξ,ρ(Θ).

    Thus

    ρI1ξα+pC11ξ,ρ(Θ).

    By Lemma 2.7, with ζ1=ξ we can write

    ρI1ξα+(p(ϑ)p(ϑ)+ρI1ξα+p(α)Γ(ξ)(ϑραρρ)ξ1)=˜α,ρI1ξα+p(α)Γ(ξ) ρI1ξα+(ϱραρρ)ξ1(ϑ)=˜α.

    By Lemma 2.5 we have

    ρI1ξα+p(α)=˜α,

    which is the condition (3.9).

    Further, from Eq (3.6), for ϑ(ϑν,ϑν+1], we have

    ρI1ξα+p(ϑ)=1Γ(ξ)(˜α+νi=1χi) ρI1ξα+(ϱραρρ)ξ1(ϑ)+ ρI1ξ+ζ1α+Ψ(ϑ)=˜α+νi=1χi+ ρI1ξ+ζ1α+Ψ(ϑ), (3.18)

    and for ϑ(ϑν1,ϑν], we have

    ρI1ξα+p(ϑ)=1Γ(ξ)(˜α+ν1i=1χi) ρI1ξα+(ϱραρρ)ξ1(ϑ)+ ρI1ξ+ζ1α+Ψ(ϑ)=˜α+ν1i=1χi+ ρI1ξ+ζ1α+Ψ(ϑ). (3.19)

    Therefore, from (3.18) and (3.19), we obtain

    ρI1ξα+p(ϑ+ν) ρI1ξα+p(ϑν)=νi=1χiν1i=1χi=χν,

    which condition (3.8). We have proved that p satisfies the problems (3.7)(3.9).

    As a consequence of Lemma 3.2, we have the following lemma.

    Lemma 3.3. Let ξ=ζ1+ζ2ζ1ζ2 where 0<ζ1<1 and 0ζ21, let ψ:(α,μ]×R×RR be a continuous function where ψ(,p(),x())PC1ξ,ρ(Θ) for any p,xPC1ξ,ρ(Θ). If pPCξ1ξ,ρ(Θ), then p verifies (1.1)(1.3) if and only if p is the fixed point of the operator S:PC1ξ,ρ(Θ)PC1ξ,ρ(Θ) given by

    Sp(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+α<ϑν<ϑχν)+(ρIζ1α+ϰ)(ϑ), (3.20)

    where ϰ:(0,μ]R be a function verifying the functional equation

    ϰ(ϑ)=ψ(ϑ,p(ϑ),ϰ(ϑ)).

    It is obvious that ϰPC1ξ,ρ(Θ). Also, by Lemma 2.4, SpPC1ξ,ρ(Θ).

    Assume that the function ψ:(α,μ]×R×RR is continuous and verifies the following:

    (H1) ψ(,x(),y())PCζ2(1ζ1)1ξ,ρ for any x,yPC1ξ,ρ(Θ).

    (H2) There exist constants θ1>0 and 0<θ2<1 such that

    |ψ(ϑ,x,y)ψ(ϑ,ˉx,ˉy)|θ1|xˉx|+θ2|yˉy|

    for any x,y,ˉx,ˉyPC1ξ,ρ(Θ) and ϑ(α,μ].

    We can now declare and demonstrate our existence result for problems (1.1)–(1.3) based on Banach's fixed point [15].

    Theorem 3.4. If (H1) and (H2) are met, and

    B:=θ1Γ(ξ)Γ(ζ1+ξ)(1θ2)(μραρρ)ζ1<1, (3.21)

    then the problems (1.1)–(1.3) has unique solution in PCξ1ξ,ρ(Θ)PCζ1,ζ21ξ,ρ(Θ).

    Proof. The proof will be presented in two segments.

    Step 1: We demonstrate that S defined in (3.20) has a unique fixed point p in PC1ξ,ρ(Θ). Let p,xPC1ξ,ρ(Θ) and ϑ(α,μ], then, we have

    |Sp(ϑ)Sx(ϑ)|=|ρIζ1α+ϰ(ϑ)ρIζ1α+γ(ϑ)|1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|ψ(ϱ,p(ϱ),ϰ(ϱ))ψ(ϱ,x(ϱ),γ(ϱ))|dϱ,

    where ϰ,γPC1ξ,ρ(Θ) such that

    ϰ(ϑ)=ψ(ϑ,p(ϑ),ϰ(ϑ)),γ(ϑ)=ψ(ϑ,x(ϑ),γ(ϑ)).

    By (H2), we have

    |ϰ(ϑ)γ(ϑ)|=|ψ(ϑ,p(ϑ),ϰ(ϑ))ψ(ϑ,x(ϑ),γ(ϑ))|θ1|p(ϑ)x(ϑ)|+θ2|ϰ(ϑ)γ(ϑ)|.

    Then,

    |ϰ(ϑ)γ(ϑ)|θ11θ2|p(ϑ)x(ϑ)|.

    Therefore, for each ϑ(α,μ]

    |Sp(ϑ)Sx(ϑ)|θ1(1θ2)Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|p(ϱ)x(ϱ)|dϱθ1(1θ2)(Iζ1α+(ϱραρρ)ξ1)(ϑ)pxPC1ξ,ρ.

    By Lemma 2.5, we have

    |Sp(ϑ)Sx(ϑ)|θ1Γ(ξ)Γ(ζ1+ξ)(1θ2)(ϑραρρ)ζ1+ξ1pxPC1ξ,ρ,

    hence

    |(ϑραρρ)1ξ(Sp(ϑ)Sx(ϑ))|θ1Γ(ξ)Γ(ζ1+ξ)(1θ2)(ϑραρρ)ζ1pxPC1ξ,ρθ1Γ(ξ)Γ(ζ1+ξ)(1θ2)(μραρρ)ζ1pxPC1ξ,ρ,

    which implies that

    SpSxPC1ξ,ρθ1Γ(ξ)Γ(ζ1+ξ)(1θ2)(μραρρ)ζ1pxPC1ξ,ρ.

    By (3.21), the operator S is a contraction. Hence, by Banach's contraction principle, S has a unique fixed point pPC1ξ,ρ(Θ).

    Step 2: We show that such a fixed point pPC1ξ,ρ(Θ) is actually in PCξ1ξ,ρ(Θ).

    Since p is the unique fixed point of operator S in PC1ξ,ρ(Θ), then, for each ϑ(α,μ], we have

    p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+α<ϑν<ϑχν)+(ρIζ1α+ψ(ϱ,p(ϱ),ϰ(ϱ)))(ϑ).

    Applying ρDξα+ to both sides and by Lemma 2.5, and Lemma 2.11, we have

    ρDξα+p(ϑ)=(ρDξα+ ρIζ1α+ψ(ϱ,p(ϱ),ϰ(ϱ)))(ϑ)=(ρDζ2(1ζ1)α+ψ(ϱ,p(ϱ),ϰ(ϱ)))(ϑ).

    Since ξζ1, by (H1), the right hand side is in PC1ξ,ρ(Θ) and thus ρDξα+pPC1ξ,ρ(Θ) which implies that pPCξ1ξ,ρ(Θ). As a consequence of Steps 1 and 2 together with Lemma 3.3, we can conclude that the problems (1.1)(1.3) has a unique solution in PCξ1ξ,ρ(Θ).

    Our second result is based on Krasnoselskii fixed point theorem [15].

    Theorem 3.5. Assume (H1) and,

    (H3) There exist constants 0<θ1<(1θ2)Γ(ζ1+ξ)2Γ(ξ)(μραρρ)ζ1 and 0<θ2<1 such that

    |ψ(ϑ,x,y)ψ(ϑ,ˉx,ˉy)|θ1|xˉx|+θ2|yˉy|

    for any x,y,ˉx,ˉyPC1ξ,ρ(Θ) and ϑ(α,μ].

    Then the problems (1.1)(1.3) has at least one solution.

    Proof. Consider the set

    Bε={pPC1ξ,ρ(Θ):||p||PC1ξ,ρε},

    where

    ε2Γ(ξ)(|˜α|+ςν=1|χν|)+2Γ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1,

    where ψ=supϑΘ|ψ(ϑ,0,0)|.

    We define the operators S1 and S2 on Bε by

    S1p(ϑ)=1Γ(ξ)(ϑραρρ)ξ1(˜α+α<ϑν<ϑχν),  ϑΘ, (3.22)
    S2p(ϑ)=1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1ϰ(ϱ)dϱ,  ϑΘ. (3.23)

    Then (3.20) can be written as

    Sp(ϑ)=S1p(ϑ)+S2p(ϑ),  pPC1ξ,ρ(Θ).

    Step 1: We demonstrate that S1p+S2xBε for any p,xBε. For operator S1, multiplying both sides of (3.22) by (ϑραρρ)1ξ, we have

    (ϑραρρ)1ξS1p(ϑ)=1Γ(ξ)(˜α+α<ϑν<ϑχν),

    then

    |(ϑραρρ)1ξS1p(ϑ)|1Γ(ξ)(|˜α|+ςν=1|χν|).

    This gives

    ||S1p||PC1ξ,ρ1Γ(ξ)(|˜α|+ςν=1|χν|). (3.24)

    By (H3), we have for each ϑ(α,μ],

    |ϰ(ϑ)|=|ψ(ϑ,x(ϑ),ϰ(ϑ))ψ(ϑ,0,0)+ψ(ϑ,0,0)||ψ(ϑ,x(ϑ),ϰ(ϑ))ψ(ϑ,0,0)|+|ψ(ϑ,0,0)|θ1|x(ϑ)|+θ2|ϰ(ϑ)|+ψ.

    Multiplying both sides of the above inequality by (ϑραρρ)1ξ, we get

    |(ϑραρρ)1ξϰ(ϑ)|(ϑραρρ)1ξψ+θ1|(ϑραρρ)1ξx(ϑ)|+θ2|(ϑραρρ)1ξϰ(ϑ)|(μραρρ)1ξψ+θ1ε+θ2|(ϑραρρ)1ξϰ(ϑ)|.

    Then, for each ϑ(α,μ], we have

    |(ϑραρρ)1ξϰ(ϑ)|(μραρρ)1ξψ+θ1ε1θ2:=θ3. (3.25)

    Thus, (3.23) and Lemma 2.5, implies

    |S2(x)(ϑ)|[Γ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)](ϑραρρ)ζ1+ξ1.

    Therefore

    |(ϑραρρ)1ξS2x(ϑ)|[Γ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)](ϑραρρ)ζ1Γ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)(μραρρ)ζ1.

    Thus

    S2xPC1ξ,ρΓ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)(μραρρ)ζ1. (3.26)

    Linking (3.24) and (3.26) for every p,xBε we obtain

    S1p+S2xPC1ξ,ρS1pPC1ξ,ρ+S2xPC1ξ,ρΓ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)(μραρρ)ζ1+1Γ(ξ)(|˜α|+ςν=1|χν|).

    Since

    ε2Γ(ξ)(|˜α|+ςν=1|χν|)+2Γ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1,

    and

    θ1<(1θ2)Γ(ζ1+ξ)2Γ(ξ)(μραρρ)ζ1,

    we have

    S1p+S2xPC1ξ,ρε.

    which infers that S1p+S2xBε.

    Step 2: Clearly S1 is a contraction.

    Step 3: S2 is compact and continuous.

    The continuity of S2 follows from the continuity of ψ. Next we prove that S2 is uniformly bounded on Bε. Let any xBε. Then by (3.26) we have

    S2xPC1ξ,ρΓ(ξ)ψ(1θ2)Γ(ζ1+ξ)(μραρρ)1ξ+ζ1+θ1Γ(ξ)ε(1θ2)Γ(ζ1+ξ)(μραρρ)ζ1.

    This means that S2 is uniformly bounded on Bε. Next, we show that S2Bε is equicontinuous. Let any xBε and 0<α<τ1<τ2μ. Then

    |S2x(τ1)S2x(τ2)|=1Γ(ζ1)|τ1α(τρ1ϱρρ)ζ11ϱρ1ϰ(ϱ)dϱτ2α(τρ2ϱρρ)ζ11ϱρ1ϰ(ϱ)dϱ|1Γ(ζ1)|τ1α(τρ1ϱρρ)ζ11ϱρ1(ϱραρρ)ξ1(ϱραρρ)1ξϰ(ϱ)dϱτ2α(τρ2ϱρρ)ζ11ϱρ1(ϱραρρ)ξ1(ϱραρρ)1ξϰ(ϱ)dϱ|,

    by using (3.25) we have

    |S2x(τ1)S2x(τ2)|θ3Γ(ζ1)|τ1α(τρ1ϱρρ)ζ11ϱρ1(ϱραρρ)ξ1dϱτ2α(τρ2ϱρρ)ζ11ϱρ1(ϱραρρ)ξ1dϱ|θ3Γ(ξ)Γ(ζ1+ξ)|(τρ1αρρ)ζ1+ξ1(τρ2αρρ)ζ1+ξ1|.

    Note that

    |S2x(τ1)S2x(τ2)|0  as  τ2τ1.

    This shows that S2 is equicontinuous on Θ. Therefore S2 is relatively compact on Bε. By PC1ξ; type Arzela-Ascoli theorem S2 is compact on Bε.

    By Krasnoselskii's fixed point theorem, S has at least a fixed point pC1ξ,ρ(Θ) and by the same way of the proof of Theorem 3.4, pCξ1ξ,ρ(Θ). Using Lemma 3.3, we conclude that the problems (1.1)(1.3) has at least one solution in the space Cξ1ξ,ρ(Θ).

    In what follows, we give the following result on Ulam-Hyers-Rassias stability.

    Theorem 4.1. Assume that (H1), (H2), (3.21) hold and,

    (H4) There exists a nondecreasing function ωPCξ1(Θ) and there exists λω>0 such that for any ϑ(α,μ]:

    ρIζ1α+(ω(ϱ))(ϑ)λωω(ϑ).

    Then, the Eq (1.1) is Ulam-Hyers-Rassias stable with respect to (ω,ϖ).

    Proof. Let wPC1ξ,ρ(Θ) be a solution of the inequality (2.2). Denote by p the unique solution of the problem:

    (ρDζ1,ζ2α+p)(ϑ)=ψ(ϑ,p(ϑ),(ρDζ1,ζ2α+p)(ϑ)), foreach , ϑ(ϑν,ϑν+1], ν=0,,ς,
    ΔρI1ξα+p|ϑ=ϑν=χνR, ν=1,,ς,
    (ρI1ξα+w)(α)=(ρI1ξα+p)(α)=˜αR.

    Using Lemma 3.3, we obtain

    p(ϑ)={˜αΓ(ξ)(ϑραρρ)ξ1+(ρIζ1α+ϰ(ϱ))(ϑ),    ϑ(α,ϑ1]1Γ(ξ)(ϑραρρ)ξ1(˜α+νi=1χi)+(ρIζ1α+ϰ(ϱ))(ϑ),   ϑ(ϑν,ϑν+1],

    where ϰ:(0,μ]R be a function satisfying the functional equation

    ϰ(ϑ)=ψ(ϑ,p(ϑ),ϰ(ϑ)).

    Since w solution of the inequality (2.2) and by Remark 2.20, we have

    (ρDζ1,ζ2α+w)(ϑ)=ψ(ϑ,w(ϑ),(ρDζ1,ζ2α+w)(ϑ))+η(ϑ), foreach, ϑ(ϑν,ϑν+1], (4.1)
    ΔρI1ξα+w|ϑ=ϑν=χν+ην,ν=1,,ς. (4.2)

    Clearly, the solution of the problems (4.1) and (4.2) is given by

    w(ϑ)={˜αΓ(ξ)(ϑραρρ)ξ1+(ρIζ1α+(γ(ϱ)+η(ϱ)))(ϑ),    ϑ(α,ϑ1]1Γ(ξ)(ϑραρρ)ξ1(˜α+νi=1(χi+ηi))+(ρIζ1α+(γ(ϱ)+η(ϱ)))(ϑ),   ϑ(ϑν,ϑν+1],

    where γ:(0,μ]R be a function satisfying the functional equation

    γ(ϑ)=ψ(ϑ,w(ϑ),γ(ϑ)).

    If ϑ(α,ϑ1], it follows that

    |w(ϑ)p(ϑ)|1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|η(ϱ)|dϱ+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱκΓ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1ω(ϱ)dϱ+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱκλωω(ϑ)+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱ,

    where ϰ,γC1ξ,ρ(Θ) such that

    ϰ(ϑ)=ψ(ϑ,p(ϑ),ϰ(ϑ)),γ(ϑ)=ψ(ϑ,w(ϑ),γ(ϑ)).

    By (H2), we have

    |γ(ϑ)ϰ(ϑ)|=|ψ(ϑ,w(ϑ),γ(ϑ))ψ(ϑ,p(ϑ),ϰ(ϑ))|θ1|w(ϑ)p(ϑ)|+θ2|γ(ϑ)ϰ(ϑ)|.

    Then,

    |γ(ϑ)ϰ(ϑ)|θ11θ2|w(ϑ)p(ϑ)|.

    Therefore, for each ϑ(α,ϑ1]

    |w(ϑ)p(ϑ)|κλωω(ϑ)+θ1(1θ2)Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|w(ϱ)p(ϱ)|dϱ.

    Applying Corollary 2.15, we get

    |w(ϑ)p(ϑ)|κλωω(ϑ)Eζ1(θ11θ2(ϑραρρ)ζ1)κλωω(ϑ)Eζ1(θ11θ2(μραρρ)ζ1)κλω(ψ+ω(ϑ))Eζ1(θ11θ2(μραρρ)ζ1)=κc1(ψ+ω(ϑ)),

    where

    c1=λωEζ1(θ11θ2(μραρρ)ζ1).

    If ϑ(ϑν,ϑν+1], ν=1,,ς, then we have

    |w(ϑ)p(ϑ)|1Γ(ξ)(ϑραρρ)ξ1ςν=1|ην|+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|η(ϱ)|dϱ+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱςκψΓ(ξ)(ϑρ1αρρ)ξ1+κΓ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1ω(ϱ)dϱ+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱςκψΓ(ξ)(ϑρ1αρρ)ξ1+κλωω(ϑ)+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱ,

    where ϰ,γC1ξ,ρ(Θ) such that

    ϰ(ϑ)=ψ(ϑ,p(ϑ),ϰ(ϑ)),γ(ϑ)=ψ(ϑ,w(ϑ),γ(ϑ)).

    By (H2), we have

    |γ(ϑ)ϰ(ϑ)|=|ψ(ϑ,w(ϑ),γ(ϑ))ψ(ϑ,p(ϑ),ϰ(ϑ))|θ1|w(ϑ)p(ϑ)|+θ2|γ(ϑ)ϰ(ϑ)|.

    Then,

    |γ(ϑ)ϰ(ϑ)|θ11θ2|w(ϑ)p(ϑ)|.

    Therefore, for each ϑ(ϑν,ϑν+1], ν=1,,ς,

    |w(ϑ)p(ϑ)|ςκϖΓ(ξ)(ϑρ1αρρ)ξ1+κλωω(ϑ)+θ1(1θ2)Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|w(ϱ)p(ϱ)|dϱ.

    Applying Corollary 2.15, we get

    |w(ϑ)p(ϑ)|[ςκψΓ(ξ)(ϑρ1αρρ)ξ1+κλωω(ϑ)]Eζ1(θ11θ2(ϑραρρ)ζ1)[ςκψΓ(ξ)(ϑρ1αρρ)ξ1+κλωω(ϑ)]Eζ1(θ11θ2(μραρρ)ζ1)κ[ςΓ(ξ)(ϑρ1αρρ)ξ1+λω](ω(ϑ)+ψ)Eζ1(θ11θ2(μραρρ)ζ1)=κc2(ω(ϑ)+ψ),

    where

    c2=[ςΓ(ξ)(ϑρ1αρρ)ξ1+λω]Eζ1(θ11θ2(μραρρ)ζ1).

    Thus, the Eq (1.1) is Ulam-Hyers-Rassias stable with respect to (ω,ϖ). The proof is complete.

    The following theorem gives Ulam-Hyers stable result.

    Theorem 4.2. Assume that (H1), (H2) and (3.21) hold. Then, the Eq (1.1) is Ulam-Hyers stable.

    Proof. Let wC1ξ,ρ(Θ) be a solution of the inequality (2.1). Denote by p the unique solution of the problem:

    (ρDζ1,ζ2α+p)(ϑ)=ψ(ϑ,p(ϑ),(ρDζ1,ζ2α+p)(ϑ)), foreach , ϑ(ϑν,ϑν+1], ν=0,,ς,
    ΔρI1ξα+p|ϑ=ϑν=χνR, ν=1,,ς,
    (ρI1ξα+w)(α)=(ρI1ξα+p)(α)=˜αR.

    By the same way of the proof of Theorem 4.1, we can easily show that:

    If ϑ(α,ϑ1], it follows that

    |w(ϑ)p(ϑ)|1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|η(ϱ)|dϱ+1Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|γ(ϱ)ϰ(ϱ)|dϱκΓ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1dϱ+θ1(1θ2)Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|w(ϱ)p(ϱ)|dϱκΓ(ζ1+1)(μραρρ)ζ1+θ1(1θ2)Γ(ζ1)ϑα(ϑρϱρρ)ζ11ϱρ1|w(ϱ)p(ϱ)|dϱ.

    Applying Lemma 2.14, we get

    |w(ϑ)p(ϑ)|κΓ(ζ1+1)(μραρρ)ζ1[1+ϑαβ=1(θ11θ2)βΓ(βζ1)(ϑρϱρρ)βζ11ϱρ1dϱ]κΓ(ζ1+1)(μραρρ)ζ1[1+β=1(θ11θ2)βΓ(βζ1+1)(μραρρ)βζ1]=κΓ(ζ1+1)(μραρρ)ζ1[1+β=11Γ(βζ1+1)[(θ11θ2)(μραρρ)ζ1]β]=κΓ(ζ1+1)(μραρρ)ζ1[1+Eζ1((θ11θ2)(μραρρ)ζ1)]=b1κ.

    If {\vartheta} \in ({\vartheta}_{\nu},{\vartheta}_{{\nu}+1}],\ {\nu}=1, \ldots,{\varsigma}, then we have

    \begin{eqnarray*} |{\mathfrak{w}}({\vartheta})-{\mathfrak{p}}({\vartheta})|&\leq& \frac {1}{\Gamma({\xi})}\left( \frac {{\vartheta}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}\sum\limits_{{\nu}=1}^{{\varsigma}}|\eta_{{\nu}}|+ \frac {1}{\Gamma({\zeta_1})}{ \int}_{{\alpha}}^{{\vartheta}}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\zeta_1}-1}{\varrho}^{\rho-1}|\eta({\varrho})|d{\varrho}\\ &+& \frac {1}{\Gamma({\zeta_1})}{ \int}_{{\alpha}}^{{\vartheta}}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\zeta_1}-1}{\varrho}^{\rho-1}|{\gamma}({\varrho})-{\varkappa}({\varrho})|d{\varrho}\\ &\leq& \frac {{\varsigma}\kappa}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {\kappa}{\Gamma({\zeta_1})}{ \int}_{{\alpha}}^{{\vartheta}}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\zeta_1}-1}{\varrho}^{\rho-1}d{\varrho}\\ &+& \frac {1}{\Gamma({\zeta_1})}{ \int}_{{\alpha}}^{{\vartheta}}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\zeta_1}-1}{\varrho}^{\rho-1}|{\gamma}({\varrho})-{\varkappa}({\varrho})|d{\varrho}\\ &\leq& \frac {{\varsigma}\kappa}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {\kappa}{\Gamma({\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\\ &+& \frac {{\theta_1}}{(1-{\theta_2})\Gamma({\zeta_1})}{ \int}_{{\alpha}}^{{\vartheta}}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\zeta_1}-1}{\varrho}^{\rho-1}|{\mathfrak{w}}({\varrho})-{\mathfrak{p}}({\varrho})|d{\varrho}. \end{eqnarray*}

    Applying Lemma 2.14, we get

    \begin{eqnarray*} |{\mathfrak{w}}({\vartheta})-{\mathfrak{p}}({\vartheta})|&\leq&\kappa\left[ \frac {{\varsigma}}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {1}{\Gamma({\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right]\\ &&\left[1+{ \int}_{{\alpha}}^{{\vartheta}}\sum\limits_{{\beta}=1}^{\infty} \frac {\left( \frac {{\theta_1}}{1-{\theta_2}}\right)^{{\beta}}}{\Gamma({\beta}{\zeta_1})}\left( \frac {{\vartheta}^{\rho}-{\varrho}^{\rho}}{\rho}\right)^{{\beta}{\zeta_1}-1}{\varrho}^{\rho-1}d{\varrho}\right]\\ &\leq&\kappa\left[ \frac {{\varsigma}}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {1}{\Gamma({\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right]\\ &&\left[1+\sum\limits_{{\beta}=1}^{\infty} \frac {\left( \frac {{\theta_1}}{1-{\theta_2}}\right)^{{\beta}}}{\Gamma({\beta}{\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\beta}{\zeta_1}}\right]\\ &=&\kappa\left[ \frac {{\varsigma}}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {1}{\Gamma({\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right]\\ &&\left[1+\sum\limits_{{\beta}=1}^{\infty} \frac {1}{\Gamma({\beta}{\zeta_1}+1)}\left[\left( \frac {{\theta_1}}{1-{\theta_2}}\right)\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right]^{{\beta}}\right]\\ &=&\kappa\left[ \frac {{\varsigma}}{\Gamma({\xi})}\left( \frac {{\vartheta}_{1}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\xi}-1}+ \frac {1}{\Gamma({\zeta_1}+1)}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right]\\ &&\left[1+E_{{\zeta_1}}\left(\left( \frac {{\theta_1}}{1-{\theta_2}}\right)\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\right)\right]= b_{2}\kappa, \end{eqnarray*}

    which completes the proof of the theorem. Moreover, if we set {\varpi}_{{\psi},{\varsigma}}(\kappa)=(b_{1}+b_{2})\kappa; {\varpi}_{{\psi},{\varsigma}}(0)=0 , then, the Eq (1.1) is generalized Ulam-Hyers stable.

    Consider the following initial value problem with impulse

    \begin{equation} ^{\frac{1}{2}}D^{\frac{1}{2},0}_{1^{+}}{\mathfrak{p}}({\vartheta})=\frac{2+|{\mathfrak{p}}({\vartheta})|+\left|^{\frac{1}{2}}D^{\frac{1}{2},0}_{0^{+}}{\mathfrak{p}}({\vartheta})\right|}{108e^{-{\vartheta}+3}\left(1+|{\mathfrak{p}}({\vartheta})|+\left|^{\frac{1}{2}}D^{\frac{1}{2},0}_{0^{+}}{\mathfrak{p}}({\vartheta})\right|\right)}+\frac{\ln(\sqrt{{\vartheta}}+1)}{3\sqrt{\sqrt{{\vartheta}}-1}} , \;{\vartheta}\in J_{0} \cup J_{1}, \end{equation} (5.1)
    \begin{equation} \Delta ^{\rho}I_{{\alpha}^{+}}^{1-{\xi}}{\mathfrak{p}}\left( \frac {3}{2}\right) = \eta \in {{\mathbb R}} , \end{equation} (5.2)
    \begin{equation} ^{\frac{1}{2}}I_{1^{+}}^{1-{\xi}}{\mathfrak{p}}(1)=0, \end{equation} (5.3)

    where J_{0}=\left(1, \frac{3}{2}\right], J_{1}=\left(\frac{3}{2},2\right].

    Set

    {\psi}({\vartheta},{\mathfrak{x}},{\mathfrak{y}})=\frac{2+|{\mathfrak{x}}|+|{\mathfrak{y}}|}{108e^{-{\vartheta}+3}(1+|{\mathfrak{x}}|+|{\mathfrak{y}}|)}+\frac{\ln(\sqrt{{\vartheta}}+1)}{3\sqrt{{\vartheta}}} , {\vartheta}\in (1,2] , {\mathfrak{x}},{\mathfrak{y}}\in {{\mathbb R}}.

    We have

    PC^{{\zeta_2}(1-{\zeta_1})}_{1-{\xi},\rho}([1,2])=PC^{0}_{\frac{1}{2},\frac{1}{2}}([1,2])=PC_{\frac{1}{2},\frac{1}{2}}([1,2]),

    with {\xi}={\zeta_1}=\rho=\frac{1}{2} and {\zeta_2}=0. Clearly, the function {\psi} \in PC_{\frac{1}{2},\frac{1}{2}}([1,2]).

    Hence condition (H1) is satisfied.

    For each {\mathfrak{x}},\bar{{\mathfrak{x}}},{\mathfrak{y}},\bar{{\mathfrak{y}}}\in{{\mathbb R}} and {\vartheta}\in(1,2] :

    \begin{eqnarray*} |{\psi}({\vartheta},{\mathfrak{x}},{\mathfrak{y}})-{\psi}({\vartheta},\bar{{\mathfrak{x}}},\bar{{\mathfrak{y}}})|&\leq &\frac{1}{108e^{-{\vartheta}+3}}(|{\mathfrak{x}}-\bar{{\mathfrak{x}}}|+|{\mathfrak{y}}-\bar{{\mathfrak{y}}}|)\\ &\leq & \frac {1}{108e}\left(|{\mathfrak{x}}-\bar{{\mathfrak{x}}}|+|{\mathfrak{y}}-\bar{{\mathfrak{y}}}|\right). \end{eqnarray*}

    Hence condition (H{2}) is satisfied with {\theta_1}={\theta_2}= \frac {1}{108e}.

    The condition

    \frac {{\theta_1}\Gamma({\xi})}{\Gamma({\zeta_1}+{\xi})(1-{\theta_2})}\left( \frac {{\mu}^{\rho}-{\alpha}^{\rho}}{\rho}\right)^{{\zeta_1}}\approx 0.0055 < 1,

    is satisfied with with {\alpha}=1 and {\mu}=2. It follows from Theorem 3.4 that the problems (5.1)–(5.3) has a unique solution in the space PC^{\frac{1}{2}}_{\frac{1}{2},\frac{1}{2}}([1,2]). Moreover, Theorem 4.2, implies that the Eq (1.1) is Ulam-Hyers stable.

    We have investigated the existence, uniqueness and stability of solutions for a class of nonlinear impulsive Hilfer-Katugampola problems. Our reasoning is founded on the Banach contraction principle and Krasnoselskii’s fixed point theorem. In addition, an example is provided to demonstrate the effectiveness of the main results. We plan to consider for for a futur study the same problem in infinite dimensional Banach space and make us of Darbo and Monch's fixed point theorems associated with the notion of measure of noncompactness.

    The research of J. J. Nieto was partially supported by the AEI of Spain under Grant MTM2016-75140-P and co-financed by European Community fund FEDER. The work of Y. Zhou was supported by the National Natural Science Foundation of China (12071396).

    The authors declare that there is no conflict of interest.



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