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Volume 2011, Article ID 483816, 17 pagesdoi:10.1155/2011/483816 Research Article Nonlocal Cauchy Problem for Nonautonomous Fractional Evolution Equations Fei Xiao Department of Mathemati

Volume 2011, Article ID 483816, 17 pages

doi:10.1155/2011/483816

Research Article

Nonlocal Cauchy Problem for

Nonautonomous Fractional Evolution Equations

Fei Xiao

Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China

Correspondence should be addressed to Fei Xiao,sheaf@mail.ustc.edu.cn

Received 28 November 2010; Accepted 29 January 2011

Academic Editor: Toka Diagana

Copyrightq 2011 Fei Xiao This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We investigate the mild solutions of a nonlocal Cauchy problem for nonautonomous fractional

evolution equations d q u t/dt q  −Atut  ft, K1u t, K2u t, , K n u t, t ∈ I  0, T,

u 0  A−10gu  u0, in Banach spaces, where T > 0, 0 < q < 1 New results are obtained by

using Sadovskii’s fixed point theorem and the Banach contraction mapping principle An example

is also given

1 Introduction

During the past decades, the fractional differential equations have been proved to be valuable tools in the investigation of many phenomena in engineering and physics; they attracted many researcherscf., e.g., 1 9 On the other hand, the autonomous and nonautonomous evolution equations and related topics were studied in, for example,6,7,10–20, and the nonlocal Cauchy problem was considered in, for example,2,5,18,21–26

In this paper, we consider the following nonlocal Cauchy problem for nonautonomous fractional evolution equations

d q u t

dt q  −Atut  ft, K1u t, K2u t, , Kn u t, t ∈ I  0, T,

u 0  A−10gu  u0,

1.1

in Banach spaces, where 0 < q < 1, g : CI; X → X The terms Ki u t, i  1, , n are

defined by

Ki u t 

t

0

k it, susds, 1.2

the positive functions k it, s are continuous on D  {t, s ∈ R2: 0≤ s ≤ t ≤ T} and

K i∗ sup

t ∈0,T

t

0

k it, sds < ∞. 1.3

Let us assume that u ∈ L0, T; X and At is a family linear closed operator defined

in a Banach space X The fractional order integral of the function u is understood here in the

Riemann-Liouville sense, that is,

I q u t  1

Γq

t

0

t − s q−1u sds. 1.4

In this paper, we denote that C is a positive constant and assume that a family of closed

linear{At : t ∈ 0, T} satisfying

A1 the domain DA of {At : t ∈ 0, T} is dense in the Banach space X and in-dependent of t,

A2 the operator At  λ−1exists in LX for any λ with Re λ ≤ 0 and



At  λ−1 ≤ C

|λ  1| , t ∈ 0, T. 1.5

A3 There exists constant γ ∈ 0, 1 and C such that



At1 − At2A−1s ≤ C|t

1− t2|γ , t1, t2, s ∈ 0, T. 1.6

Under conditionA2, each operator −As, s ∈ 0, T generates an analytic semigroup

exp−tAs, t > 0, and there exists a constant C such that

A n s exp−tAs ≤ C

where n  0, 1, t > 0, s ∈ 0, T 11

We study the existence of mild solution of 1.1 and obtain the existence theorem based on the measures of noncompactness An example is given to show an application of the abstract results

2 Preliminaries

Throughout this work, we set I  0, T We denote by X a Banach space, LX the space

of all linear and bounded operators on X, and CI, X the space of all X-valued continuous functions on I.

Lemma 2.1 see 9 1 I q : L10, T → L10, T.

2 For g ∈ L10, T, we have

t

0

η

0



t − ηq−1

η − sγ−1g sds dη  Bq, γ t

0

t − s q γ−1 g sds, 2.1

where Bq, γ is a Beta function.

Definition 2.2 Let B be a bounded set of seminormed linear space Y The Kuratowski’s

measure of noncompactnessfor brevity, α-measure of B is defined as

α B  infd > 0 : B has a finite cover by sets of diameter ≤ d. 2.2

From the definition, we can get some properties of α-measure immediately, see27

Lemma 2.3 see 27 Let A and B be bounded sets of X Then

1 αA ≤ αB, if A ⊆ B.

2 αA  αAcl, where Acldenotes the closure of A.

3 αA  0 if and only if A is precompact.

4 αλA  |λ|αA, λ ∈ R.

5 αA ∪ B  max{αA, αB}.

6 αA  B ≤ αA  αB, where A  B  {x  y : x ∈ A, y ∈ B}.

7 αA  x0  αA, for any x0∈ X.

For H ⊂ CI, X we define

t

0

H sds 

t

0

u sds : u ∈ H

for t ∈ I, where Hs  {us ∈ X : u ∈ H}.

The following lemma will be needed

Lemma 2.4 see 27 If H ⊂ CI, X is a bounded, equicontinuous set, then

1 αH  sup t ∈I α Ht.

2 α t

0H sds ≤ t

0α Hsds, for t ∈ I.

Lemma 2.5 see 28 If {un}∞n1⊂ L1I, X and there exists a m· ∈ L1I, R such that

then α {unt}∞n1 is integrable and

α

t

0

u nsds

∞

n1

≤ 2

t

0

α {uns}∞n1ds. 2.5

We need to use the following Sadovskii’s fixed point theorem

Definition 2.6see 29 Let P be an operator in Banach space X If P is continuous and takes bounded, sets into bounded sets, and αPH < αH for every bounded set H of X with

α H > 0, then P is said to be a condensing operator on X.

Lemma 2.7 Sadovskii’s fixed point theorem 29 Let P be a condensing operator on Banach

space X If P B ⊆ B for a convex, closed, and bounded set B of X, then P has a fixed point in B.

According to4, a mild solution of 1.1 can be defined as follows

Definition 2.8 A function u ∈ CI, X satisfying the equation

u t  A−10gu  u0

t

0

ψ

t − η, ηU

η

A0 A−10gu  u0



dη



t

0

ψ

t − η, ηf

η, K1uη

, K2uη

, , Kn uη

dη



t

0

η

0

ψ

t − η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη,

2.6

is called a mild solution of1.1, where

ψ t, s  q

∞

0

θt q−1ξ qθ exp−t q θA sdθ, 2.7

and ξq is a probability density function defined on0, ∞ such that its Laplace transform is

given by

∞

0

e −σx ξ qσdσ ∞

j0

−x j

Γ1 qj  , q ∈ 0, 1, x > 0,

ϕ t, τ ∞

k1

ϕ kt, τ,

2.8

ϕ1t, τ  At − Aτψt − τ, τ,

ϕ k1t, τ 

t

τ

ϕ kt, sϕ1s, τds, k  1, 2 ,

U t  −AtA−10 −

t

0

ϕ t, sAsA−10ds.

2.9

To our purpose, the following conclusions will be needed For the proofs refer to4

Lemma 2.9 see 4 The operator-valued functions ψt − η, η and Atψt − η, η are continuous

in uniform topology in the variables t, η, where 0 ≤ η ≤ t − ε, 0 ≤ t ≤ T, for any ε > 0 Clearly,

ψ

t − η, η ≤ Ct − η q−1. 2.10

Moreover, we have

ϕ

t, η  ≤ Ct − η γ−1. 2.11

Remark 2.10 From the proof of Theorem 2.5 in4, we can see

γ

2 For t ∈ I, t

0ψ t − η, ηUηdη is uniformly continuous in the norm of LX and







t

0

ψ

t − η, ηU

η

dη



 ≤C2t q

 1

q  t γ B

q, γ 1: Mt. 2.12

3 Existence of Solution

Assume that

B1 f : I × X × X × · · · × X → X satisfies f·, v1, v2, , v n : I → X is measurable for all

v i ∈ X, i  1, 2, , n and ft, ·, ·, , · : X × X × · · · × X → X is continuous for a.e t ∈ I, and there exist a positive function μ· ∈ L p I, R p > 1/q > 1 and a continuous nondecreasing function ω : 0, ∞ → 0, ∞ such that

f t, v1, v2, , v n ≤ μtωn

i1

i

, t, v1, v2, , v n ∈ I × X × X × · · · × X, 3.1

and set Tp,q  max{T q −1/p , T q}

B2 For any bounded sets D, D1, D2, , D n ⊂ X, and 0 ≤ τ ≤ s ≤ t ≤ T,

α

g D≤ βtαD,

α

ψ t − s, sfs, D1, D2, , D n

≤ β1t, sαD1  β2t, sαD2  · · ·  βnt, sαDn,

α

ψ t − s, sϕs, τfτ, D1, D2, , D n

≤ ζ1t, s, ταD1  ζ2t, s, ταD2  · · ·  ζnt, s, ταDn,

3.2

where βt is a nonnegative function, and sup t ∈I β t : β < ∞,

sup

t ∈I

t

0

β it, sds : βi < ∞, i  1, 2, , n,

sup

t ∈I

t

0

s

0

ζ jt, s, τdτ ds : ζj < ∞, j  1, 2, , n.

3.3

B3 g : CI; X → X is continuous and there exists

0 < α1<

CMT−1, α2≥ 0 3.4 such that

g u ≤ α1 2. 3.5

B4 The functions μ and ω satisfy the following condition:

C

1 CBq, γ

T p,q γ Ωp,q

n



i1

K i∗

μ

L plim inf

τ→ ∞

ω τ

τ < 1 − α1



CMT, 3.6

whereΩp,q p − 1/pq − 1 p−1/p , and T p,q γ  max{Tp,q , T p,q γ}

Theorem 3.1 Suppose that (B1)–(B4) are satisfied, and if C  MTβ  4Σ n

i1βi  2ζiK∗

i  < 1,

then1.1 has a mild solution on 0, T.

Proof Define the operator F : C I; X → CI; X by

F ut  A−10gu  u0

t

0

ψ

t − η, ηU

η

A0 A−10gu  u0



dη



t

0

ψ

t − η, ηf

η, K1uη

, K2uη

, , Kn uη

dη



t

0

η

0

ψ

t − η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη, t ∈ I.

3.7 Then we proceed in five steps

Step 1 We show that F is continuous.

Let ui be a sequence that ui → u as i → ∞ Since f satisfies B1, we have

f t, K1u it, K2u it, , Kn u it −→ ft, K1u t, K2u t, , Kn u t, as i −→ ∞.

3.8 Then

i

≤A−10g ui − gu t

0

ψ

t − η, ηU

ηg ui − gudη



t

0

ψ

t − η, ηf

η, K1u iη

, K2u iη

, , Kn u iη

−fη, K1uη

, K2uη

, , Kn uηdη



t

0

η

0

ψ

t − η, ηϕ

η, s

f s, K1u is, K2u is, , Kn u is

−fs, K1u s, K2u s, , Kn u sds dη.

3.9

According to the conditionA2, 2.12, and the continuity of g, we have



A−10g ui − gu −→ 0, as i −→ ∞;

t

0

ψ

t − η, ηU

ηg ui − gudη −→ 0, as i −→ ∞. 3.10

Noting that ui i

Therefore, we have

f t, K1u it, K2u it, , Kn u it − ft, K1u t, K2u t, , Kn u t

≤ μt

⎡

⎣ω

⎛

⎝n

j1

K j u i

t⎞⎠  ωn

j1

K j u

t⎤⎦

≤ μt

⎡

⎣ω

⎛

⎝n

j1

K j∗

⎞

⎠  ω

⎛

⎝n

j1

K∗j

⎞

⎠

⎤

⎦.

3.11

Using2.10 and by means of the Lebesgue dominated convergence theorem, we obtain

t

0

ψ

t − η, ηf

η, K1u iη

, K2u iη

, , Kn u iη

−fη, K1uη

, K2uη

, , Kn uηdη

≤ C

t

0



t − ηq−1f

η, K1u iη

, K2u iη

, , Kn u iη

−fη, K1uη

, K2uη

, , Kn uηdη,

−→ 0, as i −→ ∞.

3.12

Similarly, by2.10 and 2.11, we have

t

0

η

0

ψ

t − η, ηϕ

η, s

×f s, K1u it, K2u it, , Kn u it

−fs, K1u s, K2u s, , Kn u sds dη

≤ C2

t

0

η

0



t − ηq−1

η − sγ−1

×f s, K1u it, K2u it, , Kn u it

−fs, K1u s, K2u s, , Kn u sds dη

−→ 0, as i −→ ∞.

3.13

Therefore, we deduce that

lim

Step 2 We show that F maps bounded sets of C I, X into bounded sets in CI, X.

For any r > 0, we set Br r, byB1, we can see

f t, K1u t, K2u t, , Kn u t ≤ μtω⎛⎝n

j1

K∗j r

⎞

⎠. 3.15

Based on2.12, we denote that St : t

0ψ t − η, ηUηdη, we have

 1

q  t γ B

q, γ 1 0 0 3.16

Then for any u ∈ Br, byA2, 2.10, 2.11, andLemma 2.1, we have



A−10gu

0 S tgu 0



t

0

ψ

t − η, ηf

η, K1uη

, K2uη

, , Kn uηdη



t

0

η

0

ψ

t − η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη

≤C  Mtgu

 C

t

0



t − ηq−1μ

η

ω

⎛

⎝n

j1

K j∗r

⎞

⎠dη

 C2

t

0

η

0



t − ηq−1

η − sγ−1μ sω

⎛

⎝n

j1

K∗j r

⎞

⎠ds dη

≤ α1



C  Mt 2



C  Mt 0 0

 M1

C

t

0



t − ηq−1

μ

η

dη  C2B

q, γ t

0



t − ηq γ−1

μ

η

dη

,

3.17

where M1 ω n

j1K j∗r

By means of the H ¨older inequality, we have

t

0



t − ηq−1

μ

η

dη  t pq−1/p M p,qμ

L p ≤ Tp,qΩp,qμ

L p ,

t

0



t − ηγ q−1

μ

η

dη ≤ Tp,q γΩp,qγμ

L p

3.18

1



CMTr  α2



 M1Ωp,qT p,q γ C  C2B

q, γμ

L p: !r

3.19

This means FBr  ⊂ B !r

Step 3 We show that there exists m ∈ N such that FBm ⊂ Bm

Suppose the contrary, that for every m ∈ N, there exists um ∈ Bm and tm ∈ I, such that mtm

f t, K1u mt, K2u mt, , Kn u mt ≤ μtω⎛⎝n

j1

K j∗m

⎞

⎠, 3.20

we have



CMT m 2



CMT 0

M 0 1

C

t m

0



t m − ηq−1μ

η

dη  C2B

q, γ t m

0



t m − ηq γ−1 μ

η

dη

≤ α1



CMT m 2



CMT 0

M 0 1Ωp,qT p,q γ C  C2B

q, γμ

L p

≤ α1



CMTm  α2



CMT 0

M 0 1Ωp,qT p,q γ C  C2B

q, γμ

L p

3.21

Dividing both sides by m and taking the lower limit as m → ∞, we obtain

C

1 CBq, γ

T p,q γ Ωp,qn

j1

K∗jμ

L plim inf

m→ ∞

w m

m ≥ 1 − α1



CMT 3.22

which contradictsB4

Step 4 Denote

F ut  A−10gu  u0

t

0

ψ

t − η, ηU

η

A0 A−10gu  u0



dη  Gut, 3.23

G ut 

t

0

ψ

t − η, ηf

η, K1uη

, K2uη

, , Kn uη

dη



t

0

η

0

ψ

t − η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη.

3.24

We show that Gu· is equicontinuous.

Let 0 < t2< t1< T and u ∈ Bm Then

1 − Gut2 1 I2 I3 I4, 3.25 where

I1

t2

0

ψ

t1− η, η− ψt2− η, ηf

η, K1uη

, K2uη

, , Kn uηdη,

I2

t1

t2

ψ

t1− η, ηf

η, K1uη

, K2uη

, , Kn uηdη,

I3

t2

0

η

0

ψ

t1− η, η− ψt2− η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη,

I4

t1

t2

η

0

ψ

t1− η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη.

3.26

It follows fromLemma 2.9,B1, and 3.20 that I1, I3 → 0, as t2 → t1

For I2, from2.10, 3.20, and B1, we have

I2

t1

t2

ψ

t1− η, ηf

η, K1uη

, K2uη

, , Kn uηdη

≤ CM1

t1

t2



t1− ηq−1

μ

η

dη −→ 0, as t2−→ t1.

3.27

Similarly, by2.10, 2.11, B1, andLemma 2.1, we have

I4

t1

t2

η

0

ψ

t1− η, ηϕ

η, s

f s, K1u s, K2u s, , Kn u sds dη

≤ C2M1

t1

t



t1− ηq−1

η

0



η − sγ−1μ sds dη −→ 0, as t2−→ t1.

3.28

Step 5 We show that α FH < αH for every bounded set H ⊂ Bm For any ε > 0, we can

take a sequence{hv}∞

v1⊂ H such that

α H ≤ 2α{hv}  ε, 3.29

cf 30 So it follows from Lemmas2.3–2.5,2.9,2 inRemark 2.10, andB2 that

α FH ≤ Cαg H

M Tαg H 2αG{hv}  ε

≤ Cαg HMTαg H

 2 sup

t ∈I α

t

0

ψ

t − η, ηf

η, K1h vη

, K2h vη

, , Kn h vη

dη



t

0

η

0

ψ

t − η, ηϕ

η, s

×fs, K1h vs, K2h vs, , Kn h vsds dη

 ε

≤ CβαH  MTβαH

4 sup

t ∈I

t

0

α

ψ

t − η, ηf

η, K1h vη

, K2h vη

, , Kn h vη

dη

8 sup

t ∈I

t

0

η

0

α

ψ

t − η, ηϕ

η, s

f s, K1h vs, K2h vs, , Kn h vs

 ε ≤ CβαH 

M TβαH  4 sup

t ∈I

t

0

n



i1

β i



t, η

K i∗

α {hv}dη

 8 sup

t ∈I

t

0

η

0

n



i1

ζ i



t, η, s

K i∗

α {hv}ds dη

 ε

≤ CβαH  MTβαH 

4

n



i1

β i K∗i  8n

i1

ζ i K∗i

α {hv}  ε



CMTβ 4

n

i1



β i  2ζiK∗i

α H  ε.

3.30

Since ε is arbitrary, we can obtain

α FH ≤ C

M Tβ 4Σn

i1



β i  2ζiK∗i

α H < αH. 3.31

In summary, we have proven that F has a fixed point !u ∈ Bm Consequently,1.1 has

at least one mild solution

Our next result is based on the Banach’s fixed point theorem

G1 There exists a positive function l· ∈ L1I, R and a constant μ > 0 such that

g u − gu∗ ∗

f t, v1, v2, , v n − ft, w1, w2, , w n

≤ lt

n



i1

i − wi

, vi , w i ∈ X2, i  1, 2, , n.

3.32

G2 There exists a constant 0 < δ < 1 such that the function Λ : I → Rdefined by

Λt  μCMT C

n



i1

K∗i

Γq

I q l t  C2

n



i1

K∗i

Γq

Γγ

I q γ l t ≤ δ, t ∈ I.

3.33

Theorem 3.2 Assume that (G1), (G2) are satisfied, then 1.1 has a unique mild solution.

Proof Let F be defined as inTheorem 3.1 For any u, u∗∈ CI, X, we have

f t, K1u t, K2u t, , Kn u t − ft, K1u∗t, K2u∗t, , Kn u∗t

≤ lt

n



i1

i u t − Ki u∗

≤ ltn

i1

K i∗ ∗

3.34

Thus, fromA2, 2.10, 2.11,Lemma 2.1, we have

∗

0

ψ

t − η, ηU



t

0

ψ

t − η, ηf

η, K1uη

, K2uη

, , Kn uη

−fη, K1u∗η

, K2u∗η

, , Kn u∗ηdη



t

0

η

0

ψ

t − η, ηϕ

η, sf s, Kus, Hus − fs, Ku∗s, Hu∗sds dη

Step We show that α FH < αH for every bounded set H ⊂ Bm For any ε > 0, we... 3.10