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Research article

Some stronger forms of mean sensitivity

  • The equivalence between multi-transitive mean sensitivity and multi-transitive mean n-sensitivity for linear dynamical systems was demonstrated in this study. Furthermore, this paper presented examples that highlighted the disparities among mean sensitivity, multi-transitive mean sensitivity, and syndetically multi-transitive mean sensitivity.

    Citation: Quanquan Yao, Yuanlin Chen, Peiyong Zhu, Tianxiu Lu. Some stronger forms of mean sensitivity[J]. AIMS Mathematics, 2024, 9(1): 1103-1115. doi: 10.3934/math.2024054

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  • The equivalence between multi-transitive mean sensitivity and multi-transitive mean n-sensitivity for linear dynamical systems was demonstrated in this study. Furthermore, this paper presented examples that highlighted the disparities among mean sensitivity, multi-transitive mean sensitivity, and syndetically multi-transitive mean sensitivity.



    The well-known classical Lyapunov inequality [15] states that, if u is a nontrivial solution of the Hill's equation

    u(t)+q(t)u(t)=0, a<t<b, (1.1)

    subject to Dirichlet-type boundary conditions:

    u(a)=u(b)=0, (1.2)

    then

    ba|q(t)|dt>4ba, (1.3)

    where q:[a,b]R is a real and continuous function.

    Later, in 1951, Wintner [24], obtained the following inequality:

    baq+(t)dt>4ba, (1.4)

    where q+(t)=max{q(t),0}.

    A more general inequality was given by Hartman and Wintner in [12], that is known as Hartman Wintner-type inequality:

    ba(ta)(bt)q+(t)dt>ba, (1.5)

    Since maxt[a,b](ta)(bt)=(ba)24, then, (1.5) implies (1.4).

    The Lyapunov inequality and its generalizations have many applications in different fields such in oscillation theory, asymptotic theory, disconjugacy, eigenvalue problems.

    Recently, many authors have extended the Lyapunov inequality (1.3) for fractional differential equations [1,2,3,4,5,6,7,8,9,10,11,12,13,15,18,20,22,23,24]. For this end, they substituted the ordinary second order derivative in (1.1) by a fractional derivative or a conformable derivative. The first result in which a fractional derivative is used instead of the ordinary derivative in equation (1.1), is the work of Ferreira [6]. He considered the following two-point Riemann-Liouville fractional boundary value problem

    Dαa+u(t)+q(t)u(t)=0, a<t<b, 1<α2
    u(a)=u(b)=0.

    And obtained the Lyapunov inequality:

    ba|q(t)|dt>Γ(α)(4ba)α1.

    Then, he studied in [7], the Caputo fractional differential equation

    CDαa+u(t)+q(t)u(t)=0, a<t<b, 1<α2

    under Dirichlet boundary conditions (1.2). In this case, the corresponding Lyapunov inequality has the form

    ba|q(t)|dt>ααΓ(α)((α1)(ba))α1.

    Later Agarwal and Özbekler in [1], complimented and improved the work of Ferreira [6]. More precisely, they proved that if u is a nontrivial solution of the Riemann-Liouville fractional forced nonlinear differential equations of order α(0,2]:

    Dαa+u(t)+p(t)|u(t)|μ1u(t)+q(t)|u(t)|γ1u(t)=f(t), a<t<b,

    satisfying the Dirichlet boundary conditions (1.2), then the following Lyapunov type inequality

    (ba[p+(t)+q+(t)]dt)(ba[μ0p+(t)+γ0q+(t)+|f(t)|]dt)>42α3Γ2(α)(ba)2α2.

    holds, where p, q, f are real-valued functions, 0<γ<1<μ<2, μ0=(2μ)μμ/(2μ)22/(μ2) and γ0=(2γ)γγ/(2γ)22/(γ2).

    In 2017, Guezane-Lakoud et al. [11], derived a new Lyapunov type inequality for a boundary value problem involving both left Riemann-Liouville and right Caputo fractional derivatives in presence of natural conditions

    CDαbDβa+u(t)+q(t)u(t)=0, a<t<b, 0<α,β1
    u(a)=Dβa+u(b)=0, 

    then, they obtained the following Lyapunov inequality:

    ba|q(t)|dt>(α+β1)Γ(α)Γ(β)(ba)α+β1.

    Recently, Ferreira in [9], derived a Lyapunov-type inequality for a sequential fractional right-focal boundary value problem

    CDαa+Dβa+u(t)+q(t)u(t)=0, a<t<b
    u(a)=Dγa+u(b)=0, 

    where 0<α,β,γ1, 1<α+β2, then, they obtained the following Lyapunov inequality:

    ba(bs)α+βγ1|q(t)|dt>1C,

    where

    C=(ba)γmax{Γ(βγ+1)Γ(α+βγ)Γ(β+1),1αβΓ(α+β)(Γ(βγ+1)Γ(α+β1)Γ(α+βγ)Γ(β))α+β1α1, with α<1}

    Note that more generalized Lyapunov type inequalities have been obtained for conformable derivative differential equations in [13]. For more results on Lyapunov-type inequalities for fractional differential equations, we refer to the recent survey of Ntouyas et al. [18].

    In this work, we obtain Lyapunov type inequality for the following mixed fractional differential equation involving both right Caputo and left Riemann-Liouville fractional derivatives

    CDαbDβa+u(t)+q(t)u(t)=0, a<t<b, (1.6)

    satisfying the Dirichlet boundary conditions (1.2), here 0<βα1, 1<α+β2, CDαb denotes right Caputo derivative, Dβa+ denotes the left Riemann-Liouville and q is a continuous function on [a,b].

    So far, few authors have considered sequential fractional derivatives, and some Lyapunov type inequalities have been obtained. In this study, we place ourselves in a very general context, in that in each fractional operator, the order of the derivative can be different. Such problems, with both left and right fractional derivatives arise in the study of Euler-Lagrange equations for fractional problems of the calculus of variations [2,16,17]. However, the presence of a mixed left and right Caputo or Riemann-Liouville derivatives of order 0<α<1 leads to great difficulties in the study of the properties of the Green function since in this case it's given as a fractional integral operator.

    We recall the concept of fractional integral and derivative of order p>0. For details, we refer the reader to [14,19,21]

    The left and right Riemann-Liouville fractional integral of a function g are defined respectively by

    Ipa+g(t)=1Γ(p)tag(s)(ts)1pds,Ipbg(t)=1Γ(p)btg(s)(st)1pds.

    The left and right Caputo derivatives of order p>0, of a function g are respectively defined as follows:

    CDpa+g(t)=Inpa+g(n)(t),CDpbg(t)=(1)nInpbg(n)(t),

    and the left and right Riemann-Liouville fractional derivatives of order p>0, of a function g\ are respectively defined as follows:

    Dpa+g(t)=dndtn(Inpa+g)(t),Dpbg(t)=(1)ndndtnInpbg(t),

    where n is the smallest integer greater or equal than p.

    We also recall the following properties of fractional operators. Let 0<p<1, then:

    1- IpCa+Dpa+f(t)=f(t)f(a).

    2- IpCbDpbf(t)=f(t)f(b).

    3- (Ipa+c)(t)=c(ta)pΓ(p+1),cR

    4- Dpa+u(t)=CDpa+u(t), when u(a)=0.

    5- Dpbu(t)=CDpbu(t), when u(b)=0.

    Next we transform the problem (1.6) with (1.2) to an equivalent integral equation.

    Lemma 1. Assume that 0<α,β1. The function u is a solution to the boundary value problem (1.6) with (1.2) if and only if u satisfies the integral equation

    u(t)=baG(t,r)q(r)u(r)dr, (2.1)

    where

    G(t,r)=1Γ(α)Γ(β)(inf{r,t}a(ts)β1(rs)α1ds
    (ta)β(ba)βra(bs)β1(rs)α1ds) (2.2)

    is the Green's function of problem (1.6) with (1.2).

    Proof. Firstly, we apply the right side fractional integral Iαb to equation (1.6), then the left side fractional integral Iβa+ to the resulting equation and taking into account the properties of Caputo and\Riemann-Liouville fractional derivatives and the fact that Dβa+u(t)=CDβa+u(t), we get

    u(t)=Iβa+Iαbq(t)u(t)+c(ta)βΓ(β+1). (2.3)

    In view of the boundary condition u(b)=0, we get

    c=Γ(β+1)(ba)βIβa+Iαbq(t)u(t)t=b.

    Substituting c in (2.3), it yields

    u(t)=Iβa+Iαbq(t)u(t)(ta)β(ba)βIβa+Iαbq(t)u(t)t=b=1Γ(α)Γ(β)ta(ts)β1(bs(rs)α1q(r)u(r)dr)ds(ta)βΓ(α)Γ(β)(ba)βba(bs)β1(bs(rs)α1q(r)u(r)dr)ds.

    Finally, by exchanging the order of integration, we get

    u(t)=1Γ(α)Γ(β)ta(ra(ts)β1(rs)α1ds)q(r)u(r)dr+1Γ(α)Γ(β)bt(ta(ts)β1(rs)α1ds)q(r)u(r)dr(ta)βΓ(α)Γ(β)(ba)βba(ra(bs)β1(rs)α1ds)q(r)u(r)dr,

    thus

    u(t)=baG(t,r)q(r)u(r)dr,

    with

    G(t,r)=1Γ(α)Γ(β){ra(ts)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds,artb,ta(ts)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds,atrb.

    that can be written as

    G(t,r)=1Γ(α)Γ(β)(inf{r,t}a(ts)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds).

    Conversely, we can verify that if u satisfies the integral equation (2.1), then u is a solution to the boundary value problem (1.6) with (1.2). The proof is completed.

    In the next Lemma we give the property of the Green function G that will be needed in the sequel.

    Lemma 2. Assume that 0<βα1,1<α+β2, then the Green function G(t,r) given in (2.2) of problem (1.6) with (1.2) satisfies the following property:

    |G(t,r)|1Γ(α)Γ(β)(α+β1)(α+β)(α(ba)(β+α))α+β1,

    for all artb.

    Proof. Firstly, for artb, we have G(t,r)0. In fact, we have

    G(t,r)=1Γ(α)Γ(β)(ra(ts)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds)1Γ(α)Γ(β)(ra(bs)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds)
    =1Γ(α)Γ(β)(1(ta)β(ba)β)ra(bs)β1(rs)α1ds0 (2.4)

    in addition,

    G(t,r)1Γ(α)Γ(β)(ra(rs)β1(rs)α1ds(ra)β(ba)βra(bs)β1(rs)α1ds)1Γ(α)Γ(β)((ra)α+β1(α+β1)(ra)β(ba)βra(ba)β1(rs)α1ds)
    =1Γ(α)Γ(β)((ra)α+β1(α+β1)(ra)β+αα(ba)). (2.5)

    Thus, from (2.4) and (2.5), we get

    0G(t,r)h(r), artb, (2.6)

    where

    h(s):=1Γ(α)Γ(β)((sa)α+β1(α+β1)(sa)β+αα(ba)),

    it is clear that h(s)0, for all s[a,b].

    Now, for atrb, we have

    G(t,r)=1Γ(α)Γ(β)(ta(ts)β1(rs)α1ds(ta)β(ba)βra(bs)β1(rs)α1ds)1Γ(α)Γ(β)(ta(ts)β1(ts)α1ds(ta)β(ba)ra(rs)α1ds)=1Γ(α)Γ(β)((ta)α+β1(α+β1)(ta)β(ra)αα(ba))
    1Γ(α)Γ(β)((ta)α+β1(α+β1)(ta)β+αα(ba))=h(t). (2.7)

    On the other hand,

    G(t,r)1Γ(α)Γ(β)(ra)α1ta(ts)β1ds(ta)β(ba)βra(rs)β1(rs)α1ds)1Γ(α)Γ(β)((ta)α(ta)ββ(ba)(ta)β(ba)β(ra)α+β1(α+β1))1Γ(α)Γ(β)((ta)α+ββ(ba)(ta)β(ra)α1(α+β1))1Γ(α)Γ(β)((ta)α+ββ(ba)(ta)α+β1(α+β1)),

    since βα, we get

    G(t,r)h(t), atrb. (2.8)

    From (2.7) and (2.8) we obtain

    |G(t,r)|h(t), atrb. (2.9)

    Finally, by differentiating the function h, it yields

    h(s)=1Γ(α)Γ(β)(sa)α+β2(1(β+α)(sa)α(ba)).

    We can see that h(s)=0 for s0=a+α(ba)(β+α)(a,b), h(s)<0 for s>s0 and h(s)>0 for s<s0. Hence, the function h(s) has a unique maximum given by

    maxs[a,b]h(s)=h(s0)=1Γ(α)Γ(β)((α(ba)(β+α))α+β1(α+β1)(α(ba)(β+α))β+αα(ba))=1Γ(α)Γ(β)(α+β1)(α+β)(α(ba)(β+α))α+β1.

    From (2.6) and (2.9), we get |G(t,r)|h(s0), from which the intended result follows.

    Next, we state and prove the Lyapunov type inequality for problem (1.6) with (1.2).

    Theorem 3. Assume that 0<βα1 and 1<α+β2. If the fractional boundary value problem (1.6) with (1.2) has a nontrivial continuous solution, then

    ba|q(r)|drΓ(α)Γ(β)(α+β1)(α+β)α+β(α(ba))α+β1. (2.10)

    Proof. Let X=C[a,b] be the Banach space endowed with norm ||u||=maxt[a,b]|u(t)|. It follows from Lemma 1 that a solution uX to the boundary value problem (1.6) with (1.2) satisfies

    |u(t)|ba|G(t,r)||q(r)||u(r)|druba|G(t,r)|q(r)dr,

    Now, applying Lemma 2 to equation (2.1), it yields

    |u(t)|1Γ(α)Γ(β)(α+β1)(α+β)(α(ba)(β+α))α+β1uba|q(r)|dr

    Hence,

    u(α(ba))α+β1Γ(α)Γ(β)(α+β1)(α+β)α+βuba|q(r)|dr,

    from which the inequality (2.10) follows. Note that the constant in (2.10) is not sharp. The proof is completed.

    Remark 4. Note that, according to boundary conditions (1.2), the Caputo derivatives CDαb and  CDβa+ coincide respectively with the Riemann-Liouville derivatives Dαb and Dβa+. So, equation (1.6) is reduced to the one containing only Caputo derivatives or only Riemann-Liouville derivatives, i.e.,

    CDαCbDβa+u(t)+q(t)u(t)=0, a<t<b

    or

    DαbDβa+u(t)+q(t)u(t)=0, a<t<b

    Furthermore, by applying the reflection operator (Qf)(t)=f(a+bt) and taking into account that QCDαa+=CDαbQ and QCDβb=CDβa+Q (see [21]), we can see that, the boundary value problem (1.6) with (1.2) is equivalent to the following problem

    CDαa+Dβbu(t)+q(t)u(t)=0, a<t<b,
    u(a)=u(b)=0.

    Remark 5. If we take α=β=1, then the Lyapunov type inequality (2.3) is reduced to

    ba|q(t)|dt4ba.

    The authors thank the anonymous referees for their valuable comments and suggestions that improved this paper.

    All authors declare no conflicts of interest in this paper.



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