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        Existence of Positive Solutions for a Kind of Nonlinear Fractional Differential Equation with Nonlinear Boundary Value Conditions

        2020-01-10 05:48:32LIMengting李夢婷ZHANGKemei張克梅
        應(yīng)用數(shù)學(xué) 2020年1期

        LI Mengting(李夢婷),ZHANG Kemei(張克梅)

        (School of Mathematical Sciences,Qufu Normal University,Qufu 273165,China)

        Abstract: In this paper,we use mixed monotone operator method to deal with the existence and uniqueness of positive solutions to a kind of boundary value problem of the nonlinear fractional order differential equation.Our analysis is based on one new fixed point theorem for the mixed monotone operator.As an application,we also give a simple example to prove our main conclusions.

        Key words: Positive solution; Mixed monotone operator; Riemann-Liouville fractional derivative; Existence and uniqueness

        1.Introduction

        In this paper,we investigate the existence and uniqueness of positive solution for the fractional boundary value problem(BVP for short)of the form:

        where 3< α ≤4,β >1,is the Riemann-Liouville fractional derivative,is the Riemann-Liouville fractional integral,f:[0,1]×[0,∞)×[0,∞)→[0,∞),g:[0,∞)→(?∞,0]are continuous functions.

        Recently,the theory of boundary value problems has significant application in many areas,and many conclusions were obtained dealing with the existence and uniqueness of solution for the boundary value problems[1?10].

        In[1],the authors considered the following fourth-order two point problem with nonlinear boundary conditions:

        wheref,gare functions given,the results are given by using the Monotone iterative operator method.

        In[5],the authors studied the following fourth-order two point boundary value problem:

        wheref ∈C([0,1]×[0,∞)×(?∞,0],[0,∞)).By using the Krasnoselskii fixed point theorem,the author obtained the existence results of at least one positive solution.

        Inspired by the work of the above mentioned papers,the aim of this paper is to establish the existence and uniqueness of positive solutions of the BVP (1.1).

        To prove our conclusions,we put forward some related definitions and lemmas in Section 2,and we provide some new properties of the Green function that we compute.In Section 3,by using a mixed monotone operator method,the results for the existence and uniqueness of positive solutions of BVP (1.1)are built.To explain our results,we give a simple example in Section 4.

        2.Preliminaries and Correlative Lemmas

        We first give some essential definitions and lemmas that are significant for the paper.

        Definition 2.1[11]The Riemann-Liouville fractional integral of orderα>0 of a functionu:R+→R is given by

        whereΓ(α)is the Euler gamma function defined by

        Definition 2.2[11]The Riemann-Liouville fractional derivative of orderα >0 of a functionu:R+→R is given by

        wheren=[α]+1,provided the right side is pointwise defined on R+.

        Remark 2.1[12]In this way,we need the following conclusions:

        Lemma 2.1[11]Ifu ∈L1(0,1),α>β >0,then

        Lemma 2.2[11]Ifu ∈C(R+)and1(R+),then

        whereci ∈R,i=1,2,...,n,n=[α]+1.

        Definition 2.3[13]Let(E,||·||)be a real Banach Space.A nonempty,closed and convex setP ?Eis a cone if the following two conditions are satisfied:

        1)Ifx ∈Pandμ≥0,thenμx ∈P;

        2)Ifx ∈Pand?x ∈P,thenx=θ.

        Every coneP ?Einduces a partial order inEgiven byx1≤x2if and only ifx2?x1∈P.

        Definition 2.4[13]If there exists a constantC >0 such that,for anyx,y ∈E,0≤x ≤yimplies||x||≤C||y||,then the conePis said to be normal.In this case,the smallest constantCsatisfying the last inequality is called the normality constant ofP.

        Definition 2.5[13]Forx,y ∈E,we denotex~ywhen there exist constantsλ,μ >0 such thatλy ≤x ≤μy.Clearly~is an equivalence relation.

        Definition 2.6[13]Forh >0,we definePh={x ∈E:x~h}.It is easily seen thatPh ?P.

        Definition 2.7[13]An operatorT:E→Eis said to be increasing,if,for anyx,y ∈E,x ≤yimpliesTx ≤Ty.

        Definition 2.8[13]An operatorA:P ×P→Pis called mixed monotone,ifA(x,y)is increasing inxand decreasing iny.Hence,for any (x,y),(u,v)∈P ×P,x ≤uandy ≥v ?A(x,y)≤A(u,v).

        Definition 2.9An operatorB:P→Pis called sub-homogeneous ifB(tx)≥tBx,for anyt ∈(0,1)andx ∈P.

        Lemma 2.3Leth ∈C[0,1],3<α ≤4.The unique solution of the following boundary value problem:

        can be expressed by

        where

        ProofBy Lemma 2.2,the solution of (2.1)can be written as

        Fromu(0)=u′(0)=0,we know thatc3=c4=0,hence

        By the b oundary conditionsu′′(1)=0 and(1)=g(u(1)),we can get

        then

        So

        whereG(t,s)are defined by (2.3).The proof is completed.

        Lemma 2.4The functionG(t,s)defined by (2.3)satisfies

        1)G(t,s)≥0 is continuous for allt,s ∈[0,1];

        2)G(t,s)≥wherey1(s)=1?(1?s)α?3?(α?3)s,for allt,s ∈[0,1];

        3)G(t,s)≤wherey2(s)=1?(1?s)α?3,for allt,s ∈[0,1].

        Proof1)From (2.3),it is easy to see thatG(t,s)is continuous on [0,1]×[0,1],and obviouslyG(t,s)≥0;

        2)When 0≤s ≤t ≤1,

        When 0≤t ≤s ≤1,

        3)When 0≤s ≤t ≤1,

        When 0≤t ≤s ≤1,

        The proof is completed.

        Lemma 2.5[14]Suppose thatγ ∈(0,1),h ∈E,θE

        for anyλ ∈(0,1)andu,v ∈P.LetB:P→Pbe an increasing sub-homogeneous operator.Assume that

        (i)there existsh0∈Phsuch thatA(h0,h0)∈PhandBh0∈Ph;

        (ii)there exists a constantδ0>0 such thatA(u,v)≥δ0Bu,for anyu,v ∈P.Then

        (a)A:Ph×Ph→PhandB:Ph→Ph;

        (b)There existsu0,v0∈Phandr ∈(0,1)such thatrv0≤u0≤v0and

        (c)There exists a uniquex?∈Phsuch thatx?=A(x?,x?)+Bx?;

        (d)For any initial valuesx0,y0∈Ph,the sequences defined by

        Lemma 2.6[15]The following operator equations

        whereA:P×P→Pis a mixed monotone operator and the following conditions are satisfied:

        (A1)There existsh ∈Pwithh≠θsuch thatA(h,h)∈Ph;

        (A2)For anyu,v ∈Pandt ∈(0,1),there existsφ(t)∈(t,1)such that

        (A3)Pbe a normal cone ofC[0,1].

        Then operator equationA(x,x)=xhas a unique positive solutionx?inPh.Moreover,for any initial valuesx0,y0∈Ph,the sequences

        satisfy||xn?x?||→0 and||yn?x?||→0 asn→∞.

        Lemma 2.7[15]Suppose the above (A1),(A2)and (A3)are satisfied,andxλ(λ>0)is the unique solution of parameter equationA(x,x)=λxinPh.Then the following conclusions hold:

        (B1)Ifφ(t)fort ∈(0,1),thenxλis strictly decreasing inλ,that is,0< λ1< λ2impliesxλ1>xλ2;

        (B2)If there existsη ∈(0,1)such thatφ(t)≥tηfort ∈(0,1),thenxλis continuous inλ,that is,λ→λ0(λ0>0)implies||xλ?xλ0||→0;

        3.Main Results

        In this paper,we consider the spaceE=C[0,1]with the standard norm defined by,forx ∈C[0,1],||x||=max {|x(t)|:t ∈[0,1]}.InC[0,1],we consider the conePdefined by,P= {x ∈C[0,1]:x(t)≥0,t ∈[0,1]}.

        We give the following conditions:

        (H1)f(t,x,y)=φ(t,x,y)+ψ(t,x,y),whereφ,ψ:[0,1]×[0,∞)×[0,∞)→[0,∞)are continuous;

        (H2)For the fixedt ∈[0,1],φ(t,x,y)is nondecreasing inx,y,ψ(t,x,y)is nonincreasing inx,y,andφ(t,0,0)0;

        (H3)There exists a constantσ ∈(0,1)such that,forx,y ≥0,and for anyt ∈[0,1]andλ ∈(0,1),we have

        (H4)g:[0,∞)→(?∞,0)is decreasing continuous function andg(κx)≤κg(x),for anyκ ∈(0,1)andx ∈[0,∞);

        Theorem 3.1Assume that (H1)-(H5)hold.Then

        1)BVP (1.1)has a unique positive solutionx?∈Ph;

        2)For anyx0,y0∈Ph,there exist iteration sequencesxn(t),yn(t),forn=1,2,...,such that||xn?x?||→0 and||yn?x?||→0 asn→∞,whereh(t)=tα?2,t ∈[0,1].

        ProofFirstly,from Lemma 2.3,BVP (1.1)has the following integral formulation:

        wheret ∈[0,1],G(t,s)is Green’s function appearing in Lemma 2.3.

        Now,we consider the operators defined by:

        fort ∈[0,1]andu,v ∈P.

        From (H1),(H2),we can infer thatA:P ×P→PandB:P→P.

        Hence,usatisfies the equation (3.1)if and only ifu=A(u,u)+Bu.In the sequel,we check that operatorsA,Bsatisfy all the assumptions of Theorem 2.1.

        Step 1 It is clear thatPis a normal cone inC[0,1]and the normality constant is 1.

        Step 2 We explain thatAis a mixed monotone operator.In fact,foru1,u2,v ∈P,withu1≤u2,andt ∈[0,1],from (H2),we have

        Similarly,foru,v1,v2∈P,v1≤v2,we can get

        That is,Ais a mixed monotone operator.

        Step 3 We will prove that forγ ∈(0,1),anyλ ∈(0,1)andu,v ∈P,we have

        From (H3),there existσ1,σ2∈(0,1)such that

        Hence,forλ ∈(0,1)andu,v ∈P,we have

        whereγ=max {σ1,σ2}.

        Step 4 We are going to prove thatBis an increasing operator.Foru,v ∈P,withu ≤v,sincegis decreasing (the assumption (H4)),g(u(1))≥g(v(1)).Fort ∈[0,1],

        and this proves thatBis an increasing operator.

        Step 5 In order to prove thatBis a sub-homogeneous operator,we takeκ ∈(0,1),u ∈Pandt ∈[0,1],from the assumption (H4),we have

        Therefore,Bis a sub-homogeneous operator.

        Step 6 We will prove that assumption (i)of Lemma 2.5 holds.

        Now,we consider the functionh(t)=tα?2,for anyt ∈[0,1].Hence,0≤h(t)≤1,for anyt ∈[0,1].

        Before continuing the proof,fort ∈[0,1],we notice

        So,from (H2),we can get

        On the other hand,

        where we have used the assumption (H2)and Lemma 2.4.

        Put

        Then,we have

        Next,we will prove thata1,a2>0.In fact,it is sufficient to prove thata1>0.By the assumption (H2),

        Hence,A(h,h)∈Ph.Now,we are going to prove thatBh ∈Ph.Fort ∈[0,1],fromg(1)<0,we have

        On the other hand,

        Put

        Then,

        By (H4),sinceg(1)<0,0

        Therefore,Bh ∈Ph.

        Step 7 In order to prove assumption (ii)of Lemma 2.5,for anyu,v ∈P,by Lemma 2.4 and the assumption (H5),we have

        This proves that,foru,v ∈P,A(u,v)≥δ0Bu,withδ0=1.

        Finally,let’s apply Lemma 2.5.There exists a uniquex?∈Phsuch thatx?=A(x?,x?)+Bx?.Hence,BVP (1.1)has a unique solutionx?∈Ph.

        Moreover,for any initial valuesx0,y0∈Ph,the sequences are defined by

        forn=1,2,...Whenn→∞,we have||xn?x?||→0,||yn?x?||→0.

        The proof is finished.

        Remark 3.1If we change the range ofgin the assumption (H4)intog:[0,∞)→(?∞,0],and add the conditiong(1)<0 to Theorem 3.1,then the conclusions still hold.

        Corollary 3.1Assume that (H1),(H2),(H3)hold.Then the following fractional boundary value problem

        has a unique positive solutionx?inPh.Moreover,for any initial valuesx0,y0∈Ph,the sequences

        ProofFrom (3.2),forγ ∈(0,1),anyλ ∈(0,1)andu,v ∈P,we have

        Hence,λ < λγ <1 forγ ∈(0,1).We apply Lemma 2.6.The proof is similar to the Theorem 3.1,so we omit it.

        Corollary 3.2Assume that(H1),(H2),(H3)hold.Then the fractional boundary value problem

        has a unique positive solutionxλinPh.And we have the following conclusions:

        1)Fort ∈(0,1),ifthenxλis strictly decreasing inλ,whereγis what appears in (3.2);

        2)xλis continuous inλ.

        ProofApplying Lemma 2.6 and Lemma 2.7,we can obtain the conclusions easily.

        (H3)′:There existssuch that forx,y ≥0,and for anyt ∈[0,1],λ ∈(0,1),φ(t,λx,λy)≥λσφ(t,x,y)andψ(t,λ?1x,λ?1y)≥λσψ(t,x,y).

        Corollary 3.3Assume that (H1),(H2),(H3)′hold.Then the solution of BVP (3.4)xλsatisfies

        4.Example

        Example 4.1Consider the boundary value problem

        wherea,b ∈(0,1),φ1,φ2∈C[0,1]withφi(t)≥0,i=1,2 andφ2(t)satisties the condition

        g:[0,∞)→(?∞,0]is the function defined as

        Notice that Problem (4.1)is a particular case of BVP (1.1),with

        Obviously,the assumption(H1),(H2)of Theorem 3.1 hold.And by Remark 3.1,gsatisfies the conditions.

        Now,let’s prove that assumption (H3)holds,where

        For anyt ∈[0,1],λ ∈(0,1),x,y ∈[0,∞),we have

        Similarly,

        Next,we will prove that assumption (H5)holds.For anyx,y,z ∈[0,∞),we have

        Hence,the assumption (H5)holds.

        Finally,by using Theorem 3.1 and Remark 3.1,we obtain that Problem(4.1)has a unique positive solutionu?∈Ph,such that

        for anyt ∈[0,1]and certain positive constantsm,n.

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