Jianing Xu,Xue Gong,Chengmin Hou

(Dept.of Math.,Yanbian University,Yanji 133002,Jilin)

SOLVABILITY OF DISCRETE FRACTIONAL BOUNDARY VALUE PROBLEMS??

Jianing Xu,Xue Gong,Chengmin Hou?

(Dept.of Math.,Yanbian University,Yanji 133002,Jilin)

In this paper,by introducing a new approach,we investigate a discrete fractional boundary value problem.We transform a fractional nonlinear difference equation on a finite discrete segment with boundary conditions into a system,and obtain some conditions for the existence of solutions to the equation,based on coincidence degree theory and matrix theory but not on Green’s function.

fractional difference;boundary value problem;coincidence degree theory;matrix theory

2000 Mathematics Subject Classification 39A12;44A25;26A33

Ann.of Appl.Math.

31:2(2015),225-235

1　Introduction

In this paper we study the existence of solutions to the following fractional boundary value problem(FBVP)

The boundary value problems for differential equation and fractional differential equation have been extensively studied.Many excellent results have been obtained based on some theories such as coincidence degree theory,cone theory,etc.For details,see[1-7]and the references therein.Discrete fractional calculus has gained great attention.There are some literatures dealing with the discrete fractional difference equation subject to various boundary value conditions or initial value conditions.The study of the existence of the solutions to fractional difference equations boundary value problems is hard,and some results for these please see[8-22].For example,in[13-16],the authors investigated the existence of solutions to some boundary value problems by fixed point theorems on a cone,based on a fixed-point theorem for the completely continuous operators in cones and the Green’s function.It is necessary to study the sign of Green’s function for the corresponding linear equations.For practical purposes,serious difficulties arise in the search for a Green’s function and the judgement of its sign.In[18],it is the first time that the coincidence degree theory was introduced to investigate the boundary value problem for fractional difference equations.In this paper,we further develop the technique to study(1)with(2)by substituting(1) with(2)into a continuous operator equation with boundary value conditions,and then apply matrix theory and coincidence degree theory to establish the existence of solutions to(1) with(2).

Throughout this paper,RTstands for the T-dimensional real vector space.A-1denotes the inverse matrix of the matrix A,R(A)denotes the rank of A,A?denotes transpose of a matrix or a vector,and(·,·)denotes the inner product of two elements.

2　Preliminaries

We first introduce some definitions about discrete fractional operators,which can be found in[13,16].

First,for any integer β,let Nβ={β,β+1,β+2,···}.We define,for any t and ν,where the right-hand side is defined.We also assume that if t+1-ν is a pole of the Gamma function and t+1 is not a pole,then t(ν)=0.

Definition 2.1 The νth fractional sum of f for ν＞0 is defined by

for t∈Na+ν.We also define the νth fractional difference for ν＞0 by Δνf(t):=ΔNΔν-Nf(t), where t∈Na+N-νand N∈N is chosen so that 0≤N-1＜ν≤N.

Definition 2.2 Let f be any real-valued function and ν∈(1,2).The discrete fractional difference operator is defined as

Lemma 2.1[20]Let y:Na-→R and ν＞0 with N-1＜ν≤N,then

for t∈Na+N-ν,

Lemma 2.2[20]Let f:Na-→R be given and suppose k∈N0and ν＞0.Then for t∈Na+ν,

Moreover,ifμ＞0 with M-1≤μ≤M,then for t∈Na+M-μ+ν,

Lemma 2.3[20]Let f:Na-→R be given and supposeμ,ν＞0 with N-1＜μ≤N. Then for t∈Na+ν+N-μ,

Lemma 2.4[20]Let f:Na-→R be given and suppose ν＞0 with N-1＜ν＜N.Then for t∈Na-ν+N,

Next we recall some notations,definitions and theorems on coincidence degree theory.

Let X and Z be real normed spaces.If a linear mapping L:domL?X-→ Z is a Fredholm mappings of index zero,then there exist continuous pro jectors P:X-→X, Q:Z-→Z such that ImP=KerL,KerQ=ImL and X=KerL⊕KerP,Z=ImL⊕ImQ as topological direct sums.Consequently,the restriction LPof L to domL∩KerP is one-to-one and onto ImL,thus its pseudo inverse KP:ImL-→domL∩KerP is defined.

Let ? be an open subset of X and satisfy domL∩?≠?.The mapping N:X-→Z is called L-compact ifis bounded and KP(I-Q)N:is compact.

Lemma 2.5[21](Mawhin continuation theorem)Let X and Z be two Banach spaces and L be a Fredholm mapping of index zero.Assume that N:X-→Z is L-compact onwith? being open bounded in X and the following hypotheses are satisfied.

(i)Lx≠λNx,for any(x,λ)∈[(domLKerL)∩??]×(0,1);

(ii)Nx?ImL,for any x∈KerL∩??;

(iii)deg(QN|KerL,?∩KerL,0)≠0,where Q:Z-→ Z is a continuous projector satisfying ImL=KerQ.

Then the equation Lx=Nx has at least a solution in domL∩?.

RT+1is endowed with the normis a Banach space.We define the norm of the matrix A as,where|·|denotes the usual absolute value in R,λmax(A?A)and λmin(A?A)denotes the maximal eigenvalue and minimal eigenvalue of A?A,respectively.So‖A‖and‖x‖are consistent.

3　Main Results

Lemma 3.1 C=

is a reversible matrix.

Proof detC=1+detD.detD

Therefore,

The proof is complete.

Lemma 3.2 The set of solution to the FBVP(1)with(2)is equal to the set of solutions to the operator equation

where x=(x(ν-1),x(ν),···,x(T+ν-1))?,B-1=C,More precisely,x is a solution to the FBVP(1)with(2)if and only if x is a solution to(3).

Proof Necessity Assume that x is a solution to the FBVP(1)with(2).Let y(t)=,then by Lemmas 2.1 and 2.3,we have

and the FBVP(1)with(2)is equal to

By(4)and(6),we have

that is,y=Bx and

On the other hand,equation(5)can be expressed by

Thus,Ay=ABx=Fx.

Sufficiency If x satisfies(3).Let y=Bx,then y satisfies(9).Thus,(8)implies that x is a solution to(1)with(2).The proof is complete.

Next,we can see that A is a semi-positive definite and R(A)=T.Let λ1,λ2,···,λT+1be eigenvalues of A.We order them as

Since A is real and symmetric,there must exist T+1 linearly independent eigenvector η1,η2,···,ηT+1satisfying

It is clear that η={η1,η2,···,ηT+1}is a normal orthogonal basis of.Let

Since B is a nonsinguar matrix,ξ={ξ1,ξ2,···,ξT+1}is a basis of RT+1.Define a linear operator L:domL? RT+1-→as Lx=ABx.Define N:RT+1-→as Nx=Fx.Then a solution to problem(1)with(2)is equivalent to that to the operator equation Lx=Nx.Thus,we obtain

By(10)and(11),it is easily shown that{λ2η2,λ3η3,···,λT+1ηT+1}is a basis of ImL.

We extend it as a basis of the T+1-dimensional real vector space as follows:

We can define continuous projectors P:RT+1-→RT+1andas

Thus the Moore-Penrose(Pseudo-inverse)Kp:ImL-→(I-P)x is defined as

for i∈[2,T+1]N2.And ImP=KerL,KerQ=ImL,?RT+1=ImL⊕ImQ,RT+1=KerL⊕ImP.

Theorem 3.1 Let

Assume that:

(G1)There exist constants α,β＞0,such thatand

(G2)there exists a constant m1＞0,for x∈RT+1,such that when||x||＞m1,either (Bξi,Fx)＞0 for i=1,or(Bξi,Fx)＜0 for i=1;

(G3)for x∈RT+1,there exists a constant m2＞0 such that||x||＞m2either(Bx,Fx)＞0 or(Bx,Fx)＜0. Then,the FBVP(1)with(2)has at least one solution.

4　Proof of Main Result

Lemma 4.1 The continuous projector Q defined in Section 3 is a semi-positive definite matrix.

Proof From the definitions of Q andit follows thatthen

By the matrix theory,it is clear that Q is semi-positive definite.

Proof of Theorem 3.1 Set

Since for any x∈?1,, we see that ?1?(I-P)X.

For R(AB)=T,(I-P)X is a T-dimensional space whose normal is denoted by‖x‖, and we denote

According to what described in Section 3,{ξ2,ξ3,···,ξT+1}is a basis of(I-P)X.Then for any x∈?1,x?KerL.We haveand

By(2),

Since

we get|x(ν-2)|≤C1‖x‖.

By(7),

we see that

This shows that ?1is bounded.

Set ?2={x∈KerL:Nx∈ImL},for each x∈?2,then we have x∈KerL,and QNx=0.

Since

which is equivalent to

Then by(G2),‖x‖≤m1,hence ?2is bounded.

Next,according to(G3),there are two cases:

Case 1 Assume that(Bx,Fx)＞0 holds.Set ?3={x∈KerL:λBx+(1-λ)QNx= 0,λ∈[0,1]},where B:KerL-→ImQ is a linear isomorphism.For each x∈?3,we haveBx=-(1-λ)QNx.If λ=1,then x=0.If λ=0,then ?3becomes ?2,or λ∈(0,1)and if‖x‖＞m2,then it follows from Lemma 4.1 that

This contradicts the condition(G3).So ?3?{x∈KerL:‖x‖≤m2},that is ?3is bounded.

Case 2 Assume that(Bx,Fx)＜0 holds.Set ?3={x∈KerL:-λBx+(1-λ)QNx= 0,λ∈[0,1]},then the same result is true for ?3.

By the definitions of L and N,it is easily shown that L is a Fredholm mapping of index 0 and N is L-compact.Let ? be an open subset of RT+1satisfyingFrom the above discussion,(i)and(ii)of Lemma 2.5 are verified.

Next,let us check that condition(iii)of Lemma 2.5 is satisfied.Set

The above discussion implies

Since B:?∩KerL-→RT+1can be seen as a linear mapping,then equation Bx=0 has a unique solution in ?∩KerL.

and

So,deg(B,?∩KerL,0)≠0.

It follows from Homotopy invariance that

So,by Lemma 2.5,Lx=Nx has at least one solution inthat is,equation(1) with(2)has at least one solution.The proof is complete.

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(edited by Liangwei Huang)

?This project was supported by the National Natural Science Foundation of China(11161049).

?Manuscript December 8,2014

?.E-mail:cmhou@foxmail.com