PIECEWISE CONTINUOUS SOLUTIONS OF INITIAL VALUE PROBLEMS OF SINGULAR FRACTIONAL DIFFERENTIAL EQUATIONS WITH IMPULSE EFFECTS∗

Yuji LIUDepartment of Mathematics,Guangdong University of Finance and Economics, Guangzhou 510320,China E-mail:yujiliu@sohu.com

PIECEWISE CONTINUOUS SOLUTIONS OF INITIAL VALUE PROBLEMS OF SINGULAR FRACTIONAL DIFFERENTIAL EQUATIONS WITH IMPULSE EFFECTS∗

Results on the existence of piecewise continuous solutions for two classes of initial value problems of impulsive singular fractional differential equations are obtained.

singular fractional differential equation;impulsive effect;piecewise continuous solution;fixed point theorem

2010 MR Subject Classification92D25;34A37;34K15

1　Introduction

Theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments.Processes with such a character arise naturally and often,for example,phenomena studied in physics,chemical technology,population dynamics, biotechnology and economics.For an introduction of the basic theory of impulsive differential equation,we refer the reader to[17].

Fractional differential equations were found numerous applications in the field of viscoelasticity,feedback amplifiers,electrical circuits,electro analytical chemistry,fractional multipoles,neuron modelling encompassing different branches of physics,chemistry and biological sciences[18,19].

In recent years,many authors studied the existence and uniqueness of solutions of the different kinds of initial value problems,two-point boundary value problems or multi-point boundary value problems for the impulsive fractional differential equations on finite intervals see papers[1–7,9–16,2o–34]and the references therein.

In[9,1o,28,3o],the concept of solutions for fractional differential equations with impulse effects was argued extensively,while the concept presented in these papers could be controversial and deserved a further argument and mending.

Motivated by[9,1o,28,3o],in this paper,we discuss the existence of piecewise continuous solutions of the following initial value problems of nonlinear singular fractional differential equations with impulse effects

where

(a)n is a positive integer and α satisfies n−1<α

(b)o=to

that there exist constants k>−α and l≤o with α+k+l−n+1>o and α+l−n+1>o such that|m1(t)|≤(t−ti)k(ti+1−t)lfor all t∈(ti,ti+1](m may be singular at t=ti),

(d)m2:(o,1)→R satisfies that m2|(ti,ti+1]∈Co(ti,ti+1](s=o,1,2,···,p)and that there exist constants k>−α and l≤o with α+k+l−n+1>o and α+l−n+1>o such that|m2(t)|≤tk(1−t)lfor all t∈(o,1)(m may be singular at t=o,1),

(e)f,Ij:(o,1)×Rn→R are Caratheodory functions(j=o,1,2,···,n−1).

A functions x:(o,1]→R is said to be a piecewise continuous solution of(1.1)if x(j)|(ti,ti+1)∈Co(ti,ti+1)(i=o,1,2,···,p),(j=o,1,2,···,n−1)and the limits

exist and all equations in(1.1)are satisfied.Similarly we can define the piecewise continuous solution of(1.2).

We establish the existence results of solutions for impulsive singular fractional differential systems(1.1)and(1.2),respectively.Two example are given to illustrate the efficiency of the main theorems.

The remainder of this paper is as follows:in Section 2,we present preliminary results.In Section 3,the main theorems on the existence of solutions of(1.1)and(1.2)are presented, respectively.

2　Preliminary Results

For the convenience of the readers,we present the necessary definitions from the fractional calculus theory.These definitions and results can be found in the monograph[19].For φ∈ L1(o,1),denoteLet the Gamma and beta functions Γ(α)and B(p,q)be defined by

Definition 2.1(see[19])Let a≥o.The Riemann-Liouville fractional integral of order α>o of a function g:(a,∞)→R is given by

provided that the right-hand side exists.

Definition 2.2(see[19])Let a≥o.The Caputo fractional derivative of order α>o of a n-times differentiable function g:(a,∞)→R is given by

where n−1≤α

Remark 2.1Let n−1≤µ

where Ci∈R,i=o,1,2,···,n−1.

For x∈X,define the norm by

It is easy to show that X is a real Banach space.

Remark 2.2Define the matrix Miby

Then|M|/=o and the inverse of M is denoted by

Remark 2.3Define the matrix Ni(i=1,2,···,p)by

By direct computation,we get that

We know that aj,v,s,o=1 for j=v and aj,v,s,o=o for j/=v.

Lemma 2.1u∈X is a solution of

if and only if

for t∈(ti,ti+1],i=o,1,2,···,p.Furthermore,we have

ProofSuppose that u is a solution of(2.3).Thenis continuous and the limitsexist.Then by Remark 2.1,we get that there exists cj,i∈R (i=o,1,2,···,p,j=o,1,2,···,n−1)such that

t∈(ti,ti+1],i=o,1,2,···,p,j=1,2,···,n−1.

We know for t∈(ti,ti+1](using(c))that

By Δu(j)(ti)=Ij,i(i=1,2,···,p,j=o,1,2,···,n−1),we get that

where Ni,Biwere defined by Remark 2.3.Using Remark 2.3,we have that

Hence we get for j=o,1,2,···,n−1 and i=1,2,···,p that

Substituting cj,i(j=o,1,2,···,n−1,i=o,1,2,···,p)into(2.6),we get(2.4)and(2.5).Now we prove that x∈X.It is easy to see that x(j)|(ti,ti+1)is continuous and the limitsexists for all j=o,1,2,···,n−1 and i=1,2,···,p+1.exists for all j=o,1,2,···,n−1 and i=o,1,2,···,p. So x∈X.

Now suppose that x satisfies(2.4).It is easy to show that x∈X and x is a solution of (2.3).The proof is completed.

Lemma 2.2x∈X is a solution of

for i=o,1,2,···,p. ProofSuppose that u is a solution of(2.7).One sees from Remark 2.1 thatu(t)= m2(t)implies that there exist constants cj,o∈R such that

Now we will prove by using the mathematical induction method that there exist constants cj,i∈R such that

From(2.9),we see that(2.1o)holds for i=o.Now,we suppose that(2.1o)holds for i= o,1,···,s(s≤p−1).We will prove that

By mathematical induction method,we have that(2.1o)holds for all i=o,1,2,···,p. By(2.1o),we get that

where i=o,1,2,···,p,v=o,1,2,···,n−1.By(d)we have

By Δu(v)(ti)=Iv,i(v=o,1,2,···,n−1,i=1,2,···,p)and

That is as follows:

It follows from Remark 2.2 for i=1,2,···,p that

Hence we get for i=o,1,2,···,p and v=o,1,2,···,n−1 that

Substituting cj,iinto(2.1o),we get for i=o,1,2,···,p thatThis is just(2.8).We can prove that u∈X easily.On the other hand,if u satisfies(2.8),it is easy to show that u∈X and u is a solution of(2.7).The proof is completed.□

Definition 2.3We call K:(o,1)×Rn→R a Caratheodory function if it satisfies the followings:

(i)t→K(t,x1,···,xn)is measurable on(ti,ti+1](i=o,1,2,···,p),respectively,

(ii)(x,y)→K(t,x1,···,xn)is continuous on Rnfor all most all t∈(o,1),

(iii)for each r>o there exists a constant Mr>o such that

Now,we define the operator T1,T2on X by

Remark 2.4By Lemma 2.1,x∈X is a solution of(1.1)if and only if x∈X is a fixed point of the operator T1.By Lemma 2.2,x∈X is a solution of(1.2)if and only if x∈X is a fixed point of the operator T2.

Lemma 2.3Suppose that(a)–(e)hold.Then both T1and T2:X→X are well defined and completely continuous.

ProofFirst,we prove that T1is well defined;second,we prove that T1is continuous and finally,we prove that T1is compact.So T1is completely continuous.Similarly we can prove that T2is well defined and completely continuous.Thus the proof is divided into three steps.

Step(i)Prove that T:X→X is well defined.

For x∈X,we have‖x‖=r>o.From f,Ijare Caratheodory functions,then there exist constants Mf≥o,MI≥o such that

By the definition of T1and Lemma 2.1,we have T1x∈X.Then T1:X→X is well defined.

Step(ii)We prove that T1is continuous.Let xκ∈X with xκ→xoas n→∞.We will show that T1xκ→T1xoas κ→∞.

In fact,we have r>o such that‖xκ‖≤r>o(κ=o,1,2,···).Since f,Ijare Caratheodory functions,then there exists Mf≥o,MI≥o such that(2.14)holds with x=xκ.By

Hence for v=o,1,2,···,n−1,we have

It is easy to show that{(T1xκ)(v)}is uniformly bounded.From the Lebesgue dominated convergence theorem,we get that both‖T1xκ−T1xo‖→o as κ→∞.It follows that T1is continuous.

Step(iii)We prove that T1is compact,i.e.,for each nonempty open bounded subset Ω of X,prove thatis uniformly bounded, equi-continuous on each interval(ti,ti+1](i=o,1,2,···,p).

Let Ω be a bounded open subset of X.We have r>o such that then‖x‖≤r for all x∈Ω.Since f,Ijare Caratheodory functions,then there exists Mf≥o,MI≥o such that (2.14)holds.

Sub-step(iii1)Prove that is uniformly bounded.

This follows similarly from the method used in Step(ii)and the details are omitted.

Sub-step(iii2)Prove that T1(Ω)is equi-continuous on each interval(ti,ti+1](i=o,1,2, ···,p).

3　Existence of Solutions

In this section we shall establish the existence of at least one solution of(1.1)and(1.2) respectively.

For easy referencing,we list the conditions needed as follows:

(A)there exist numbers σ≥o(i=1,2,···,n,j=1,···,m)with

σi,j(s=1,2,···,m)and bounded function ψ:(o,1)→R,numbers λi(i=1,2,···,p),and numbers aj≥o,bj≥o(j=1,2,···,m)such that

Theorem 3.1Suppose that(a)–(e)and(A)σhold,f,Ijare Caratheodory functions. Then,system(1.1)has at least one solution if

(i)σ=max{σs:s=1,2,···,m}>1 and

(ii)σ=max{σs:s=1,2,···,m}∈[o,1),or

(iii)σ=max{σs:s=1,2,···,m}=1 and Mo<1.

ProofLet the Banach space X and the operator T1be defined as in Section 2.We know that

(i)T1:X→X is well defined;

(ii)For x∈X is a fixed point of T1if and only if x∈X is a solution of(1.1);

(iii)T1:X→X is completely continuous.

It is easy to show from Ψ∈X.Let r>o and define Mr={x∈X:‖x−Ψ‖≤r}. For x∈Mr,we find

Then for t∈(ti,ti+1]and v=o,1,2,···,n−1 we get that

Then,for x∈Mrowe have

Hence,we have a bounded subset Mro⊆X such that T1(Mro)⊆Mro.Then,Schauder fixed point theorem implies that T1has a fixed point x∈Mro.Hence,x is a bounded solution of BVP(1.1).

Case(ii)σ∈[o,1).Choose r>o sufficiently large such that Mo(r+‖Ψ‖)σ≤r.Then, for x∈Mrwe have

So T(Mr)⊆Mrand Schauder fixed point theorem implies that T1has a fixed point x∈Mr. This x is a bounded solution of BVP(1.1).

From above discussion,the proof is complete.

Remark 3.1Suppose that(a)–(e)hold and f,Ijare Caratheodory functions.It follows from Theorem 3.1 that(1.1)has at least one solution if f and Ijare bounded.

Now,let

Theorem 3.2Suppose that(a)–(e)and(B)σhold,f,Ijare Caratheodory functions. Then,system(1.2)has at least one solution if

(i)σ=max{σs:s=1,2,···,m}>1 and

(ii)σ=max{σs:s=1,2,···,m}∈(o,1),or

(iii)σ=max{σs:s=1,2,···,m}=1 and No<1.

ProofLet the Banach space X and the operator T2be defined as in Section 2.We know that

(i)T2:X→X is well defined;

(ii)For x∈X is a fixed point of T2if and only if x∈X is a solution of(1.2);

(iii)T2:X→X is completely continuous.

It is easy to show from Ψ∈X.Let r>o and define Mr={x∈X:‖x−Ψ‖≤r}.For x∈Mr,we have(3.15).Then for t∈(ti,ti+1]and v=o,1,2,···,n−1,we get that

The remainder of the proof is similar to that of the proof of Theorem 3.1 and is omitted.

Remark 3.2Suppose that(a)–(e)hold and f,Ijare Caratheodory functions.It follows from Theorem 3.2 that(1.2)has at least one solution if f and Ijare bounded.

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∗July 25,2014.Supported by the Natural Science Foundation of Guangdong Province (S2011010001900)and the Guangdong Higher Education Foundation for High-Level Talents.