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Hindawi Publishing Corporation
Advances in Difference Equations
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,
Received 28 November 2010; Accepted 29 January 2011
Academic Editor: Toka Diagana
Copyright q 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
 −Atutft, K
1
ut, K
2
ut, ,K
n
ut,t∈ I 0,T,
u0A
−1
0guu
0


, 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 researchers cf., 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,andthe
nonlocal C auchy 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
 −A

t

u

t

 f

t,


K
1
u

t

,

K
2
u

t

, ,

K
n
u

t

,t∈ I 

0,T

,
u


0

 A
−1

0

g

u

 u
0
,
1.1
in Banach spaces, where 0 <q<1, g : CI; X → X.ThetermsK
i
ut, i  1, ,n are
2AdvancesinDifference Equations
defined by

K
i
u

t



t

0
k
i

t, s

u

s

ds, 1.2
the positive functions k
i
t, s are continuous on D  {t, s ∈ R
2
:0≤ s ≤ t ≤ T} and
K

i
 sup
t∈0,T

t
0
k
i

t, s

ds < ∞. 1.3

Let us assume that u ∈ L0,T; X and At 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
Γ

q


t
0

t − s

q−1
u

s

ds. 1.4
In this paper, we denote that C is a positive constant and assume that a family of closed
linear {At : t ∈ 0,T} satisfying
A1 the domain DA of {At : t ∈ 0 ,T} is dense in the Banach space X and in-

dependent of t,
A2 the operator Atλ
−1
exists in LX for any λ with Re λ ≤ 0and




A

t

 λ

−1




C
|
λ  1
|
,t∈

0,T

. 1.5
A3 There exists constant γ ∈ 0, 1 and C such that





A

t
1

− A

t
2

A
−1

s




≤ C
|
t
1
− t
2
|
γ
,t

1
,t
2
,s∈

0,T

. 1.6
Under condition A2,eachoperator−As, s ∈ 0,T generates an analytic semigroup
exp−tAs, t>0, and there exists a constant C such that


A
n

s

exp

−tA

s




C
t
n
, 1.7

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.
Advances in Difference Equations 3
2. Preliminaries
Throughout this work, we set I 0,T.WedenotebyX a Banach space, LX the space
of all linear and bounded operators on X,andCI, X the space of all X-valued continuous
functions on I.
Lemma 2.1 see 9. 1 I
q
: L
1
0,T → L
1
0,T.
2 For g ∈ L
1
0,T,wehave

t
0

η
0

t − η

q−1


η − s

γ−1
g

s

ds dη  B

q, γ


t
0

t − s

qγ−1
g

s

ds, 2.1
where Bq, γ is a Beta function.
Definition 2.2. Let B be a bounded set of seminormed linear space Y. The Kuratowski’s
measure of noncompactness for brevity, α-measure of B is defined as
α

B


 inf

d>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, see 27.
Lemma 2.3 see 27. Let A and B be bounded sets of X.Then
1 αA ≤ αB,ifA ⊆ B.
2 αAαA
cl
,whereA
cl
denotes the closure of A.
3 αA0 if and only if A is precompact.
4 αλA|λ|αA, λ ∈ R.
5 αA ∪ Bmax{αA,αB}.
6 αA  B ≤ αAαB,whereA  B  {x  y : x ∈ A, y ∈ B}.
7 αA  x
0
αA, for any x
0
∈ X.
For H ⊂ CI, X we define

t
0
H

s


ds 


t
0
u

s

ds : u ∈ H

, 2.3
for t ∈ I,whereHs{us ∈ X : u ∈ H}.
The following lemma will be needed.
Lemma 2.4 see 27. If H ⊂ CI,X is a bounded, equicontinuous set, then
1 αHsup
t∈I
αHt.
2 α

t
0
Hsds ≤

t
0
αHsds,fort ∈ I.
4AdvancesinDifference Equations
Lemma 2.5 see 28. If {u
n

}

n1
⊂ L
1
I, X and there exists a m· ∈ L
1
I, R

 such that

u
n

t

≤ m

t

, a.e t ∈ I, 2.4
then α{u
n
t}

n1
 is integrable and
α



t
0
u
n

s

ds


n1

≤ 2

t
0
α
{
u
n

s
}

n1

ds. 2.5
We need to use the following Sadovskii’s fixed point theorem.
Definition 2.6 see 29.LetP be an operator in Banach space X.IfP is continuous and takes
bounded, sets into bounded sets, and αPH <αH 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 o perator on Banach
space X.IfP B ⊆ B for a convex, closed, and bounded set B of X,thenP has a fixed point in B.
According to 4, a mild solution of 1.1 can be defined as follows.
Definition 2.8. Afunctionu ∈ CI, X satisfying the equation
u

t

 A
−1

0

g

u

 u
0


t
0
ψ

t − η, η

U


η

A

0


A
−1

0

g

u

 u
0




t
0
ψ

t − η, η

f


η,

K
1
u


η

,

K
2
u


η

, ,

K
n
u


η





t
0

η
0
ψ

t − η, η

ϕ

η, s

f

s,

K
1
u

s

,

K
2
u

s


, ,

K
n
u

s

ds dη,
2.6
is called a mild solution of 1.1,where
ψ

t, s

 q


0
θt
q−1
ξ
q

θ

exp

−t

q
θA

s

dθ, 2.7
and ξ
q
is a probability density function defined on 0, ∞ such that its Laplace transform is
given by


0
e
−σx
ξ
q

σ

dσ 


j0

−x

j
Γ


1  qj

,q∈

0, 1

,x>0,
ϕ

t, τ




k1
ϕ
k

t, τ

,
2.8
Advances in Difference Equations 5
where
ϕ
1

t, τ




A

t

− A

τ

ψ

t − τ, τ

,
ϕ
k1

t, τ



t
τ
ϕ
k

t, s

ϕ
1


s, τ

ds, k  1, 2 ,
U

t

 −A

t

A
−1

0



t
0
ϕ

t, s

A

s

A

−1

0

ds.
2.9
To our purpose, the following conclusions will be needed. For the proofs refer to 4.
Lemma 2.9 see 4. The operator-valued functions ψt − η, η and Atψt − η, η are continuous
in uniform topology in the variables t, η,where0 ≤ η ≤ t − ε, 0 ≤ t ≤ T, for any ε>0. Clearly,


ψ

t − η, η



≤ C

t − η

q−1
. 2.10
Moreover, we have


ϕ

t, η




≤ C

t − η

γ−1
. 2.11
Remark 2.10. From the proof of Theorem 2.5 in 4,wecansee
1 Ut≤C  Ct
γ
.
2 For t ∈ I,

t
0
ψt − η, ηUηdη is uniformly continuous in the norm of LX and






t
0
ψ

t − η, η

U


η







≤ C
2
t
q

1
q
 t
γ
B

q, γ  1


: M

t

. 2.12
3. Existence of Solution
Assume that

B1 f : I × X × X ×···×X → X satisfies f·,v
1
,v
2
, ,v
n
 : I → X is measurable for all
v
i
∈ X, i  1, 2, ,n and ft, ·, ·, ,· : 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, v
1
,v
2
, ,v
n



≤ μ


t

ω

n

i1

v
i


,

t, v
1
,v
2
, ,v
n

∈ I × X × X ×···×X, 3.1
and set T
p,q
 max{T
q−1/p
,T
q
}.

6AdvancesinDifference Equations
B2 For any bounded sets D, D
1
,D
2
, ,D
n
⊂ X,and0≤ τ ≤ s ≤ t ≤ T,
α

g

D


≤ β

t

α

D

,
α

ψ

t − s, s


f

s, D
1
,D
2
, ,D
n


≤ β
1

t, s

α

D
1

 β
2

t, s

α

D
2


 ··· β
n

t, s

α

D
n

,
α

ψ

t − s, s

ϕ

s, τ

f

τ, D
1
,D
2
, ,D
n



≤ ζ
1

t, s, τ

α

D
1

 ζ
2

t, s, τ

α

D
2

 ··· ζ
n

t, s, τ

α

D
n


,
3.2
where βt is a nonnegative function, and sup
t∈I
βt : β<∞,
sup
t∈I

t
0
β
i

t, s

ds : β
i
< ∞,i 1, 2, ,n,
sup
t∈I

t
0

s
0
ζ
j


t, s, τ

dτ ds :  ζ
j
< ∞,j 1, 2, ,n.
3.3
B3 g : CI; X → X is continuous and there exists
0 <α
1
<

C  M

T


−1

2
≥ 0 3.4
such that


g

u



≤ α

1

u

 α
2
. 3.5
B4 The functions μ and ω satisfy the following c ondition:
C

1  CB

q, γ

T
γ
p,q
Ω
p,q

n

i1
K

i



μ



L
p
lim inf
τ →∞
ω

τ

τ
< 1 − α
1

C 
M

T


, 3.6
where Ω
p,q
p − 1/pq − 1
p−1/p
,andT
γ
p,q
 max{T
p,q

,T
p,qγ
}.
Theorem 3.1. Suppose that (B1)–(B4) are satisfied, and if C 
MTβ  4Σ
n
i1
β
i
 2ζ
i
K

i
 < 1,
then 1.1 has a mild solution on 0,T.
Advances in Difference Equations 7
Proof. Define the operator F : CI; X → CI; X by
F

u

t

 A
−1

0

g


u

 u
0


t
0
ψ

t − η, η

U

η

A

0


A
−1

0

g

u


 u
0




t
0
ψ

t − η, η

f

η,

K
1
u


η

,

K
2
u



η

, ,

K
n
u


η




t
0

η
0
ψ

t − η, η

ϕ

η, s

f


s,

K
1
u

s

,

K
2
u

s

, ,

K
n
u

s

ds dη, t ∈ I.
3.7
Then we proceed in five steps.
Step 1. We show that F is continuous.
Let u
i

be a sequence that u
i
→ u as i →∞.Sincef satisfies B1,wehave
f

t,

K
1
u
i

t

,

K
2
u
i

t

, ,

K
n
u
i


t

−→ f

t,

K
1
u

t

,

K
2
u

t

, ,

K
n
u

t

, as i −→ ∞ .
3.8

Then

F

u
i

t

− F

u

t





A
−1

0






g


u
i

− g

u





t
0


ψ

t − η, η

U

η





g


u
i

− g

u






t
0


ψ

t − η, η

f

η,

K
1
u
i



η

,

K
2
u
i


η

, ,

K
n
u
i


η

−f

η,

K
1
u



η

,

K
2
u


η

, ,

K
n
u


η






t
0

η

0


ψ

t − η, η

ϕ

η, s

f

s,

K
1
u
i

s

,

K
2
u
i

s


, ,

K
n
u
i

s

−f

s,

K
1
u

s

,

K
2
u

s

, ,


K
n
u

s




ds dη.
3.9
According to the condition A2, 2.12, and the continuity of g,wehave



A
−1

0






g

u
i


− g

u



−→ 0, as i −→ ∞ ;

t
0


ψ

t − η, η

U

η





g

u
i

− g


u



dη −→ 0, as i −→ ∞ .
3.10
8AdvancesinDifference Equations
Noting that u
i
→ u in CI,X,thereexistsε>0suchthatu
i
− u≤ε for i sufficiently large.
Therefore, we have



f

t,

K
1
u
i

t

,


K
2
u
i

t

, ,

K
n
u
i

t

− f

t,

K
1
u

t

,

K
2

u

t

, ,

K
n
u

t




≤ μ

t



ω


n

j1




K
j
u
i


t





 ω
n

j1



K
j
u


t






≤ μ

t



ω


n

j1
K

j

u

 ε



 ω


n

j1
K


j

u





.
3.11
Using 2.10 and by means of the Lebesgue dominated convergence theorem, we obtain

t
0


ψ

t − η, η

f

η,

K
1
u
i



η

,

K
2
u
i


η

, ,

K
n
u
i


η

−f

η,

K
1
u



η

,

K
2
u


η

, ,

K
n
u


η




≤ C

t
0

t − η


q−1



f

η,

K
1
u
i


η

,

K
2
u
i


η

, ,

K

n
u
i


η

−f

η,

K
1
u


η

,

K
2
u


η

, ,

K

n
u


η



dη,
−→ 0, as i −→ ∞ .
3.12
Similarly, by 2.10 and 2.11,wehave

t
0

η
0


ψ

t − η, η

ϕ

η, s

×


f

s,

K
1
u
i

t

,

K
2
u
i

t

, ,

K
n
u
i

t

−f


s,

K
1
u

s

,

K
2
u

s

, ,

K
n
u

s




ds dη
≤ C

2

t
0

η
0

t − η

q−1

η − s

γ−1
×


f

s,

K
1
u
i

t

,


K
2
u
i

t

, ,

K
n
u
i

t

−f

s,

K
1
u

s

,

K

2
u

s

, ,

K
n
u

s



ds dη
−→ 0, as i −→ ∞ .
3.13
Therefore, we deduce that
lim
i →∞

F

u
i

− F

u


 0. 3.14
Advances in Difference Equations 9
Step 2. We show that F maps bounded sets of CI, X into bounded sets in CI, X.
For any r>0, we set B
r
 {u ∈ CI, X : u≤r}. Now, for u ∈ B
r
,byB1,wecansee


f

t,

K
1
u

t

,

K
2
u

t

, ,


K
n
u

t



≤ μ

t

ω


n

j1
K

j
r


. 3.15
Based on 2.12,wedenotethatSt :

t
0

ψt − η, ηUηdη,wehave

S

t

A

0

u
0

≤ C
2
t
q

1
q
 t
γ
B

q, γ  1



A


0

u
0


M

t

A

0

u
0

. 3.16
Then for any u ∈ B
r
,byA2, 2.10, 2.11,andLemma 2.1,wehave

Fu

t






A
−1

0

g

u






u
0




S

t

g

u






S

t

A

0

u
0



t
0


ψ

t − η, η

f

η,

K
1
u



η

,

K
2
u


η

, ,

K
n
u


η






t
0


η
0


ψ

t − η, η

ϕ

η, s

f

s,

K
1
u

s

,

K
2
u

s


, ,

K
n
u

s



ds dη


C 
Mt



g

u





u
0



Mt

A

0

u
0

 C

t
0

t − η

q−1
μ

η

ω


n

j1
K

j

r



 C
2

t
0

η
0

t − η

q−1

η − s

γ−1
μ

s

ω


n

j1

K

j
r


ds dη
≤ α
1

C 
Mt


u

 α
2

C 
M

t




u
0



Mt

A

0

u
0

 M
1

C

t
0

t − η

q−1
μ

η

dη  C
2
B

q, γ



t
0

t − η

qγ−1
μ

η



,
3.17
where M
1
 ω

n
j1
K

j
r.
By means of the H
¨
older inequality, we have


t
0

t − η

q−1
μ

η

dη  t
pq−1/p
M
p,q


μ


L
p
≤ T
p,q
Ω
p,q


μ



L
p
,

t
0

t − η

γq−1
μ

η

dη ≤ T
p,qγ
Ω
p,qγ


μ


L
p
.
3.18
10 Advances in Difference Equations
Thus


Fu

t

≤ α
1

C 
M

T


r  α
2

C 
M

T




u
0


M


T

A

0

u
0

 M
1
Ω
p,q
T
γ
p,q

C  C
2
B

q, γ




μ


L

p
: r.
3.19
This means FB
r
 ⊂ B
r
.
Step 3. We show that there exists m ∈ N such that FB
m
 ⊂ B
m
.
Suppose the contrary, that for every m ∈ N,thereexistsu
m
∈ B
m
and t
m
∈ I,suchthat
Fu
m
t
m
 >m. However, on the other hand


f

t,


K
1
u
m

t

,

K
2
u
m

t

, ,

K
n
u
m

t



≤ μ


t

ω


n

j1
K

j
m


, 3.20
we have
m<

Fu
m

t
m

≤ α
1

C 
M


T



u
m

 α
2

C 
M

T




u
0


M

T

A

0


u
0

 M
1

C

t
m
0

t
m
− η

q−1
μ

η

dη  C
2
B

q, γ


t
m

0

t
m
− η

qγ−1
μ

η



≤ α
1

C 
M

T



u
m

 α
2

C 

M

T




u
0


M

T

A

0

u
0

 M
1
Ω
p,q
T
γ
p,q


C  C
2
B

q, γ




μ


L
p
≤ α
1

C 
M

T


m  α
2

C 
M

T





u
0


M

T

A

0

u
0

 M
1
Ω
p,q
T
γ
p,q

C  C
2
B


q, γ




μ


L
p
.
3.21
Dividing both sides by m and taking the lower limit as m →∞,weobtain
C

1  CB

q, γ

T
γ
p,q
Ω
p,q
n

j1
K


j


μ


L
p
lim inf
m →∞
w

m

m
≥ 1 − α
1

C 
M

T


3.22
which contradicts B4.
Step 4. Denote
F

u


t

 A
−1

0

g

u

 u
0


t
0
ψ

t − η, η

U

η

A

0



A
−1

0

g

u

 u
0

dη  G

u

t

, 3.23
Advances in Difference Equations 11
where
G

u

t




t
0
ψ

t − η, η

f

η,

K
1
u


η

,

K
2
u


η

, ,

K
n

u


η




t
0

η
0
ψ

t − η, η

ϕ

η, s

f

s,

K
1
u

s


,

K
2
u

s

, ,

K
n
u

s

ds dη.
3.24
We show that Gu· is equicontinuous.
Let 0 <t
2
<t
1
<Tand u ∈ B
m
.Then


Gu


t
1



Gu

t
2

≤I
1
 I
2
 I
3
 I
4
, 3.25
where
I
1


t
2
0




ψ

t
1
− η, η

− ψ

t
2
− η, η

f

η,

K
1
u


η

,

K
2
u



η

, ,

K
n
u


η



dη,
I
2


t
1
t
2


ψ

t
1
− η, η


f

η,

K
1
u


η

,

K
2
u


η

, ,

K
n
u


η




dη,
I
3


t
2
0

η
0



ψ

t
1
− η, η

− ψ

t
2
− η, η

ϕ


η, s

f

s,

K
1
u

s

,

K
2
u

s

, ,

K
n
u

s




ds dη,
I
4


t
1
t
2

η
0


ψ

t
1
− η, η

ϕ

η, s

f

s,

K
1

u

s

,

K
2
u

s

, ,

K
n
u

s



ds dη.
3.26
It follows from Lemma 2.9, B1,and3.20 that I
1
,I
3
→ 0, as t
2

→ t
1
.
For I
2
,from2.10, 3.20,andB1,wehave
I
2


t
1
t
2


ψ

t
1
− η, η

f

η,

K
1
u



η

,

K
2
u


η

, ,

K
n
u


η




≤ CM
1

t
1
t

2

t
1
− η

q−1
μ

η

dη −→ 0, as t
2
−→ t
1
.
3.27
Similarly, by 2.10, 2.11, B1,andLemma 2.1,wehave
I
4


t
1
t
2

η
0



ψ

t
1
− η, η

ϕ

η, s

f

s,

K
1
u

s

,

K
2
u

s

, ,


K
n
u

s



ds dη
≤ C
2
M
1

t
1
t
2

t
1
− η

q−1

η
0

η − s


γ−1
μ

s

ds dη −→ 0, as t
2
−→ t
1
.
3.28
12 Advances in Difference Equations
Step 5. We show that αFH <αH for every bounded set H ⊂ B
m
.Foranyε>0, we can
take a sequence {h
v
}

v1
⊂ H such that
α

H

≤ 2α
{
h
v

}
 ε, 3.29
cf. 30. So it follows f rom Lemmas 2.3–2.5, 2.9, 2 in Remark 2.10,andB2 that
α

F

H

≤ Cα

g

H



M

T

α

g

H


 2α


G
{
h
v
}
 ε
≤ Cα

g

H



M

T

α

g

H


 2sup
t∈I
α



t
0
ψ

t − η, η

f

η,

K
1
h
v


η

,

K
2
h
v


η

, ,


K
n
h
v


η






t
0

η
0
ψ

t − η, η

ϕ

η, s

×f

s,


K
1
h
v

s

,

K
2
h
v

s

, ,

K
n
h
v

s

ds dη

 ε
≤ Cβα


H


M

T

βα

H

4sup
t∈I


t
0
α

ψ

t − η, η

f

η,

K
1
h

v


η

,

K
2
h
v


η

, ,

K
n
h
v


η



8sup
t∈I



t
0

η
0
α

ψ

t − η, η

ϕ

η, s

f

s,

K
1
h
v

s

,

K

2
h
v

s

, ,

K
n
h
v

s



 ε ≤ Cβα

H


M

T

βα

H


 4sup
t∈I


t
0

n

i1
β
i

t, η

K

i

α
{
h
v
}


 8sup
t∈I



t
0

η
0

n

i1
ζ
i

t, η, s

K

i

α
{
h
v
}
ds dη

 ε
≤ Cβα

H



M

T

βα

H



4
n

i1
β
i
K

i
 8
n

i1
ζ
i
K

i


α
{
h
v
}
 ε



C 
M

T


β  4

n

i1

β
i
 2ζ
i

K

i


α

H

 ε.
3.30
Since ε is arbitrary, we can obtain
α

F

H



C 
M

T


β  4

Σ
n
i1

β
i
 2ζ

i

K

i


α

H



H

. 3.31
Advances in Difference Equations 13
In summary, we have proven that F has a fixed point u ∈ B
m
.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· ∈ L
1
I, R

 and a constant μ>0suchthat


g


u

− g

u




≤ μ

u − u


,


f

t, v
1
,v
2
, ,v
n

− f

t, w

1
,w
2
, ,w
n



≤ l

t


n

i1

v
i
− w
i


,

v
i
,w
i


∈ X
2
,i 1, 2, ,n.
3.32
G2 There exists a constant 0 <δ<1 such that the function Λ : I → R

defined by
Λ

t

 μ

C 
M

T


 C

n

i1
K

i

Γ


q

I
q
l

t

 C
2

n

i1
K

i

Γ

q

Γ

γ

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 in Theorem 3.1.Foranyu, u

∈ CI,X,wehave


f

t,

K
1
u

t

,

K
2
u

t

, ,


K
n
u

t

− f

t,

K
1
u


t

,

K
2
u


t

, ,

K
n

u


t



≤ l

t


n

i1

K
i
u

t



K
i
u


t



≤ l

t

n

i1
K

i

u − u


.
3.34
Thus, from A2, 2.10, 2.11, Lemma 2.1,wehave

Fu

t



Fu


t


≤ μC

u − u


 μ

t
0


ψ

t − η, η

U

η




u − u





t

0


ψ

t − η, η





f

η,

K
1
u


η

,

K
2
u


η


, ,

K
n
u


η

−f

η,

K
1
u



η

,

K
2
u




η

, ,

K
n
u



η






t
0

η
0


ψ

t − η, η

ϕ


η, s





f

s,

Ku

s

,

Hu

s

− f

s,

Ku


s

,


Hu


s



ds dη
14 Advances in Difference Equations


u − u



μ

C 
M

T


 C

n

i1
K


i


t
0

t − η

q−1
l

η


C
2

n

i1
K

i


t
0

η

0

t − η

q−1

η − s

γ−1
l

s

ds dη



μ

C  M

T


 C

n

i1
K


i

Γ

q

I
q
l

t

 C
2

n

i1
K

i

Γ

q

Γ

γ


I
qγ
l

t



u − u


Λ

t

u − u


.
3.35
We get

F

u

− F

u



≤ δ

u − u


. 3.36
By the Banach contraction mapping principle, F has a unique fixed point, which is a mild
solution of 1.1.
4. An Example
To illustrate the usefulness of our main result, we consider the following fractional differential
equation:

q
∂t
q
u

t, ξ

 b

t, ξ


2
∂ξ
2
u


t, ξ


t
n
n

t
0

t − s

u

s, ξ

ds 
t
n
n

t
0
e
−ts
u

s, ξ


ds, ξ ∈

0, 1

,
u

t, 0

 u

t, 1

 0,
u

0,ξ

 −

ξ
0

y
0
b
−1

0,x


sin



u
λ



dx dy,
4.1
where 0 <q<1, 0 ≤ t ≤ 1, λ>C
M1, n ∈ N, bt, ξ is continuous function and is uniformly
H
¨
older continuous in t, that is, there exist C>0andγ ∈ 0, 1 such that

b

t
1


− b

t
2


≤ C

|
t
1
− t
2
|
γ
, 0 ≤ t
1
≤ t
2
≤ 1. 4.2
Let X  L
2
0, 1,R and define At by
D

A

t

 H
2

0, 1

∩ H
1
0


0, 1



H
2

0, 1

: z

0

 z

1

 0

,
−A

t

z

 b

t, ξ


z

.
4.3
Then −As generates an analytic semigroup exp−tAs.
Advances in Difference Equations 15
For t ∈ 0, 1, ξ ∈ 0, 1,weset
u

t

ξ

 u

t, ξ

,
g

u

 sin



u
λ




,
A
−1

0

g

u

 −

ξ
0

y
0
b
−1

0,x

sin



u
λ




dx dy,
f

t,

K
1
u

t

,

K
2
u

t

ξ


t
n
n

t
0


t − s

u

s, ξ

ds 
t
n
n

t
0
e
−ts
u

s, ξ

ds,
4.4
where

K
1
u

t


ξ



t
0

t − s

u

s, ξ

ds,

K
2
u

t

ξ



t
0
e
−ts
u


s, ξ

ds,
K

1
 sup
t∈I

t
0

t − s

ds <
1
2
< ∞,
K

2
 sup
t∈I

t
0
e
−ts
ds 

1
4
< ∞.
4.5
Moreover, we can get


g

u




1
λ

u

,
α

g

D



1
λ

α

D

4.6
for any D ⊂ X. Then the above equation 4.1 can be written in the abstract form as 1.1.On
the other hand,


f

t,

Ku

t

,

Hu

t

ξ




t
n

n

K
1
u

t, ξ



K
2
u

t, ξ


t
n
n

K

1

u

 K

2


u


 μ

t

ω

K

1

u

 K

2

u


,
4.7
16 Advances in Difference Equations
where μtt
n
, ωzz/n satisfying B1.Foranyu
1

,u
2
∈ X,


ψ

t − s, s

f

s,

K
1
u
1

s

,

K
2
u
1

s

ξ


− ψ

t − s, s

f

s,

K
1
u
2

s

,

K
2
u
2

s

ξ





Cs
n
n

t − s

q−1

K
1
u
1

s

ξ



K
1
u
2

s

ξ




K
2
u
1

s

ξ



K
2
u
2

s

ξ

.
4.8
Therefore, for a ny bounded sets D
1
,D
2
⊂ X,wehave
α

ψ


t − s, s

f

s, D
1
,D
2



Cs
n
n

t − s

q−1

α

D
1

 α

D
2


.
4.9
Moreover,
C
n
sup
t∈0,1

t
0

t − s

q−1
s
n
ds 
C
n
sup
t∈0,1
t
nq
B

q, n  1


C
n

B

q, n  1

: β
1
 β
2
. 4.10
Similarly, we obtain
α

ψ

t − s, s

ϕ

s, τ

f

τ, D
1
,D
2



C

2
n

t − s

q−1

s − τ

γ−1
τ
n

α

D
1

 α

D
2

,
C
2
n
sup
t∈0,1


t
0

s
0

t − s

q−1

s − τ

γ−1
τ
n
dτ ds ≤
C
2
n
B

q, γ

B

q  γ,n 1

: ζ
1
 ζ

2
.
4.11
Suppose further that
13/4nC1  CBq, γp − 1/pq − 1
p−1/p
μ
L
p
< 1 − C  M1/λ,
21/λC 
M1  3β
1
 2ζ
1
 < 1.
Then 4.1 has a mild solution by Theorem 3.1.
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