Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 903518, 15 pages
doi:10.1155/2012/903518
Research Article
Existence of Mild Solutions for a
Class of Fractional Evolution Equations
with Compact Analytic Semigroup
He Yang
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to He Yang, yanghe256@163.com
Received 2 May 2012; Revised 9 September 2012; Accepted 27 September 2012
Academic Editor: Dumitru Bǎleanu
Copyright q 2012 He Yang. 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.
This paper deals with the existence of mild solutions for a class of fractional evolution equations
with compact analytic semigroup. We prove the existence of mild solutions, assuming that the
nonlinear part satisfies some local growth conditions in fractional power spaces. An example is
also given to illustrate the applicability of abstract results.
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,
physics, economics, and science. Numerous applications can be found in electrochemistry,
control, porous media, electromagnetic, see for example, 1–5 and references therein. Hence
the study of such equations has become an object of extensive study during recent years, see
6–23 and references therein.
In this paper, we consider the existence of the following fractional evolution equation:
Dq ut Aut ft, ut, Gut,
u0 x0 ,
t ∈ J 0, T ,
1.1
where Dq is the Caputo fractional derivative of order q ∈ 0, 1, −A is the infinitesimal
generator of a compact analytic semigroup S· of uniformly bounded linear operators, f
is the nonlinear term and will be specified later, and
2
Abstract and Applied Analysis
Gut t
Kt, susds
1.2
0
is a Volterra integral operator with integral kernel K ∈ CΔ, R , Δ {t, s : 0 ≤ s ≤ t ≤
T }, R 0, ∞. Throughout this paper, we denote by K ∗ : maxt,s∈Δ Kt, s
In some existing articles, the fractional differential equations were treated under the
hypothesis that nonlinear term satisfies Lipschitz conditions or linear growth conditions. It
is obvious that these conditions are not easy to be verified sometimes. To make the things
more applicable, in this work, we will prove the existence of mild solutions for 1.1 under
some new conditions. More precisely, the nonlinear term only satisfies some local growth
conditions see conditions H1 and H2 . These conditions are much weaker than Lipschitz
conditions and linear growth conditions. The main techniques used here are fractional
calculus, theory of analytic semigroup, and Schauder fixed point theorem.
The rest of this paper is organized as follows. In Section 2, some preliminaries are given
on the fractional power of the generator of a compact analytic semigroup and the definition
of mild solutions of 1.1. In Section 3, we study the existence of mild solutions for 1.1. In
Section 4, an example is given to illustrate the applicability of abstract results obtained in
Section 3.
2. Preliminaries
In this section, we introduce some basic facts about the fractional power of the generator of a
compact analytic semigroup and the fractional calculus that are used throughout this paper.
Let X be a Banach space with norm · . Throughout this paper, we assume that −A :
DA ⊂ X → X is the infinitesimal generator of a compact analytic semigroup St t ≥ 0
of uniformly bounded linear operator in X, that is, there exists M ≥ 1 such that St ≤ M
for all t ≥ 0. Without loss of generality, let 0 ∈ ρ−A, where ρ−A is the resolvent set of −A.
Then for any α > 0, we can define A−α by
A−α : 1
Γα
∞
tα−1 Stdt.
2.1
0
It follows that each A−α is an injective continuous endomorphism of X. Hence we
can define Aα by Aα : A−α −1 , which is a closed bijective linear operator in X. It can be
shown that each Aα has dense domain and that DAβ ⊂ DAα for 0 ≤ α ≤ β. Moreover,
Aαβ x Aα Aβ x Aβ Aα x for every α, β ∈ R and x ∈ DAμ with μ : max{α, β, α β}, where
A0 I, I is the identity in X. For proofs of these facts, we refer to the literature 24–26.
We denote by Xα the Banach space of DAα equipped with norm xα Aα x for
x ∈ DAα , which is equivalent to the graph norm of Aα . Then we have Xβ → Xα for 0 ≤ α ≤
β ≤ 1 with X0 X, and the embedding is continuous. Moreover, Aα has the following basic
properties.
Lemma 2.1 see 24. Aα has the following properties.
i St : X → Xα for each t > 0 and α ≥ 0.
ii Aα Stx StAα x for each x ∈ DAα and t ≥ 0.
Abstract and Applied Analysis
3
iii For every t > 0, Aα St is bounded in X and there exists Mα > 0 such that
Aα St ≤ Mα t−α .
2.2
iv A−α is a bounded linear operator for 0 ≤ α ≤ 1 in X.
In the following, we denote by CJ, Xα the Banach space of all continuous functions
from J into Xα with supnorm given by uC supt∈J utα for u ∈ CJ, Xα . From
Lemma 2.1iv, since A−α is a bounded linear operator for 0 ≤ α ≤ 1, there exists a constant
Cα such that A−α ≤ Cα for 0 ≤ α ≤ 1.
For any t ≥ 0, denote by Sα t the restriction of St to Xα . From Lemma 2.1i and ii,
for any x ∈ Xα , we have
Stxα Aα · Stx St · Aα x ≤ St · Aα x St · xα ,
Stx − xα Aα · Stx − Aα x St · Aα x − Aα x −→ 0
2.3
as t → 0. Therefore, St t ≥ 0 is a strongly continuous semigroup in Xα , and Sα tα ≤
St for all t ≥ 0. To prove our main results, the following lemma is also needed.
Lemma 2.2 see 27. Sα t t ≥ 0 is an immediately compact semigroup in Xα , and hence it is
immediately norm-continuous.
Let us recall the following known definitions in fractional calculus. For more details,
see 16–20, 23.
Definition 2.3. The fractional integral of order σ > 0 with the lower limits zero for a function
f is defined by
1
I ft Γσ
σ
t
t − sσ−1 fsds,
t > 0,
2.4
0
where Γ is the gamma function.
The Riemann-Liouville fractional derivative of order n − 1 < σ < n with the lower
limits zero for a function f can be written as
L
Dσ ft dn
1
Γn − σ dtn
t
t − sn−σ−1 fsds,
t > 0, n ∈ N.
2.5
0
Also the Caputo fractional derivative of order n − 1 < σ < n with the lower limits zero for a
function f ∈ Cn 0, ∞ can be written as
1
D ft Γn − σ
σ
t
0
t − sn−σ−1 f n sds,
t > 0, n ∈ N.
2.6
4
Abstract and Applied Analysis
Remark 2.4. 1 The Caputo derivative of a constant is equal to zero.
2 If f is an abstract function with values in X, then integrals which appear in
Definition 2.3 are taken in Bochner’s sense.
Lemma 2.5 see 12. A measurable function h : J → X is Bochner integrable if h is Lebesgue
integrable.
For x ∈ X, we define two families {Ut}t≥0 and {V t}t≥0 of operators by
Utx ∞
ηq θStq θxdθ,
V tx q
0
∞
θηq θStq θxdθ,
0 < q < 1,
2.7
0
where
1
ηq θ θ−1−1/q ρq θ−1/q ,
q
∞
1
n−1 −qn−1 Γ nq 1
sin nπq ,
ρq θ −1 θ
π n1
n!
2.8
θ ∈ 0, ∞,
ηq is a probability density function defined on 0, ∞, which has properties ηq θ ≥ 0 for all
θ ∈ 0, ∞ and
∞
ηq θdθ 1,
0
∞
0
1
θηq θdθ .
Γ q1
2.9
The following lemma follows from the results in 7, 11–13.
Lemma 2.6. The operators U and V have the following properties.
i For fixed t ≥ 0 and any x ∈ Xα , we have
Utxα ≤ Mxα ,
qM
M
xα xα .
V txα ≤ Γ 1q
Γ q
2.10
ii The operators Ut and V t are strongly continuous for all t ≥ 0.
iii Ut and V t are norm-continuous in X for t > 0.
iv Ut and V t are compact operators in X for t > 0.
v For every t > 0, the restriction of Ut to Xα and the restriction of V t to Xα are normcontinuous.
vi For every t > 0, the restriction of Ut to Xα and the restriction of V t to Xα are compact
operators in Xα .
Based on an overall observation of the previous related literature, in this paper, we
adopt the following definition of mild solution of 1.1.
Abstract and Applied Analysis
5
Definition 2.7. By a mild solution of 1.1, we mean a function u ∈ CJ, Xα satisfying
ut Utx0 t
2.11
t − sq−1 V t − sfs, us, Gusds
0
for all t ∈ J.
3. Existence of Mild Solutions
In this section, we give the existence theorems of mild solutions of 1.1. The discussions
are based on fractional calculus and Schauder fixed point theorem. Our main results are as
follows.
Theorem 3.1. Assume that the following condition on f is satisfied.
H1 There exists a constant β ∈ α, 1 such that f : J × Xα × Xα → Xβ satisfies:
i for each x, y ∈ Xα × Xα , the function f·, x, y : J → Xβ is measurable;
ii for each t ∈ J, the function ft, ·, · : Xα × Xα → Xβ is continuous;
iii for any r > 0, there exists a function gr ∈ L∞ J, R such that
f t, x, y ≤ gr t,
sup
β
xα ≤r, y α ≤K ∗ T r
t ∈ J,
3.1
and there is a constant γ > 0 such that
lim inf
r−→∞
1
r
t
0
gr s
t − s1−q
ds ≤ γ < ∞.
3.2
If x0 ∈ Xα and MCβ−α γ < Γq, then 1.1 has at least one mild solution.
Proof. Define an operator Q by
Qut Utx0 t
t − sq−1 V t − sfs, us, Gusds,
t ∈ J.
3.3
0
It is not difficult to verify that Q : CJ, Xα → CJ, Xα . We will use Schauder fixed point
theorem to prove that Q has fixed points in CJ, Xα .
For any r > 0, let Br : {u ∈ CJ, Xα : utα ≤ r, t ∈ J}. We first show that there
is a positive number r such that QBr ⊂ Br . If this were not the case, then for each r > 0,
6
Abstract and Applied Analysis
there would exist ur ∈ Br and tr ∈ J such that Qur tr α > r. Thus, from Lemma 2.6i and
H1 iii, we see that
tr
r < Qur tr α ≤ Utr x0 α ≤ Mx0 α ≤ Mx0 α tr
0
tr − sq−1 V tr − sfs, ur s, Gur sα ds
tr − sq−1 Aα−β V tr − s · Aβ fs, ur s, Gur sds 3.4
0
MCβ−α
Γ q
tr
tr − sq−1 gr sds.
0
Dividing on both sides by r and taking the lower limit as r → ∞, we have
MCβ−α γ ≥ Γ q ,
3.5
which is a contradiction. Hence QBr ⊂ Br for some r > 0.
To complete the proof, we separate the rest of proof into the following three steps.
Step 1. Q : Br → Br is continuous.
Let {un } ⊂ Br with un → u ∈ Br as n → ∞. From the assumption H1 ii, for each
s ∈ J, we have
fs, un s, Gun s −→ fs, us, Gus
3.6
as n → ∞. Since fs, un s, Gun s − fs, us, Gusβ ≤ 2gr s, by the Lebesgue
dominated convergence theorem, for each t ∈ J, we have
Qun t − Qutα ≤
≤
t
0
t
t − sq−1 V t − s fs, un s, Gun s − fs, us, Gus α ds
t − sq−1 Aα−β V t − s
0
MCβ−α
≤
Γ q
·Aβ fs, un s, Gun s − fs, us, Gus ds
t
0
t − sq−1 fs, un s, Gun s
−fs, us, Gusβ ds −→ 0
3.7
as n → ∞, which implies that Q : Br → Br is continuous.
Abstract and Applied Analysis
7
Step 2. QBr t : {Qut : u ∈ Br } is relatively compact in Xα for all t ∈ J.
It follows from 2.9 and 3.3 that QBr 0 {Qu0 : u ∈ Br } {x0 } is compact in
Xα . Hence it is only necessary to consider the case of t > 0. For each t ∈ 0, T , ∈ 0, t, and
any δ > 0, we define a set Q,δ Br t by
Q,δ Br t : {Q,δ ut : u ∈ Br },
3.8
where
Q,δ ut ∞
ηq θStq θdθx0
δ
q
t−
0
S δ
q
t − s
∞
q−1
∞
θηq θS t − sq θ dθ · fs, us, Gusds
δ
3.9
ηq θSt θ − δdθx0
q
q
δ
q
t−
0
t − s
q−1
∞
q
θηq θS t − s θ − δ dθ · fs, us, Gusds .
q
δ
Then the set Q,δ Br t is relatively compact in Xα since by Lemma 2.2, the operator
Sα q δ is compact in Xα . For any u ∈ Br and t ∈ 0, T , from the following inequality:
δ
Qut − Q,δ utα ≤ ηq θStq θdθx0 0
α
δ
t
q t − sq−1
θηq θS t − sq θ dθ
0
0
·fs, us, Gusds
α
∞
t
q−1
q t − s
θηq θS t − sq θ dθ · fs, us, Gusds
0
δ
−q
t−
t − s
q−1
∞
0
θηq θS t − sq θ dθ
δ
·fs, us, Gusds
≤ Mx0 α
δ
0
α
ηq θdθ qMCβ−α gr L∞
qMCβ−α gr L∞
t
t−
q−1
t − s
∞
ds
0
t
t − sq−1 ds
0
θηq θdθ
δ
0
θηq θdθ
8
Abstract and Applied Analysis
≤ Mx0 α
×
δ
0
δ
0
ηq θdθ MCβ−α T q gr L∞
MCβ−α gr L∞ q
θηq θdθ .
Γ q1
3.10
One can obtain that the set QBr t is relatively compact in Xα for all t ∈ 0, T . And since it
is compact at t 0, we have the relatively compactness of QBr t in Xα for all t ∈ J.
Step 3. QBr : {Qu ∈ CJ, Xα : u ∈ Br } is equicontinuous.
For τ ∈ 0, T , by 3.3, we have
Quτ − Qu0α ≤ Uτx0 − x0 α
τ
q−1
V
−
sfs,
us,
Gusds
−
s
τ
τ
0
α
MCβ−α gr L∞ q
≤ Uτ − I · x0 α τ .
Γ q1
3.11
Hence it is only necessary to consider the case of t > 0. For 0 < t1 < t2 ≤ T , by Lemma 2.1 and
Lemma 2.6i, we have
Qut2 − Qut1 α ≤ Ut2 x0 − Ut1 x0 α
t2
t2 − sq−1 V t2 − sfs, us, Gusds
0
−
t1
0
t1 − s
q−1
V t1 − sfs, us, Gusds
α
≤ Ut2 − Ut1 · x0 α
t2
q−1
t2 − s V t2 − sfs, us, Gusds
t1
α
t1 q−1
q−1
V t2 − sfs, us, Gusds
t2 − s − t1 − s
0
t1
q−1
t1 − s fs, us, GusV t2 − s − V t1 − sds
0
MCβ−α gr L∞
≤ Ut2 − Ut1 · x0 α t2 − t1 q
Γ q1
MCβ−α gr L∞ q q
t2 − t1 − t2 − t1 q
Γ q1
α
α
Abstract and Applied Analysis
t1
t1 − sq−1 V t2 − s − V t1 − s · fs, us, Gusds
0
9
α
I1 I2 I3 I4 .
3.12
From Lemma 2.6v, we see that I1 → 0 as t2 → t1 independently of u ∈ Br . From the
expressions of I2 and I3 , it is clear that I2 → 0 and I3 → 0as t2 → t1 independently of u ∈ Br .
For any ∈ 0, t1 , we have
t1 −
q−1
I4 ≤ t1 − s V t2 − s − V t1 − s · fs, us, Gusds
0
α
t1
t1 − sq−1 V t2 − s − V t1 − s · fs, us, Gusds
t1 −
3.13
α
2MCβ−α gr L∞ q
1
q
q
.
≤ Cβ−α gr L∞ T sup V t2 − s − V t1 − s q
Γ q1
0≤s≤t1 −
It follows from Lemma 2.6v that I4 → 0 as t2 → t1 and → 0 independently of
u ∈ Br . Therefore, we prove that QBr is equicontinuous.
Thus, the Arzela-Ascoli theorem guarantees that Q is a compact operator. By the
Schauder fixed point theorem, the operator Q has at least one fixed point u∗ in Br , which
is a mild solution of 1.1. This completes the proof.
Remark 3.2. In assumption H1 iii, if the function gr t is independent of t, then we can
easily obtain a constant γ > 0 satisfying 3.2. For example, if there is a constant af > 0 such
that
f t, x, y ≤ af 1 x y
α
α
β
3.14
for all x, y ∈ Xα and t ∈ J, then for any r > 0, x, y ∈ Xα with xα ≤ r, yα ≤ K ∗ T r,
we have ft, x, yβ ≤ af af 1 K ∗ T r gr t, where gr t is independent of t. Thus,
γ : 1/qaf T q 1 K ∗ T > 0 is the constant in 3.2.
More generally, if f satisfies the following condition:
H2 there is a constant β ∈ α, 1 such that f : J × Xα × Xα → Xβ satisfies:
i for each x, y ∈ Xα × Xα , the function f·, x, y : J → Xβ is measurable,
ii for any r > 0, there exists a function ∈ L∞ J, R such that
f t, x1 , y1 − f t, x2 , y2 ≤ t x1 − x2 y1 − y2 α
α
β
3.15
for any xi , yi ∈ Xα with xi α ≤ r, yi α ≤ K ∗ T r i 1, 2 and t ∈ J, then we have the
following existence and uniqueness theorem.
10
Abstract and Applied Analysis
Theorem 3.3. Assume that the condition H2 is satisfied. If x0 ∈ Xα and MCβ−α T q 1 K ∗ T L∞ < Γq 1, then 1.1 has a unique mild solution.
Proof. For any r > 0, if x, y ∈ Xα with xα ≤ r, yα ≤ K ∗ T r, then from H2 ii, we have
f t, x, y ≤ t1 K ∗ T r bt gr t,
β
3.16
where bt ft, 0, 0β . Therefore, the condition H1 iii is satisfied with γ 1 K ∗ T T q L∞ /q. By Theorem 3.1, 1.1 has at least one mild solution u∗ ∈ Br .
Let u1 , u2 ∈ Br be the solutions of 1.1. We show that u1 ≡ u2 . Since u1 t Qu1 t
and u2 t Qu2 t for all t ∈ J, we have
u1 t − u2 tα Qu1 t − Qu2 tα
t
≤
t − sq−1 V t − s fs, u1 s, Gu1 s − fs, u2 s, Gu2 s α ds
0
t
t − sq−1 Aα−β V t − s · Aβ fs, u1 s, Gu1 s − fs, u2 s, Gu2 s ds
0
≤
≤
≤
MCβ−α
Γ q
MCβ−α
Γ q
t
0
t
0
t − sq−1 · fs, u1 s, Gu1 s − fs, u2 s, Gu2 sβ ds
t − sq−1 · su1 s − u2 sα Gu1 s − Gu2 sα ds
MCβ−α L∞
1 K ∗ T Γ q
t
0
t − sq−1 · u1 s − u2 sα ds.
3.17
By using the Gronwall-Bellman inequality see 14, Theorem 1, we can deduce that u1 t −
u2 tα 0 for all t ∈ J, which implies that u1 ≡ u2 . Hence 1.1 has a unique mild solution
u∗ ∈ Br . This completes the proof.
Remark 3.4. In Theorem 3.3, we only assume that f satisfies a local Lioschitz condition see
condition H2 , and an existence and uniqueness result is obtained. If ft, u, v ≡ ft, u :
J × Xα → X, then the assumption H2 deletes the linear growth condition 3 of assumption
Hf in 12. Therefore, the Theorem 3.3 extends and improves the main result in 12.
4. An Example
Assume that X L2 0, π equipped with its natural norm and inner product defined,
respectively, for all u, v ∈ L2 0, π, by
uX π
0
1/2
2
|ux| dx
,
u, v π
uxvxdx.
0
4.1
Abstract and Applied Analysis
11
Consider the following fractional partial differential equation:
t
∂1/2
∂2
ux,
t
−
ux,
t
g
x,
t,
ux,
t,
Kt,
sux,
sds
,
0
2
∂x
∂t1/2
u0, t uπ, t 0, t ∈ 0, T ,
ux, 0 u0 x, x ∈ 0, π,
t ∈ 0, T , x ∈ 0, π
4.2
where T > 0 is a constant.
Let the operator A : DA ⊂ X → X be defined by
DA : v ∈ X : v ∈ X, v0 vπ 0 ,
Au −
∂2 u
.
∂x2
4.3
It is well known that A has a discrete spectrum with eigenvalues
of the form n2 , n ∈ N,
and corresponding normalized eigenfunctions given by zn 2/π sinnx. Moreover, −A
generates a compact analytic semigroup St t ≥ 0 in X, and
Stu ∞
e−n t u, zn zn .
2
4.4
n1
It is not difficult to verify that StLX ≤ e−t for all t ≥ 0. Hence, we take M 1.
The following results are also well known.
I The operator A can be written as
Au ∞
n2 u, zn zn
4.5
∞
nu, zn zn
4.6
n1
for every u ∈ DA.
II The operator A1/2 is given by
A1/2 u n1
for each u ∈ DA1/2 : {v ∈ X :
∞
n < v, zn > zn ∈ X} and A−1/2 LX 1.
n1
Lemma 4.1 see 28. If m ∈ DA1/2 , then m is absolutely continuous, m ∈ X and m X A1/2 mX .
Let X1/2 DA1/2 , · 1/2 , where x1/2 : A1/2 xX for all x ∈ DA1/2 . Assume
that g : 0, π × 0, T × R × R → R satisfies the following conditions.
i For each x, t ∈ 0, π × 0, T , the function gx, t, ·, · is continuous.
ii For each ξ, η ∈ R2 , the function g·, ·, ξ, η is measurable.
12
Abstract and Applied Analysis
iii For each t ∈ 0, T and ξ, η ∈ R, g·, t, ξ, η is differentiable, and ∂/∂xgx, t, ξ, η ∈
X.
iv g0, ·, ·, · gπ, ·, ·, · 0.
v There exist the functions 1 , 0 ∈ L∞ 0, T , R such that
∂ g x, t, ξ, η ≤ 1 t |ξ| η 0 t
∂x
4.7
for all x, t, ξ, η ∈ 0, π × 0, T × R × R.
t
Define ft, ut, Gutx gx, t, ux, t, 0 Kt, sux, sds. Then, for each φ ∈ X1/2 ,
from assumptions iii and iv, we have
f t, φ, Gφ , zn t
π 2
sinnxdx
g x, t, φx, t, Kt, sφx, sds ·
π
0
0
1
n
π
0
t
∂
2
g x, t, φx, t, Kt, sφx, sds
cosnxdx.
·
∂x
π
0
4.8
This implies from II that f : 0, T × X1/2 × X1/2 → X1/2 . Moreover, for any r > 0, by
Minkowski inequality, assumption v and Lemma 4.1, we have
t
∂
sup f t, φ, Gφ 1/2 sup g x, t, φx, t, Kt, sφx, sds ∂x
0
φ1/2 ≤r
φ1/2 ≤r
X
⎛ 2 ⎞1/2
π t
∂
sup ⎝ g x, t, φx, t, Kt, sφx, sds dx⎠
∂x
0
0
φ1/2 ≤r
π
t
1 t φx, t Kt, sφx, sds
≤ sup
0
0
φ1/2 ≤r
!2
1/2
0 t dx
sup 1 t φX K ∗ T φX 0 t
φ1/2 ≤r
sup 1 K ∗ T 1 tA−1/2 · A1/2 φ 0 t
X
φ1/2 ≤r
≤
≤ 1 K ∗ T r1 t 0 t gr t.
4.9
Therefore, f satisfies the condition H1 with γ 2T 1/2 1 K ∗ T 1 L∞ . Thus, 4.2
√
has at least one mild solution provided that γ < π due to Theorem 3.1.
Abstract and Applied Analysis
13
Assume furthermore that the function g satisfies the following:
vi for any r > 0, there exists a function 2 ∈ L∞ 0, T , R such that
∂ g x, t, ξ1 , η1 − ∂ g x, t, ξ2 , η2 ≤ 2 t |ξ1 − ξ2 | η1 − η2 ∂x
∂x
4.10
for x, t, ξ1 , η1 , x, t, ξ2 , η2 ∈ 0, π × 0, T × R × R with |ξi | ≤ r and |ηi | ≤ K ∗ T r, i 1, 2.
Then for each φ1 , φ2 ∈ X1/2 , by Lemma 4.1, we have
1/2
f t, φ1 , Gφ1 − f t, φ2 , Gφ2 ,
Gφ
,
Gφ
f
t,
φ
−
f
t,
φ
A
1
1
2
2
1/2
X
t
∂
g x, t, φ1 x, t, Kt, sφ1 x, sds
∂x
0
t
∂
− g x, t, φ2 x, t, Kt, sφ2 x, sds ∂x
0
X
π
t
∂
g x, t, φ1 x, t, Kt, sφ1 x, sds
0 ∂x
0
2 ⎞1/2
t
∂
− g x, t, φ2 x, t, Kt, sφ2 x, sds dx⎠
∂x
0
π
2 t φ1 x, t − φ2 x, t
≤
0
t
Kt, sφ1 x, sds
0
2 ⎞1/2
− Kt, sφ2 x, sds dx⎠
0
t
⎡
1/2
π
⎢
φ1 x, t − φ2 x, t2 dx
≤ 2 t⎣
0
π t
Kt, sφ1 x, sds
0 0
⎤
2 ⎞1/2
⎥
− Kt, sφ2 x, sds dx⎠ ⎦
0
t
14
Abstract and Applied Analysis
2 t φ1 − φ2 X Gφ1 − Gφ2 X
2 t A−1/2 · A1/2 φ1 − φ2 X
A−1/2 · A1/2 Gφ1 − Gφ2 X
≤ 2 t A1/2 φ1 − φ2 A1/2 Gφ1 − Gφ2 X
2 t φ1 − φ2 1/2 Gφ1 − Gφ2 1/2 .
X
4.11
This shows that f satisfies the condition H2 . Hence by Theorem 3.3, the mild solution of
4.2 is unique.
Acknowledgments
This research was supported by the National Natural Science Foundation of China Grant
no. 11261051, the Fundamental Research Funds for the Gansu Universities and The Project
of NWNU-LKQN-11-3.
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