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,[email protected] Received 28 November 2010; Accepted 29 January 2011
Academic Editor: Toka Diagana
Copyrightq2011 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 equationsdqut/dtq−Atut ft,K1ut,K2ut, . . . ,Knut, t∈I 0, T, u0 A−10gu u0, in Banach spaces, whereT >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
dqut
dtq −Atut ft,K1ut,K2ut, . . . ,Knut, t∈I 0, T, u0 A−10gu u0,
1.1
in Banach spaces, where 0 < q < 1,g : CI;X → X. The termsKiut,i 1, . . . , nare
defined by
Kiut t
0
kit, susds, 1.2
the positive functionskit, sare continuous onD{t, s∈R2: 0≤s≤t≤T}and
Ki∗ sup
t∈0,T
t
0
kit, sds <∞. 1.3
Let us assume thatu∈L0, T;XandAtis a family linear closed operator defined in a Banach spaceX. The fractional order integral of the functionuis understood here in the Riemann-Liouville sense, that is,
Iqut 1 Γ
q t
0
t−sq−1usds. 1.4
In this paper, we denote thatCis a positive constant and assume that a family of closed linear{At:t∈0, T}satisfying
A1the domain DA of {At : t ∈ 0, T} is dense in the Banach space X and in- dependent oft,
A2the operatorAt λ−1exists inLXfor anyλwith Reλ≤0 and
At λ−1≤ C
|λ1|, t∈0, T. 1.5
A3There exists constantγ∈0,1andCsuch that
At1−At2A−1s≤C|t1−t2|γ, t1, t2, s∈0, T. 1.6 Under conditionA2, each operator−As,s∈0, Tgenerates an analytic semigroup exp−tAs,t >0, and there exists a constantCsuch that
Ansexp−tAs≤ C
tn, 1.7
wheren0,1,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 setI 0, T. We denote by X a Banach space,LXthe space of all linear and bounded operators onX, andCI, Xthe space of allX-valued continuous functions onI.
Lemma 2.1see9. 1Iq :L10, T → L10, T. 2Forg∈L10, T, we have
t
0
η
0
t−ηq−1
η−sγ−1
gsds dηB q, γ
t 0
t−sqγ−1gsds, 2.1
whereBq, γis 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-measureofBis defined as
αB inf
d >0 :Bhas a finite cover by sets of diameter≤d
. 2.2
From the definition, we can get some properties ofα-measure immediately, see27.
Lemma 2.3see27. LetAandBbe bounded sets ofX. Then 1αA≤αB, ifA⊆B.
2αA αAcl, whereAcldenotes the closure ofA.
3αA 0 if and only ifAis precompact.
4αλA |λ|αA,λ∈R.
5αA∪B max{αA, αB}.
6αAB≤αA αB, whereAB{xy:x∈A, y∈B}.
7αAx0 αA, for anyx0∈X.
ForH⊂CI, Xwe define t
0
Hsds
t
0
usds:u∈H , 2.3
fort∈I, whereHs {us∈X :u∈H}.
The following lemma will be needed.
Lemma 2.4see27. IfH⊂CI, Xis a bounded, equicontinuous set, then 1αH supt∈IαHt.
2αt
0Hsds≤t
0αHsds, fort∈I.
Lemma 2.5see28. If{un}∞n1⊂L1I, Xand there exists am·∈L1I, Rsuch that
unt ≤mt, a.et∈I, 2.4
thenα{unt}∞n1is integrable and
α t
0
unsds
∞ n1
≤2 t
0
α{uns}∞n1ds. 2.5
We need to use the following Sadovskii’s fixed point theorem.
Definition 2.6see29. LetPbe an operator in Banach spaceX. IfPis continuous and takes bounded, sets into bounded sets, andαPH < αHfor every bounded setHofXwith αH>0, thenPis said to be a condensing operator onX.
Lemma 2.7Sadovskii’s fixed point theorem29. Let P be a condensing operator on Banach spaceX. IfPB⊆Bfor a convex, closed, and bounded setBofX, thenPhas a fixed point inB.
According to4, a mild solution of1.1can be defined as follows.
Definition 2.8. A functionu∈CI, Xsatisfying the equation ut A−10gu u0
t
0
ψ
t−η, η U
η
A0
A−10gu u0
dη
t
0
ψ
t−η, η f
η,K1u η
,K2u η
, . . . ,Knu η
dη
t
0
η
0
ψ
t−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη,
2.6
is called a mild solution of1.1, where ψt, s q
∞
0
θtq−1ξqθexp−tqθAsdθ, 2.7
andξq is a probability density function defined on0,∞such that its Laplace transform is given by
∞
0
e−σxξqσdσ∞
j0
−xj Γ
1qj, q∈0,1, x >0,
ϕt, τ ∞
k1
ϕkt, τ,
2.8
where
ϕ1t, τ At−Aτψt−τ, τ,
ϕk1t, τ t
τ
ϕkt, sϕ1s, τds, k1,2. . . ,
Ut −AtA−10− t
0
ϕt, sAsA−10ds.
2.9
To our purpose, the following conclusions will be needed. For the proofs refer to4.
Lemma 2.9see4. The operator-valued functionsψt−η, ηandAtψt−η, ηare continuous in uniform topology in the variablest,η, where 0≤η≤t−ε, 0≤t≤T, for anyε >0. Clearly,
ψ
t−η, η≤C
t−ηq−1
. 2.10
Moreover, we have
ϕ
t, η≤C
t−ηγ−1
. 2.11
Remark 2.10. From the proof of Theorem 2.5 in4, we can see 1Ut ≤CCtγ.
2Fort∈I,t
0ψt−η, ηUηdηis uniformly continuous in the norm ofLXand
t
0
ψ
t−η, η U
η dη
≤C2tq 1
qtγB
q, γ1
:Mt. 2.12
3. Existence of Solution
Assume that
B1f :I×X×X× · · · ×X → Xsatisfiesf·, v1, v2, . . . , vn:I → Xis measurable for all vi ∈ X,i 1,2, . . . , nand ft,·,·, . . . ,· : X×X× · · · ×X → X is continuous for a.et ∈ I, and there exist a positive functionμ· ∈ LpI, R p > 1/q > 1and a continuous nondecreasing functionω:0,∞ → 0,∞such that
ft, v1, v2, . . . , vn≤μtω n
i1
vi
, t, v1, v2, . . . , vn∈I×X×X× · · · ×X, 3.1
and setTp,qmax{Tq−1/p, Tq}.
B2For any bounded setsD, D1, D2, . . . , Dn⊂X, and 0≤τ≤s≤t≤T, α
gD
≤βtαD,
α
ψt−s, sfs, D1, D2, . . . , Dn
≤β1t, sαD1 β2t, sαD2 · · ·βnt, sαDn, α
ψt−s, sϕs, τfτ, D1, D2, . . . , Dn
≤ζ1t, s, ταD1 ζ2t, s, ταD2 · · ·ζnt, s, ταDn,
3.2
whereβtis a nonnegative function, and supt∈Iβt:β <∞,
sup
t∈I
t
0
βit, sds:βi<∞, i1,2, . . . , n,
sup
t∈I
t
0
s
0
ζjt, s, τdτ ds:ζj<∞, j 1,2, . . . , n.
3.3
B3g:CI;X → Xis continuous and there exists
0< α1<
CMT−1
, α2≥0 3.4
such that
gu≤α1uα2. 3.5
B4The functionsμandωsatisfy the following condition:
C
1CB q, γ
Tp,qγ Ωp,q
n i1
Ki∗ μ
Lplim inf
τ→ ∞
ωτ
τ <1−α1
CMT
, 3.6
whereΩp,q p−1/pq−1p−1/p, andTp,qγ max{Tp,q, Tp,qγ}.
Theorem 3.1. Suppose that (B1)–(B4) are satisfied, and ifCMTβ4Σni1βi2ζiKi∗<1, then1.1has a mild solution on0, T.
Proof. Define the operatorF:CI;X → CI;Xby
Fut A−10gu u0 t
0
ψ
t−η, η U
η
A0
A−10gu u0 dη
t
0
ψ
t−η, η f
η,K1u η
,K2u η
, . . . ,Knu η
dη
t
0
η
0
ψ
t−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη, t∈I.
3.7
Then we proceed in five steps.
Step 1. We show thatFis continuous.
Letuibe a sequence thatui → uasi → ∞. SincefsatisfiesB1, we have
ft,K1uit,K2uit, . . . ,Knuit−→ft,K1ut,K2ut, . . . ,Knut, asi−→ ∞.
3.8
Then
Fuit−Fut
≤A−10gui−gu t
0
ψ
t−η, η U
ηgui−gudη
t
0
ψ
t−η, η f
η,K1ui η
,K2ui η
, . . . ,Knui η
−f
η,K1u η
,K2u η
, . . . ,Knu
ηdη
t
0
η
0
ψ
t−η, η ϕ
η, s
fs,K1uis,K2uis, . . . ,Knuis
−fs,K1us,K2us, . . . ,Knusds dη.
3.9
According to the conditionA2,2.12, and the continuity ofg, we have A−10gui−gu−→0, asi−→ ∞;
t
0
ψ
t−η, η U
ηgui−gudη−→0, asi−→ ∞.
3.10
Noting thatui → uinCI, X, there existsε >0 such thatui−u ≤εforisufficiently large.
Therefore, we have
ft,K1uit,K2uit, . . . ,Knuit−ft,K1ut,K2ut, . . . ,Knut
≤μt
⎡
⎣ω
⎛
⎝n
j1
Kjui t⎞
⎠ω n j1
Kju t⎤
⎦
≤μt
⎡
⎣ω
⎛
⎝n
j1
Kj∗uε
⎞
⎠ω
⎛
⎝n
j1
K∗ju
⎞
⎠
⎤
⎦.
3.11
Using2.10and by means of the Lebesgue dominated convergence theorem, we obtain t
0
ψ
t−η, η f
η,K1ui η
,K2ui η
, . . . ,Knui η
−f
η,K1u η
,K2u η
, . . . ,Knu
ηdη
≤C t
0
t−ηq−1f
η,K1ui η
,K2ui η
, . . . ,Knui η
−f
η,K1u η
,K2u η
, . . . ,Knu
ηdη,
−→0, asi−→ ∞.
3.12
Similarly, by2.10and2.11, we have t
0
η
0
ψ
t−η, η ϕ
η, s
×
fs,K1uit,K2uit, . . . ,Knuit
−fs,K1us,K2us, . . . ,Knusds dη
≤C2 t
0
η
0
t−ηq−1
η−sγ−1
×fs,K1uit,K2uit, . . . ,Knuit
−fs,K1us,K2us, . . . ,Knusds dη
−→0, asi−→ ∞.
3.13
Therefore, we deduce that
ilim→ ∞Fui−Fu0. 3.14
Step 2. We show thatFmaps bounded sets ofCI, Xinto bounded sets inCI, X.
For anyr >0, we setBr {u∈CI, X:u ≤r}. Now, foru∈Br, byB1, we can see ft,K1ut,K2ut, . . . ,Knut≤μtω
⎛
⎝n
j1
K∗jr
⎞
⎠. 3.15
Based on2.12, we denote thatSt:t
0ψt−η, ηUηdη, we have StA0u0 ≤C2tq
1 qtγB
q, γ1
A0u0MtA0u0. 3.16
Then for anyu∈Br, byA2,2.10,2.11, andLemma 2.1, we have Fut ≤A−10guu0StguStA0u0
t
0
ψ
t−η, η f
η,K1u η
,K2u η
, . . . ,Knu ηdη
t
0
η
0
ψ
t−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη
≤
CMtguu0MtA0u0 C
t
0
t−ηq−1 μ
η ω
⎛
⎝n
j1
Kj∗r
⎞
⎠dη
C2 t
0
η
0
t−ηq−1
η−sγ−1 μsω
⎛
⎝n
j1
K∗jr
⎞
⎠ds dη
≤α1
CMt
uα2
CMt
u0MtA0u0
M1
C
t
0
t−ηq−1 μ
η
dηC2B q, γ
t 0
t−ηqγ−1 μ
η dη
,
3.17 whereM1ω nj1Kj∗r.
By means of the H ¨older inequality, we have t
0
t−ηq−1 μ
η
dηtpq−1/pMp,qμ
Lp ≤Tp,qΩp,qμ
Lp, t
0
t−ηγq−1 μ
η
dη≤Tp,qγΩp,qγμ
Lp.
3.18
Thus
Fut ≤α1
CMT rα2
CMT
u0MTA0u0 M1Ωp,qTp,qγ
CC2B
q, γμ
Lp:r.!
3.19
This meansFBr⊂Br!.
Step 3. We show that there existsm∈Nsuch thatFBm⊂Bm.
Suppose the contrary, that for everym∈N, there existsum∈Bmandtm∈I, such that Fumtm> m. However, on the other hand
ft,K1umt,K2umt, . . . ,Knumt≤μtω
⎛
⎝n
j1
Kj∗m
⎞
⎠, 3.20
we have
m <Fumtm ≤α1
CMT
umα2
CMT u0 MTA0u 0M1
C
tm
0
tm−ηq−1 μ
η
dηC2B q, γ
tm
0
tm−ηqγ−1 μ
η dη
≤α1
CMT
umα2
CMT u0 MTA0u 0M1Ωp,qTp,qγ
CC2B
q, γμ
Lp
≤α1
CMT mα2
CMT u0 MTA0u 0M1Ωp,qTp,qγ
CC2B
q, γμ
Lp.
3.21
Dividing both sides bymand taking the lower limit asm → ∞, we obtain
C
1CB q, γ
Tp,qγ Ωp,q
n j1
K∗jμ
Lplim inf
m→ ∞
wm
m ≥1−α1
CMT
3.22
which contradictsB4.
Step 4. Denote
Fut A−10gu u0 t
0
ψ
t−η, η U
η
A0
A−10gu u0
dηGut, 3.23
where Gut
t
0
ψ
t−η, η f
η,K1u η
,K2u η
, . . . ,Knu η
dη
t
0
η
0
ψ
t−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη.
3.24
We show thatGu·is equicontinuous.
Let 0< t2< t1< T andu∈Bm. Then
Gut1−Gut2 ≤I1I2I3I4, 3.25
where
I1 t2
0
ψ
t1−η, η
−ψ
t2−η, η f
η,K1u η
,K2u η
, . . . ,Knu
ηdη, I2
t1
t2
ψ
t1−η, η f
η,K1u η
,K2u η
, . . . ,Knu
ηdη, I3
t2
0
η
0
ψ
t1−η, η
−ψ
t2−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη, I4
t1
t2
η
0
ψ
t1−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη.
3.26 It follows fromLemma 2.9,B1, and3.20thatI1, I3 → 0,ast2 → t1.
ForI2, from2.10,3.20, andB1, we have
I2 t1
t2
ψ
t1−η, η f
η,K1u η
,K2u η
, . . . ,Knu ηdη
≤CM1
t1
t2
t1−ηq−1 μ
η
dη−→0, ast2−→t1.
3.27
Similarly, by2.10,2.11,B1, andLemma 2.1, we have
I4 t1
t2
η
0
ψ
t1−η, η ϕ
η, s
fs,K1us,K2us, . . . ,Knusds dη
≤C2M1 t1
t2
t1−ηq−1η
0
η−sγ−1
μsds dη−→0, ast2−→t1.
3.28
Step 5. We show thatαFH< αHfor every bounded setH⊂Bm. For anyε >0, we can take a sequence{hv}∞v1 ⊂Hsuch that
αH≤2α{hv} ε, 3.29
cf.30. So it follows from Lemmas2.3–2.5,2.9,2inRemark 2.10, andB2that
αFH≤Cα
gH
MTα
gH
2αG{hv} ε
≤Cα
gH
MTα
gH
2 sup
t∈I α t
0
ψ
t−η, η f
η,K1hv η
,K2hv η
, . . . ,Knhv η
dη
t
0
η
0
ψ
t−η, η ϕ
η, s
×fs,K1hvs,K2hvs, . . . ,Knhvsds dη ε
≤CβαH MTβαH
4 sup
t∈I
t
0
α ψ
t−η, η f
η,K1hv η
,K2hv η
, . . . ,Knhv
η
dη
8 sup
t∈I
t
0
η
0
α ψ
t−η, η ϕ
η, s
fs,K1hvs,K2hvs, . . . ,Knhvs
ε≤CβαH MTβαH 4 sup
t∈I
t
0
n
i1
βi
t, η Ki∗
α{hv}dη
8 sup
t∈I
t
0
η
0
n
i1
ζi
t, η, s Ki∗
α{hv}ds dη
ε
≤CβαH MTβαH
4
n i1
βiK∗i 8 n
i1
ζiK∗i
α{hv} ε
CMT β4
n
i1
βi2ζi K∗i
αH ε.
3.30
Sinceεis arbitrary, we can obtain
αFH≤
CMT β4
Σni1 βi2ζi
K∗i
αH< αH. 3.31
In summary, we have proven thatFhas a fixed pointu!∈Bm. Consequently,1.1has at least one mild solution.
Our next result is based on the Banach’s fixed point theorem.
G1There exists a positive functionl·∈L1I, Rand a constantμ >0 such that gu−gu∗≤μu−u∗,
ft, v1, v2, . . . , vn−ft, w1, w2, . . . , wn
≤lt n
i1
vi−wi
, vi, wi∈X2, i1,2, . . . , n.
3.32
G2There exists a constant 0< δ <1 such that the functionΛ:I → Rdefined by
Λt μ
CMT C
n i1
K∗i
Γ q
Iqlt C2 n
i1
K∗i
Γ q
Γ γ
Iqγlt≤δ, t∈I.
3.33
Theorem 3.2. Assume that (G1), (G2) are satisfied, then1.1has a unique mild solution.
Proof. LetFbe defined as inTheorem 3.1. For anyu, u∗∈CI, X, we have
ft,K1ut,K2ut, . . . ,Knut−ft,K1u∗t,K2u∗t, . . . ,Knu∗t
≤lt n
i1
Kiut−Kiu∗t
≤ltn
i1
Ki∗u−u∗.
3.34
Thus, fromA2,2.10,2.11,Lemma 2.1, we have Fut−Fu∗t
≤μCu−u∗μ t
0
ψ
t−η, η U
ηu−u∗dη
t
0
ψ
t−η, ηf
η,K1u η
,K2u η
, . . . ,Knu η
−f
η,K1u∗ η
,K2u∗ η
, . . . ,Knu∗ ηdη
t
0
η
0
ψ
t−η, η ϕ
η, sfs,Kus,Hus−fs,Ku∗s,Hu∗sds dη
≤ u−u∗
μ
CMT C
n
i1
Ki∗
t 0
t−ηq−1 l
η dη
C2 n
i1
K∗i
t 0
η
0
t−ηq−1
η−sγ−1
lsds dη
μ
CMT C
n
i1
K∗i
Γ q
Iqlt C2 n
i1
K∗i
Γ q
Γ γ
Iqγlt
u−u∗ Λtu−u∗.
3.35
We get
Fu−Fu∗ ≤δu−u∗. 3.36
By the Banach contraction mapping principle,F has a unique fixed point, which is a mild solution of1.1.
4. An Example
To illustrate the usefulness of our main result, we consider the following fractional differential equation:
∂q
∂tqut, ξ bt, ξ∂2
∂ξ2ut, ξ tn n
t
0
t−sus, ξdstn n
t
0
e−tsus, ξds, ξ∈0,1, ut,0 ut,1 0,
u0, ξ − ξ
0
y
0
b−10, xsin"""u λ
""
"dx dy,
4.1
where 0< q <1, 0≤t≤1,λ > CM1, n∈N,bt, ξis continuous function and is uniformly H ¨older continuous int, that is, there existC >0 andγ∈0,1such that
bt1, ξ−bt2, ξ ≤C|t1−t2|γ, 0≤t1≤t2≤1. 4.2
LetXL20,1, Rand defineAtby
DAt H20,1∩H010,1 #
H20,1:z0 z1 0
$ ,
−Atz bt, ξz.
4.3
Then−Asgenerates an analytic semigroup exp−tAs.
Fort∈0,1,ξ∈0,1, we set
utξ ut, ξ, gu sin"""u
λ
""
", A−10gu −
ξ
0
y
0
b−10, xsin"""u λ
""
"dx dy, ft,K1ut,K2utξ tn
n t
0
t−sus, ξdstn n
t
0
e−tsus, ξds,
4.4
where
K1utξ t
0
t−sus, ξds,
K2utξ t
0
e−tsus, ξds,
K1∗sup
t∈I
t
0
t−sds < 1 2 <∞, K∗2sup
t∈I
t
0
e−tsds1 4 <∞.
4.5
Moreover, we can get
gu≤ 1 λu, α
gD
≤ 1 λαD
4.6
for anyD⊂X. Then the above equation4.1can be written in the abstract form as1.1. On the other hand,
ft,Kut,Hutξ≤ tn
nK1ut, ξK2ut, ξ
≤ tn n
K∗1uK2∗u
μtω
K∗1uK2∗u ,
4.7
whereμt tn,ωz z/nsatisfyingB1. For anyu1, u2∈X,
ψt−s, sfs,K1u1s,K2u1sξ−ψt−s, sfs,K1u2s,K2u2sξ
≤ Csn
n t−sq−1K1u1sξ−K1u2sξK2u1sξ−K2u2sξ.
4.8
Therefore, for any bounded setsD1, D2⊂X, we have α
ψt−s, sfs, D1, D2
≤ Csn
n t−sq−1αD1 αD2. 4.9
Moreover, C n sup
t∈0,1
t
0
t−sq−1snds C n sup
t∈0,1tnqB
q, n1 C
nB
q, n1
:β1β2. 4.10
Similarly, we obtain
α
ψt−s, sϕs, τfτ, D1, D2
≤ C2
n t−sq−1s−τγ−1τnαD1 αD2, C2
n sup
t∈0,1
t
0
s
0
t−sq−1s−τγ−1τndτ ds≤ C2 n B
q, γ B
qγ, n1
:ζ1ζ2.
4.11
Suppose further that
1 3/4nC1CBq, γp−1/pq−1p−1/pμLp <1−CM1/λ, 2 1/λCM1 3β12ζ1<1.
Then4.1has a mild solution byTheorem 3.1.
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