Processing math: 100%
Research article

A class of impulsive vibration equation with fractional derivatives

  • Received: 25 October 2020 Accepted: 30 November 2020 Published: 07 December 2020
  • MSC : 26A33, 34A08, 34A37, 34B15

  • In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives. By using monotone iterative technique, some new results on the multiplicity for solutions of the equations under nonlinear boundary conditions are obtained, and the properties of the solutions are discussed. Finally, the practicability of our results is discussed through a concrete example.

    Citation: Xue Wang, Xiping Liu, Mei Jia. A class of impulsive vibration equation with fractional derivatives[J]. AIMS Mathematics, 2021, 6(2): 1965-1990. doi: 10.3934/math.2021120

    Related Papers:

    [1] M. Mossa Al-Sawalha, Azzh Saad Alshehry, Kamsing Nonlaopon, Rasool Shah, Osama Y. Ababneh . Approximate analytical solution of time-fractional vibration equation via reliable numerical algorithm. AIMS Mathematics, 2022, 7(11): 19739-19757. doi: 10.3934/math.20221082
    [2] Said Mesloub, Hassan Altayeb Gadain, Lotfi Kasmi . On the well posedness of a mathematical model for a singular nonlinear fractional pseudo-hyperbolic system with nonlocal boundary conditions and frictional damping terms. AIMS Mathematics, 2024, 9(2): 2964-2992. doi: 10.3934/math.2024146
    [3] Hui Huang, Kaihong Zhao, Xiuduo Liu . On solvability of BVP for a coupled Hadamard fractional systems involving fractional derivative impulses. AIMS Mathematics, 2022, 7(10): 19221-19236. doi: 10.3934/math.20221055
    [4] Ravi Agarwal, Snezhana Hristova, Donal O'Regan . Integral presentations of the solution of a boundary value problem for impulsive fractional integro-differential equations with Riemann-Liouville derivatives. AIMS Mathematics, 2022, 7(2): 2973-2988. doi: 10.3934/math.2022164
    [5] Muneerah Al Nuwairan, Ahmed Gamal Ibrahim . Solutions and anti-periodic solutions for impulsive differential equations and inclusions containing Atangana-Baleanu fractional derivative of order $ \zeta \in (1, 2) $ in infinite dimensional Banach spaces. AIMS Mathematics, 2024, 9(4): 10386-10415. doi: 10.3934/math.2024508
    [6] Weerawat Sudsutad, Wicharn Lewkeeratiyutkul, Chatthai Thaiprayoon, Jutarat Kongson . Existence and stability results for impulsive $ (k, \psi) $-Hilfer fractional double integro-differential equation with mixed nonlocal conditions. AIMS Mathematics, 2023, 8(9): 20437-20476. doi: 10.3934/math.20231042
    [7] Reny George, Sina Etemad, Ivanka Stamova, Raaid Alubady . Existence of solutions for $ [\mathtt{p},\mathtt{q}] $-difference initial value problems: application to the $ [\mathtt{p},\mathtt{q}] $-based model of vibrating eardrums. AIMS Mathematics, 2025, 10(2): 2321-2346. doi: 10.3934/math.2025108
    [8] Thabet Abdeljawad, Pshtiwan Othman Mohammed, Hari Mohan Srivastava, Eman Al-Sarairah, Artion Kashuri, Kamsing Nonlaopon . Some novel existence and uniqueness results for the Hilfer fractional integro-differential equations with non-instantaneous impulsive multi-point boundary conditions and their application. AIMS Mathematics, 2023, 8(2): 3469-3483. doi: 10.3934/math.2023177
    [9] Pinghua Yang, Caixia Yang . The new general solution for a class of fractional-order impulsive differential equations involving the Riemann-Liouville type Hadamard fractional derivative. AIMS Mathematics, 2023, 8(5): 11837-11850. doi: 10.3934/math.2023599
    [10] Abdelatif Boutiara, Mohammed S. Abdo, Manar A. Alqudah, Thabet Abdeljawad . On a class of Langevin equations in the frame of Caputo function-dependent-kernel fractional derivatives with antiperiodic boundary conditions. AIMS Mathematics, 2021, 6(6): 5518-5534. doi: 10.3934/math.2021327
  • In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives. By using monotone iterative technique, some new results on the multiplicity for solutions of the equations under nonlinear boundary conditions are obtained, and the properties of the solutions are discussed. Finally, the practicability of our results is discussed through a concrete example.



    As we all know, the vibration equation is one of the important research topics in mechanics, physics and other disciplines. In the recent decades, researchers have been paying more and more attentions to the fractional differential equations due to its wide applications on mechanics, physical science, biological sciences and engineering disciplines, etc., see [3,6,7,12,13,17,18,19,20,25] and the references therein. The development of fractional differential equations provides some new theoretical bases for the study of vibration problems. In [22], the vibration equations with fractional derivatives are used to describe the vibration behavior of viscoelastic polymers and good results are obtained. The theoretical study of vibration equations with fractional derivative has also been widely concerned. These studies include the mechanical properties, dynamic characteristics of the system and the correlation functions with various influences for the vibration equations with fractional derivatives, and so on see[15,16,21,22]. In addition, mutation often occurs in vibration system. These abrupt changes can be simulated by the impulsive vibration equation. And this simulation are effective in describing the behavior of real system. There have been a large number of references for the study of fractional impulsive differential equations, see [1,4,5,9,11,23,24,26].

    In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives under the nonlinear boundary conditions

    {x(t)λcDα0+x(t)=f(t,x(t),x(t)),tJ,Δx(t)|t=tk=Ik(x(tk)),k=1,2,,m,Δx(t)|t=tk=Qk(x(tk)),k=1,2,,m,g0(x(0),x(1))=0,g1(x(0),x(1))=0, (1.1)

    where 0<α<1, λ>0 and cDα0+ is the Caputo derivative, 0=t0<t1<<tm<tm+1=1. Set J=[0,1],J=J{t1,t2,,tm},J0=[0,t1],Jk=(tk,tk+1],k=1,2,,m. Δx(tk)=x(t+k)x(tk),Δx(tk)=x(t+k)x(tk). x(tk),x(t+k) denote the left limit and the right limit of x(t) at t=tk. x(tk),x(t+k) denote the left limit and the right limit of x(t) at t=tk. Let x(tk)=x(tk). fC(J×R2,R),Ik,QkC(R,R),k=1,2,, m. g0,g1C(R2,R) are given nonlinear functions. By using monotone iterative technique, some new results on multiplicity of boundary value problems are obtained, and the properties of the solutions are discussed. Finally, an example is given out to illustrate the applicability of our main results.

    In this section, we present some basic definitions and lemmas, which will be used to prove our main results.

    Definition 2.1. (See[8], P67) Let α,β>0. The function Eα,β is defined by

    Eα,β(z)=j=0zjΓ(jα+β),

    whenever the series converge is called the two parameters Mittag-Leffler function with parameters α and β.

    Lemma 2.1. (See[8], P68) Consider the two parameters Mittag-Leffler function Eα,β for some α,β>0. The power series defining Eα,β is convergent for all zC. In other words, Eα,β is an entire function.

    Lemma 2.2. Let α,β>0,k=0,1,2,,zR. Then

    E(k)α,β(z)=j=0zjΓ(k+j+1)Γ(j+1)Γ(α(k+j)+β).

    Proof. By Definition 2.1 and Lemma 2.1, we get

    E(k)α,β(z)=dkdzkEα,β(z)=j=0dkdzk(zjΓ(jα+β))=j=kΓ(j+1)zjkΓ(jk+1)Γ(αj+β)=j=0zjΓ(k+j+1)Γ(j+1)Γ(α(k+j)+β).

    Lemma 2.3. (See[14], P314) Let 0<α<β,n1<αn,l1<βl(n,lN,nl,λR). Then

    cDβ0+x(t)λcDα0+x(t)=0(t>0)

    has its linearly independent solutions given by

    xj(t)=tjEβα,j+1(λtβα)λtβα+jEβα,βα+j+1(λtβα)(j=0,1,,n1), (2.1)
    xj(t)=tjEβα,j+1(λtβα)(j=n,,l1). (2.2)

    Lemma 2.4. (See[14], P324) Let l1<βl(lN),0<α<β be such that βl+1α, λR, and h(t) be a given real function defined on R+. The general solution to the nonhomogeneous linear differential equation

    cDβ0+x(t)λcDα0+x(t)=h(t)(t>0)

    is given by

    x(t)=t0(ts)β1Eβα,β(λ(ts)βα)h(s)ds+l1j=0cjxj(t),

    where xj(t) are given by (2.1) and (2.2), cj are arbitrary real constants (j=0,1,,l1).

    Lemma 2.5. Let L[0,1] denote the space of Lebesgue integrable functions on [0, 1], hL[0,1] and 0<α<1, then the Cauchy problem of the second order vibration equation with fractional derivative

    {x(t)λcDα0+x(t)=h(t),tJ,x(ξ)=x0,x(ξ)=x1,ξJ

    has a unique solution, which is given by

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)dsξ0(ξs)E2α,2(λ(ξs)2α)h(s)ds+x0+tE2α,2(λt2α)ξE2α,2(λξ2α)E2α,1(λξ2α)(x1ξ0E2α,1(λ(ξs)2α)h(s)ds), (2.3)

    and x(t) is derivable while its derivative is given by

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+E2α,1(λt2α)E2α,1(λξ2α)(x1ξ0E2α,1(λ(ξs)2α)h(s)ds). (2.4)

    Proof. In view of Lemma 2.4, for β=2,0<α<1, the general solution of the equation

    x(t)λcDα0+x(t)=h(t)

    is given by

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)ds+c1tE2α,2(λt2α)+c0,

    where c0,c1R, and

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+c1E2α,1(λt2α).

    Therefore,

    x(ξ)=ξ0(ξs)E2α,2(λ(ξs)2α)h(s)ds+c1ξE2α,2(λξ2α)+c0=x0,x(ξ)=ξ0E2α,1(λ(ξs)2α)h(s)ds+c1E2α,1(λξ2α)=x1.

    We get

    c1=1E2α,1(λξ2α)(x1ξ0E2α,1(λ(ξs)2α)h(s)ds),c0=x0ξ0(ξs)E2α,2(λ(ξs)2α)h(s)dsc1ξE2α,2(λξ2α).

    So

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)dsξ0(ξs)E2α,2(λ(ξs)2α)h(s)ds+x0+c1(tE2α,2(λt2α)ξE2α,2(λξ2α))=t0(ts)E2α,2(λ(ts)2α)h(s)dsξ0(ξs)E2α,2(λ(ξs)2α)h(s)ds+x0+tE2α,2(λt2α)ξE2α,2(λξ2α)E2α,1(λξ2α)(x1ξ0E2α,1(λ(ξs)2α)h(s)ds),

    and

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+E2α,1(λt2α)E2α,1(λξ2α)(x1ξ0E2α,1(λ(ξs)2α)h(s)ds).

    The proof is completed.

    Let PC1(J)={x:JR|x,xC(J,R),x(t+k),x(tk),x(t+k),x(tk)exist,andx(tk)=x(tk),k=1,2,,m} and endowed with the normal x=max{supt[0,1]|x(t)|,supt[0,1]|x(t)|}. Then PC1(J) is a Banach space.

    For xPC1(J), by the Lagrange mean value theorem, there exists ξk[tkε,tk] such that

    x(tk)x(tkε)=x(ξk)ε,

    and

    x(tk)=limε0+x(tk)x(tkε)ε=limε0+x(ξk)εε=x(tk),k=1,2,,m.

    Thus, for xPC1(J), we denote

    x(tk)=x(tk)=x(tk),k=1,2,,m. (2.5)

    Let P={xPC1(J)|x(t)0,x(t)0,tJ}. It is obvious that PPC1(J) is a normal solid cone. We denote x_yPC1(J) if and only if x(t)y(t) and x(t)y(t) on t[0,1], i.e. yxP. We denote xy if x_yPC1(J) and xy, and x≺≺y if yx˚P.

    Lemma 2.6. (See[10], P220, [2], P666) Let E be a Banach space, and PE be a normal solid cone. Suppose that there exist α1,β1,α2,β2E with α1β1α2β2 and A:[α1,β2]E is a completely continuous strongly increasing operator such that

    α1_Aα1,Aβ1β1,α2Aα2,Aβ2_β2.

    Then the operator A has at least three distinct fixed points x1,x2,x3 on [α1,β2] such that

    α1_x1≺≺β1,α2≺≺x2_β2,α2_x3_β1.

    In this section, we obtain the solution of the linear impulsive vibration equation and discuss the properties of its kernel function.

    Lemma 3.1. For any pk,qkR (k=1,2,,m), mi,niR (i=1,2) and hL[0,1], the following boundary value problem of the second order impulsive vibration equation with fractional derivative

    {x(t)λcDα0+x(t)=h(t),tJ,Δx(t)|t=tk=pk,k=1,2,,m,Δx(t)|t=tk=qk,k=1,2,,m,m1x(0)+m2x(1)=γ0,n1x(0)+n2x(1)=γ1 (3.1)

    has a unique solution, which is given by

    x(t)=10G(t,s)h(s)ds+φ(t)+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γ1+γ0m1+m2,tJ, (3.2)

    where

    G(t,s)={(ts)E2α,2(λ(ts)2α)n2tE2α,2(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)+m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ))m2(1s)E2α,2(λ(1s)2α)m1+m2,0st1,m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ))n2tE2α,2(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)m2(1s)E2α,2(λ(1s)2α)m1+m2,0t<s1, (3.3)
    φ(t)=0<ti<tpim2m1+m2mi=1pi+0<ti<ttE2α,2(λt2α)tiE2α,2(λt2αi)E2α,1(λt2αi)qi+mi=1(m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λt2αi)m2(E2α,2(λ)tiE2α,2(λt2αi))(m1+m2)E2α,1(λt2αi)n2E2α,1(λ)tE2α,2(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi))qi,tJ. (3.4)

    Furthermore,

    x(t)=10Gt(t,s)h(s)ds+φ(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γ1,tJ, (3.5)
    Gt(t,s)={E2α,1(λ(ts)2α)n2E2α,1(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ),0st1,n2E2α,1(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ),0t<s1, (3.6)

    and

    φ(t)=0<ti<tE2α,1(λt2α)E2α,1(λt2αi)qimi=1n2E2α,1(λ)E2α,1(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi)qi,tJ. (3.7)

    Proof. For t[0,t1], let ξ=0,x(0)=c0,x(0)=c1, by Lemma 2.5, Cauchy problem

    {x(t)λcDα0+x(t)=h(t),x(0)=c0,x(0)=c1

    has a unique solution

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)ds+c1tE2α,2(λt2α)+c0,

    and

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+c1E2α,1(λt2α),c0,c1R.

    So

    x(t1)=t10(t1s)E2α,2(λ(t1s)2α)h(s)ds+c1t1E2α,2(λt2α1)+c0,x(t1)=t10E2α,1(λ(t1s)2α)h(s)ds+c1E2α,1(λt2α1).

    For t(t1,t2], let ξ=t1,x(t+1)=x(t1)+p1,x(t+1)=x(t1)+q1. By Lemma 2.5, we can obtain that

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)dst10(t1s)E2α,2(λ(t1s)2α)h(s)ds+x(t+1)+tE2α,2(λt2α)t1E2α,2(λt2α1)E2α,1(λt2α1)(x(t+1)t10E2α,1(λ(t1s)2α)h(s)ds)=t0(ts)E2α,2(λ(ts)2α)h(s)dst10(t1s)E2α,2(λ(t1s)2α)h(s)ds+x(t1)+p1+tE2α,2(λt2α)t1E2α,2(λt2α1)E2α,1(λt2α1)(x(t1)+q1t10E2α,1(λ(t1s)2α)h(s)ds)=t0(ts)E2α,2(λ(ts)2α)h(s)dst10(t1s)E2α,2(λ(t1s)2α)h(s)ds+p1+t10(t1s)E2α,2(λ(t1s)2α)h(s)ds+c1t1E2α,2(λt2α1)+c0+tE2α,2(λt2α)t1E2α,2(λt2α1)E2α,1(λt2α1)(t10E2α,1(λ(t1s)2α)h(s)ds+c1E2α,1(λt2α1)+q1t10E2α,1(λ(t1s)2α)h(s)ds)=t0(ts)E2α,2(λ(ts)2α)h(s)ds+tE2α,2(λt2α)c1+c0+p1+tE2α,2(λt2α)t1E2α,2(λt2α1)E2α,1(λt2α1)q1,

    and

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+E2α,1(λt2α)E2α,1(λt2α1)(x(t1)+q1t10E2α,1(λ(t1s)2α)h(s)ds)=t0E2α,1(λ(ts)2α)h(s)ds+E2α,1(λt2α)c1+E2α,1(λt2α)E2α,1(λt2α1)q1.

    For t(tk,tk+1], let ξ=tk,x(t+k)=x(tk)+pk,x(t+k)=x(tk)+qk,k=2,3,,m. In the same way, we have

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)ds+tE2α,2(λt2α)c1+c0+0<ti<tpi+0<ti<ttE2α,2(λt2α)tiE2α,2(λt2αi)E2α,1(λt2αi)qi,

    and

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds+E2α,1(λt2α)c1+0<ti<tE2α,1(λt2α)E2α,1(λt2αi)qi.

    Hence,

    x(1)=10(1s)E2α,2(λ(1s)2α)h(s)ds+E2α,2(λ)c1+c0+mi=1pi+mi=1E2α,2(λ)tiE2α,2(λt2αi)E2α,1(λt2αi)qi,x(1)=10E2α,1(λ(1s)2α)h(s)ds+E2α,1(λ)c1+mi=1E2α,1(λ)E2α,1(λt2αi)qi.

    By the boundary conditions m1x(0)+m2x(1)=γ0,n1x(0)+n2x(1)=γ1, we can get that

    {(m1+m2)c0m2E2α,2(λ)c1=m210(1s)E2α,2(λ(1s)2α)h(s)ds+m2mi=1pi+m2mi=1E2α,2(λ)tiE2α,2(λt2αi)E2α,1(λt2αi)qiγ0,(n1+n2E2α,1(λ))c1=n210E2α,1(λ(1s)2α)h(s)ds+mi=1n2E2α,1(λ)E2α,1(λt2αi)qiγ1.

    So

    c0=10(m2(1s)E2α,2(λ(1s)2α)m1+m2+m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ)))h(s)dsm2m1+m2mi=1pi+mi=1(m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λt2αi)m2(E2α,2(λ)tiE2α,2(λt2αi))(m1+m2)E2α,1(λt2αi))qim2E2α,2(λ)γ1(m1+m2)(n1+n2E2α,1(λ))+γ0m1+m2,
    c1=n2n1+n2E2α,1(λ)(10E2α,1(λ(1s)2α)h(s)ds+mi=1E2α,1(λ)E2α,1(λt2αi)qi)+γ1n1+n2E2α,1(λ).

    Therefore,

    x(t)=t0(ts)E2α,2(λ(ts)2α)h(s)ds+10(m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ))n2tE2α,2(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)m2(1s)E2α,2(λ(1s)2α)m1+m2)h(s)ds+0<ti<tpim2m1+m2mi=1pi+0<ti<ttE2α,2(λt2α)tiE2α,2(λt2αi)E2α,1(λt2αi)qi+mi=1(m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λt2αi)m2(E2α,2(λ)tiE2α,2(λt2αi))(m1+m2)E2α,1(λt2αi)n2E2α,1(λ)tE2α,2(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi))qi+γ0m1+m2+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γ1=10G(t,s)h(s)ds+φ(t)+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γ1+γ0m1+m2,t[0,1].

    And

    x(t)=t0E2α,1(λ(ts)2α)h(s)ds10n2E2α,1(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)h(s)ds+0<ti<tE2α,1(λt2α)E2α,1(λt2αi)qimi=1n2E2α,1(λ)E2α,1(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi)qi+E2α,1(λt2α)n1+n2E2α,1(λ)γ1=10Gt(t,s)h(s)ds+φ(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γ1,t[0,1].

    Therefore, boundary value problem (3.1) has a unique solution x=x(t) which is given by (3.2), and G(t,s),φ(t) are given by (3.3) and (3.4), respectively. Furthermore, x(t) is also established.

    For convenience, we give out the following hypothesis:

    (H1) The constants mi,niR(i=1,2) satisfy m2(m1+m2)<0 and n2(n1+n2E2α,1(λ))<0.

    Lemma 3.2. Suppose that (H1) holds. Then functions G and φ defined by (3.3) and (3.4) satisfy the following properties:

    (1) G(t,s) is continuous for t,s[0,1].

    (2) G(t,s)>0 for t,s[0,1] and maxt[0,1]G(t,s)=G(1,s),mint[0,1]G(t,s)=G(0,s).

    (3) Gt(t,s)>0 for t,s[0,1] and maxt[0,1]Gt(t,s)=Gt(1,s),mint[0,1]Gt(t,s)=Gt(0,s).

    (4) If pk0,qk0,k=1,2,,m, then φ(t)0,φ(t)0, for tJk.

    Proof. (1) By the definition of G(t,s),GC([0,1]×[0,1]) is obvious.

    (2) By (H1), for 0st1,

    Gt(t,s)=G(t,s)t=E2α,1(λ(ts)2α)n2E2α,1(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)>0.

    Then G(s,s)G(t,s)G(1,s), for s[0,1] and t[s,1]. And

    G(s,s)=m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ))n2sE2α,2(λs2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)m2(1s)E2α,2(λ(1s)2α)m1+m2>0.

    Hence, G(t,s)>0 for 0st1 and G(t,s)G(1,s) for t[s,1].

    For 0t<s1,

    G(t,s)t=n2E2α,1(λt2α)E2α,1(λ(1s)2α)n1n2E2α,1(λ)>0,

    we can get that G(0,s)G(t,s)<G(s,s), for s[0,1] and t[0,s). And

    G(0,s)=m2n2E2α,2(λ)E2α,1(λ(1s)2α)(m1+m2)(n1+n2E2α,1(λ))m2(1s)E2α,2(λ(1s)2α)m1+m2>0.

    Hence, G(t,s)>0 for 0t<s1 and G(t,s)<G(s,s) for t[0,s).

    Therefore, G(t,s)>0 for any t,s[0,1]. And G(t,s) is monotone increasing with respect to t[0,1], so maxt[0,1]G(t,s)=G(1,s),mint[0,1]G(t,s)=G(0,s).

    (3) Since

    (E2α,1(λt2α))=(k=0(λt2α)kΓ((2α)k+1))=k=1(2α)kλkt(2α)k1Γ((2α)k+1)0,t[0,1].

    By (H1), for 0st1,

    2G(t,s)t2=(E2α,1(λ(ts)2α))n2E2α,1(λ(1s)2α)(E2α,1(λt2α))n1+n2E2α,1(λ)0.

    Then Gt(s,s)Gt(t,s)Gt(1,s), for any s[0,1],t[s,1]. Because

    Gt(s,s)=1n2E2α,1(λs2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ)>0,

    we can get that Gt(t,s)>0 for 0st1.

    For 0t<s1,

    G2(t,s)t2=n2E2α,1(λ(1s)2α)(E2α,1(λt2α))n1n2E2α,1(λ)0,

    we can get that Gt(0,s)Gt(t,s)Gt(s,s) for any s[0,1] and t[0,s). Since

    Gt(0,s)=n2E2α,1(λ(1s)2α)n1n2E2α,1(λ)>0,

    we have Gt(t,s)>0 for 0t<s1.

    Therefore, Gt(t,s)>0 for any t,s[0,1] and maxt[0,1]Gt(t,s)=Gt(1,s),mint[0,1]Gt(t,s)=Gt(0,s).

    (4) If pk,qk0, k=1,2,,m, by (H1), (3.4) and (3.7), we can easily get that

    φ(t)0,φ(t)0,tJk.

    The proof is completed.

    In this section, we will establish the existence results of the solutions for the boundary value problem (1.1).

    For any uPC1(J), we consider the following boundary value problem

    {x(t)λcDα0+x(t)=f(t,u(t),u(t)),tJ,Δx(t)|t=tk=Ik(u(tk)),k=1,2,,m,Δx(t)|t=tk=Qk(u(tk)),k=1,2,,m,m1x(0)+m2x(1)=g0(u(0),u(1))+m1u(0)+m2u(1):=γu,0,n1x(0)+n2x(1)=g1(u(0),u(1))+n1u(0)+n2u(1):=γu,1. (4.1)

    By Lemma 3.1, we can get that boundary value (4.1) is equivalent to the following integral equation

    x(t)=10G(t,s)f(s,u(s),u(s))ds+φu(t)+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γu,1+γu,0m1+m2,tJ,

    and

    x(t)=10Gt(t,s)f(s,u(s),u(s))ds+φu(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γu,1,tJ,

    where

    φu(t)=0<ti<tIi(u(ti))m2m1+m2mi=1Ii(u(ti))+0<ti<ttE2α,2(λt2α)tiE2α,2(λt2αi)E2α,1(λt2αi)Qi(u(ti))+mi=1(m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λt2αi)m2(E2α,2(λ)tiE2α,2(λt2αi))(m1+m2)E2α,1(λt2αi)n2E2α,1(λ)tE2α,2(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi))Qi(u(ti)),tJ (4.2)

    and

    φu(t)=0<ti<tE2α,1(λt2α)E2α,1(λt2αi)Qi(u(ti))mi=1n2E2α,1(λ)E2α,1(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi)Qi(u(ti)),tJ. (4.3)

    We define an operator T:PC1(J)PC1(J) by

    Tu(t)=10G(t,s)f(s,u(s),u(s))ds+φu(t)+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γu,1+γu,0m1+m2,tJ.

    By Lemma 3.1 and (3.5),

    (Tu)(t)=10Gt(t,s)f(s,u(s),u(s))ds+φu(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γu,1.

    We can easily get that the following Lemma 4.1 holds.

    Lemma 4.1. The function x=x(t) is the solution of boundary value problem (1.1) if and only if x is a fixed point of the operator T in PC1(J).

    Lemma 4.2. If (H1) holds, then T:PC1(J)PC1(J) is completely continuous.

    Proof. Step 1: T is a continuous operator.

    Suppose that {un}PC1(J) and there exists uPC1(J) such that unu0(n). Then there exists a constant M>0 such that unM,uM.

    Since fC(J×R2,R),Ik,QkC(R,R),k=1,2,,m, g0,g1C(R2,R) and γun,0,γun,1,γu,0,γu,1,R, then

    limn|f(t,un(t),un(t))f(t,u(t),u(t))|=0,tJ,
    limn|Ik(un(tk))Ik(u(tk))|=0,limn|Qk(un(tk))Qk(u(tk))|=0,k=1,2,,m,
    limn|γun,0γu,0|=0,limn|γun,1γu,1|=0.

    By (H1), Lemma 3.2 and Lebesgue dominated convergence theorem, for any tJ, we have

    |Tun(t)Tu(t)|=|10G(t,s)(f(s,un(s),un(s))f(s,u(s),u(s)))ds+(φun(t)φu(t))+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))(γun,1γu,1)+γun,0γu,0m1+m2|10G(1,s)|f(s,un(s),un(s))f(s,u(s),u(s))|ds+|m1+m2|+|m2||m1+m2|mi=1|Ii(un(ti))Ii(u(ti))|+(|m1|+2|m2|)(|n1|+2|n2|E2α,1(λ))E2α,2(λ)|m1+m2||n1+n2E2α,1(λ)|mi=1|Qi(un(ti))Qi(u(ti))|+(|m1|+2|m2|)E2α,2(λ)|m1+m2||n1+n2E2α,1(λ)||γun,1γu,1|+|γun,0γu,0||m1+m2|0(n),

    and

    |(Tun)(t)(Tu)(t)|=|10Gt(t,s)(f(s,un(s),un(s))f(s,u(s),u(s)))ds+(φun(t)φu(t))+E2α,1(λt2α)n1+n2E2α,1(λ)(γun,1γu,1)|10Gt(1,s)|f(s,un(s),un(s))f(s,u(s),u(s))|ds+mi=1(|n1|+2|n2|E2α,1(λ))E2α,1(λ)|n1+n2E2α,1(λ)||Qi(un(ti))Qi(u(ti))|+E2α,1(λ)|n1+n2E2α,1(λ)||γun,1γu,1|0(n).

    So ||TunTu||0 as n, which means T is continuous.

    Step 2: T is relatively compact.

    Let ΩPC1(J) be a bounded set. By the continuity of the functions f,Ik,Qk(k=1,2,,m), g0 and g1, there exists a constant L>0, for any uΩ and tJ,

    |f(t,u(t),u(t))|L,|γu,0|L,|γu,1|L,
    |Ik(u(t))|L,|Qk(u(t))|L,k=1,2,,m.

    Then

    |φu(t)|m(|m1|+2|m2|)(|n1|(1+E2α,2(λ))+|n2|E2α,1(λ)(1+2E2α,2(λ)))|m1+m2||n1+n2E2α,1(λ)|L,
    |φu(t)|m(|n1|+2|n2|E2α,1(λ))E2α,1(λ)|n1+n2E2α,1(λ)|L.

    By Lemma 3.2, for any tJ,

    |Tu(t)|=|10G(t,s)f(s,u(s),u(s))ds+φu(t)+γu,0m1+m2+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γu,1|(10G(1,s)ds+(|m1|+2|m2|)E2α,2(λ)+|n1|+|n2|E2α,1(λ)|m1+m2||n1+n2E2α,1(λ)|+m(|m1|+2|m2|)(|n1|(1+E2α,2(λ))+|n2|E2α,1(λ)(1+2E2α,2(λ)))|m1+m2||n1+n2E2α,1(λ)|)L,

    and

    |(Tu)(t)|=|10Gt(t,s)f(s,u(s),u(s))ds+φu(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γu,1|(10Gt(1,s)ds+m(|n1|+2|n2|E2α,1(λ))E2α,1(λ)+E2α,1(λ)|n1+n2E2α,1(λ)|)L.

    Therefore, the operator T(Ω) is uniformly bounded.

    Because G(t,s) is continuous on [0,1]×[0,1], then it is uniformly continuous on [0,1]×[0,1]. Thus, for any ε>0, there exists a constant δ1>0 such that for any s[0,1], t1,t2Jk,k=0,1,,m, whenever |t1t2|<δ1, we can get that

    |G(t1,s)G(t2,s)|<ε2L.

    Denote

    G0(t,s)=n2tE2α,2(λt2α)E2α,1(λ(1s)2α)n1+n2E2α,1(λ).

    By (3.6),

    Gt(t,s)={E2α,1(λ(ts)2α)+G0(t,s),0st1,G0(t,s),0t<s1.

    Similarly, for the ε>0, there exists a constant δ2>0 such that

    |G0(t1,s)G0(t2,s)|<ε3L

    whenever |t1t2|<δ2 and t1,t2Jk,k=0,1,,m.

    By the uniformly continuity of functions tE2α,2(λt2α) and E2α,1(λt2α) on t[0,1], we can show that for the ε>0, there exists a constant δ3>0 such that

    |t1E2α,2(λt12α)t2E2α,2(λt22α)|<|n1+n2E2α,1(λ)|ε2(m(|n1|+2|n2|E2α,1(λ))+1)L,

    and

    |E2α,1(λt2α1)E2α,1(λt22α)|<|n1+n2E2α,1(λ)|ε3(m(|n1|+2|n2|E2α,1(λ))+1)L,

    whenever |t1t2|<δ3 and t1,t2Jk,k=0,1,,m. Hence, by (4.2) and (4.3), we have

    |φu(t1)φu(t2)|mi=1|t1E2α,2(λt12α)t2E2α,2(λt22α)||Qi(u(ti))|+mi=1|n2|E2α,1(λ)|t1E2α,2(λt12α)t2E2α,2(λt22α)||n1+n2E2α,1(λ)||Qi(u(ti))|mL(|n1|+2|n2|E2α,1(λ))|n1+n2E2α,1(λ)||t1E2α,2(λt12α)t2E2α,2(λt22α)|,

    and

    |φu(t1)φu(t2)|mi=1|E2α,1(λt12α)E2α,1(λt22α)||Qi(u(ti))|+mi=1|n2|E2α,1(λ)|E2α,1(λt12α)E2α,1(λt22α)||n1+n2E2α,1(λ)||Qi(u(ti))|mL(|n1|+2|n2|E2α,1(λ))|n1+n2E2α,1(λ)||E2α,1(λt12α)E2α,1(λt22α)|.

    By the uniformly continuity of E2α,1(λ(ts)2α) on D={(t,s)|0st1}, for the ε>0, there exists a constant 0<δ4<ε6LE2α,1(λ), for (t1,s),(t2,s)D, |t1t2|<δ4 and t1,t2Jk,k=0,1,,m, we have

    |E2α,1(λ(t1s)2α)E2α,1(λ(t2s)2α)|<ε6L.

    Then

    |t10E2α,1(λ(t1s)2α)f(s,u(s),u(s))dst20E2α,1(λ(t2s)2α)f(s,u(s),u(s))ds||t10E2α,1(λ(t1s)2α)f(s,u(s),u(s))dst10E2α,1(λ(t2s)2α)f(s,u(s),u(s))dst2t1E2α,1(λ(t2s)2α)f(s,u(s),u(s))ds|Lt10|(E2α,1(λ(t1s)2α)E2α,1(λ(t2s)2α))|ds+L|t2t1E2α,1(λ(t2s)2α)ds|L|(E2α,1(λ(t1s)2α)E2α,1(λ(t2s)2α))|+L|E2α,1(λ)||t2t1|<ε3.

    We take δ=min{δ1,δ2,δ3,δ4}. Therefore, for any ε>0, there exists a constant δ>0 such that for t1,t2Jk,k=0,1,,m whenever |t1t2|<δ and any uΩ, we can get that

    |Tu(t1)Tu(t2)|10|G(t1,s)G(t2,s)||f(s,u(s),u(s))|ds+|φu(t1)φu(t2)|+|t1E2α,2(λt2α1)t2E2α,2(λt2α2)||n1+n2E2α,1(λ)|γu,1L10|G(t1,s)G(t2,s)|ds+|φu(t1)φu(t2)|+L|n1+n2E2α,1(λ)||t1E2α,2(λt12α)t2E2α,2(λt22α)|<ε2+(m(|n1|+2|n2|E2α,1(λ))+1)L|n1+n2E2α(λ)||t1E2α,2(λt12α)t2E2α,2(λt22α)|<ε2+ε2=ε,

    and

    |(Tu)(t1)(Tu)(t2)|=|t10E2α,1(λ(t1s)2α)f(s,u(s),u(s))dst20E2α,1(λ(t2s)2α)f(s,u(s),u(s))ds+10(G0(t1,s)G0(t1,s))f(s,u(s),u(s))ds+φu(t1)φu(t2)+γu,1n1+n2E2α,1(λ)(E2α,1(λt2α1)E2α,1(λt22α))||t10E2α,1(λ(t1s)2α)f(s,u(s),u(s))dst20E2α,1(λ(t2s)2α)f(s,u(s),u(s))ds|+L|G0(t1,s)G0(t2,s)|+(m(|n1|+2|n2|E2α,1(λ))+1)L|n1+n2E2α(λ)||E2α,1(λt12α)E2α,1(λt22α)|<ε3+ε3+ε3=ε.

    Thus, the operator T(Ω) is equicontinuous on every interval Jk.

    According to the Arzela-Ascoli theorem, T(Ω) is relatively compact.

    Therefore, T:PC1(J)PC1(J) is completely continuous.

    In the following, we give out some hypotheses.

    (H2) f(t,x1,y1)f(t,x2,y2) if tJ, x1x2, y1y2R. Furthermore, f(t,x1,y1)<f(t,x2,y2) if x1<x2 and y1y2.

    And for x1x2R, Ik(x1)Ik(x2),Qk(x1)Qk(x2),k=1,2,,m.

    (H3.1) If m1+m2>0,m2<0,n1+n2E2α,1(λ)>0,n2<0, then for y1y2, z1z2R,

    g0(z2,y2)g0(z1,y1)m1(z2z1)m2(y2y1),

    and

    g1(z2,y2)g1(z1,y1)n1(z2z1)n2(y2y1).

    (H3.2) If m1+m2<0,m2>0,n1+n2E2α,1(λ)>0,n2<0, then for y1y2, z1z2R,

    g0(z2,y2)g0(z1,y1)m1(z2z1)m2(y2y1),

    and

    g1(z2,y2)g1(z1,y1)n1(z2z1)n2(y2y1).

    (H3.3) If m1+m2>0,m2<0,n1+n2E2α,1(λ)<0,n2>0, then for y1y2, z1z2R,

    g0(z2,y2)g0(z1,y1)m1(z2z1)m2(y2y1),

    and

    g1(z2,y2)g1(z1,y1)n1(z2z1)n2(y2y1).

    (H3.4) If m1+m2<0,m2>0,n1+n2E2α,1(λ)<0,n2>0, then for y1y2, z1z2R,

    g0(z2,y2)g0(z1,y1)m1(z2z1)m2(y2y1),

    and

    g1(z2,y2)g1(z1,y1)n1(z2z1)n2(y2y1).

    Remark. It is easy to see that if one of (H3.1)–(H3.4) holds, then (H1) holds.

    Lemma 4.3. For mi,niR(i=1,2), if one of the following conditions holds, then x(t)0 and x(t)0 for tJ.

    (1) m1+m2>0,m2<0,n1+n2E2α,1(λ)>0,n2<0 and xPC1(J) satisfies

    {x(t)λcDα0+x(t)0,tJ,Δx(t)|t=tk0,k=1,2,,m,Δx(t)|t=tk0,k=1,2,,m,m1x(0)+m2x(1)0,n1x(0)+n2x(1)0.

    (2) m1+m2<0,m2>0,n1+n2E2α,1(λ)>0,n2<0 and xPC1(J) satisfies

    {x(t)λcDα0+x(t)0,tJ,Δx(t)|t=tk0,k=1,2,,m,Δx(t)|t=tk0,k=1,2,,m,m1x(0)+m2x(1)0,n1x(0)+n2x(1)0.

    (3) m1+m2>0,m2<0,n1+n2E2α,1(λ)<0,n2>0 and xPC1(J) satisfies

    {x(t)λcDα0+x(t)0,tJ,Δx(t)|t=tk0,k=1,2,,m,Δx(t)|t=tk0,k=1,2,,m,m1x(0)+m2x(1)0,n1x(0)+n2x(1)0.

    (4) m1+m2<0,m2>0,n1+n2E2α,1(λ)<0,n2>0 and xPC1(J) satisfies

    {x(t)λcDα0+x(t)0,tJ,Δx(t)|t=tk0,k=1,2,,m,Δx(t)|t=tk0,k=1,2,,m,m1x(0)+m2x(1)0,n1x(0)+n2x(1)0.

    Proof. (1) Denote x(t)λcDα0+x(t)=h(t),Δx(t)|t=tk=pk,Δx(t)|t=tk=qk,k=1,2,,m,m1x(0)+m2x(1)=γ0,n1x(0)+n2x(1)=γ1. Then h(t)0,pk0,qk0,k=1,2,,m.

    By Lemma 3.1, the following boundary value problem

    {x(t)λcDα0+x(t)=h(t),tJ,Δx(t)|t=tk=pk,k=1,2,,m,Δx(t)|t=tk=qk,k=1,2,,m,m1x(0)+m2x(1)=γ0,n1x(0)+n2x(1)=γ1,

    has a unique solution

    x(t)=t0G(t,s)h(s)ds+φ(t)+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))γ1+γ0m1+m2,tJ,

    and

    x(t)=10Gt(t,s)h(s)ds+φ(t)+E2α,1(λt2α)n1+n2E2α,1(λ)γ1,tJ.

    It's easy to see that x(t)0 and x(t)0 for tJ.

    Similarly, (2)–(4) are easy to be proved.

    Lemma 4.4. Suppose (H2) and one of (H3.1)-(H3.4) holds, then T is a strongly increasing operator.

    Proof. Here, we will only prove the conclusion when (H3.1) holds, and other situations are similar.

    For any u1,u2PC1(J), and u1u2 which implies that u1(t)u2(t),u1(t)u2(t) and u1(t)u2(t) for tJ. By (H2), for any tJ,

    f(t,u2(t),u2(t))f(t,u1(t),u1(t))0,Ik(u2(tk))Ik(u1(t2))0,Qk(u2(tk))Qk(u2(tk))0.

    Since u1(t)u2(t), there exists an interval [a,b]J such that u1(t)<u2(t) for t[a,b]. It follows from (H2)

    f(t,u2(t),u2(t))f(t,u1(t),u1(t))>0,t[a,b]. (4.4)

    By (H3.1), we can get that

    γu2,0γu1,0=g0(u2(0),u2(1))g0(u1(0),u1(1))+(m1u2(0)+m2u2(1))(m1u1(0)+m2u1(1))0,γu2,1γu1,1=g1(u2(0),u2(1))g1(u1(0),u1(1))+(n1u2(0)+n2u2(1))(n1u1(0)n2u1(1))0.

    By (4.2), (4.3), (H3.1) and (H2), we can get that for tJ,

    φu2(t)φu1(t)=0<ti<t(Ii(u2(ti))Ii(u1(ti)))m2m1+m2mi=1(Ii(u2(ti))Ii(u1(ti)))+0<ti<ttE2α,2(λt2α)tiE2α,2(λt2αi)E2α,1(λt2αi)(Qi(u2(ti))Qi(u1(ti)))+mi=1(m2(E2α,2(λ)tiE2α,1(λt2αi))(m1+m2)E2α,1(λt2αi)n2E2α,1(λ)(tE2α,2(λt2α))(n1+n2E2α,1(λ))E2α,1(λt2αi)+m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λt2αi))(Qi(u2(ti))Qi(u1(ti)))0<ti<t(Ii(u2(ti))Ii(u1(ti)))m2m1+m2mi=1(Ii(u2(ti))Ii(u1(ti)))+mi=1m2n2E2α,2(λ)E2α,1(λ)(m1+m2)(n1+n2E2α,1(λ))E2α,1(λ)(Qi(u2(ti))Qi(u1(ti)))0,

    and

    φu2(t)φu1(t)=0<ti<tE2α,1(λt2α)E2α,1(λt2αi)(Qi(u2(ti))Qi(u1(ti)))mi=1n2E2α,1(λ)E2α,1(λt2α)(n1+n2E2α,1(λ))E2α,1(λt2αi)(Qi(u2(ti))Qi(u1(ti)))0<ti<t1E2α,1(λ)(Qi(u2(ti))Qi(u1(ti)))mi=1n2E2α,1(λ)(n1+n2E2α,1(λ))E2α,1(λ)(Qi(u2(ti))Qi(u1(ti)))0.

    By Lemma 3.2, (H2) and (4.4), for any u1,u2PC1(J) and u1u2, as tJ, we have

    Tu2(t)Tu1(t)=10G(t,s)(f(s,u2(s),u2(s))f(s,u1(s),u1(s)))ds+(φu2(t)φu1(t))+(tE2α,2(λt2α)n1+n2E2α,1(λ)m2E2α,2(λ)(m1+m2)(n1+n2E2α,1(λ)))(γu2,1γu1,1)+γu2,0γu1,0m1+m2baG(0,s)(f(s,u2(s),u2(s))f(s,u1(s),u1(s)))ds>0,

    and

    (Tu2)(t)(Tu1)(t)=10Gt(t,s)(f(s,u2(s),u2(s))f(s,u1(s),u1(s)))ds+(φu2(t)φu1(t))+E2α,1(λt2α)n1+n2E2α,1(λ)(γu2,1γu1,1)baGt(0,s)(f(s,u2(s),u2(s))f(s,u1(s),u1(s)))ds>0.

    Thus,

    Tu2Tu1˚P

    which implies that T is a strongly increasing operator.

    The proof is completed.

    Theorem 4.5. If (H2) and (H3.1) hold, there exist zi,yiPC1(J),i=1,2, with z1y1z2y2 such that

    {zi(t)λcDα0+zi(t)f(t,zi(t),zi(t)),tJ,Δzi(t)|t=tkIk(zi(tk)),k=1,2,,m,Δzi(t)|t=tkQk(zi(tk)),k=1,2,,m,g0(zi(0),zi(1))0,g1(zi(0),zi(1))0, (4.5)

    and

    {yi(t)λcDα0+yi(t)f(t,yi(t),yi(t)),tJ,Δyi(t)|t=tkIk(yi(tk)),k=1,2,,m,Δyi(t)|t=tkQk(yi(tk)),k=1,2,,m,g0(yi(0),yi(1))0,g1(yi(0),yi(1))0, (4.6)

    where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then boundary value problem (1.1) has at least three distinct solutions x1,x2,x3[z1,y2] and satisfies

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x2(t)y1(t),tJ.

    Proof. By Lemma 4.2 and Lemma 4.4, we can get that T:[z1,y2]PC1(J) is a completely continuous strongly increasing operator.

    By the definition of operator T, we can show that

    {(Tz1)(t)λcDα0+(Tz1)(t)=f(t,z1(t),z1(t)),tJ,Δ(Tz1)(t)|t=tk=Ik(z1(tk)),k=1,2,,m,Δ(Tz1)(t)|t=tk=Qk(z1(tk)),k=1,2,,m,m1(Tz1)(0)+m2(Tz1)(1)=g0(z1(0),z1(1))+m1z1(0)+m2z1(1),n1(Tz1)(0)+n2(Tz1)(1)=g1(z1(0),z1(1))+n1z1(0)+n2z1(1).

    Let x=Tz1z1. By (4.5),

    x(t)λcDα0+x(t)=((Tz1)(t)z1(t))λ(cDα0+(Tz1)(t)cDα0+z1(t))=((Tz1)(t)λcDα0+(Tz1)(t))(z1(t)λcDα0+z1(t))=f(t,z1(t),z1(t))(z1(t)λcDα0+z1(t))0,tJ.
    Δx(t)|t=tk=ΔTz1(t)|t=tkΔz1(t)|t=tk=Ik(z1(tk))Δz1(t)|t=tk0.Δx(t)|t=tk=Δ(Tz1)(t)|t=tkΔz1(t)|t=tk=Qk(z1(tk))Δz1(t)|t=tk0.
    m1x(0)+m2x(1)=m1((Tz1)(0)z1(0))+m2((Tz1)(1)z1(1))=m1((Tz1)(0)+m2(Tz1)(1))(m1z1(0)+m2z1(1))=g0(z1(0),z1(1))+(m1z1(0)+m2z1(1))(m1z1(0)+m2z1(1))=g0(z1(0),z1(1))0.

    Similarly, we can get that

    n1x(0)+n2x(1)0.

    By (1) in Lemma 4.3, we have

    x(t)=(Tz1)(t)z1(t)0,x(t)=(Tz1)(t)z1(t)0,tJ.

    Therefore, z1_Tz1.

    It is similar that we can obtain z2_Tz2. Because z2 is not the solution of boundary value problem (1.1), then Tz2z2. Thus, z2Tz2.

    By using the same method, we can easily get Ty1y1,Ty2_y2.

    By using Lemma 2.6, we can get that the operator T has at least three distinct fixed points x1,x2,x3[z1,y2], and satisfies

    z1_x1≺≺y1,z2≺≺x2_y2,z2_x3_y1.

    Therefore, by Lemma 4.1, we can obtain that boundary value problem (1.1) has at least three distinct solutions x1,x2,x3[z1,y2], and

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x2(t)y1(t),tJ.

    The proof is completed.

    In the similar way, the following three theorems can be established.

    Theorem 4.6. If (H2) and (H3.2) hold, there exist zi,yiPC1(J),i=1,2, with z1y1z2y2 such that

    {zi(t)λcDα0+zi(t)f(t,zi(t),zi(t)),tJ,Δzi(t)|t=tkIk(zi(tk)),k=1,2,,m,Δzi(t)|t=tkQk(zi(tk)),k=1,2,,m,g0(zi(0),zi(1))0,g1(zi(0),zi(1))0,

    and

    {yi(t)λcDα0+yi(t)f(t,yi(t),yi(t)),tJ,Δyi(t)|t=tkIk(yi(tk)),k=1,2,,m,Δyi(t)|t=tkQk(yi(tk)),k=1,2,,m,g0(yi(0),yi(1))0,g1(yi(0),yi(1))0,

    where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3[z1,y2] and satisfies

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x2(t)y1(t),tJ.

    Theorem 4.7. If (H2) and (H3.3) hold, there exist zi,yiPC1(J),i=1,2, with z1y1z2y2 such that

    {zi(t)λcDα0+zi(t)f(t,zi(t),zi(t)),tJ,Δzi(t)|t=tkIk(zi(tk)),k=1,2,,m,Δzi(t)|t=tkQk(zi(tk)),k=1,2,,m,g0(zi(0),zi(1))0,g1(zi(0),zi(1))0,

    and

    {yi(t)λcDα0+yi(t)f(t,yi(t),yi(t)),tJ,Δyi(t)|t=tkIk(yi(tk)),k=1,2,,m,Δyi(t)|t=tkQk(yi(tk)),k=1,2,,m,g0(yi(0),yi(1))0,g1(yi(0),yi(1))0,

    where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3[z1,y2] and satisfies

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x2(t)y1(t),tJ.

    Theorem 4.8. If (H2) and (H3.4) hold, there exist zi,yiPC1(J),i=1,2, with z1y1z2y2 such that

    {zi(t)λcDα0+zi(t)f(t,zi(t),zi(t)),tJ,Δzi(t)|t=tkIk(zi(tk)),k=1,2,,m,Δzi(t)|t=tkQk(zi(tk)),k=1,2,,m,g0(zi(0),zi(1))0,g1(zi(0),zi(1))0,

    and

    {yi(t)λcDα0+yi(t)f(t,yi(t),yi(t)),tJ,Δyi(t)|t=tkIk(yi(tk)),k=1,2,,m,Δyi(t)|t=tkQk(yi(tk)),k=1,2,,m,g0(yi(0),yi(1))0,g1(yi(0),yi(1))0,

    where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3[z1,y2] and satisfies

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x2(t)y1(t),tJ.

    In this section, we discuss the applicability of our main results.

    Consider the following the boundary value problem

    {x(t)1100cD120+x(t)=2t3πarctan(x(t)+x(t)),t(0,12)(12,1),Δx(t)|t=12=12x(12),Δx(t)|t=12=1209x(12),cos(2πx(0))+0.02x(1)0.52=0,6πcos(πx(0))+πx(1)6π=0, (5.1)

    where α=12,λ=1100,f(t,x,y)=2t3πarctan(x+y),I1(x)=12x,Q1(x)=1209x.

    g0(x,y)=cos(2πx)+0.02y0.52,g1(x,y)=6πcos(πx)+πy6π.

    Obviously, fC([0,1]×R2,R),I1,Q1C(R,R),g0,g1C(R2,R) are nonlinear functions. And f,I1,Q1 satisfy (H2).

    Let m1=2π,m2=0.02,n1=6π,n2=π. Since

    g0(x2,y2)g0(x1,y1)=cos(2πx2)+cos(2πx1)+0.02(y2y1)2π(x2x1)+0.02(y2y1),g1(x2,y2)g1(x1,y1)=6πcos(πx2)+6πcos(πx1)+π(y2y1)6π(x2x1)+π(y2y1),

    then g0,g1 satisfy (H3.1).

    For t[0,1], we take

    z1(t)={32+t,t[0,12],2+t,t(12,1],y1(t)={4+32t+18t2+14t4,t[0,12],9+32t+16t2+14t4,t(12,1],
    z2(t)={232+3t,t[0,12],12+3t,t(12,1],y2(t)={16+72t+25t2+16t4,t[0,12],33+72t+34t2+16t4,t(12,1].

    We can easily get that

    z1(t)1100cD120+z1(t)2t3πarctan(z1(t)+z1(t))={t50π2t3arctan(52+t)π<0,t(0,12),t50π2t3arctan(3+t)π<0,t(12,1),
    y1(t)1100cD120+y1(t)2t3πarctan(y1(t)+y1(t))={14+3t2t(315+35t+96t3)10500π2t3arctan(112+7t4+t28+t3+t44)π>0,t[0,12],84000πt(1+9t2)7560t840t22304t4352(t(2t1)+4t3(2t1))252000πt2t3arctan(212+11t6+t26+t3+t44)π>0,t(12,1],
    z2(t)1100cD120+z2(t)2t3πarctan(z2(t)+z2(t))={3t50π2t3arctan(292+3t)π<0,t(0,12),3t50π2t3arctan(15+3t)π<0,t(12,1),

    and

    y2(t)1100cD120+y2(t)2t3πarctan(y2(t)+y2(t))={45+2t2t(735+112t+64t3)10500π2t3arctan(392+43t10+2t25+2t33+t46)π>0,t[0,12],21000πt(3+4t2)2940t448t2256t4492(t(2t1)+4t3(2t1))42000πt2t3arctan(732+5t+3t24+2t33+t46)π>0,t(12,1].

    So

    {z1(t)1100cD120+z1(t)<2t3πarctan(z1(t)+z1(t)),t(0,12)(12,1),Δz1(t)|t=12=z1(12+)z1(12)=12<12z1(12)=1,Δz1(t)|t=12=z1(12+)z1(12)=0<1209z1(12)=2209,cos(2πz1(0))+0.02z1(1)0.52>0,6πcos(πz1(0))+πz1(1)6π>0,
    {y1(t)1100cD120+y1(t)>2t3πarctan(y1(t)+y1(t)),t(0,12)(12,1),Δy1(t)|t=12=y1(12+)y1(12)=48196>12y1(12)=307128,Δy1(t)|t=12=y1(12+)y1(12)=124>1209y1(12)=30713376,cos(2πy1(0))+0.02y1(1)0.52<0,6πcos(πy1(0))+πy1(1)6π<0,
    {z2(t)1100cD120+z2(t)<2t3πarctan(z2(t)+z2(t)),t(0,12)(12,1),Δz2(t)|t=12=z2(12+)z2(12)=12<12z2(12)=132,Δz2(t)|t=12=z2(12+)z2(12)=0<1209z2(12)=13209,cos(2πz2(0))+0.02z2(1)0.52>0,6πcos(πz2(0))+πz2(1)6π>0

    and

    {y2(t)1100cD120+y2(t)>2t3πarctan(y2(t)+y2(t)),t(0,12)(12,1),Δy2(t)|t=12=y2(12+)y2(12)=136780>12y2(12)=8573960,Δy2(t)|t=12=y2(12+)y2(12)=720>1209y2(12)=8573100320,cos(2πy2(0))+0.02y2(1)0.52<0,6πcos(πy2(0))+πy2(1)6π<0.

    We can easily get that z1(t),z2(t) satisfy (4.5) and y1(t),y2(t) satisfy (4.6) with z1y1z2y2.

    The conditions of Theorem 4.5 are all satisfied. So by Theorem 4.5, the boundary value problem (5.1) has at least three distinct solutions x1,x2,x3[z1,y2], and moreover,

    z1(t)x1(t)<y1(t),z2(t)<x2(t)y2(t),z2(t)x3(t)y1(t),t[0,1].

    In this work, we investigate the existence of solutions for a class of second order impulsive vibration equation with fractional derivatives. Some sufficient conditions for existence of the multiplicity solutions are established by applying monotone iterative technique. Finally, a concrete example is given to illustrate the wide range of potential applications of our main results.

    Further extensions of this paper are to study the motion state of the vibrator in the system described by boundary value problem (1.1) and the existence of solutions to the boundary value problems with other boundary conditions. Moreover, fractional differential equation models such as the rheological model of the fractional derivative and singular systems model of fractional differential equations have real world applications. So, we also can consider using the boundary value problem of impulsive differential equations to simulate the abrupt changes in the systems described by these models.

    The authors would like to thank college mathematics characteristic pilot team of University of Shanghai for science and technology for its support to this project.

    The authors declare that they have no conflicts of interest.



    [1] A. Arshad, S. Kamal, J. Fahd, Existence and stability analysis to a coupled system of implicit type impulsive boundary value problems of fractional-order differential equations, Adv. Differ. Equ., 2019 (2019), 1–21. doi: 10.1186/s13662-018-1939-6
    [2] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Rev., 18 (1976), 620–709. doi: 10.1137/1018114
    [3] Z. Bai, Z. Du, S. Zhang, Iterative method for a class of fourth-order p-Laplacian beam equation, J. Appl Anal. Comput., 9 (2019), 1443–1453.
    [4] C. Promsakon, E. Suntonsinsoungvon, S. K. Ntouyas, J. Tariboon, Impulsive boundary value problems containing Caputo fractional derivative of a function with respect to another function, Adv. Differ. Equ., 2019 (2019), 1–17. doi: 10.1186/s13662-018-1939-6
    [5] J. Cao, H. Chen, Impulsive fractional differential equations with nonlinear boundary conditions, Math. Comp. Model., 55 (2012), 303–311. doi: 10.1016/j.mcm.2011.07.037
    [6] I. Dassios, D. Baleanu, Caputo and related fractional derivatives in singular systems, Appl. Math. Comput., 337 (2018), 591–606.
    [7] S. Dhar, Q. Kong, M. McCabe, Fractional boundary value problems and Lyapunov-type inequalities with fractional integral boundary conditions, Electron. J. Qual. Theo., 2016 (2016),
    [8] K. Diethelm, The analysis of fractional differential equations, Spring-Verlag Berlin, Heidelberg, 2010.
    [9] M. Feckan, Y. Zhou, J. R. Wang, On the concept and existence of solution for impulsive fractional differential equations, Commun. Nonlinear Sci., 17 (2012), 3050–3060. doi: 10.1016/j.cnsns.2011.11.017
    [10] D. Guo, J. Sun, Z. Liu, Nonlinear ordinary differential equation functional method, Shandong Science and Technology Press, Jinan (in Chinese), 2005.
    [11] Z. Gao, T. Hu, H. Pang, Existence and uniqueness theorems for a fractional differential equation with impulsive effect under Band-Like integral boundary conditions, Adv. Math. Phys., 2020 (2020), 1–8.
    [12] M. Jia, X. Liu, Multiplicity of solutions for integral boundary value problems of fractional differential equations with upper and lower solutions, Appl. Math. Comput., 232 (2014), 313–323.
    [13] M. Jia, L. Li, X. Liu, J. Song, Z. Bai, A class of nonlocal problems of fractional differential equations with composition of derivative and parameters, Adv. Differ. Equ., 2019 (2019), 1–26. doi: 10.1186/s13662-018-1939-6
    [14] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and applications of fractional differential equations, North-Holland Mathematics Studies, Elsevier Science, 2006.
    [15] M. Lázaro, J. L. Pérez-Aparicio, Dynamic analysis of frame structures with free viscoelastic layers: New closed-form solutions of eigenvalues and a viscous approach, Eng. Struct., 54 (2013), 69–81.
    [16] R. Lewandowski, Z. Pawlak, Dynamic analysis of frames with viscoelastic dampers modelled by rheological models with fractional derivatives, J. Sound Vib., 330 (2011), 923–936. doi: 10.1016/j.jsv.2010.09.017
    [17] X. Liu, M. Jia, Solvability and numerical simulations for BVPs of fractional coupled systems involving left and right fractional derivatives, Appl. Math. Comput., 353 (2019), 230–242.
    [18] X. Liu, M. Jia, W. Ge, The method of lower and upper solutions for mixed fractional four-point boundary value problem with p-Laplacian operator, Appl. Math. Lett., 65 (2017), 56–62. doi: 10.1016/j.aml.2016.10.001
    [19] X. Liu, M. Jia, The method of lower and upper solutions for the general boundary value problems of fractional differential equations with p-Laplacian, Adv. Differ. Equ., 2018 (2018), 1–15. doi: 10.1186/s13662-017-1452-3
    [20] I. Podluny, Fractional differential equtions, Academic press, San Diego, 1999.
    [21] T. Sandev, R. Metzler, Z. Tomovski, Correlation functions for the fractional generalized Langevin equation in the presence of internal and external noise, J. Math. Phys., 55 (2014), 1–23.
    [22] P. J. Torvik, R. L. Bagley, On the appearance of the fractional derivative in the behavior of real materials, J. Appl. Mech., 51 (1984), 294–298. doi: 10.1115/1.3167615
    [23] A. Ullah, K. Shah, T. Abdeljawad, R. A. Khan, I. Mahariq, Study of impulsive fractional differential equation under Robin boundary conditions by topological degree method, Bound. Value Probl., 2020 (2020), 1–17. doi: 10.1186/s13661-019-01311-5
    [24] B. S. Vadivoo, R. Ramachandran, J. Cao, H. Zhang, X. Li, Controllability analysis of nonlinear neutral-type fractional-order differential systems with state delay and impulsive effects, Int. J. Control. Autom., 16 (2018), 659–669. doi: 10.1007/s12555-017-0281-1
    [25] B. Zhu, L. Liu, Y. Wu, Existence and uniqueness of global mild solutions for a class of nonlinear fractional reaction-diffusion equations with delay, Comput. Math. Appl., 78 (2019), 1811–1818. doi: 10.1016/j.camwa.2016.01.028
    [26] T. Zhang, L. Xiong, Periodic motion for impulsive fractional functional differential equations with piecewise Caputo derivative, Appl. Math. Lett., 101 (2020), 1–7.
  • Reader Comments
  • © 2021 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)
通讯作者: 陈斌, bchen63@163.com
  • 1. 

    沈阳化工大学材料科学与工程学院 沈阳 110142

  1. 本站搜索
  2. 百度学术搜索
  3. 万方数据库搜索
  4. CNKI搜索

Metrics

Article views(2557) PDF downloads(237) Cited by(0)

Other Articles By Authors

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog