Existence of positive solutions for integral boundaryvalue problem of fractional differential equations

Xiping Liu, Guiyun Wu

(College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China)

1 Introduction

Fractional differential equations have been widely applied in various modern scientific fields,such as physics,chemistry,electrodynamics of complex medium,electrical circuits and biology,etc (see [1-5]).Recently,there are many papers discussing the existence of solutions of fractional differential equations (see [6-10] and the references therein) or fractional differential equations with the fractional order linear derivative operator (see [11-13] and the references therein).

In [11],the authors used contraction mapping principle to show the existence of solutions of the following initial value problem of fractional differential equations

In [12],the existence and uniqueness for solution of the following initial problem was discussed:

whereL(D)=Dα-rtnDβ,nis a non-negative integer,r∈,0<β≤α< 1 andDα,Dβare the standard Riemann-Liouville fractional derivatives,f:+×→is a continuous function.

In this paper,we study the following integral boundary value problem

(1)

whereL(D) =Dα-rtnDβ,nis a positive integer,r∈+,0<β<1 <α< 2 withα-β≥1 andDα,Dβare the standard Riemann-Liouville fractional derivatives,fandgare given functions.

2 Preliminaries

In this section,we introduce some notations,definitions and basic lemmas about the Riemann-Liouville fractional derivative,which are used to prove our main results.

Definition1 (see [1]) The Riemann-Liouville fractional integral of orderα> 0 of a functionx:[a,b]→is given by

provided the integral exists,whereΓis the Gamma function.

Definition2 (see [1]) The Riemann-Liouville fractional derivative of orderα>0 of a functionx:[a,b]→is given by

provided the right side is pointwise defined on (0,+∞),whereN∈+andN-1<α

Lemma3 (see [1]) Suppose thatx∈C[a,b] and 0<β<1,β≤α,then fractional derivativeDβx(t) is integrable and

(2)

Dβ(Iαx(t))=Iα-βx(t).

Furthermore,if (I1-βx(t))|t=0= 0,then equation (2) will be reduced to

Iα(Dβx(t))=Iα-βx(t).

Lemma4 (see [2]) Letx∈C[a,b] and 0<β<1,β≤α.If (I1-βx(t))|t=0=0,then

Lemma5 (see [3]) Letα> 0,then the fractional differential equationDαx(t) = 0 has solutions

x(t)=c1tα-1+c2tα-2+…+cNtα-N,ci∈,i= 1,2,…,N,

whereNis the smallest integer greater than or equal toα.

Lemma6 (see [3]) Letα> 0, assume thatx∈C(0,1)∩L[0,1], then

IαDαx(t) =x(t)+c1tα-1+… +cNtα- N,ci∈,i= 1,2,…,N,

whereNis the smallest integer greater than or equal toα.

(i) {x∈P(θ,b,d) |θ(x)>b}≠Ø andθ(Ax)>bforx∈P(θ,b,d);

(ii) ‖Ax‖

(iii)θ(Ax)>bforx∈P(θ,b,c) with ‖Ax‖>d.

Then A has at least three fixed pointsx1,x2,x3with

‖x1‖

Remark1 If there holdsd=c,then the condition of (i) of Lemma 8 implies condition (iii) of that one.

3 Expression and Properties of Integral Kernel

We denote

(H1)f:[0,1]×+→+is a continuous function.

Lemma9 Suppose (H1) holds,then the boundary value problem (1) is equivalent to the following integral equation

(3)

where

(4)

(5)

and

(6)

ProofBy Lemma 5 and Lemma 6,we obtain that

IαL(D)x(t)=-Iαf(t,x(t))

and

x(t)=-Iαf(t,x(t))+Iα(rtnDβx)(t)+c1tα-1+c2tα-2,c1,c2∈.

(7)

c2= 0.

Therefore,

whereG(t,s),Hk(t,s) andφ(t,s) are defined by (4),(5) and (6),respectively.

Lemma10 The functionG(t,s) defined by (4) satisfies

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

(ii)G(t,s) is decreasing with respect totfor 0≤s≤t≤1 and increasing with respect totfor 0≤t≤s≤1.And

(8)

(iii) There exists a positive function

(9)

such that

(10)

Particularly,s0=0.5 ifα=2;s0→0.5 asα→2 ands0→0.75 asα→1.

Proof(i) It is easy to show thatG(t,s) is continuous on [0,1]×[0,1] from the expression ofG(t,s).

For 0≤t≤s≤1,it is obvious thatG(t,s)=(t(1-s))α-1≥0.For 0≤s≤t≤1,

(t(1-s))α-1-(t-s)α-1=(t-ts)α-1-(t-s)α-1≥0.

Therefore,G(t,s)≥0,for allt,s∈[0,1] ,andG(t,s) > 0,for allt,s∈(0,1).

(ii) For 0≤s

So,G(t,s) is decreasing with respect totfor 0≤s

(iii) The proof is the same as that in [8].

Lemma11 The functionHk(t,s) defined by (6) satisfies

(i)Hk(t,s) is continuous for (t,s)∈[0,1]×[0,1];

(ii)Hk(t,s)< 0 fort,s∈(0,1) and |Hk(t,s)|≤tα-1(1-s)α-β+k-1,k=0,1,2,…;

Proof(i) It is obvious that (i) holds.

(ii) For 0

Hk(t,s)=tn-k(t-s)α-β+k-1-tα-1(1-s)α-β+k-1,k= 0,1,…,n.

Ifn-k≥1,it also means thatk≠n,thentn-k

thus

Hk(t,s)=tn-k(t-s)α-β+k-1-tα-1(1-s)α-β+k-1< 0.

Ifk=n,we have

Hn(t,s)=(t-s)α-β+n-1-tα-1(1-s)α-β+n-1<

(t-s)α-β+n-1-tα-β+n-1(1-s)α-β+n-1=

(t-s)α-β+n-1-(t-ts)α-β+n-1<0.

Then,Hk(t,s)<0 for 0

It is obvious thatHk(t,s)=-tα-1(1-s)α-β+k-1<0 for 0

Above all,Hk(t,s)<0.For allt,s∈[0,1] andk=0,1,2,….

Hence,|Hk(t,s)|≤tα-1(1-s)α-β+k-1,k=0,1,2,….

(iii) It follows from (ii) that

Since

by Lemma 11 (iii),we can assume that the following condition is satisfied:

LetE=C[0,1] be endowed with the norm

and

P={x∈E|x(t)≥0,t∈[0,1]}.

ThenPis a cone on the Banach spaceE.

We defineA:P→Eby

(11)

ProofIt is obvious that operatorAis linear.

By (H2),we have |Ax(t)|≤M‖x‖,soAis bounded,that is to say the linear operatorAis bounded.

It follows from (H2) that ‖Ax‖≤M‖x‖,therefore,‖A‖≤M<1,which impliesI-Ais reversible and

In the following,we will give the expression of (I-A)-1.

By using the theory of Fredholm integral equations,we havex(t)=(I-A)-1y(t) if and only ifx(t)=y(t)+Ax(t) fort∈[0,1].

The definition of the operatorAimplies that

(12)

(13)

where the resolvent kernelR(t,s) is given by

and

whereφ1(t,s)=φ(t,s).

It follows from (H2) that fort,s∈[0,1]×[0,1],we havemj≤φj(t,s)≤Mj,j=1,2,3,….Then we can obtain the following lemma.

Lemma13 Suppose (H2) holds,then

(14)

4 Main results

LetT:P→C[0,1],

So the solutionx(t) of boundary value problem (1) satisfiesx(t)=Tx(t)+Ax(t),that is to say

x(t)=(I-A)-1Tx(t).

(15)

We defineS:P→C[0,1] by

According to (12),(15) and Lemma 12,we know thatx(t) is the solution of (1) if and only ifx(t) is the solution ofx(t)=(I-A)-1Tx(t)=Sx(t),andx(t) is the fixed point ofS.

Lemma14 If (H1) and (H2) hold,thenS:P→Pis completely continuous.

ProofBy Lemma 10,(H1) and (H2),we can easily showS:P→P.

SinceG(t,s) andf(t,x) are continuous,by using Ascoli-Arzela theorem,we can show getT:P→Pis completely continuous.

On the other hand,(I-A)-1is linear bounded operator.It followsSis completely continuous.

Denote

(i)f(t,x)≤(1-M)L1r1for (t,x)∈[0,1]×[0,r1],

Then the boundary value problem (1) has at least one positive solutionxsuch thatr2≤‖x‖≤r1.

ProofBy Lemma 14,we knowS:P→Pis completely continuous.

Let Ω1={x∈P|‖x‖

r1=‖x‖.

Let Ω2={x∈P|‖x‖

r2=‖x‖.

So,‖Sx‖≥‖x‖,forx∈∂Ω2.

By using Lemma 7,we can get the boundary value problem (1) has at least one positive solutionxsuch thatr2≤‖x‖≤r1.

(i)f(t,x)<(1-M)L1afor (t,x)∈[0,1]×[0,a];

(ii)f(t,x)>(1-m)L2bfor (t,x)∈[1/4,3/4]×[b,c];

(iii)f(t,x)≤(1-M)L1cfor (t,x)∈[0,1]×[0,c].

Then the boundary value problem (1) has at least three positive solutionsx1,x2andx3with

ProofLet

thenθ:P→[0,+∞) is a nonnegative continuous concave functional defined on the coneP.

That is,

θ(Sx)>b, for allx∈P(θ,b,c).

This shows that the condition (i) of Lemma 8 is also satisfied.

By Lemma 8,the boundary value problem (1) has at least three positive solutionsx1,x2andx3satisfies

5 Illustrations

Example1 We consider the boundary value problem

(16)

By a simple calculation,we can shows0≈0.711325,L1≈2.25676,L2≈13.6649,m=0 andM=0.122.

f(t,x) =tsinx+ 1≤1.8415≤(1 -M)L1r1≈1.98143,for (t,x)∈[0,1]×[0,1].

f(t,x) =tsinx+ 1≥1≥(1 -m)L2r2≈0.488031,for (t,x)∈[0,1]×[0,1/28].

Example2 Consider the boundary value problem

(17)

where we take that

f(t,x) =14 +t+x≤35(1 -M)L1c≈39.63,for (t,x)∈[0,1]×[0,20].

By using Theorem 16,we can obtain that the boundary value problem (17) has at least three positive solutionsx1,x2andx3with

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