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Existence of solutions for impulsive hybrid boundary value problems to fractional differential systems

  • In this paper, the existence of solutions for impulsive mixed boundary value problems involving Caputo fractional derivatives is obtained. And then, our conclusions are based on Krasnoselskii's fixed point theorem and Arzela-Ascoli theorem. Finally, some examples are given to illustrate the main results.

    Citation: Jie Yang, Guoping Chen. Existence of solutions for impulsive hybrid boundary value problems to fractional differential systems[J]. AIMS Mathematics, 2021, 6(8): 8895-8911. doi: 10.3934/math.2021516

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  • In this paper, the existence of solutions for impulsive mixed boundary value problems involving Caputo fractional derivatives is obtained. And then, our conclusions are based on Krasnoselskii's fixed point theorem and Arzela-Ascoli theorem. Finally, some examples are given to illustrate the main results.



    Fractional differential equations have used in many engineering and scientific disciplines, such as physics, chemistry, aerodynamics, electrodynamics of complex media, polymer rheology, and other fields [1,2,3]. Many authors obtained solutions for fractional differential equations boundary value problems (BVP) by a fixed point theorem [4,5,6,7,8]. Many established mathmatics methods were applied to the existence solutions of BVP. For example, the numerical method [9,10,11], the Mawhin continuation method [12,13,14], the upper and lower solution method[15,16,17], the critical point theory [18,19,20].

    In the past, impulsive differential and partial differential equations have become more and more crucial in mathematical models of real phenomena, especially in the fields of control, biological and medical [21,22,23].

    In 2016, Bai and Dong [4] studied the existence of solutions for a class of hybrid BVP for fractional impulsive differential equation:

    {CDq0+u(t)=f(t,u(t)),1<q<2,tJ{t1,t2,,tp},Δut=tk=Ik(u(tk)),Δut=tk=Jk(u(tk)),k=1,2,,p,u(0)+u(1)=0,u(0)+u(1)=0,

    where CDq0+ is the Caputo fractional derivative of order q(1,2), J=[0,1], f:[0,1]×RR is a given function, tk satisfy 0=t0<t1<t2<<tp<tp+1=1, the right and left limits of u(t) at t=t+k are represented by u(t+k) and u(tk).

    In 2017, Mahmudov et al. [5] using the same way to investigate the existence and uniqueness of solutions for the following mixed impulsive BVP:

    {CDq0+u(t)=f(t,u(t)),1<q<2,J=[0,1],tJ{t1,t2,,tp},Δut=tk=Ik(u(tk)),Δut=tk=Jk(u(tk)),k=1,2,,p,u(0)+μ1u(1)=σ1,u(0)+μ2u(1)=σ2,

    where CDq0+ is the Caputo fractional derivative of order q(1,2).

    Motivated by the above works, in this paper, we will apply Arzela-Ascoli theorem, Krasnoselskii's fixed point theorem and contraction mapping principle to study the existence of solution for a class of hybrid boundary value problem under impulse conditions. Precisely, we consider the existence and uniqueness of solutions for an impulsive mixed BVP of fractional differential equation:

    {CDα0+u(t)=f(t,u(t)),1<α2,tJ,Δu(tk)=Ik(u(tk)),ΔCDβ0+u(tk)=Jk(u(tk)),0<β1,u(0)+κ1CDβ0+u(1)=θ1,CDβ0+u(0)+κ2u(1)=θ2, (1.1)

    where CDα0+ and CDβ0+ are Caputo fractional derivatives of order α(1<α2) and β(0<β1) respectively; f:J×RR,Ik,Jk:RR are continuous functions; J=[0,1],J=J{t1,t2,,tp} and tk satisfy 0=t0<t1<t2<<tp<tp+1=1,Δu(tk)=u(t+k)u(tk),ΔCDβ0+u(tk)=CDβ0+u(t+k)CDβ0+u(tk). Here, respectively, the left and the right limits of u(t) at t=tk(k=1,2,,p) are represented by u(tk) and u(t+k); κ1,κ2,θ1,θ2 are constants and κ1,κ2 are different from zero.

    When β=1, the results of (1.1) will be degenerate to Lemma 6 in [5], and κ1=κ2=1, the results degenerate to Lemma 2.4 in [4]. Therefore, this conclusion further expands the research results of [4] and [5].

    The structure of this article is as follows. In Sect 2, The definitions and theorems related to Caputo's fractional integral and derivative are given. In Sect 3, the existence and uniqueness of solutions to mixed impulsive boundary value problems are proved by using Arzela-Ascoli theorem and Krasnoselskii's fixed point theorem. In Sect 4, some examples are provided to illustrate the main research results.

    In this section, we mainly introduce related definitions, theorems, lemmas and necessary symbol descriptions.

    Let PC(J)={u:[0,1]R|uC(J), u(t+k),u(tk) exist, and u(tk)=u(tk),1kp}. Thus, PC(J) is a Banach space with the norm uPC=sup0t1|u(t)|.

    Definition 2.1 ([5]) The fractional integral of order α of a function f:[0,+)R is defined as

    Iα0+f(t)=t0(ts)α1Γ(α)f(s)ds,t>0,α>0, (2.1)

    provided that the right hand side of the integral is point-wise defined on (0,).

    Definition 2.2 ([5]) The Caputo fractional-order derivative of order α>0 for a function f:[0,+)R is defined by

    CDα0+f(t)=t0(ts)nα1Γ(nα)f(n)(s)ds,t>0,n=[α]+1, (2.2)

    where [α] denotes the integer part of real number α, and Γ() is the gamma function.

    Lemma 2.3. ([3]) For α>0, the general solution of the fractional differential equation CDαu(t)=0 is given by

    u(t)=k0+k1t+k2t2++kn1tn1,kiR, (2.3)

    and

    Iα0+(Dα0+u)(t)=u(t)+k0+k1t+k2t2++kn1tn1, (2.4)

    where n=[α],[α] denotes the integer part of the real number α.

    Now, we state two known results due to Krasnoselskii and Arzela-Ascoli which are used to prove the existence and uniqueness of solutions of (1.1), respectively.

    Lemma 2.4. (Krasnoselskii's fixed point Theorem[24]) Assume C is a closed, convex and non-empty subset of a Banach space H, and the operators A and B be such that: Ax+ByC, whenever x,yC; A is compact and continuous; and B is a contraction mapping. Therefore, there exists zC such that z=Az+Bz.

    Lemma 2.5. (Arzela-Ascoli Theorem[3]) The set GPC([0,1],Rn) is relatively compact set if and only if G is bounded, therefore, xM for each xG and some M>0; the G is quasi-equicontinuous in [0,1], in other words, for any ε>0 there exists γ>0 such that if xG,kN, χ1,χ2(tk1,tk] and |χ1χ2|<γ, we have |x(χ1)x(χ2)|<ε.

    Lemma 2.6. Let α(1,2],β(0,1], and g:JR be continuous. A functional u is a solution of the following impulsive hybrid BVP:

    {CDα0+u(t)=g(t),1<α2,tJ,Δu(tk)=Ik(u(tk)),ΔCDβ0+u(tk)=Jk(u(tk)),0<β1,u(0)+κ1CDβ0+u(1)=θ1,CDβ0+u(0)+κ2u(1)=θ2, (2.5)

    if u is a unique solution of the following impulsive fractional integral equation:

    (P)u(t)={t0(ts)α1Γ(α)g(s)ds+ν1(t)θ1+ν2(t)θ2κ2ν1(t)10(1s)α1Γ(α)g(s)dsκ1ν1(t)10(1s)αβ1Γ(αβ)g(s)ds,t[0,t1);t0(ts)α1Γ(α)g(s)ds+ν1(t)θ1+ν2(t)θ2κ2ν1(t)10(1s)α1Γ(α)g(s)dsκ1ν1(t)10(1s)αβ1Γ(αβ)g(s)dsΓ(2β)ν1(t)pj=1tβjJj(u(tj))+ν1(t)pj=1Ij(u(tj))+Γ(2β)ν2(t)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβ1j(tjt)Jj(u(tj))pj=k+1Ij(u(tj)),t[tk,tk+1],k=1,2,,p1;t0(ts)α1Γ(α)g(s)ds+ν1(t)θ1+ν2(t)θ2κ2ν1(t)10(1s)α1Γ(α)g(s)dsκ1ν1(t)10(1s)αβ1Γ(αβ)g(s)ds+ν1(t)pj=1Ij(u(tj))Γ(2β)ν1(t)pj=1tβjJj(u(tj))+Γ(2β)ν2(t)pj=1tβ1jJj(u(tj)),t(tp,tp+1],

    where κ21+1κ1 and

    ν1(t)=Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2,ν2(t)=Γ(2β)tκ1Γ(2β)(κ2+1)κ1κ2.

    Proof. With the Lemma 2.3, a general solution u of the equationCDα0+u(t)=g(t) on each interval (tk,tk+1](k=0,1,2,,p) is given by

    u(t)=Iα0+g(t)+dk+wkt=t0(ts)α1Γ(α)g(s)ds+dk+wkt,t[0,t1), (2.6)

    for some dk,wkR, where t0=0 and tp+1=1.

    If 0<β<1, we get

    CDβ0+u(t)=t0(ts)αβ1Γ(αβ)g(s)ds+t1βΓ(2β)wk,t(tk,tk+1], (2.7)

    if β=1, we obtain

    CDβ0+u(t)=u(t)=t0(ts)α2Γ(α1)g(s)ds+wk,t(tk,tk+1], (2.8)

    where CDβ0+dk=0(0<β1),CDβ0+t=t1βΓ(2β)(0<β<1). When β=1, (2.7) and (2.8) are equivalent, thus, 0<β1, get

    CDβ0+u(t)=t0(ts)αβ1Γ(αβ)g(s)ds+t1βΓ(2β)wk,t(tk,tk+1]. (2.9)

    By the (2.6) and (2.9), we obtain u(0)=d0,CDβ0+u(0)=w0, and

    u(1)=10(1s)α1Γ(α)g(s)ds+dp+wp,CDβ0+u(1)=10(1s)αβ1Γ(αβ)g(s)ds+wpΓ(2β).

    Using the boundary conditions in (2.5) to get

    d0+κ1[10(1s)αβ1Γ(αβ)g(s)ds+wpΓ(2β)]=θ1, (2.10)

    and

    w0+κ2[10(1s)α1Γ(α)g(s)ds+dp+wp]=θ2. (2.11)

    Next, using the condition of ΔCDβ0+u(tk)=CDβ0+u(t+k)CDβ0+u(tk)=Jk(u(tk)), we get

    t1βkΓ(2β)wk=t1βkΓ(2β)wk1+Jk(u(tk)),wk=wk1+Γ(2β)tβ1kJk(u(tk)),wk+1=wk+Γ(2β)tβ1k+1Jk+1(u(tk)),wk+2=wk+1+Γ(2β)tβ1k+2Jk+2(u(tk)),wp=wk1+Γ(2β)tβ1pJp(u(tp)),wp=wk1+Γ(2β)pj=ktβ1jJj(u(tj)),wk1=wpΓ(2β)pj=ktβ1jJj(u(tj)),wk=wpΓ(2β)pj=k+1tβ1jJj(u(tj)). (2.12)

    In the same way, using the condition of Δut=tk=Ik(u(tk))=u(t+k)u(tk), we obtain

    dk+wktk=dk1+wk1tk+Ik(u(tk)),

    which by (2.12) implies that

    dk+(wk1+Γ(2β)tβ1kJk(u(tk)))tk=dk1+wk1tk+Ik(u(tk)),dk+wk1tk+Γ(2β)tβ1kJk(u(tk))tk=dk1+wk1tk+Ik(u(tk)),dk+Γ(2β)tβkJk(u(tk))=dk1+Ik(u(tk)),dk=dk1Γ(2β)tβkJk(u(tk))+Ik(u(tk)),dk=dp+Γ(2β)pj=k+1tβjJj(u(tj))pj=k+1Ij(u(tj)). (2.13)

    By combining (2.10), (2.11), (2.12) and (2.13), we have

    dp+Γ(2β)pj=1tβjJj(u(tj))pj=1Ij(u(tj))+κ1wpΓ(2β)+κ110(1s)αβ1Γ(αβ)g(s)ds=θ1,wpΓ(2β)pj=1tβ1jJj(u(tj))+κ2wp+κ2dp+κ210(1s)α1Γ(α)g(s)ds=θ2.

    Then

    dp=(Γ(2β)(κ2+1)Γ(2β)(κ2+1)κ1κ2)θ1(κ1Γ(2β)(κ2+1)κ1κ2)θ2+(κ1κ2Γ(2β)(κ2+1)κ1κ2)10(1s)α1Γ(α)g(s)ds(Γ(2β)(κ1κ2+κ1)Γ(2β)(κ2+1)κ1κ2)10(1s)αβ1Γ(αβ)g(s)ds+(Γ(2β)(κ2+1)Γ(2β)(κ2+1)κ1κ2)pj=1Ij(u(tj))((Γ(2β))2(κ2+1)Γ(2β)(κ2+1)κ1κ2)pj=1tβjJj(u(tj))(Γ(2β)κ1Γ(2β)(κ2+1)κ1κ2)pj=1tβ1jJj(u(tj)), (2.14)

    and

    wp=(Γ(2β)(κ2)Γ(2β)(κ2+1)κ1κ2)θ1+(Γ(2β)Γ(2β)(κ2+1)κ1κ2)θ2(κ2Γ(2β)Γ(2β)(κ2+1)κ1κ2)10(1s)α1Γ(α)g(s)ds+(Γ(2β)(κ1κ2)Γ(2β)(κ2+1)κ1κ2)10(1s)αβ1Γ(αβ)g(s)ds(Γ(2β)κ2Γ(2β)(κ2+1)κ1κ2)pj=1Ij(u(tj))+((Γ(2β))2κ2Γ(2β)(κ2+1)κ1κ2)pj=1tβjJj(u(tj))+((Γ(2β))2Γ(2β)(κ2+1)κ1κ2)pj=1tβ1jJj(u(tj)). (2.15)

    Combining (2.12), (2.13), (2.14) and (2.15), we obtain

    dk=dp+Γ(2β)pj=k+1tβjJj(u(tj))pj=k+1Ij(u(tj))=(Γ(2β)(κ2+1)Γ(2β)(κ2+1)κ1κ2)θ1(κ1Γ(2β)(κ2+1)κ1κ2)θ2+(κ1κ2Γ(2β)(κ2+1)κ1κ2)10(1s)α1Γ(α)g(s)ds(Γ(2β)(κ1κ2+κ1)Γ(2β)(κ2+1)κ1κ2)10(1s)αβ1Γ(αβ)g(s)ds+(Γ(2β)(κ2+1)Γ(2β)(κ2+1)κ1κ2)pj=1Ij(u(tj))((Γ(2β))2(κ2+1)Γ(2β)(κ2+1)κ1κ2)pj=1tβjJj(u(tj))(Γ(2β)κ1Γ(2β)(κ2+1)κ1κ2)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβjJj(u(tj))pj=k+1Ij(u(tj)), (2.16)

    and

    wk=wpΓ(2β)pj=k+1tβ1jJj(u(tj))=(Γ(2β)(κ2)Γ(2β)(κ2+1)κ1κ2)θ1+(Γ(2β)Γ(2β)(κ2+1)κ1κ2)θ2(κ2Γ(2β)Γ(2β)(κ2+1)κ1κ2)10(1s)α1Γ(α)g(s)ds+(Γ(2β)(κ1κ2)Γ(2β)(κ2+1)κ1κ2)10(1s)αβ1Γ(αβ)g(s)ds(Γ(2β)κ2Γ(2β)(κ2+1)κ1κ2)pj=1Ij(u(tj))+((Γ(2β))2κ2Γ(2β)(κ2+1)κ1κ2)pj=1tβjJj(u(tj))+((Γ(2β))2Γ(2β)(κ2+1)κ1κ2)pj=1tβ1jJj(u(tj))Γ(2β)pj=k+1tβ1jJj(u(tj)), (2.17)

    for k=0,1,,p1. By using (2.16) and (2.17), we get

    dk+wkt=(Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2)θ1+(Γ(2β)tκ1Γ(2β)(κ2+1)κ1κ2)θ2+(κ2(Γ(2β)tκ1)Γ(2β)(κ2+1)κ1κ2)10(1s)α1Γ(α)g(s)ds+(κ1Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2)10(1s)αβ1Γ(αβ)g(s)ds+(Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2)pj=1Ij(u(tj))+((Γ(2β))2(κ2tκ21)Γ(2β)(κ2+1)κ1κ2)pj=1tβjJj(u(tj))+(Γ(2β)(Γ(2β)tκ1)Γ(2β)(κ2+1)κ1κ2)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβ1j(tjt)Jj(u(tj))pj=k+1Ij(u(tj)).

    Therefore, by the (2.6), we get

    u(t)=t0(ts)α1Γ(α)g(s)ds+ν1(t)θ1+ν2(t)θ2κ2ν2(t)10(1s)α1Γ(α)g(s)dsκ1ν1(t)10(1s)αβ1Γ(αβ)g(s)ds+ν1(t)pj=1Ij(u(tj))Γ(2β)ν1(t)pj=1tβjJj(u(tj))+Γ(2β)ν2(t)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβ1j(tjt)Jj(u(tj))pj=k+1Ij(u(tj)),

    where

    ν1(t)=Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2,ν2(t)=Γ(2β)tκ1Γ(2β)(κ2+1)κ1κ2.

    Thus, we obtain (P) for solution of (2.5). Conversely, assume that u is a solution of the impulsive fractional integral equation (2.5), then by a direct computation, it follows that the solution given by (P) satisfies (2.5). This completes the proof.

    In this section, we state and prove the existence and uniqueness results of the fractional boundary value of (1.1) by using the fixed point theorem. We use the following notations throughout this paper:

    ν1(t)=Γ(2β)(1+κ2κ2t)Γ(2β)(κ2+1)κ1κ2ν1:=Γ(2β)(1+2|κ2|)|Γ(2β)(κ2+1)κ1κ2|,
    ν2(t)=Γ(2β)tκ1Γ(2β)(κ2+1)κ1κ2ν2:=Γ(2β)+|κ1||Γ(2β)(κ2+1)κ1κ2|.

    In this paper, we following conditions of (H1) and (H2), and then we state and prove our first result.

    (H1) The function f:J×RR is continuous.

    (H2) There exist positive constants M1,M2,M3,L1,L2 such that

    |f(t,u)f(t,v)|M1|uv|,t[0,1],u,vR,
    |Ik(u)Ik(v)|M2|uv|,|Jk(u)Jk(v)|M3|uv|,
    |Ik(u)|L1,|Jk(u)|L2.

    Also it is clear that

    |f(t,u)||f(t,u)f(t,0)|+|f(t,0)|M1|u|+L,

    where supt[0,1]|f(t,0)|=L.

    Theorem 3.1. Assume (H1) and (H2) holds. If

    M1(1Γ(α+1)+|κ2|ν2Γ(α+1)+|κ1|ν1Γ(αβ+1))+Γ(2β)p(ν1+ν2+2)M3+p(ν1+1)M2<1 (3.1)

    then BVP of (1.1) has a unique solution on [0, 1].

    Proof. By using (3.1), r can be chosen as follows :

    r>{1M1Γ(α+1)(1+ν2|κ2|)M1Γ(αβ+1)ν1|κ1|}1(LΓ(α+1)+ν1|θ1|+ν2|θ2|+ν1|κ2|LΓ(α+1)+ν2|κ1|LΓ(αβ+1)+Γ(2β)p(ν1+ν2+2)L2+p(ν1+1)L1).

    Define an operator ϝ:PC([0,1],R)PC([0,1],R) to transform (1.1) into the fixed point problem

    (ϝu)(t)=t0(ts)α1Γ(α)f(s,u(s))ds+ν1(t)θ1+ν2(t)θ2κ2ν2(t)10(1s)α1Γ(α)f(s,u(s))dsκ1ν1(t)10(1s)αβ1Γ(αβ)f(s,u(s))ds+ν1(t)pj=1Ij(u(tj))Γ(2β)ν1(t)pj=1tβjJj(u(tj))+Γ(2β)ν2(t)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβ1j(tjt)Jj(u(tj))pj=k+1Ij(u(tj)),

    where tk<t<tk+1,k=0,,p. Then

    |ϝu(t)|t0(ts)α1Γ(α)|f(s,u(s))|ds+|ν1(t)||θ1|+|ν2(t)||θ2|+|κ2||ν2(t)|10(1s)α1Γ(α)|f(s,u(s))|ds+|κ1||ν1(t)|10(1s)αβ1Γ(αβ)|f(s,u(s))|ds+|ν1(t)|pj=1|Ij(u(tj))|+Γ(2β)|ν1(t)|pj=1|Jj(u(tj))|+Γ(2β)|ν2(t)|pj=1|Jj(u(tj))|+2Γ(2β)pj=k+1|Jj(u(tj))|+pj=k+1|Ij(u(tj))|,

    and then

    |ϝu|(t)t0(ts)α1Γ(α)|f(s,u(s))f(s,0)|ds+t0(ts)α1Γ(α)|f(s,0)|ds+|ν1(t)||θ1|+|ν2(t)||θ2|+|κ2||ν2(t)|[10(1s)α1Γ(α)|f(s,u(s))f(s,0)|ds+10(1s)α1Γ(α)|f(s,0)|ds]+|κ1||ν1(t)|[10(1s)αβ1Γ(αβ)|f(s,u(s))f(s,0)|ds+10(1s)α1Γ(αβ)|f(s,0)|ds]+|ν1(t)|pj=1|Ij(u(tj))|+Γ(2β)|ν1(t)|pj=1|Jj(u(tj))|+Γ(2β)|ν2(t)|pj=1|Jj(u(tj))|+2Γ(2β)pj=k+1|Jj(u(tj))|+pj=k+1|Ij(u(tj))|.

    Using the upper bound Ik(u)L1, we have

    pj=1|Ij(u(tj))|=|I1(u(t1))|+|I2(u(t2))|++|Ip1(u(tp1))|+|Ip(u(tp))|pL1+L1++L1+L1=pL1,

    and

    pj=1|Jj(u(tj))|=|J1(u(t1))|+|J2(u(t2))|++|Jp1(u(tp1))|+|Jp(u(tp))|pL2+L2++L2+L2=pL2.

    By the same way, we get

    pj=k+1|Ij(u(tj))|=|Ik+1(u(tk+1))|+|Ik+2(u(tk+2))|++|Ipk1(u(tpk1))|+|Ipk(u(tpk))|pkL1+L1++L1+L1=(pk)L1pL1,

    and

    pj=k+1|Jj(u(tj))|=|Jk+1(u(tk+1))|+|Jk+2(u(tk+2))|++|Jpk1(u(tpk1))|+|Jpk(u(tpk))|pkL2+L2++L2+L2=(pk)L2pL2.

    Thus

    |ϝu|(t)M1rΓ(α+1)+LΓ(α+1)+ν1|θ1|+ν2|θ2|+ν2|κ2|[M1rΓ(α+1)+LΓ(α+1)]+ν1|κ1|[M1rΓ(αβ+1)+LΓ(αβ+1)]+Γ(2β)p(ν1+ν2+2)L2+p(ν1+1)L1<r.

    Therefore, for all t[0,1], the fixed point of the operator ϝ is the solution of our BVP (1.1). Next, the fixed point theorem is used and then it is shown that ϝ is a contraction mapping and we have

    |(ϝu)(t)(ϝv)(t)|t0(ts)α1Γ(α)|f(s,u(s))f(s,v(s))|ds+|κ2||ν2(t)|10(1s)α1Γ(α)|f(s,u(s))f(s,v(s))|ds+|κ1||ν1(t)|10(1s)αβ1Γ(αβ)|f(s,u(s))f(s,v(s))|ds+|ν1(t)|pj=1|Ij(u(tj))Ij(v(tj))|+Γ(2β)|ν1(t)|pj=1|Jj(u(tj))Jj(v(tj))|+Γ(2β)|ν2(t)|pj=1|Jj(u(tj))Jj(v(tj))|+2Γ(2β)pj=k+1|Jj(u(tj))Jj(v(tj))|+pj=k+1|Ij(u(tj))Ij(v(tj))|,

    Thus

    |(ϝu)(t)(ϝv)(t)|[M1(1Γ(α+1)+|κ2||ν2(t)|Γ(α+1)+|κ1||ν1(t)|Γ(αβ+1))+Γ(2β)p(|ν1(t)|+|ν2(t)|+2)M3+p(|ν1(t)|+1)M2]uv,

    ϝ is contraction mapping. By condition (3.1), we have

    |ϝuϝv|[M1(1Γ(α+1)+|κ2|ν2Γ(α+1)+|κ1|ν1Γ(αβ+1))+Γ(2β)p(ν1+ν2+2)M3+p(ν1+1)M2]uv

    Therefore, ϝ is a contraction mapping. The conclusion follows the principle of contraction mapping, that is the unique solution of impulsive mixed BVP can be obtained by using the fixed point theorem.

    Theorem 3.2. Assume that |f(t,u)|ζ(t) for all (t,u)J×R where ζL1μ(J×R),μ(0,α1) and the (H2) holds. If

    Γ(2β)p(ν1+ν2+2)M3+p(ν1+1)M2<1. (3.2)

    Then problem (1.1) has at least one solution on J.

    Proof. Let

    rζL1μ(J)((1+|κ2|ν2)Γ(α)(αμ1μ)1μ+|κ1|ν1Γ(αβ)(αβμ1μ)1μ)+Γ(2β)p(ν1+ν2+2)L2+p(ν1+1)L1,

    and denote Sr={uPC(J,R)|uPCr}.

    Define the operators B and N on Sr as

    (Bu)(t)=t0(ts)α1Γ(α)f(s,u(s))dsκ2ν2(t)10(1s)α1Γ(α)f(s,u(s))dsκ1ν1(t)10(1s)αβ1Γ(αβ)f(s,u(s))ds,(Nu)(t)=ν1(t)pj=1Ij(u(tj))Γ(2β)ν1(t)pj=1tβjJj(u(tj))+Γ(2β)ν2(t)pj=1tβ1jJj(u(tj))+Γ(2β)pj=k+1tβ1j(tjt)Jj(u(tj))pj=k+1Ij(u(tj)).

    For any u,vSr(tJ), using the condition that |f(t,u)ζ(t)| and the Hölder inequality, we get

    t0|(ts)α1Γ(α)f(s,u(s))|ds(t0(ts)α11μds)1μ(t0(ζ(s))1μds)μΓ(α)ζL1μ(J)Γ(α)(αμ1μ)1μ,
    10|(1s)α1Γ(α)f(s,u(s))|ds(10(1s)α11μds)1μ(10(ζ(s))1μds)μΓ(α)ζL1μ(J)Γ(α)(αμ1μ)1μ,

    and

    10|(1s)αβ1Γ(αβ)f(t,u(s))|ds(10(1s)αβ11μds)1μ(10(ζ(s))1μds)μΓ(αβ)ζL1μ(J)Γ(αβ)(αβμ1μ)1μ.

    Thus

    (Bu)+(Nv)(1+|κ2|ν2)ζL1μ(J)Γ(α)(αμ1μ)1μ+(|κ1|ν1)ζL1μ(J)Γ(αβ)(αβμ1μ)1μ+Γ(2β)p(ν1+ν2+2)L2+p(ν1+1)L1=ζL1μ(J)((1+|κ2|ν2)Γ(α)(αμ1μ)1μ+|κ1|ν1Γ(αβ)(αβμ1μ)1μ)+Γ(2β)p(ν1+ν2+2)L2+p(ν1+1)L1.

    Therefore, Bu+NvSr. By the (3.2), it is obvious that N is a contraction mapping. And the continuity of f implies that the operator B is continuous. Thus, B is uniformly bounded on Sr where

    (Bu)ζL1μ(J)((1+|κ2|ν2)Γ(α)(αμ1μ)1μ+|κ1|ν1Γ(αβ)(αβμ1μ)1μ)r.

    Next the quasi-equicontinuity of the operator B is proved. Let Ω=J×Sr,fsup=sup(t,u)Ω|f(t,u)|. For any tk<χ2<χ1tk+1, we have

    (Bu)(χ2)(Bu)(χ1)=|χ20(χ2s)α1Γ(α)f(s,u(s))ds10κ2ν2(χ2)(1s)α1Γ(α)f(s,u(s))ds10κ1ν1(χ2)(1s)αβ1Γ(αβ)f(s,u(s))dsχ10(χ1s)α1Γ(α)f(s,u(s))ds+10κ2ν2(χ1)(1s)α1Γ(α)f(s,u(s))ds+10κ1ν1(χ1)(1s)αβ1Γ(αβ)f(s,u(s))ds|fsupΓ(α)|χ20(χ2s)α1(χ1s)α1ds+χ1χ2(χ1s)α1ds|+|κ2ν2(χ2χ1)fsupΓ(α)10(1s)α1ds|+|κ1ν1(χ1χ2)fsupΓ(αβ)10(1s)αβ1ds|fsup[(χ2χ1)α+χα1χα2Γ(α+1)+|κ2|(χα1χα2)Γ(α+1)+|κ1|(χα2χα1)Γ(αβ+1)],

    which tends to zero as χ1χ2. This proves that B is quasi-equicontinuous on the (tk,tk+1]. So that B is compact by Lemma 2.5, and B is relatively compact on Sr.

    Thus all the assumptions of Lemma 2.4 are satisfied and problem (1.1) has at least one solution on J.

    In this section, two examples are given to verify the feasibility of the results.

    Example 4.1. Let us consider the hybrid fractional BVP:

    {CD320+u(t)=sin3u(t)(t+20)2(1+u(t)),t[0,1],t14,Δu(14)=|u(14|)100+|u(14)|,ΔCD120+u(14)=|u(14|)100+|u(14)|,t=14,u(0)+CD120+u(1)=0,CD120+u(0)+u(1)=0, (4.1)

    where Ik(u(t))=Jk(u(t))=|u(14|)100+|u(14)|, and α=32,β=12,p=1,f(t,u(t))=sin3u(t)(t+20)2(1+u(t)). Let θ1=θ2=0,M1=M2=M3=L1=L2=1100.

    By the f(t,u(t))=sin3u(t)(t+20)2(1+u(t)), (H1) is obviously established. Next, consider the (H2), we obtain

    |f(t,u)f(t,v)|=|sin3u(t)(t+20)2(1+u(t))sin3v(t)(t+12)2(1+v(t))|1400|sin3u(t)(1+u(t))sin3v(t)(1+v(t))|1400(|sin3u(t)sin3v(t)|+|u(t)v(t)|)1400(3|u(t)v(t)|+|u(t)v(t)|)=1100|uv|.

    Therefore, the |f(t,u)f(t,v)| is established. By the same way, we have |Ik(u)Ik(v)|M2|uv|,|Jk(u)Jk(v)|M3|uv|,|Ik(u)|L1 and |Jk(u)|L2 are established. Thus, (H2) is completed. Last, consider the condition of (3.1). Because the θ1=θ2=0,M1=M2=M3=1100, we have

    ν1:=2Γ(2β)1Γ(2β)15.5747,ν2:=1+Γ(2β)1Γ(2β)16.5747,
    ν3:=2Γ(2β)1.7724.

    Thus

    M1Γ(α+1)+ν1(1Γ(αβ+1)+1Γ(α+1))M1+ν1pM2+Γ(2β)ν1pM3+2pM2+2ν3pM3+ν1θ1+ν2θ20.6292<1.

    Therefore, all assumptions in the Theorem 3.1 are satisfied. Hence, the fractional impulsive hybrid BVP of (4.1) has a unique solution on [0,1].

    Example 4.2. We consider the hybrid fractional BVP of following:

    {CD320+u(t)=sint|u(t)|(t+10)2(1+|u(t)|),t[0,1],t13,Δu(13)=|u(13|)100+|u(13)|,ΔCD120+u(15)=|u(13|)100+|u(13)|,t=13u(0)+CD120+u(1)=0,CD120+u(0)+u(1)=0, (4.2)

    where Ik(u(t))=Jk(u(t))=|u(13|)100+|u(13)|, and p=1,α=32,β=12,f(t,u(t))=sint|u(t)|(t+10)2(1+|u(t)|). The

    ν1:=2Γ(2β)1Γ(2β)15.5747,ν2:=1+Γ(2β)1Γ(2β)16.5747,

    and

    f(t,u)=sint|u|(t+10)2(1+|u|),(t,u)[0,1]×[0,+).

    Clearly, we obtain

    |f(t,u)|sint|u|(t+10)2,ζ(t)=sint(t+10)2L3([0,1],R),

    and

    |f(t,u)f(t,v)|=|sint|u|(t+10)2sint|v|(t+10)2|=|sint(t+10)2(|u||v|)|sint(t+10)2|uv|.

    Let M2=M3=1100, we get

    Γ(2β)p(ν1+ν2+2)M3+p(ν1+1)M20.4683<1.

    Thus, all assumptions in the Theorem 3.2 are satisfied. Hence, the fractional impulsive mixed BVP of (4.2) has a least one solution on [0,1].

    The authors declare there is no conflict of interest.



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