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Research article Special Issues

Regularity criteria for a two dimensional Erying-Powell fluid flowing in a MHD porous medium


  • Received: 28 July 2022 Revised: 23 August 2022 Accepted: 25 August 2022 Published: 05 September 2022
  • The intention and novelty in the presented study were to develop the regularity analysis for a parabolic equation describing a type of Eyring-Powell fluid flow in two dimensions. We proved that, under certain general conditions involving the space of bounded mean oscillation (BMO) and the Lebesgue space L2, there exist bounded and regular velocity solutions under the L2 space scope. This conclusion was additionally supplemented by the condition of a finite square integrable initial data (also some of the obtained expressions involved the gradient and the laplacian of the initial velocity distribution). To make our results further general, the proposed analysis was extended to cover regularity results in Lp(p>2) spaces. As a remarkable conclusion, we highlight that the solutions to the two dimensional Eyring-Powell fluid flow did not exhibit blow up behaviour.

    Citation: José Luis Díaz Palencia, Saeed Ur Rahman, Saman Hanif. Regularity criteria for a two dimensional Erying-Powell fluid flowing in a MHD porous medium[J]. Electronic Research Archive, 2022, 30(11): 3949-3976. doi: 10.3934/era.2022201

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  • The intention and novelty in the presented study were to develop the regularity analysis for a parabolic equation describing a type of Eyring-Powell fluid flow in two dimensions. We proved that, under certain general conditions involving the space of bounded mean oscillation (BMO) and the Lebesgue space L2, there exist bounded and regular velocity solutions under the L2 space scope. This conclusion was additionally supplemented by the condition of a finite square integrable initial data (also some of the obtained expressions involved the gradient and the laplacian of the initial velocity distribution). To make our results further general, the proposed analysis was extended to cover regularity results in Lp(p>2) spaces. As a remarkable conclusion, we highlight that the solutions to the two dimensional Eyring-Powell fluid flow did not exhibit blow up behaviour.



    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 by

    By Lemma 3.1 and (3.5),

    We can easily get that the following Lemma 4.1 holds.

    Lemma 4.1. The function is the solution of boundary value problem (1.1) if and only if is a fixed point of the operator in .

    Lemma 4.2. If (H1) holds, then is completely continuous.

    Proof. Step 1: is a continuous operator.

    Suppose that and there exists such that . Then there exists a constant such that

    Since , and , then

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

    and

    So as , which means is continuous.

    Step 2: is relatively compact.

    Let be a bounded set. By the continuity of the functions , and , there exists a constant , for any and ,

    Then

    By Lemma 3.2, for any ,

    and

    Therefore, the operator is uniformly bounded.

    Because is continuous on , then it is uniformly continuous on . Thus, for any , there exists a constant such that for any , , whenever , we can get that

    Denote

    By (3.6),

    Similarly, for the , there exists a constant such that

    whenever and .

    By the uniformly continuity of functions and on , we can show that for the , there exists a constant such that

    and

    whenever and . Hence, by (4.2) and (4.3), we have

    and

    By the uniformly continuity of on , for the , there exists a constant , for , and , we have

    Then

    We take . Therefore, for any , there exists a constant such that for whenever and any , we can get that

    and

    Thus, the operator is equicontinuous on every interval .

    According to the Arzela-Ascoli theorem, is relatively compact.

    Therefore, is completely continuous.

    In the following, we give out some hypotheses.

    (H2) if , , . Furthermore, if and .

    And for ,

    (H3.1) If , then for , ,

    and

    (H3.2) If , then for , ,

    and

    (H3.3) If , then for , ,

    and

    (H3.4) If , then for , ,

    and

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

    Lemma 4.3. For , if one of the following conditions holds, then and for .

    (1) and satisfies

    (2) and satisfies

    (3) and satisfies

    (4) and satisfies

    Proof. (1) Denote Then .

    By Lemma 3.1, the following boundary value problem

    has a unique solution

    and

    It's easy to see that and for .

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

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

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

    For any , and which implies that and for . By (H2), for any ,

    Since , there exists an interval such that for . It follows from (H2)

    (4.4)

    By (H3.1), we can get that

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

    and

    By Lemma 3.2, (H2) and (4.4), for any and , as , we have

    and

    Thus,

    which implies that is a strongly increasing operator.

    The proof is completed.

    Theorem 4.5. If (H2) and (H3.1) hold, there exist , with such that

    (4.5)

    and

    (4.6)

    where , and are not the solutions of boundary value problem (1.1). Then boundary value problem (1.1) has at least three distinct solutions and satisfies

    Proof. By Lemma 4.2 and Lemma 4.4, we can get that is a completely continuous strongly increasing operator.

    By the definition of operator , we can show that

    Let . By (4.5),

    Similarly, we can get that

    By (1) in Lemma 4.3, we have

    Therefore, .

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

    By using the same method, we can easily get .

    By using Lemma 2.6, we can get that the operator has at least three distinct fixed points , and satisfies

    Therefore, by Lemma 4.1, we can obtain that boundary value problem (1.1) has at least three distinct solutions , and

    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 , with such that

    and

    where , and are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions and satisfies

    Theorem 4.7. If (H2) and (H3.3) hold, there exist , with such that

    and

    where , and are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions and satisfies

    Theorem 4.8. If (H2) and (H3.4) hold, there exist , with such that

    and

    where , and are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions and satisfies

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

    Consider the following the boundary value problem

    (5.1)

    where .

    Obviously, are nonlinear functions. And satisfy (H2).

    Let . Since

    then satisfy (H3.1).

    For , we take

    We can easily get that

    and

    So

    and

    We can easily get that satisfy (4.5) and satisfy (4.6) with .

    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 and moreover,

    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.



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