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, sheaf@mail.ustc.edu.cn
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 dq ut/dtq −Atut f t, K1 ut, K2 ut, . . . , Kn ut, t ∈ I 0, T,
u0 A−1 0gu 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 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, 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
dq ut
−Atut ft, K1 ut, K2 ut, . . . , Kn ut,
dtq
t ∈ I 0, T,
1.1
−1
u0 A 0gu u0 ,
in Banach spaces, where 0 < q < 1, g : CI; X → X. The terms Ki ut, i 1, . . . , n are
2
Advances in Difference Equations
defined by
t
Ki ut ki t, susds,
1.2
0
the positive functions ki t, s are continuous on D {t, s ∈ R2 : 0 ≤ s ≤ t ≤ T} and
Ki∗ sup
t
t∈0,T ki t, sds < ∞.
1.3
0
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,
1
I ut Γ q
t
q
t − sq−1 usds.
1.4
0
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 independent of t,
A2 the operator At λ−1 exists 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 − At2 A−1 s ≤ C|t1 − t2 |γ ,
t1 , t2 , s ∈ 0, T.
1.6
Under condition A2, each operator −As, s ∈ 0, T generates an analytic semigroup
exp−tAs, t > 0, and there exists a constant C such that
n
A s exp−tAs ≤ C ,
tn
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.
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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 : L1 0, T → L1 0, T.
2 For g ∈ L1 0, T, we have
t η
0
q−1 γ −1
t−η
η−s
gsds dη B q, γ
t
0
t − sqγ −1 gsds,
2.1
0
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 noncompactness for brevity, α-measure of B is defined as
αB 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 αA ≤ αB, if A ⊆ B.
2 αA αAcl , where Acl denotes 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
Hsds t
0
usds : u ∈ H ,
0
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 supt∈I αHt.
t
t
2 α 0 Hsds ≤ 0 αHsds, for t ∈ I.
2.3
4
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1
1
Lemma 2.5 see 28. If {un }∞
n1 ⊂ L I, X and there exists a m· ∈ L I, R such that
un t
≤ mt,
a.e t ∈ I,
2.4
then α{un t}∞
n1 is integrable and
t
α
∞ ≤2
un sds
0
t
0
n1
α{un s}∞
n1 ds.
2.5
We need to use the following Sadovskii’s fixed point theorem.
Definition 2.6 see 29. Let P be an operator in Banach space X. If P is continuous and takes
bounded, sets into bounded sets, and αP H < αH for every bounded set H of X with
αH > 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 to 4, a mild solution of 1.1 can be defined as follows.
Definition 2.8. A function u ∈ CI, X satisfying the equation
ut A−1 0gu u0 t
ψ t − η, η U η A0 A−1 0gu u0 dη
0
t
ψ t − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη
0
t η
0
2.6
ψ t − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη,
0
is called a mild solution of 1.1, where
ψt, s q
∞
θtq−1 ξq θ exp−tq θAsdθ,
2.7
0
and ξq is a probability density function defined on 0, ∞ such that its Laplace transform is
given by
∞
0
e−σx ξq σdσ ∞
−xj
,
j0 Γ 1 qj
ϕt, τ ∞
k1
q ∈ 0, 1, x > 0,
ϕk t, τ,
2.8
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5
where
ϕ1 t, τ At − Aτψt − τ, τ,
ϕk1 t, τ t
ϕk t, sϕ1 s, τds,
k 1, 2 . . . ,
τ
−1
Ut −AtA 0 −
t
2.9
ϕt, sAsA−1 0ds.
0
To our purpose, the following conclusions will be needed. For the proofs refer to 4.
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 ≤ C t − η q−1 .
2.10
ϕ t, η ≤ C t − η γ −1 .
2.11
Moreover, we have
Remark 2.10. From the proof of Theorem 2.5 in 4, we can see
1 Ut
≤ C Ctγ .
t
2 For t ∈ I, 0 ψt − η, ηUηdη is uniformly continuous in the norm of LX and
t 1
tγ B q, γ 1
: Mt.
ψ t − η, η U η dη ≤ C2 tq
0
q
2.12
3. Existence of Solution
Assume that
B1 f : I × X × X × · · · × X → X satisfies f·, v1 , v2 , . . . , vn : I → X is measurable for all
vi ∈ 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· ∈ Lp I, R p > 1/q > 1 and a
continuous nondecreasing function ω : 0, ∞ → 0, ∞ such that
n
ft, v1 , v2 , . . . , vn ≤ μtω
vi ,
i1
and set Tp,q max{T q−1/p , T q }.
t, v1 , v2 , . . . , vn ∈ I × X × X × · · · × X,
3.1
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B2 For any bounded sets D, D1 , D2 , . . . , Dn ⊂ X, and 0 ≤ τ ≤ s ≤ t ≤ T,
α gD ≤ βtαD,
α ψt − s, sfs, D1 , D2 , . . . , Dn ≤ β1 t, sαD1 β2 t, sαD2 · · · βn t, sαDn ,
3.2
α ψt − s, sϕs, τfτ, D1 , D2 , . . . , Dn ≤ ζ1 t, s, ταD1 ζ2 t, s, ταD2 · · · ζn t, s, ταDn ,
where βt is a nonnegative function, and supt∈I βt : β < ∞,
t
sup
t s
sup
t∈I
0
βi t, sds : βi < ∞,
i 1, 2, . . . , n,
0
t∈I
3.3
ζj t, s, τdτ ds : ζj < ∞,
j 1, 2, . . . , n.
0
B3 g : CI; X → X is continuous and there exists
−1
0 < α1 < C MT ,
α2 ≥ 0
3.4
such that
gu ≤ α1 u
α2 .
3.5
B4 The functions μ and ω satisfy the following condition:
n
γ
ωτ
∗ C 1 CB q, γ Tp,q Ωp,q
Ki
μ Lp lim inf
< 1 − α1 C MT ,
τ →∞ τ
i1
3.6
where Ωp,q p − 1/pq − 1p−1/p , and Tp,q max{Tp,q , Tp,qγ }.
γ
Theorem 3.1. Suppose that (B1)–(B4) are satisfied, and if C MTβ 4Σni1 βi 2ζi Ki∗ < 1,
then 1.1 has a mild solution on 0, T.
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7
Proof. Define the operator F : CI; X → CI; X by
−1
Fut A 0gu u0 t
ψ t − η, η U η A0 A−1 0gu u0 dη
0
t
ψ t − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη
0
t η
0
ψ t − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη,
t ∈ I.
0
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
ft, K1 ui t, K2 ui t, . . . , Kn ui t −→ ft, K1 ut, K2 ut, . . . , Kn ut,
as i −→ ∞.
3.8
Then
Fui t − Fut
t
−1 ≤ A 0 gui − gu ψ t − η, η U η gui − gudη
0
t
ψ t − η, η f η, K1 ui η , K2 ui η , . . . , Kn ui η
0
−f η, K1 u η , K2 u η , . . . , Kn u η dη
t η
0
3.9
ψ t − η, η ϕ η, s fs, K1 ui s, K2 ui s, . . . , Kn ui s
0
−fs, K1 us, K2 us, . . . , Kn us ds dη.
According to the condition A2, 2.12, and the continuity of g, we have
−1 A 0gui − gu −→ 0,
t
0
as i −→ ∞;
ψ t − η, η U η gui − gudη −→ 0,
3.10
as i −→ ∞.
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Noting that ui → u in CI, X, there exists ε > 0 such that ui − u
≤ ε for i sufficiently large.
Therefore, we have
ft, K1 ui t, K2 ui t, . . . , Kn ui t − ft, K1 ut, K2 ut, . . . , Kn ut ⎡ ⎛
⎤
⎞
n n ≤ μt⎣ω⎝ Kj ui t⎠ ω Kj u t⎦
j1
j1
3.11
⎡ ⎛
⎞
⎛
⎞⎤
n
n
≤ μt⎣ω⎝ Kj∗ u
ε⎠ ω⎝ Kj∗ u
⎠⎦.
j1
j1
Using 2.10 and by means of the Lebesgue dominated convergence theorem, we obtain
t
ψ t − η, η f η, K1 ui η , K2 ui η , . . . , Kn ui η
0
−f η, K1 u η , K2 u η , . . . , Kn u η dη
≤C
t
q−1 f η, K1 ui η , K2 ui η , . . . , Kn ui η
t−η
3.12
0
−f η, K1 u η , K2 u η , . . . , Kn u η dη,
−→ 0,
as i −→ ∞.
Similarly, by 2.10 and 2.11, we have
t η
0
0
ψ t − η, η ϕ η, s
× fs, K1 ui t, K2 ui t, . . . , Kn ui t
−fs, K1 us, K2 us, . . . , Kn us ds dη
t η
q−1 γ −1
2
t−η
η−s
≤C
0
3.13
0
× fs, K1 ui t, K2 ui t, . . . , Kn ui t
−fs, K1 us, K2 us, . . . , Kn usds dη
−→ 0,
as i −→ ∞.
Therefore, we deduce that
lim Fui − Fu
0.
i→∞
3.14
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Step 2. We show that F maps bounded sets of CI, X into bounded sets in CI, X.
For any r > 0, we set Br {u ∈ CI, X : u
≤ r}. Now, for u ∈ Br , by B1, we can see
⎛
⎞
n
∗
ft, K1 ut, K2 ut, . . . , Kn ut ≤ μtω⎝ K r ⎠.
j
3.15
j1
Based on 2.12, we denote that St :
StA0u0 ≤ C t
2 q
t
0
ψt − η, ηUηdη, we have
1
γ t B q, γ 1 A0u0 Mt
A0u0 .
q
3.16
Then for any u ∈ Br , by A2, 2.10, 2.11, and Lemma 2.1, we have
Fut
≤ A−1 0gu u0 Stgu StA0u0 t
ψ t − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη
0
t η
0
ψ t − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη
0
≤ C Mt gu u0 Mt
A0u0 C
t
0
C
⎛
⎞
n
q−1 ∗
t−η
μ η ω⎝ Kj r ⎠dη
j1
t η
2
0
0
⎛
⎞
n
q−1 γ −1
∗
μsω⎝ Kj r ⎠ds dη
t−η
η−s
j1
≤ α1 C Mt u
α2 C Mt u0 Mt
A0u0 t
q−1 qγ −1 t
2 M1 C
t−η
t−η
μ η dη C B q, γ
μ η dη ,
0
0
3.17
where M1 ω nj1 Kj∗ r.
By means of the Hölder inequality, we have
t
0
q−1 μ η dη tpq−1/p Mp,q μLp ≤ Tp,q Ωp,q μLp ,
t−η
t
0
γ q−1 μ η dη ≤ Tp,qγ Ωp,qγ μLp .
t−η
3.18
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Thus
Fut
≤ α1 C MT r α2 C MT u0 MT
A0u0 γ
M1 Ωp,q Tp,q C C2 B q, γ μLp : r!.
3.19
This means FBr ⊂ Br! .
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
Fum tm > m. However, on the other hand
⎛
⎞
n
ft, K1 um t, K2 um t, . . . , Kn um t ≤ μtω⎝ K ∗ m⎠,
j
3.20
j1
we have
m < Fum tm ≤ α1 C MT um α2 C MT u0 t
m
q−1 tm qγ −1 2 tm − η
tm − η
MT
A0u0 M1 C
μ η dη C B q, γ
μ η dη
0
0
≤ α1 C MT um α2 C MT u0 γ
MT
A0u0 M1 Ωp,q Tp,q C C2 B q, γ μLp
≤ α1 C MT m α2 C MT u0 γ
MT
A0u0 M1 Ωp,q Tp,q C C2 B q, γ μLp .
3.21
Dividing both sides by m and taking the lower limit as m → ∞, we obtain
n
γ
wm
≥ 1 − α1 C MT
C 1 CB q, γ Tp,q Ωp,q Kj∗ μLp lim inf
m→∞ m
j1
3.22
which contradicts B4.
Step 4. Denote
−1
Fut A 0gu u0 t
0
ψ t − η, η U η A0 A−1 0gu u0 dη Gut,
3.23
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11
where
Gut t
ψ t − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη
0
t η
0
ψ t − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη.
3.24
0
We show that Gu· is equicontinuous.
Let 0 < t2 < t1 < T and u ∈ Bm . Then
Gut1 − Gut2 ≤ I1 I2 I3 I4 ,
3.25
where
I1 t2
ψ t1 − η, η − ψ t2 − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη,
0
I2 t1
ψ t1 − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη,
t2
I3 t2 η
0
0
I4 ψ t1 − η, η − ψ t2 − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη,
t1 η
ψ t1 − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη.
0
t2
3.26
It follows from Lemma 2.9, B1, and 3.20 that I1 , I3 → 0, as t2 → t1 .
For I2 , from 2.10, 3.20, and B1, we have
I2 t1
ψ t1 − η, η f η, K1 u η , K2 u η , . . . , Kn u η dη
t2
≤ CM1
t1
q−1 μ η dη −→ 0,
t1 − η
3.27
as t2 −→ t1 .
t2
Similarly, by 2.10, 2.11, B1, and Lemma 2.1, we have
I4 t1 η
t2
ψ t1 − η, η ϕ η, s fs, K1 us, K2 us, . . . , Kn usds dη
0
≤ C2 M1
t1
t2
q−1
t1 − η
η
0
γ −1
η−s
μsds dη −→ 0,
3.28
as t2 −→ t1 .
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Step 5. We show that αFH < αH for every bounded set H ⊂ Bm . For any ε > 0, we can
take a sequence {hv }∞
v1 ⊂ H such that
αH ≤ 2α{hv } ε,
3.29
cf. 30. So it follows from Lemmas 2.3–2.5, 2.9, 2 in Remark 2.10, and B2 that
αFH ≤ Cα gH MTα gH 2αG{hv } ε
≤ Cα gH MTα gH
t
2 sup α
ψ t − η, η f η, K1 hv η , K2 hv η , . . . , Kn hv η dη
0
t∈I
t η
0
ψ t − η, η ϕ η, s
0
×fs, K1 hv s, K2 hv s, . . . , Kn hv sds dη
ε
≤ CβαH MTβαH
t
dη
4 sup
α ψ t − η, η f η, K1 hv η , K2 hv η , . . . , Kn hv η
0
t∈I
t η
8 sup
t∈I
0
α ψ t − η, η ϕ η, s fs, K1 hv s, K2 hv s, . . . , Kn hv s
0
t n
∗
βi t, η Ki α{hv }dη
ε ≤ CβαH MTβαH 4 sup
t∈I
0
i1
t η n
∗
ζi t, η, s Ki α{hv }ds dη ε
8 sup
t∈I
0
0
i1
≤ CβαH MTβαH C MT β 4
n
n
∗
∗
4 βi Ki 8 ζi Ki α{hv } ε
n
i1
βi 2ζi Ki∗
i1
αH ε.
i1
3.30
Since ε is arbitrary, we can obtain
αFH ≤
C MT β 4 Σni1 βi 2ζi Ki∗ αH < αH.
3.31
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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· ∈ L1 I, R and a constant μ > 0 such that
gu − gu∗ ≤ μ
u − u∗ ,
ft, v1 , v2 , . . . , vn − ft, w1 , w2 , . . . , wn n
≤ lt
vi − wi , vi , wi ∈ X 2 , i 1, 2, . . . , n.
3.32
i1
G2 There exists a constant 0 < δ < 1 such that the function Λ : I → R defined by
n
n
q
∗
2
∗
Λt μ C MT C
Ki Γ q I lt C
Ki Γ q Γ γ I qγ lt ≤ δ,
i1
i1
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. For any u, u∗ ∈ CI, X, we have
ft, K1 ut, K2 ut, . . . , Kn ut − ft, K1 u∗ t, K2 u∗ t, . . . , Kn u∗ t
n
∗
≤ lt
Ki ut − Ki u t
i1
3.34
n
≤ lt Ki∗ u − u∗ .
i1
Thus, from A2, 2.10, 2.11, Lemma 2.1, we have
Fut − Fu∗ t
t
≤ μC
u − u∗ μ ψ t − η, η U η u − u∗ dη
t
0
ψ t − η, η f η, K1 u η , K2 u η , . . . , Kn u η
0
−f η, K1 u∗ η , K2 u∗ η , . . . , Kn u∗ η dη
t η
0
0
ψ t − η, η ϕ η, s fs, Kus, Hus − fs, Ku∗ s, Hu∗ sds dη
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t
n
q−1 ∗
t−η
Ki
l η dη
≤ u − u μ C MT C
∗
C
2
n
Ki∗
i1
0
i1
t η
0
q−1 γ −1
t−η
η−s
lsds dη
0
n
n
q
qγ
∗
2
∗
μ C MT C
Ki Γ q I lt C
Ki Γ q Γ γ I lt u − u∗ i1
i1
Λt
u − u∗ .
3.35
We get
Fu − Fu∗ ≤ 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:
∂2
tn
∂q
ut, ξ bt, ξ 2 ut, ξ q
∂t
n
∂ξ
t
t − sus, ξds 0
tn
n
t
e−ts us, ξds,
ut, 0 ut, 1 0,
u0, ξ −
ξ y
0
0
ξ ∈ 0, 1,
0
4.1
"u"
" "
b−1 0, x sin" "dx dy,
λ
where 0 < q < 1, 0 ≤ t ≤ 1, λ > C M1, n ∈ N, bt, ξ is continuous function and is uniformly
Hölder continuous in t, that is, there exist C > 0 and γ ∈ 0, 1 such that
bt1 , ξ − bt2 , ξ
≤ C|t1 − t2 |γ ,
0 ≤ t1 ≤ t2 ≤ 1.
4.2
Let X L2 0, 1, R and define At by
#
$
DAt H 2 0, 1 ∩ H01 0, 1 H 2 0, 1 : z0 z1 0 ,
−Atz bt, ξz .
Then −As generates an analytic semigroup exp−tAs.
4.3
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15
For t ∈ 0, 1, ξ ∈ 0, 1, we set
utξ ut, ξ,
"u"
" "
gu sin" ",
λ
A−1 0gu −
ξ y
0
tn
ft, K1 ut, K2 utξ n
0
"u"
" "
b−1 0, x sin" "dx dy,
λ
t
tn
t − sus, ξds n
0
t
4.4
e−ts us, ξds,
0
where
K1 utξ t
t − sus, ξds,
0
K2 utξ t
e−ts us, ξds,
0
K1∗
4.5
t
1
sup t − sds < < ∞,
2
0
t∈I
K2∗ sup
t∈I
t
e−ts ds 0
1
< ∞.
4
Moreover, we can get
gu ≤ 1 u
,
λ
1
α gD ≤ αD
λ
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,
n
ft, Kut, Hutξ ≤ t K1 ut, ξ
K2 ut, ξ
n
tn ∗
K1 u
K2∗ u
n
μtω K1∗ u
K2∗ u
,
≤
4.7
16
Advances in Difference Equations
where μt tn , ωz z/n satisfying B1. For any u1 , u2 ∈ X,
ψt − s, sfs, K1 u1 s, K2 u1 sξ − ψt − s, sfs, K1 u2 s, K2 u2 sξ
≤
Csn
t − sq−1 K1 u1 sξ − K1 u2 sξ
K2 u1 sξ − K2 u2 sξ
.
n
4.8
Therefore, for any bounded sets D1 , D2 ⊂ X, we have
Csn
α ψt − s, sfs, D1 , D2 ≤
t − sq−1 αD1 αD2 .
n
4.9
Moreover,
C
sup
n t∈0,1
t
t − sq−1 sn ds 0
C C
sup tnq B q, n 1 B q, n 1 : β1 β2 .
n t∈0,1
n
4.10
Similarly, we obtain
C2
α ψt − s, sϕs, τfτ, D1 , D2 ≤
t − sq−1 s − τγ −1 τ n αD1 αD2 ,
n
t s
C2
C2 sup
B q, γ B q γ, n 1 : ζ1 ζ2 .
t − sq−1 s − τγ −1 τ n dτ ds ≤
n t∈0,1 0 0
n
Suppose further that
4.11
1 3/4nC1 CBq, γ p − 1/pq − 1p−1/p μ
Lp < 1 − C M1/λ,
2 1/λC M1 3β1 2ζ1 < 1.
Then 4.1 has a mild solution by Theorem 3.1.
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