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Open Access Research

Existence of mild solutions for fractional evolution equations with nonlocal conditions

He Yang

Author Affiliations

Department of Mathematics, Northwest Normal University, Lanzhou, 730070, People’s Republic of China

Boundary Value Problems 2012, 2012:113  doi:10.1186/1687-2770-2012-113

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/113


Received:8 August 2012
Accepted:28 September 2012
Published:17 October 2012

© 2012 Yang; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

This paper deals with the existence and uniqueness of mild solutions for a class of fractional evolution equations with nonlocal initial conditions. We present some local growth conditions on a nonlinear part and a nonlocal term to guarantee the existence theorems. An example is given to illustrate the applicability of our results.

MSC: 34A12, 35F25.

Keywords:
fractional evolution equations; nonlocal initial conditions; existence; uniqueness

1 Introduction

The differential equations involving fractional derivatives in time have recently been proved to be valuable tools in the modeling of many phenomena in various fields of engineering and science. Indeed, we can find numerous applications in electrochemistry, control, porous media, electromagnetic processing etc. (see [1-5]). Hence, the research on fractional differential equations has become an object of extensive study during recent years; see [6-11] and references therein.

On the other hand, the nonlocal initial condition, in many cases, has much better effect in applications than the traditional initial condition. As remarked by Byszewski and Lakshmikantham (see [12,13]), the nonlocal initial value problems can be more useful than the standard initial value problems to describe many physical phenomena.

Let X be a Banach space with norm , and let T > 0 be a constant. Consider the existence and uniqueness of mild solutions of fractional evolution equation with nonlocal condition in the form

{ D q u ( t ) + A u ( t ) = f ( t , u ( t ) ) + 0 t K ( t , s ) h ( s , u ( s ) ) d s , t J = [ 0 , T ] , u ( 0 ) + g ( u ) = x 0 , (1)

where D q is the Caputo fractional derivative of order q ( 0 , 1 ) , the linear operator −A is the infinitesimal generator of an analytic semigroup { S ( t ) } t 0 in X, the functions f, h and g will be specified later. K C ( Δ , R + ) , where Δ = { ( t , s ) | 0 s t T } , R + = [ 0 , + ) . Throughout this paper, we always assume that K = max ( t , s ) Δ K ( t , s ) .

In some existing articles, the fractional differential equations with nonlocal initial conditions were treated under the hypothesis that the nonlocal term is completely continuous or global Lipschitz continuous. It is obvious that these conditions are not easy to verify in many cases. To make the things more applicable, in [6] the authors studied the existence and uniqueness of mild solutions of Eq. (1) under the case K ( t , s ) 0 . In their main results, they did not assume the complete continuity of the nonlocal term, but they needed the following assumptions:

(F1) there exist a constant q 1 [ 0 , q ) and m L 1 q 1 ( J , R + ) such that f ( t , x ) m ( t ) for all x X and almost all t J ;

(F2) there exists a constant L > 0 such that g ( u ) g ( v ) L u v for u , v C ( J , X ) ;

and some other conditions.

In this paper, we will improve the conditions (F1) and (F2). We only assume that f and h satisfy local growth conditions (see assumption ( H 3 )) and g is local Lipschitz continuous (see assumption ( H 4 )). We will carry out our investigation in the Banach space X α : = ( D ( A α ) , α ) , 0 < α < 1 , where D ( A α ) is the domain of the fractional power of A. Finally, an example is given to illustrate the applicability of our main results. We can see that the main results in [6] cannot be applied to our example.

The rest of this paper is organized as follows. In Section 2, some preliminaries are given on the fractional power of the generator of an analytic semigroup and on the mild solutions of Eq. (1). In Section 3, we study the existence and uniqueness of mild solutions of Eq. (1). In Section 4, we give an example to illustrate the applicability of our results.

2 Preliminaries

Let X be a Banach space with norm , and let A : D ( A ) X X be the infinitesimal generator of an analytic semigroup S ( t ) ( t 0 ) of a uniformly bounded linear operator in X, that is, there exists M 1 such that S ( t ) M for all t 0 . Without loss of generality, let 0 ρ ( A ) . Then for any α > 0 , we can define A α by

A α = 1 Γ ( α ) 0 t α 1 S ( t ) d t .

A α L ( X ) is injective, and A α can be defined by A α = ( A α ) 1 with the domain D ( A α ) = A α ( X ) . For α = 0 , let A α = I .

Lemma 1 ([14])

A α has the following properties:

(i) D ( A α ) is a Banach space with the norm x α : = A α x for x D ( A α ) ;

(ii) S ( t ) : X X α for each t > 0 ;

(iii) A α S ( t ) x = S ( t ) A α x for each x D ( A α ) and t 0 ;

(iv) A α is a bounded linear operator onXwith D ( A α ) = I m ( A α ) ;

(v) If 0 < α β , then D ( A β ) D ( A α ) .

Let X α be the Banach space of D ( A α ) endowed with the norm α . Denote by C ( J , X α ) the Banach space of all continuous functions from J into X α with the supnorm given by u C = sup t J u ( t ) α for u C ( J , X α ) . From Lemma 1(iv), since A α is a bounded linear operator for α > 0 , we denote by C α the operator norm of A α in X, that is, C α : = A α . For any t 0 , denote by S α ( t ) the restriction of S ( t ) to X α . From Lemma 1(ii) and (iii), for any x X α , we have

S ( t ) x α = A α S ( t ) x = S ( t ) A α x S ( t ) A α x = S ( t ) x α ,

and

S ( t ) x x α = A α S ( t ) x A α x = S ( t ) A α x A α x 0

as t 0 . Therefore, S ( t ) ( t 0 ) is a strongly continuous semigroup in X α , and S α ( t ) α S ( t ) for all t 0 . To prove our main results, the following lemma is needed.

Lemma 2 ([15])

If S ( t ) ( t 0 ) is a compact semigroup inX, then S α ( t ) ( t 0 ) is an immediately compact semigroup in X α , and hence it is immediately norm-continuous.

For x X , define two families { U ( t ) } t 0 and { V ( t ) } t 0 of operators by

U ( t ) x = 0 η q ( θ ) S ( t q θ ) x d θ , V ( t ) x = q 0 θ η q ( θ ) S ( t q θ ) x d θ , 0 < q < 1 ,

where

η q ( θ ) = 1 q θ 1 1 q ρ q ( θ 1 q ) , ρ q ( θ ) = 1 π n = 1 ( 1 ) n 1 θ q n 1 Γ ( n q + 1 ) n ! sin ( n π q ) , θ ( 0 , ) ,

where η q is the probability density function defined on ( 0 , ) , which has properties η q ( θ ) 0 for all θ ( 0 , ) and

0 η q ( θ ) d θ = 1 , 0 θ η q ( θ ) d θ = 1 Γ ( 1 + q ) . (2)

The following lemma follows from the results in [6-8,10].

Lemma 3The following properties are valid:

(i) For fixed t 0 and any x X α , we have

U ( t ) x α M x α , V ( t ) x α q M Γ ( 1 + q ) x α = M Γ ( q ) x α .

(ii) The operators U ( t ) and V ( t ) are strongly continuous for all t 0 .

(iii) If S ( t ) ( t 0 ) is a compact semigroup inX, then U ( t ) and V ( t ) are norm-continuous inXfor t > 0 .

(iv) If S ( t ) ( t 0 ) is a compact semigroup inX, then U ( t ) and V ( t ) are compact operators inXfor t > 0 .

In this paper, we adopt the following definition of a mild solution of Eq. (1).

Definition 1 By a mild solution of Eq. (1), we mean a function u C ( J , X α ) satisfying

u ( t ) = U ( t ) ( x 0 g ( u ) ) + 0 t ( t s ) q 1 V ( t s ) [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s

for all t J .

To prove our main results, we also need the following two lemmas.

Lemma 4A measurable function H : [ 0 , T ] X is Bochner integrable if | H | is Lebesgue integrable.

Lemma 5 (Krasnoselskii’s fixed point theorem)

LetXbe a Banach space, letBbe a bounded closed and convex subset ofXand let Q 1 and Q 2 be mappings fromBintoXsuch that Q 1 x + Q 2 y B for every pair x , y B . If Q 1 is a contraction and Q 2 is completely continuous, then the operator equation Q 1 x + Q 2 x = x has a solution onB.

Lemmas 4 and 5, which can be found in many books, are classical.

The following are the basic assumptions of this paper.

( H 1 ) S ( t ) ( t 0 ) is a compact operator semigroup in X.

( H 2 ) There exists a constant β [ α , 1 ] such that the functions f , h : J × X α X β satisfy the following conditions:

(i) For each x X α , the functions f ( , x ) , h ( , x ) are measurable.

(ii) For each t J , the functions f ( t , ) , h ( t , ) are continuous.

( H 3 ) For t J and r > 0 , there exist positive functions φ r satisfying φ r ( ) ( t ) 1 q L 1 ( [ 0 , t ] , R + ) and ϕ r L 1 ( J , R + ) such that

sup x α r f ( t , x ) β φ r ( t ) , sup x α r h ( t , x ) β ϕ r ( t ) ,

and there are positive constants σ 1 and σ 2 such that

lim inf r + 1 r 0 t φ r ( s ) ( t s ) 1 q d s σ 1 < + , lim inf r + 1 r 0 t 0 s ϕ r ( τ ) d τ ( t s ) 1 q d s σ 2 < + ,

( H 4 ) g : C ( J , X α ) X α and for r > 0 , there exists a positive constant L such that

g ( u ) g ( v ) α L u v C

for all u , v B r : = { u C ( J , X α ) : u ( t ) α r , t J } .

3 Main results

In this section, we introduce the existence and uniqueness theorems of mild solutions of Eq. (1). The discussions are based on fractional calculus and Krasnoselskii’s fixed point theorem. Our main results are as follows.

Theorem 1Suppose that the assumptions ( H 1 )-( H 4 ) hold. If x 0 X α and the following inequality holds:

M L + M C β α Γ ( q ) ( σ 1 + K σ 2 ) < 1 , (3)

then Eq. (1) has at least one mild solution onJ.

Proof Define two operators Q 1 and Q 2 as follows:

( Q 1 u ) ( t ) = U ( t ) ( x 0 g ( u ) ) , t J , ( Q 2 u ) ( t ) = 0 t ( t s ) q 1 V ( t s ) [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s , t J .

Obviously, u is a mild solution of Eq. (1) if and only if u is a solution of the operator equation u = Q 1 u + Q 2 u on J. To prove the operator equation u = Q 1 u + Q 2 u has solutions, we first show that there is a positive number r such that Q 1 u + Q 2 v B r for every pair u , v B r . If this were not the case, then for each r > 0 , there would exist u r , v r B r and t r J such that ( Q 1 u r ) ( t r ) + ( Q 2 v r ) ( t r ) α > r . Thus, from Lemma 3, ( H 3 ) and ( H 4 ), we have

r < ( Q 1 u r ) ( t r ) + ( Q 2 v r ) ( t r ) α U ( t r ) ( x 0 g ( u r ) ) α + 0 t r ( t r s ) q 1 V ( t r s ) [ f ( s , v r ( s ) ) + 0 s K ( s , τ ) h ( τ , v r ( τ ) ) d τ ] α d s M x 0 [ g ( u r ) g ( 0 ) ] g ( 0 ) α + M Γ ( q ) 0 t r ( t r s ) q 1 A α β A β [ f ( s , v r ( s ) ) + 0 s K ( s , τ ) h ( τ , v r ( τ ) ) d τ ] d s M x 0 α + M L r + M g ( 0 ) α + M C β α Γ ( q ) 0 t r ( t r s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s .

Dividing on both sides by r and taking the lower limit as r + , we have

M L + M C β α Γ ( q ) ( σ 1 + K σ 2 ) 1 ,

which contradicts (3). Hence, for some r > 0 , Q 1 u + Q 2 v B r for every pair u , v B r .

The next proof will be given in two steps.

Step 1. Q 1 is a contraction on B r .

For any u , v B r and t J , according to Lemma 3 and assumption ( H 4 ), we have

( Q 1 u ) ( t ) ( Q 1 v ) ( t ) α = U ( t ) [ g ( v ) g ( u ) ] α M g ( u ) g ( v ) α M L u v C ,

which implies that Q 1 u Q 1 v C M L u v C . It follows from (3) that M L < 1 , hence Q 1 is a contraction on B r .

Step 2. Q 2 is a completely continuous operator on B r .

We first prove that Q 2 is continuous on B r . Let { u n } B r with u n u B r as n . Then for any t J , s [ 0 , t ] , by assumption ( H 2 ), we have

f ( s , u n ( s ) ) f ( s , u ( s ) ) , h ( s , u n ( s ) ) h ( s , u ( s ) )

as n , and from assumption ( H 3 ), we have

f ( s , u n ( s ) ) f ( s , u ( s ) ) β 2 φ r ( s ) , h ( s , u n ( s ) ) h ( s , u ( s ) ) β 2 ϕ r ( s ) , 0 s K ( s , τ ) h ( τ , u n ( τ ) ) h ( τ , u ( τ ) ) β d τ 2 K ϕ r L 1 .

This together with the Lebesgue dominated convergence theorem gives that

( Q 2 u n ) ( t ) ( Q 2 u ) ( t ) α 0 t ( t s ) q 1 V ( t s ) [ f ( s , u n ( s ) ) f ( s , u ( s ) ) ] α d s + 0 t ( t s ) q 1 V ( t s ) 0 s K ( s , τ ) [ h ( τ , u n ( τ ) ) h ( τ , u ( τ ) ) ] d τ α d s M Γ ( q ) 0 t ( t s ) q 1 A α β A β [ f ( s , u n ( s ) ) f ( s , u ( s ) ) ] d s + M Γ ( q ) 0 t ( t s ) q 1 A α β 0 s K ( s , τ ) A β [ h ( τ , u n ( τ ) ) h ( τ , u ( τ ) ) ] d τ d s M C β α Γ ( q ) 0 t ( t s ) q 1 f ( s , u n ( s ) ) f ( s , u ( s ) ) β d s + M C β α Γ ( q ) 0 t ( t s ) q 1 0 s K ( s , τ ) h ( τ , u n ( τ ) ) h ( τ , u ( τ ) ) β d τ d s 0

as n . Hence, lim n Q 2 u n Q 2 u C = 0 . This means that Q 2 is continuous on B r .

Next, we will show that the set { Q 2 u : u B r } is relatively compact. It suffices to show that the family of functions { Q 2 u : u B r } is uniformly bounded and equicontinuous, and for any t J , the set { ( Q 2 u ) ( t ) : u B r } is relatively compact.

For any u B r , we have Q 2 u C r for some r > 0 , which means that { Q 2 u : u B r } is uniformly bounded. In what follows, we show that { Q 2 u : u B r } is a family of equicontinuous functions. For t [ 0 , T ) , we have

( Q 2 u ) ( t ) ( Q 2 u ) ( 0 ) α M C β α Γ ( q ) 0 t ( t s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s .

Hence, it is only necessary to consider t > 0 . For 0 < t 1 < t 2 T , from Lemma 3 and assumption ( H 3 ), we have

( Q 2 u ) ( t 2 ) ( Q 2 u ) ( t 1 ) α 0 t 1 ( t 2 s ) q 1 [ V ( t 2 s ) V ( t 1 s ) ] [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s α + 0 t 1 [ ( t 2 s ) q 1 ( t 1 s ) q 1 ] V ( t 1 s ) × [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s α + t 1 t 2 ( t 2 s ) q 1 V ( t 2 s ) [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s α 0 t 1 ( t 2 s ) q 1 A α β [ V ( t 2 s ) V ( t 1 s ) ] × A β [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s + M Γ ( q ) 0 t 1 | ( t 2 s ) q 1 ( t 1 s ) q 1 | × A α β A β [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s + M Γ ( q ) t 1 t 2 ( t 2 s ) q 1 A α β A β [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s C β α 0 t 1 ( t 2 s ) q 1 V ( t 2 s ) V ( t 1 s ) [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s + M C β α Γ ( q ) 0 t 1 | ( t 2 s ) q 1 ( t 1 s ) q 1 | [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s + M C β α Γ ( q ) t 1 t 2 ( t 2 s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s I 1 + I 2 + I 3 ,

where

I 1 = C β α 0 t 1 ( t 2 s ) q 1 V ( t 2 s ) V ( t 1 s ) [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s , I 2 = M C β α Γ ( q ) 0 t 1 | ( t 2 s ) q 1 ( t 1 s ) q 1 | [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s , I 3 = M C β α Γ ( q ) t 1 t 2 ( t 2 s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s .

For any ϵ ( 0 , t 1 ) , we have

I 1 C β α 0 t 1 ϵ ( t 2 s ) q 1 V ( t 2 s ) V ( t 1 s ) [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s + C β α t 1 ϵ t 1 ( t 2 s ) q 1 V ( t 2 s ) V ( t 1 s ) [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s C β α 0 t 1 ϵ ( t 1 s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s sup s [ 0 , t 1 ϵ ] V ( t 2 s ) V ( t 1 s ) + 2 M C β α Γ ( q ) t 1 ϵ t 1 ( t 1 s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s .

It follows from Lemma 3 that I 1 0 as t 2 t 1 and ϵ 0 independently of u B r . From the expressions of I 2 and I 3 , it is clear that I 2 0 and I 3 0 as t 2 t 1 independently of u B r . Therefore, we prove that { Q 2 u : u B r } is a family of equicontinuous functions.

It remains to prove that for any t J , the set W ( t ) : = { ( Q 2 u ) ( t ) : u B r } is relatively compact.

Obviously, W ( 0 ) is relatively compact in X α . Let 0 < t T be fixed. For each δ ( 0 , t ) , ϵ > 0 and u B r , we define an operator Q 2 δ , ϵ by

( Q 2 δ , ϵ u ) ( t ) = 0 t δ ( t s ) q 1 ϵ q θ η q ( θ ) S ( ( t s ) q θ ) f ( s , u ( s ) ) d θ d s + 0 t δ ( t s ) q 1 ϵ q θ η q ( θ ) S ( ( t s ) q θ ) 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ d θ d s = S ( δ q ϵ ) [ 0 t δ ( t s ) q 1 ϵ q θ η q ( θ ) S ( ( t s ) q θ δ q ϵ ) f ( s , u ( s ) ) d θ d s + 0 t δ ( t s ) q 1 ϵ q θ η q ( θ ) S ( ( t s ) q θ δ q ϵ ) × 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ d θ d s ] .

Then the sets W δ , ϵ ( t ) : = { ( Q 2 δ , ϵ u ) ( t ) : u B r } are relatively compact in X α since by Lemma 2, the operator S α ( t ) is compact for t > 0 in X α . Moreover, for every u B r , from Lemma 3 and assumption ( H 3 ), we have

( Q 2 u ) ( t ) ( Q 2 δ , ϵ u ) ( t ) α 0 t ( t s ) q 1 0 ϵ q θ η q ( θ ) S ( ( t s ) q θ ) × [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d θ d s α + t δ t ( t s ) q 1 ϵ q θ η q ( θ ) S ( ( t s ) q θ ) × [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d θ d s α 0 t ( t s ) q 1 A α β 0 ϵ q θ η q ( θ ) S ( ( t s ) q θ ) d θ [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s + t δ t ( t s ) q 1 A α β ϵ q θ η q ( θ ) S ( ( t s ) q θ ) d θ [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s q M C β α 0 t ( t s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s 0 ϵ θ η q ( θ ) d θ + M C β α Γ ( q ) t δ t ( t s ) q 1 [ φ r ( s ) + K 0 s ϕ r ( τ ) d τ ] d s .

Therefore, there are relatively compact sets arbitrarily close to the set W ( t ) for t ( 0 , T ] and since it is compact at t = 0 , we have the relative compactness of W ( t ) in X α for all t J .

Therefore, the set { Q 2 u : u B r } is relatively compact by the Ascoli-Arzela theorem. Thus, the continuity of Q 2 and relative compactness of the set { Q 2 u : u B r } imply that Q 2 is a completely continuous operator. Hence, Krasnoselskii’s fixed point theorem shows that the operator equation u = Q 1 u + Q 2 u has a solution on B r . Therefore, Eq. (1) has at least one mild solution. The proof is completed. □

The following existence and uniqueness theorem for Eq. (1) is based on the Banach contraction principle. We will also need the following assumptions.

( H 5 ) There exists a constant β [ α , 1 ] such that the functions f , h : J × X α X β are strongly measurable.

( H 6 ) For r > 0 , there exist functions ρ 1 , ρ 2 L 1 ( J , R + ) such that

f ( t , x ) f ( t , y ) β ρ 1 ( t ) x y α , h ( t , x ) h ( t , y ) β ρ 2 ( t ) x y α

for any x , y B r and t J .

Theorem 2Let the assumptions ( H 3 )-( H 6 ) be satisfied. If x 0 X α and the inequalities (3) and

μ M L + M T q C β α Γ ( q + 1 ) ( ρ 1 L 1 + K ρ 2 L 1 ) < 1 (4)

hold, then Eq. (1) has a unique mild solution.

Proof From Lemma 4 and assumption ( H 5 ), it is easy to see that ( t s ) q 1 V ( t s ) [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] is Bochner integrable with respect to s [ 0 , t ] for all t J . For any u B r , we define an operator Q by

( Q u ) ( t ) = U ( t ) ( x 0 g ( u ) ) + 0 t ( t s ) q 1 V ( t s ) × [ f ( s , u ( s ) ) + 0 s K ( s , τ ) h ( τ , u ( τ ) ) d τ ] d s , t J .

According to the proof of Theorem 1, we know that Q ( B r ) B r for some r > 0 . For any u , v B r and t J , from Lemma 3, assumptions ( H 4 ) and ( H 6 ), we have

( Q u ) ( t ) ( Q v ) ( t ) α U ( t ) [ g ( v ) g ( u ) ] α + 0 t ( t s ) q 1 V ( t s ) [ f ( s , u ( s ) ) f ( s , v ( s ) ) ] α d s + 0 t ( t s ) q 1 V ( t s ) 0 s K ( s , τ ) [ h ( τ , u ( τ ) ) h ( τ , v ( τ ) ) ] d τ α d s M L u v C + q M C β α Γ ( q + 1 ) 0 t ( t s ) q 1 ρ 1 ( s ) u ( s ) v ( s ) α d s + q M C β α K Γ ( q + 1 ) 0 t ( t s ) q 1 0 s ρ 2 ( τ ) u ( τ ) v ( τ ) α d τ d s M L u v C + q M C β α Γ ( q + 1 ) 0 t ( t s ) q 1 d s ρ 1 L 1 u v C + q M C β α K Γ ( q + 1 ) 0 t ( t s ) q 1 d s ρ 2 L 1 u v C = [ M L + M T q C β α Γ ( q + 1 ) ( ρ 1 L 1 + K ρ 2 L 1 ) ] u v C .

Thus,

Q u Q v C μ u v C ,

which means that Q is a contraction according to (4). By applying the Banach contraction principle, we know that Q has a unique fixed point on B r , which is the unique mild solution of Eq. (1). This completes the proof. □

4 An example

Let X = L 2 [ 0 , π ] equip with its natural norm 2 and inner product , . Consider the following system:

{ 1 2 t 1 2 u ( t , x ) 2 x 2 u ( t , x ) = u ( t , x ) sin t 1 2 + 0 t K ( t , s ) u ( s , x ) cos s d s , t [ 0 , T ] , x [ 0 , π ] , u ( t , 0 ) = u ( t , π ) = 0 , t [ 0 , T ] , u ( 0 , x ) + i = 1 m 0 π K 0 ( x , y ) u ( t i , y ) d y = u 0 ( x ) , x [ 0 , π ] , (5)

where T > 0 is a constant, 0 < t 1 < < t m < T , u 0 X and K 0 will be specified later.

Let the operator A : D ( A ) X X be defined by

D ( A ) : = { v X : v , v X , v ( 0 ) = v ( π ) = 0 } , A u = 2 u x 2 .

Then −A generates a compact analytic semigroup S ( ) of uniformly bounded linear operators and S ( t ) L ( X ) e t for all t 0 . Hence, we take M = 1 . Moreover, the eigenvalues of A are n 2 , n N and the corresponding normalized eigenvectors are z n ( x ) = 2 π sin ( n x ) , n = 1 , 2 ,  . The operator A 1 2 is given by

A 1 2 ξ = n = 1 n ξ , z n z n

for each ξ D ( A 1 2 ) : = { v X : n = 1 n v , z n z n X } and A 1 2 L ( X ) = 1 .

Lemma 6 ([16])

If ξ D ( A 1 2 ) , thenξis absolutely continuous, ξ X and ξ 2 = A 1 2 ξ 2 .

Let X 1 2 = ( D ( A 1 2 ) , 1 2 ) , where ξ 1 2 : = A 1 2 ξ X for all ξ D ( A 1 2 ) . Assume that

(P1) The function K 0 L 2 ( [ 0 , π ] × [ 0 , π ] , R ) , K 0 ( 0 , y ) = K 0 ( π , y ) = 0 , and the partial derivative ( x , y ) x K 0 ( x , y ) belongs to L 2 ( [ 0 , π ] × [ 0 , π ] , R ) .

Define

f ( t , u ( t ) ) ( x ) = u ( t , x ) sin t 1 2 , h ( t , u ( t ) ) ( x ) = u ( t , x ) cos t , g ( u ) ( x ) = i = 1 m 0 π K 0 ( x , y ) u ( t i , y ) d y .

Let ξ X 1 2 , it follows from

f ( t , ξ ) , z n = 0 π u ( t , x ) sin t 1 2 2 π sin ( n x ) d x = 1 n 0 π [ x u ( t , x ) ] sin t 1 2 2 π cos ( n x ) d x

and

h ( t , ξ ) , z n = 0 π u ( t , x ) cos t 2 π sin ( n x ) d x = 1 n 0 π [ x u ( t , x ) ] cos t 2 π cos ( n x ) d x

that f and h are functions from [ 0 , T ] × X 1 2 into X 1 2 and they are continuous. Moreover, a similar computation of [17] together with Lemma 6 and assumption (P1) shows that g ( u ) X 1 2 whenever u C ( [ 0 , T ] , X 1 2 ) .

Then for any r > 0 , we see that the assumptions ( H 3 )-( H 6 ) are satisfied with

φ r ( t ) = r sin t 1 2 , ϕ r ( t ) = r cos t , σ 1 = σ 2 = 2 T 1 2 , L = ( n + 1 ) ( 0 π 0 π [ x K 0 ( x , y ) ] 2 d x d y ) 1 2 , ρ 1 ( t ) = ρ 2 ( t ) = 1 .

Thus, the system (5) has at least one mild solution due to Theorem 1 provided that L + 2 T 1 2 π ( 1 + K ) < 1 . And by Theorem 2, this mild solution of the system (5) is unique on [ 0 , T ] × [ 0 , π ] .

Competing interests

The author declares that they have no competing interests.

Acknowledgements

Research was supported by the Fundamental Research Funds for the Gansu Universities.

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