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Research article Special Issues

A study on centrality measures in weighted networks: A case of the aviation network

  • Received: 10 November 2023 Revised: 03 December 2023 Accepted: 12 December 2023 Published: 09 January 2024
  • MSC : 05C82

  • Identifying influential spreaders in complex networks is a crucial issue that can help control the propagation process in complex networks. An aviation network is a typical complex network, and accurately identifying the key city nodes in the aviation network can help us better prevent network attacks and control the spread of diseases. In this paper, a method for identifying key nodes in undirected weighted networks, called weighted Laplacian energy centrality, was proposed and applied to an aviation network constructed from real flight data. Based on the analysis of the topological structure of the network, the paper recognized critical cities in this network, then simulation experiments were conducted on key city nodes from the perspectives of network dynamics and robustness. The results indicated that, compared with other methods, weighted Laplacian energy centrality can identify the city nodes with the most spreading influence in the network. From the perspective of network robustness, the identified key nodes also have the characteristics of accurately and quickly destroying network robustness.

    Citation: Shuying Zhao, Shaowei Sun. A study on centrality measures in weighted networks: A case of the aviation network[J]. AIMS Mathematics, 2024, 9(2): 3630-3645. doi: 10.3934/math.2024178

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  • Identifying influential spreaders in complex networks is a crucial issue that can help control the propagation process in complex networks. An aviation network is a typical complex network, and accurately identifying the key city nodes in the aviation network can help us better prevent network attacks and control the spread of diseases. In this paper, a method for identifying key nodes in undirected weighted networks, called weighted Laplacian energy centrality, was proposed and applied to an aviation network constructed from real flight data. Based on the analysis of the topological structure of the network, the paper recognized critical cities in this network, then simulation experiments were conducted on key city nodes from the perspectives of network dynamics and robustness. The results indicated that, compared with other methods, weighted Laplacian energy centrality can identify the city nodes with the most spreading influence in the network. From the perspective of network robustness, the identified key nodes also have the characteristics of accurately and quickly destroying network robustness.



    This article is devoted to the following nonlinear fractional differential equation with periodic boundary condition

    {cDα0+x(t)λx(t)=f(t,x(t)), 0<tω,x(0)=x(ω), (1.1)

    where λ0, 0<α1 and cDα0+ is Caputo fractional derivative

    cDα0+x(t)=1Γ(1α)t0(ts)αx(s)ds.

    Differential equations of fractional order occur more frequently on different research areas and engineering, such as physics, economics, chemistry, control theory, etc. In recent years, boundary value problems for fractional differential equation have become a hot research topic, see [2,3,4,5,6,7,9,10,11,12,13,15,16,17,18,19,20,21,24,25]. In [27], Zhang studied the boundary value problem for nonlinear fractional differential equation

    {cDα0+u(t)=f(t,u(t)), 0<t<1,u(0)+u(0)=0,u(1)+u(1)=0, (1.2)

    where 1<α2, f:[0,1]×[0,+)[0,+) is continuous and cDα0+ is Caputo fractional derivative

    cDα0+u(t)=1Γ(2α)t0(ts)1αu(s)ds.

    The author obtained the existence of the positive solutions by using the properties of the Green function, Guo-Krasnosel'skill fixed point theorem and Leggett-Williams fixed point theorem.

    Ahmad and Nieto [1] studied the anti-periodic boundary value problem of fractional differential equation

    {cDq0+u(t)=f(t,u(t)), 0tT, 1<q2,u(0)=u(T), cDp0+u(0)=cDp0+u(T), 0<p<1, (1.3)

    where f:[0,T]×RR is continuous. The authors obtained some existence and uniqueness results by applying fixed point principles. The anti-periodic boundary value condition in this article corresponds to the anti-periodic condition u(0)=u(T),u(0)=u(T) in ordinary differential equation.

    In [26], Zhang studied the following fractional differential equation

    {Dδ0+u(t)=f(t,u), 0<tT,limt0+t1αu(t)=u0, (1.4)

    where 0<δ<1, T>0, u0R and Dδ0+ is Riemann-Liouville fractional derivative

    Dδ0+u(t)=1Γ(1δ)ddtt0(ts)δu(s)ds.

    The author obtained the existence and uniqueness of the solutions by the method of upper and lower solutions and monotone iterative method.

    In [7], Belmekki, Nieto and Rodriguez-Lopez studied the following equation

    {Dδ0+u(t)λu(t)=f(t,u(t)),0<t1,limt0+t1δu(t)=u(1), (1.5)

    where 0<δ<1, λR, f is continuous. The authors obtained the existence and uniqueness of the solutions by using the fixed point theorem. Cabada and Kisela [8] studied the following equation

    {Dδ0+u(t)λu(t)=f(t,t1αu(t)),0<t1,limt0+t1δu(t)=u(1), (1.6)

    where 0<δ<1, λ0(λR), f is continuous. The authors studied the existence and uniqueness of periodic solutions by using Krasnosel'skii fixed point theorem and monotone iterative method. In [7,8], the boundary condition limt0+t1δu(t)=u(1) was called as periodic boundary value condition of Riemann-Liouville fractional differential equation, which is different from the periodic condition for ordinary differential equation. The boundary value condition u(0)=u(1) is not suitable for Riemann-Liouville fractional differential equation.

    For the ordinary differential equation, the periodic boundary value problem is closely related to the periodic solution. For the Caputo fractional differential equation, the periodic boundary value condition u(0)=u(w) is meaningful. As far as we know, few work involves the periodic boundary value problem for Caputo fractional. The aim of this paper is to show the existence of positive solutions of (1.1) by using Krasnosel'skii fixed point theorem. Meanwhile, we also use the monotone iterative method to study the extremal solutions problem

    {cDα0+u(t)=f(t,u(t)),0<tω,u(0)=u(ω). (1.7)

    The paper is organized as follows. In Section 2, we recall and derive some results on Mittag-Leffler functions. In Section 3, we use the Laplace transform to obtain the solution of a linear problem and discuss some properties of Green's function. In Section 4, the existence of positive solution is studied by using the Krasnosel'skii fixed point theorem. In Section 5, the existence of extremal solutions is proved by utilizing the monotone iterative technique. Section 6 is conclusion of the paper.

    A key role in the theory of linear fractional differential equation is played by the well-known two-parameter Mittag-Leffler function

    Eα,β(z)=Σk=0zkΓ(αk+β),  zR, α, β>0. (2.1)

    We recall and derive some of their properties and relationships summarized in the following.

    Proposition 2.1. Let α(0,1],β>0,λR and ξ>0. Then it holds

    (C1) limt0+Eα,β(λtα)=1Γ(β),limt0+Eα,1(λtα)=1.

    (C2) Eα,α+1(λtα)=λ1tα(Eα,1(λtα)1).

    (C3) Eα,α(λtα)>0, Eα,1(λtα)>0 for all t0.

    (C4) Eα,α(λtα) is decreasing in t for λ<0and increasing for λ>0 for all t>0.

    (C5) Eα,1(λtα) is decreasing in t for λ<0 and increasing forλ>0 for all t>0.

    (C6) ξ0tβ1Eα,β(λtα)dt=ξβEα,β+1(λξα).

    Proof. (C1) It is obtained by an immediate calculation from (2.1).

    (C2) By (2.1), we get

    Eα,1(λtα)=Σk=0(λtα)kΓ(αk+1)=1+λtαΓ(α+1)+(λtα)2Γ(2α+1)+(λtα)3Γ(3α+1)+,
    Eα,α+1(λtα)=Σk=0(λtα)kΓ(αk+α+1)=1Γ(α+1)+λtαΓ(2α+1)+(λtα)2Γ(3α+1)+.

    Hence,

    Eα,α+1(λtα)=λ1tα(Eα,1(λtα)1).

    (C3) It follows from [23,Lemma 2.2].

    (C4) It follows from [8,Proposition 1].

    (C5) By a direct calculation, we get

    ddtEα,1(λtα)=λtα1Eα,α(λtα),

    since α(0,1], t>0 and Eα,α(λtα) is positive by Proposition (C3), the assertion is proved.

    (C6) It follows from (1.99) of [20].

    In this section, we deal with the linear case that f(t,x)=f(t) is a continuous function by mean of the Laplace transform for caputo fractional derivative

    (LcDα0+x)(s)=sαX(s)sα1x(0), 0<α1, (3.1)

    where L denotes the Laplace transform operator, X(s) denotes the Laplace transform of x(t).

    From Lemma 3.2 of [14], we get

    (LEα(λtα))(s)=sα1sαλ,Re(s)>0,λC,|λsα|<1, (3.2)

    and

    (Ltβ1Eα,β(λtα))(s)=sαβsαλ,Re(s)>0,λC,|λsα|<1. (3.3)

    We do Laplace transform to the equation

    cDα0+x(t)λx(t)=f(t),  x(0)=x(ω). (3.4)

    By (3.1), we obtain

    sαX(s)λX(s)=F(s)+x(0)sα1,
    X(s)=F(s)sαλ+sα1sαλx(0),

    where F denotes the Laplace transform of f. By (3.2) and (3.3), we obtain that

    x(t)=t0(ts)α1Eα,α(λ(ts)α)f(s)ds+x(0)Eα,1(λtα). (3.5)

    Hence,

    x(ω)=x(0)Eα,1(λwα)+ω0(ωs)α1Eα,α(λ(ωs)α)f(s)ds=x(0),

    which implies that

    x(0)=ω0(ωs)α1Eα,α(λ(ωs)α)f(s)ds1Eα,1(λωα)

    if Eα,1(λωα)1. Therefore, if Eα,1(λωα)1, the solution of the problem (3.4) is

    x(t)=ω0(ωs)α1Eα,α(λ(ωs)α)f(s)ds1Eα,1(λωα)Eα,1(λtα)+t0Eα,α(λ(ts)α)(ts)1αf(s)ds=t0Eα,1(λtα)Eα,α(λ(ωs)α)(1Eα,1(λωα))(ωs)1αf(s)ds+t0Eα,α(λ(ts)α)(ts)1αf(s)ds+ωtEα,1(λtα)Eα,α(λ(ωs)α)(1Eα,1(λωα))(ωs)1αf(s)ds.

    Theorem 3.1. Let Eα,1(λωα)1, the periodic boundary value problem (3.4) has a unique solutiongiven by

    x(t)=ω0Gα,λ(t,s)f(s)ds,

    where

    Gα,λ(t,s)={Eα,1(λtα)Eα,α(λ(ωs)α)(1Eα,1(λωα))(ωs)1α+Eα,α(λ(ts)α)(ts)1α, 0s<tω,Eα,1(λtα)Eα,α(λ(ωs)α)(1Eα,1(λωα))(ωs)1α, 0ts<ω. (3.6)

    Remark 3.2. The unique solution x of (3.4) is continuous on [0,ω].

    Lemma 3.3. Let 0<α1,λ0 and sign(η) denotes the signumfunction. Then

    (F1) limt0+Gα,λ(t,s)=Eα,α(λ(ωs)α)(1Eα,1(λωα))(ωs)1α for anyfixed s[0,ω),

    (F2)limsωGα,λ(t,s)=sign(λ) for any fixed t[0,ω],

    (F3)limts+Gα,λ(t,s)= for any fixeds[0,ω),

    (F4)Gα,λ(t,s)>0 for λ<0 and for all t[0,ω] and s[0,ω),

    (F5)Gα,λ(t,s) changes its sign for λ>0 for t[0,ω] and s[0,ω).

    Proof. (F1) When 0ts<ω, by Proposition 2.1 (C1) we can get (F1).

    (F2) When 0ts<ω, it follows by Proposition 2.1 (C5) that 1Eα,1(λωα) is positive for λ<0 and negative for λ>0. The unboundedness is implied by continuity of Mittag-Leffler function, Proposition 2.1 (C1) and the relation limt0+tr= for r>0.

    (F3) When 0s<tω, the first term of (3.6) is finite due to the continuity of the involved functions. And by a similar argument as in the previous point of this proof we have the second term tends to infinity.

    (F4) It is obtained by the positivity of all involved functions (Proposition 2.1 (C3)) and the inequation 1Eα,1(λωα)>0 for λ<0.

    (F5) When 0s<tω, the second term of (3.6) is positive due to

    limst(ts)α1Eα,α(λ(ts)α)=+,

    and by the positivity of all involved functions (Proposition 2.1 (C3)) we get the proof. When 0ts<ω, it is obtained by 1Eα,1(λωα)<0 for λ>0 and the positivity of all involved functions (Proposition 2.1 (C3)).

    Proposition 3.4. Let α(0,1] and λ<0. Then the Green's function (3.6) satisfies

    (K1) Gα,λ(t,s)m=:Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα)>0,

    (K2)ω0Gα,λ(t,s)ds=M=:1|λ| for all t[0,ω].

    Proof. (K1) For 0ts<ω, we deduce from Proposition 2.1 (C4),(C5) that Gα,λ has the minimum on the line t=s. Hence,

    Gα,λ(t,s)Gα,λ(t,t)=Eα,1(λtα)Eα,α(λ(ωt)α)(1Eα,1(λωα))(ωt)1αEα,1(λωα)Eα,α(λωα)(1Eα,1(λωα))ω1α=Eα,1(λωα)Eα,α(λωα)[1(Eα,α+1(λωα)λωα+1)]ω1α=Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα).

    For 0s<tω, we have

    Gα,λ(t,s)Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα)+Eα,α(λωα)ω1αEα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα).

    (K2) Employing Proposition 2.1, we get

    ω0Gα,λ(t,s)ds=ω0Eα,1(λtα)Eα,α(λ(ωs)α)(ωs)1α(1Eα,1(λωα))ds+t0Eα,α(λ(ts)α)(ts)1αds=Eα,1(λtα)1Eα,1(λωα)ω0(ωs)α1Eα,α(λ(ωs)α)ds+t0(ts)α1Eα,α(λ(ts)α)ds=Eα,1(λtα)1Eα,1(λωα)ωαEα,α+1(λωα)+tαEα,α+1(λtα)=Eα,1(λtα)1Eα,1(λωα)ωαλ1ωα(Eα,1(λωα)1)+tαλ1tα(Eα,1(λtα)1)=1|λ|,

    which completes the proof.

    Let C[0,ω] be the space continuous function on [0,ω] with the norm x=sup{|x(t)|:t[0,ω]}.

    In this section, we always assume that λ<0. Clearly, x is a solution of (1.1) if and only if

    x(t)=ω0Gα,λ(t,s)f(s,x(s))ds, (4.1)

    where Gα,λ is Green's function defined in Theorem 3.1.

    The following famous Krasnosel'skii fixed point theorem, which is main tool of this section.

    Theorem 4.1. [22] Let B be a Banach space, and let PB be a cone. AssumeΩ1,Ω2 two open and bounded subsets of B with0Ω1,Ω1Ω2 and letA:P(¯Ω2Ω1)P bea completely continuous operator such that one of the followingconditions is satisfied:

    (L1) Axx, if xPΩ1, andAxx, if xPΩ2,

    (L2) Axx, if xPΩ1, andAxx, if xPΩ2.

    Then, A has at least one fixed point inP(¯Ω2Ω1).

    Proposition 4.2. Assume that there exist 0<r<R,0<c1<c2 such that

    f:[0,ω]×[mc1ωMc2r,R]R  is continuous, (4.2)
    c1f(t,u)c2, (t,u)[0,ω]×[mc1ωMc2r,R]. (4.3)

    Let PC[0,ω] be the cone

    P={xC[0,ω]:mint[0,ω]x(t)mc1ωMc2x}.

    Then the operator A:¯PRPrP givenby

    Ax(t)=ω0Gα,λ(t,s)f(s,x(s))ds (4.4)

    is completely continuous, where Pl={uP:u<l}.

    Proof. Let x¯PRPr, then

    mc1ωMc2rx(t)R  for  all  t[0,ω]. (4.5)

    We first show that A is well-defined, i.e. that A:¯PRPrP. Note that

    Ax(t)=ω0Gα,λ(t,s)f(s,x(s))ds=ω0Eα,1(λtα)Eα,α(λ(ωs)α)(ωs)1α(1Eα,1(λωα))f(s,x(s))ds       +t0Eα,α(λ(ts)α)(ts)1αf(s,x(s))ds=kq(ω)Eα,1(λtα)+q(t), (4.6)

    where k=11Eα,1(λωα) and

    q(t)={t0Eα,α(λ(ts)α)(ts)1αf(s,x(s))ds,0<tω,0,t=0. (4.7)

    Clearly, for t(0,ω]

    0<q(t)c2t0Eα,α(λ(ts)α)(ts)1αdsc2Γ(α)t0(ts)α1ds=c2Γ(α+1)tα, (4.8)

    which implies that q is continuous at t=0. On the other hand,

    q(t)=k<1α1λkt0(ts)αk+α1Γ(αk+α)f(s,x(s))ds+t0k1α1λkΓ(αk+α)(ts)αk+α1f(s,x(s))ds=k<1α1λkΓ(αk+α)t0uαk+α1f(tu,x(tu))du+t0k1α1λkΓ(αk+α)(ts)αk+α1f(s,x(s))ds=:H1(t)+H2(t),

    where

    H1(t)=k<1α1λkΓ(αk+α)t0uαk+α1f(tu,x(tu))du,H2(t)=t0k1α1λkΓ(αk+α)(ts)αk+α1f(s,x(s))ds.

    Since

    |uαk+α1f(tu,x(tu))|c2uαk+α1,u>0, t(0,ω], x¯PR/Pr,|λk(ts)αk+α1Γ(αk+α)f(s,x(s))|λkΓ(αk+α)tαk+α1c2, 0st, x¯PR/Pr,t0uαk+α1du<+, t(0,ω],k1α1λktαk+α1Γ(αk+α)<+, t(0,ω],

    we obtain that H1C[0,ω] and

    H2(t)=k1α1λkΓ(αk+α)t0(ts)αk+α1f(s,x(s))ds=:k1α1λkΓ(αk+α)uk(t).

    Noting that ukC(0,ω]

    |uk(t)|c2ωαk+ααk+α, t(0,ω],   λkΓ(αk+α)ωαk+ααk+α<+,

    we have H2C(0,ω]. Hence, qC[0,ω].

    Moreover,

    mc1ωMc2Ax=mc1ωMc2supω0Gα,λ(t,s)f(s,x(s))dsmc1ωMsupω0Gα,λ(t,s)ds=mc1ωminω0Gα,λ(t,s)f(s,x(s))ds=mint[0,ω]Ax(t), (4.9)

    which means that A:¯PRPrP.

    Next, we show that A is continuous on ¯PRPr. Let xn,x¯PRPr and xnx0. From (4.2), we have f(t,xn(t))f(t,x(t))0,

    AxnAx=supt[0,ω]|ω0Gα,λ(t,s)(f(s,x(s))f(s,y(s)))ds|supt[0,ω]ω0Gα,λ(t,s)dsf(t,xn(t))f(t,x(t))Mf(t,xn(t))f(t,x(t))0,

    which implies that A is continuous. From (4.6), we get that Ax(t) is uniformly bounded. Finally, we show that {Ax|x¯PR/Pr} is an equicontinuity in C[0,ω]. By (4.6), we have

    |Ax(t1)Ax(t2)|=|ω0(Gα,λ(t1,s)Gα,λ(t2,s))f(s,x(s))ds|kq(ω)|Eα,1(λtα1)Eα,1(λtα2)|+|q(t1)q(t2)|.

    Since Eα,1(λtα)C[0,ω], q(t)C[0,ω] are uniformly continuous, |Eα,1(λtα1)Eα,1(λtα2)| and |q(t1)q(t2)| tend to zero as |t1t2|0. Hence, {Ax(t)|x¯PRPr} is equicontinuous in C[0,ω].

    Finally, by Arzela-Ascoli theorem, we can obtain that A is compact. Hence, it is completely continuous.

    Theorem 4.3. Assume that there exist 0<r<R,0<c1<c2 such that (4.2)and (4.3) hold. Further suppose one of the followingconditions is satisfied

    (i)f(t,u)Mc2m2ω2c1u,  (t,u)[0,ω]×[mc1ωMc2r,r],

    f(t,u)|λ|u,  (t,u)[0,ω]×[mc1ωMc2R,R],

    (ii)f(t,u)|λ|u,  (t,u)[0,ω]×[mc1ωMc2r,r],

    f(t,u)Mc2m2ω2c1u,  (t,u)[0,ω]×[mc1ωMc2R,R].

    Then (1.1) has at least a positive solution x with rxR.

    Proof. Here we only consider the case (i). By Proposition 4.2, A:¯PRPrP is completely continuous. For xPr, we have

    x=r,  mc1ωMc2rx(t)r, t[0,ω]

    and

    Ax(t)mω0f(s,x(s))dsMc2mω2c1ω0x(s)dsr=x.

    Similarly, if xPR,

    mc1ωMc2Rx(t)R,  t[0,ω],
    0Ax(t)ω0Gα,λ(t,s)|λ|x(s)ds|λ|Rω0Gα,λ(t,s)ds=R=x.

    By Theorem 4.1, there exists x¯PRPr such that Ax=x and x is a solution of (1.1). Moreover,

    mc1ωMc2rx(t)R.

    Corollary 4.4. Let c1<c2 be positive reals and f(t,x) satisfy the conditions

    (i) c1f(t,x)c2 for all x0,

    (ii) f:[0,w]×(0,+)R is a continuous function.

    Then problem (1.1) has a positive solution.

    Proof. Let 0<r<c21m2ω2Mc2, R>c22M|λ|m1c1ω, then (4.2) and (4.3) are satisfied. Clearly, for (t,u)[0,ω]×[mc1ωMc2r,r],

    f(t,u)c1Mc2m2ω2c1rMc2m2ω2c1u

    and for (t,u)[0,ω]×[mc1ωMc2R,R],

    f(t,u)c2|λ|mc1ωMc2R|λ|u.

    Hence, by Theorem 4.3 (1.1) has at least a positive solution.

    Example 4.5. Consider the equation

    {cDα0+x(t)λx(t)=1+x1β(t), 0<xω,x(0)=x(ω), (4.10)

    where 0<α1,β>1 and

    Λ={λ<0,|λ|Eα,1(λωα)Eα,α(λωα)4Eα,α+1(λωα)}.

    Choosing c1=1,c2=2,r=110min{1,m2ω22M},R=1. It is easy to check that (4.2) and (4.3) hold. For λΛ,

    f(t,u)=1+u1β1Mc2m2ωc1rMc2m2ωc1u,(t,u)[0,ω]×[mc1ωMc2r,r],
    f(t,u)2|λ|mc1ωMc2=|λ|mω2M|λ|u,(t,u)[0,ω]×[mc1ωMc2R,R].

    Hence, (4.10) has at least one positive solution for λΛ.

    In this section, by using the monotone iterative method, we discuss the existence of solutions when λ=0 in (1.1). Firstly, we give the definition of the upper and lower solutions and get monotone iterative sequences with the help of the corresponding linear equation. Finally, we prove the limits of the monotone iterative sequences are solutions of (1.7).

    Definition 5.1. Let h,kC1[0,ω]. h and k are called lower solution and upper solution of problem (1.7), respectively if h and k satisfy

    cDα0+h(t)f(t,h(t)), 0<tω, h(0)h(ω), (5.1)
    cDα0+k(t)f(t,k(t)), 0<tω, k(0)k(ω), (5.2)

    Clearly, if g the lower solution or upper solution of (1.7), then cDα0+g is continuous on [0,ω].

    Lemma 5.2. Let δC[0,ω] with δ0 and pR with p0. Then

    {cDα0+z(t)λz(t)=δ(t), 0<tω,z(ω)z(0)=p, (5.3)

    has a unique solution z(t)0 for t[0,ω], where 0<α1,λ<0.

    Proof. Let z1, z2 are two solutions of (5.3) and v=z1z2, then

    {cDα0+v(t)λv(t)=0, 0<tω,v(ω)=v(0). (5.4)

    Using Theorem 3.1, (5.4) has trivial solution v=0.

    By (3.5), we can verify that problem (5.3) has a unique solution

    z=ω0Gα,λ(t,s)δ(t)ds+pEα,1(λωα)Eα,1(λωα)1.

    As consequence, by Proposition 2.1 (C3), (C5) and Lemma 3.3 (F4), we conclude that z(t)0. This completes the proof.

    Theorem 5.3. Assume that h,k are the lower and upper solutions ofproblem (1.7) and hk. Moreover, supposethat f satisfies the following properties:

    (M) there is λ<0 such that for all fixed t[0,ω], f(t,x)λx is nondecreasing in h(t)xk(t),

    (J) f:[0,ω]×[h(t),k(t)]R is a continuous function.

    Then there are two monotone sequences {hn} and {kn} are nonincreasingand nondecreasing, respectively with h0=h and k0=ksuch that limnhn=¯h(t), limnkn=¯k(t)uniformly on [0,ω], and ¯h,¯k are the minimal and the maximalsolutions of (1.7) respectively, such that

    h0h1h2...hn¯hx¯kkn...k2k1k0

    on [0,ω], where x is any solution of (1.7) such that h(t)x(t)k(t)on [0,ω].

    Proof. Let [h,k]={uC[0,ω]:h(t)u(t)k(t),t[0,ω]}. For any η[h,k], we consider the equation

    {cDα0+x(t)λx(t)=f(t,η(t))λη(t), 0<tω,x(0)=x(ω),

    Theorem 3.1 implies the above problem has a unique solution

    x(t)=ω0Gα,λ(t,s)(f(s,η(s))λη(s))ds. (5.5)

    Define an operator B by x=Bη, we shall show that

    (a) kBk,Bhh,

    (b) B is nondecreasing on [h,k].

    To prove (a). Denote θ=kBk, we have

    cDα0+θ(t)λθ(t)= cDα0+k(t) cDα0+Bk(t)λ(k(t)Bk(t))f(t,k(t))((f(t,k(t))λk(t))λk(t)=0,

    and θ(w)θ(0)0. Since kC1[0,ω],

    cDα0+kC[0,ω],   cDα0+BkC[0,ω].

    By Lemma 5.2, θ0, i.e. kBk. In an analogous way, we can show that Bhh.

    To prove (b). We show that Bη1Bη2 if hη1η2k. Let z1=Bη1, z2=Bη2 and z=z2z1, then by (M), we have

    cDα0+z(t)λz(t)= cDα0+z2(t) cDα0+z1(t)λ(z2(t)z1(t))=f(t,η2(t))λη2(t)(f(t,η1(t))λη1(t))0,

    and v(ω)=v(0). By Lemma 5.2, z(t)0, which implies Bη1Bη2.

    Define the sequence {hn}, {kn} with h0=h,k0=k such that hn+1=Bhn,kn+1=Bkn for n=0,1,2,.... From (a) and (b), we have

    h0h1h2...hnkn...k2k1k0

    on t[0,ω], and

    hn(t)=ω0Gα,λ(t,s)(f(s,hn1(s))λhn1(s))ds,
    kn(t)=ω0Gα,λ(t,s)(f(s,kn1(s))λkn1(s))ds.

    Therefore, there exist ¯h,¯k such that limnhn=¯h, limnkn=¯k.

    Similar to the proof of Proposition 4.2, we can show that B:[h,k][h,k] is a completely continuous operator. Therefore, ¯h,¯k are solutions of (1.7).

    Finally, we prove that if x[h0,k0] is one solution of (1.7), then ¯h(t)x(t)¯k(t) on [0,ω]. To this end, we assume, without loss of generality, that hn(t)x(t)kn(t) for some n. From property (b), we can get that hn+1(t)x(t)kn+1(t),t[0,ω]. Since h0(t)x(t)k0(t), we can conclude that

    hn(t)x(t)kn(t),  forall n.

    Passing the limit as n, we obtain ¯h(t)x(t)¯k(t), t[0,ω]. This completes the proof.

    Example 5.4. Consider the equation

    {cDα0+x(t)=t+1x2(t), 0<x1,x(0)=x(1). (5.6)

    It easy to check that h=1,k=2 are the low solution and upper solution of (5.6), respectively. Let λ=10. For all t[0,ω],

    f(t,u)λu=t2+1u2+10u

    is nondecreasing on u[1,2] and

    f(t,u)=t+1u2

    is continuous on [0,ω]×[1,2].

    Hence, there exist two monotone sequences {hn} and {kn}, nonincreasing and nondecreasing respectively, that converge uniformly to the extremal solutions of (5.6) on [h,k].

    This paper focuses on the existence of solutions for the Caputo fractional differential equation with periodic boundary value condition. We use Green's function to transform the problem into the existence of the fixed points of some operator, and we prove the existence of positive solutions by using the Krasnosel'skii fixed point theorem. On the other hand, the existence of the extremal solutions for the special case of the problem is obtained from monotone iterative technique and lower and upper solutions method. Since the fractional differential equation is nonlocal equation, the process of verifying the compactness of operator is very tedious, and we will search for some better conditions to prove the compactness of the operator A in the follow-up research. Meanwhile, since the existence result for 0<α1 is obtained in present paper, we will discuss the existence of solutions for the Caputo fractional differential equation when n1<αn in follow-up research.

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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