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Research article

On the generalized Gronwall inequalities involving ψ-fractional integral operator with applications

  • Received: 28 July 2022 Revised: 03 September 2022 Accepted: 09 September 2022 Published: 19 September 2022
  • MSC : 34A08

  • In this paper, a Gronwall inequality involving ψ-fractional integral operator is obtained as a generalization of [23]. An example is listed to show the applications.

    Citation: Qi Wang, Shumin Zhu. On the generalized Gronwall inequalities involving ψ-fractional integral operator with applications[J]. AIMS Mathematics, 2022, 7(11): 20370-20380. doi: 10.3934/math.20221115

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  • In this paper, a Gronwall inequality involving ψ-fractional integral operator is obtained as a generalization of [23]. An example is listed to show the applications.



    The fractional Gronwall inequalities are effective tools to study the qualitative and quantitative properties of solution for fractional differential and integral equations [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21] by giving the explicit bounds of solutions. Further detail on fractional Gronwall inequalities mainly involving the Riemann-Liouville fractional integrals [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16], the Caputo fractional integrals [17], the Hadamard fractional integrals [18], the Katugampola fractional integrals [19,20], and the generalized proportional fractional integrals [21].

    In [22], the authors produced the ψ-Hilfer fractional derivative as the Riemann-Liouville fractional derivative and the Caputo fractional derivative. In [23], considering the continuous dependence of the solution on the order and the initial condition of ψ-fractional differential equations, the authors presented the following theorem involving the ψ-fractional integral operator.

    Theorem 1.1. [23] Let u,v be two integrable functions and g continuous with domain [a,b]. Let ψC1([a,b]) be an increasing function such that ψ(t)0,t[a,b]. Assume that (1) u and v are nonnegative; (2) g is nonnegative and nondecreasing. If

    u(t)v(t)+g(t)taψ(s)(ψ(t)ψ(s))α1u(s)ds.

    Then

    u(t)v(t)+tak=1(g(t)Γ(α))kΓ(αk)ψ(s)(ψ(t)ψ(s))kα1v(s)ds, t[a,b].

    More inequalities related to ψ-fractional integral operator, see [24,25,26] for details.

    As the generalizations of the classical fractional calculus operators, the ψ-fractional operator (i.e. the fractional derivative and integral of a function f with respect to another function ψ) has wide applications and properties [27,28,29,30,31,32,33,34] according to the choice of the ψ-function, which makes the Riemann-Liouville, Hadamard, Katugampola, etc fractional integral operators and the properties of above operators can be unified and considered as a whole.

    Motivated by [23], in order to release the limitation of the number of nonlinear terms, new generalized forms of Theorem 1.1 are presented in this article, which is effective in dealing with neutral fractional differential equations involving ψ-fractional integral operator.

    The organization of this paper is: In Section 2, we give some preliminaries. In Section 3, main results are obtained. In Section 4, the applications of (1.1) are given. In Section 5, an example is given to illustrate our result. In Section 6, the paper is concluded.

    We introduce some basic definitions and properties of the calculus theory, please see the details in [27,34].

    Definition 2.1. [27,34] Let f be an integrable function defined on [a,b] and ψC1([a,b]) be an increasing function with ψ(t)0,t[a,b]. The left ψ-Riemann-Liouville fractional integral operator of order γ of a function f is defined by

    (t0Iγψf)(t)=1Γ(γ)tt0(ψ(t)ψ(s))γ1ψ(s)f(s)ds,γ>0.

    Definition 2.2. [27,34] Let γ(n1,n),fCn([a,b]) and ψCn([a,b]) be an increasing function with ψ(t)0,t[a,b]. The left ψ-Caputo fractional derivative of order γ of a function f is defined by

    (Ct0Dγψf)(t)=(t0Inγψf[n])(t)=1Γ(nγ)tt0(ψ(t)ψ(s))nγ1ψ(s)f[n](s)ds,

    where n=[α]+1,f[n](s):=(1ψ(t)ddt)nf(t) on [a,b].

    Theorem 2.1. [35] Let X be a Banach space, F:XX be a completely continuous operator. If the set E(F)={yX:y=lFy for some l[0,1]} is bounded, then F has at least a fixed point.

    Theorem 3.1. Assume that x,a are integrable and nonnegative functions and bj,j=1,2,,m are continuous integrable and nonnegative functions with t[a,b]. Let ψC1([a,b]) be an increasing function with ψ(t)0,t[a,b]. If

    x(t)a(t)+mj=1bj(t)taψ(s)(ψ(t)ψ(s))αj1x(s)ds, (3.1)

    then

    x(t)a(t)+k=1bk(t)mj1=1mj2=1mjk=1Γ(αj1)Γ(αj2)××Γ(αjk)Γ(kν=1αjν)taψ(τ)(ψ(t)ψ(τ))(kν=1αjν1)a(τ)dτ, (3.2)

    provided that there exist a constant M>0 such that

    limnMmj1=1mj2=1mjn=1mjn+1=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(αjn+1)Γ(n+1ν=1αjν)mj1=1mj2=1mjn=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(nν=1αjν)=ρ[0,1),

    where αjn{α1,α2,,αm},nN,b(t)=max{bj(t)}M,j=1,2,,m.

    Proof. Let

    Ax(t)=mj=1bj(t)taψ(s)(ψ(t)ψ(s))αj1x(s)ds. (3.3)

    Then by (3.1), we get

    x(t)a(t)+Ax(t). (3.4)

    By the monotonicity of the operators A and (3.1) and mathematical induction, for t[a,b], we have

    x(t)a(t)+Ax(t)a(t)+A(a(t)+Ax(t))=a(t)+Aa(t)+A2x(t)a(t)+Aa(t)+A2(a(t)+Ax(t))n1k=0Aka(t)+Anx(t), (3.5)

    i.e.

    x(t)n1k=0Aka(t)+Anx(t), (3.6)

    where A0a(t)=a(t).

    For t[a,b], by mathematical induction, we will show that

    Anx(t)bn(t)mj1=1mj2=1mjn=1Γ(αj1)Γ(αj2)××Γ(αjn)Γ(nν=1αjν)taψ(τ)(ψ(t)ψ(τ))(nν=1αjν1)x(τ)dτ, (3.7)

    nN, and limnAnx(t)=0.

    For n=1, the conclusion in (3.7) holds naturally. Using the change of variables θ=ψ(s)ψ(τ)ψ(t)ψ(τ) and the Beta function Γ(αj)Γ(βj)Γ(αj+βj)=B(αj,βj), we have

    A2x(t)=A(Ax(t))=mj1=1bj1(t)taψ(s)(ψ(t)ψ(s))αj11mj2=1bj2(s)saψ(τ)(ψ(s)ψ(τ))αj21x(τ)dτdsmj1=1bj1(t)mj2=1bj2(t)taψ(s)(ψ(t)ψ(s))αj11saψ(τ)(ψ(s)ψ(τ))αj21x(τ)dτds=mj1=1bj1(t)mj2=1bj2(t)taψ(τ)x(τ)tτψ(s)(ψ(t)ψ(s))αj11(ψ(s)ψ(τ))αj21dsdτ=mj1=1bj1(t)mj2=1bj2(t)taψ(τ)x(τ)tτψ(s)(ψ(t)ψ(τ))αj11×[1ψ(s)ψ(τ)ψ(t)ψ(τ)]αj11(ψ(s)ψ(τ))αj21dsdτ=mj1=1bj1(t)mj2=1bj2(t)taψ(τ)x(τ)10(1θ)αj11θαj21dθ(ψ(t)ψ(τ))αj1+αj21dτ=mj1=1mj2=1bj1(t)bj2(t)Γ(αj1)Γ(αj2)Γ(αj+αl)taψ(τ)(ψ(t)φ(τ))αj1+αj21x(τ)dτb2(t)mj=1ml=1Γ(αj1)Γ(αj2)Γ(αj1+αj2)taψ(τ)(ψ(t)φ(τ))αj1+αj21x(τ)dτ, t[a,b]. (3.8)

    For t[a,b], we can suppose

    Akx(t)bk(t)mj1=1mj2=1mjk=1Γ(αj1)Γ(αj2)××Γ(αjk)Γ(kν=1αjν)taψ(τ)(ψ(t)ψ(τ))(kν=1αjν1)x(τ)dτ. (3.9)

    For n=k+1, using the non-increasing properties of bj(t),j=1,2,,m,t[a,b], we have

    Ak+1x(t)=A(Akx(t))mjk+1=1bjk+1(t)taψ(s)(ψ(t)ψ(s))αjk+11dsbk(s)×mj1=1mjk=1Γ(αj1)Γ(αj2)××Γ(αjk)Γ(kν=1αjν)saψ(τ)(ψ(s)ψ(τ))(kν=1αjν1)x(τ)dτbk+1(t)mj1=1mjk+1=1Γ(αj1)Γ(αj2)Γ(αjk+1)Γ(k+1ν=1αjν)taψ(τ)(ψ(t)ψ(τ))(k+1ν=1αjν1)x(τ)dτ. (3.10)

    Since bj(t),j=1,2,,m are all continuous functions on [a,b], then there exist a constant M>0 such that b(t)=max{bj(t)}M,j=1,2,,m. So we have

    Anx(t)Mnmj1=1mj2=1mjn=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(nν=1αjν)taψ(τ)(ψ(t)ψ(τ))(nν=1αjν1)x(τ)dτ. (3.11)

    Consider the infinite series of number n=1Mnmj1=1mj2=1mjn=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(nν=1αjν), by virtue of the ratio test

    to the infinite series of number and the asymptotic approximation in [36], we get

    limnMmj1=1mj2=1mjn=1mjn+1=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(αjn+1)Γ(n+1ν=1αjν)mj1=1mj2=1mjn=1Γ(αj1)Γ(αj2)Γ(αjn)Γ(nν=1αjν)=ρ[0,1), (3.12)

    which implies that Anx(t) is convergent. Hence the conclusion in (3.2) holds.

    Theorem 3.2. Under the hypotheses of Theorem 3.1 and let a(t) be a nondecreasing function for t[a,b]. Then

    x(t)a(t)[1+k=1bk(t)mj1=1mjk=1Γ(αj1)Γ(αj2)××Γ(αjk)Γ(kν=1αjν)taψ(τ)(ψ(t)ψ(τ))(kν=1αjν1)dτ]. (3.13)

    Proof. Since a(t) ia a nondecreasing function, for αj,j=1,2,,m, then we get

    saψ(τ)(ψ(s)ψ(τ))(k+1ν=1αjν1)a(τ)dτa(s)saψ(τ)(ψ(s)ψ(τ))(k+1ν=1αjν1)dτ=a(s)(k+1ν=1αjν1)(ψ(s)ψ(a))(k+1ν=1αjν). (3.14)

    So from (3.2) and (3.14), we have

    x(t)a(t)+k=1bk(t)mj1=1mjk=1Γ(αj1)××Γ(αjk)Γ(k+1ν=1αjν1)taψ(τ)(ψ(t)ψ(τ))(k+1ν=1αjν1)1a(t)dτ=a(t)[1+k=1bk(t)mj1=1mjk=1Γ(αj1)××Γ(αjk)Γ(k+1ν=1αjν1)taψ(τ)(ψ(t)ψ(τ))(k+1ν=1αjν1)1dτ]. (3.15)

    Consider the following neutral fractional equations involving ψ-fractional integral operator

    Ct0Dγψ[x(t)li=1t0Iγiψgi(t,x(t))]=f(t,x(t)), t0,tJ=[a,b], (4.1)

    where γ>0,γi>0,i=1,2,,l.

    (H1) For the functions f,giC(J×R,R), there are some constants ci,c0 such that

    gi(t,ϕ)gi(t,φ)ciϕφ, f(t,ϕ)f(t,φ)cϕφ, tJ. (4.2)

    (H1) For the functions f,giC(J×R,R), there are some constants ci,c0 such that

    gi(t,ϕ)ci(1+ϕ), f(t,ϕ)c(1+ϕ), tJ.

    (H2) H=li=1ci(ψ(b)ψ(a))γiΓ(γi+1)+c(ψ(b)ψ(a))γΓ(γ+1)<1.

    By using Definitions 2.1 and 2.2, we get the following result.

    Lemma 4.1. Under the hypotheses (H1),(H2). x(t) satisfies (4.1) if and only if x(t) satisfies the equality

    x(t)=X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t)), t0,tJ, (4.3)

    where

    X(t0)=x(t0)+li=1t0Iγiψgi(t0,x(t0)).

    Theorem 4.1. Under the hypotheses (H1),(H2). Then (4.1) has a unique solution on J.

    Proof. For xC(J,R), denote by

    Br={xC1(J,R):xr},r>0

    with

    X(t0)+[li=1ci(ψ(b)ψ(a))γiΓ(γi+1)+c(ψ(b)ψ(a))γΓ(γ+1)]rr.

    On Br, we define the operator Γx as

    (Γx)(t)=X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t)), t0,tJ. (4.4)

    By (H1),(H2), we have

    (Γx)X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t))X(t0)+li=1t0Iγiψcix+t0IγψcxX(t0)+[li=1ci(ψ(t)ψ(t0))γiΓ(γi+1)+c(ψ(t)ψ(t0))γΓ(γ+1)]rX(t0)+[li=1ci(ψ(b)ψ(a))γiΓ(γi+1)+c(ψ(b)ψ(a))γΓ(γ+1)]rr, t0,tJ, (4.5)

    Then for x,yC(J,R), by (H2), we get

    ΓxΓy=li=1[t0Iγiψgi(t,x(t))t0Iγiψgi(t,y(t))]+[t0Iγψf(t,x(t))t0Iγψf(t,y(t))]li=1t0Iγiψgi(t,x(t))gi(t,y(t))+t0Iγψf(t,x(t))f(t,y(t))li=1t0Iγiψcixy+t0Iγψcxy[li=1cit0Iγiψ1+ct0Iγψ1]xy[li=1ci(ψ(t)ψ(t0))γiΓ(γi+1)+c(ψ(t)ψ(t0))γΓ(γ+1)]xyHxy<xy, (4.6)

    i.e. the operator Γ has a unique solution on J.

    Theorem 4.2. Under the hypotheses (H1),(H2). Then (4.1) has at least one solution on J.

    Proof. Consider the Cauchy problem (4.1). Define the operator Γ as in (4.4).

    Claim 1: Γ is continuous. Let xn be a sequence such that xnxC1(J,R). Then since gi,f are continuous and (H1), then we have

    (Γxn)(t)(Γx)(t)li=1t0Iγiψ[gi(t,xn(t))gi(t,x(t))+t0Iγψ[f(t,xn(t))f(t,x(t))]ε[li=1cit0Iγiψ1+ct0Iγψ1]xxn0, tJ. (4.7)

    Thus (Γxn)(Γx) in C1(J,R) and Γ is continuous.

    Claim 2: Γ maps bounded sets into bounded sets in C1(J,R). Denote by Br as in Theorem 4.1. Then as (4.5), we get that (Γx)r, tJ, which implies that Γxr and the operator Γ is uniformly bounded.

    Claim 3: Γ maps bounded sets into equi-continuous sets of C1(J,R). For any xBr, where Br is defined as in Claim 2. As t1t2 for t1,t2J, we have

    |(Γx)(t2)(Γx)(t1)|li=1|[t0Iγiψgi(t2,x(t2))t0Iγiψgi(t1,x(t1))]|+|t0Iγψf(t2,x(t2))t0Iγψf(t1,x(t1))|li=11Γ(γi)[t1t0|((ψ(t2)ψ(s))γ1(ψ(t1)ψ(s))γ1)ψ(s)gi(s,x(s))|ds+t2t1|(ψ(t2)ψ(s))γ1ψ(s)gi(s,x(s))|ds]+1Γ(γ)[t1t0|((ψ(t2)ψ(s))γ1(ψ(t1)ψ(s))γ1)ψ(s)f(s,x(s))|ds+t2t1|(ψ(t2)ψ(s))γ1ψ(s)f(s,x(s))|ds]li=1[εΓ(γi)t1t0|ψ(s)gi(s,x(s))|ds+1Γ(γi)t2t1|(ψ(t2)ψ(s))γi1ψ(s)gi(s,x(s))|ds]+εΓ(γ)[t1t0|ψ(s)f(s,x(s))|ds+1Γ(γ)t2t1|(ψ(t2)ψ(s))γ1ψ(s)f(s,x(s))|ds]li=1[εΓ(γi)t1t0ψ(s)ci(1+|x(s)|)ds+1Γ(γi)t2t1|(ψ(t2)ψ(s))γi1ψ(s)ci(1+|x(s)|ds]+[εΓ(γ)t1t0ψ(s)c(1+|x(s)|ds+1Γ(γ)t2t1(ψ(t2)ψ(s))γ1ψ(s)c(1+|x(s)|ds]li=1[ci(1+r)εΓ(γi)t1t0ψ(s)ds+ci(1+r)Γ(γi)t2t1(ψ(t2)ψ(s))γi1ψ(s)ds]+[c(1+r)εΓ(γ)t1t0ψ(s)ds+c(1+r)Γ(γ)t2t1(ψ(t2)ψ(s))γ1ψ(s)ds]li=1[ci(1+r)εΓ(γi)(ψ(t1)ψ(t0))+ci(1+r)(ψ(t2)ψ(t1))γiΓ(γi+1)]+[c(1+r)εΓ(γ)(ψ(t1)ψ(t0))+c(1+r)(ψ(t2)ψ(t1))γΓ(γ+1)]0. (4.8)

    Thus (Γx)(ˆt2)(Γx)(ˆt1)0, as ˆt1ˆt2. As a consequence of Claims 1–3, it follows that Γ:C1(J,R)C1(J,R) is continuous and completely continuous.

    Claim 4: We show that the set K={xC1(J,R):x=λΓx for some 0<λ<1} is bounded. Let xK, then x=λΓx for some 0<λ<1. Thus we have

    x(t)=λ[X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t))], t0,tJ. (4.9)

    By (H1), we have

    x(t)X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t))X(t0)+li=1t0Iγiψgi(t,x(t))+t0Iγψf(t,x(t))X(t0)+li=1cit0Iγiψ1+ct0Iγψ1+li=1t0Iγiψcix(t)+t0Iγψcx(t), t0,tJ, (4.10)

    and Theorem 3.2 implies that

    x(t)(X(t0)+li=1cit0Iγiψ1+ct0Iγψ1)×[1+k=1Ckmj1=1mjk=1Γ(αj1)××Γ(αjk)Γ(kν=1αjν)taψ(τ)(ψ(t)ψ(τ))kν=1αjν1dτ]=(X(t0)+li=1cit0Iγiψ1+ct0Iγψ1)[1+k=1Ckmj1=1×mjk=1Γ(αj1)Γ(αj2)××Γ(αjk)(kν=1αjν)Γ(kν=1αjν)(ψ(b)ψ(a))(kν=1αjν)], (4.11)

    where C=max{c1,,cl,c},αjk{γ1,,γl,γ},kN and which shows that the set K is bounded.

    By Theorem 2.1, the operator Γ has a fixed point, which is a solution of problem (4.1).

    Consider the following neutral ψ-fractional differential equation

    C1Dγψ[x(t)1Iγ1ψgi(t,x(t))]=f(t,x(t)), tJ=[1,6], (5.1)

    where γ=23,γ1=34,i=1,g1(t,x(t))=t5sinx(t),f(t,x(t))=lnt4arctanx(t). Then g1,f are continuous and satisfy the assumptions (H1),(H2) with ψ(t)=3t,c1=65,c=ln64 and

    c1(ψ(t)ψ(t0))γ1Γ(γ1+1)+c(ψ(t)ψ(t0))γΓ(γ+1)=65(361)74Γ(74)+ln64(361)23Γ(53)=0.8918<1.

    Then by Theorem 4.1, (5.1) has a unique solution x(t) on the interval [1,6].

    By Theorem 4.2, (5.1) also has at least one solution x(t) on the interval [1,6].

    In this paper, we obtained a new generalized Gronwall inequality involving ψ-ractional integral operator that include the results in [23]. Furthermore, the Riemann-Liouville, the Hadamard, the Katugampola fractional integrals etc can be considered uniformly. The feasibility of the main results is checked by considering the existence of solutions of a type of neutral fractional differential equation involving ψ-fractional derivative. In the future, we will consider the stabilities for the neutral ψ-fractional differential equation.

    We are really thankful to the reviewers for their careful reading of our manuscript and their many insightful comments and valuable suggestions that have improved the quality of our manuscript. This work is supported by the Key Natural Science Project of Anhui Provincial Education Department (KJ2018A0027).

    The authors declare that there are no conflicts of interest.



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