doi:10.1155/2010/340349
Research Article
Solutions to Fractional Differential Equations with Nonlocal Initial Condition in Banach Spaces
Zhi-Wei Lv,
1Jin Liang,
2and Ti-Jun Xiao
31Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China
2Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
3Shanghai Key Laboratory for Contemporary Applied Mathematics, School of Mathematical Sciences, Fudan University, Shanghai 200433, China
Correspondence should be addressed to Jin Liang,[email protected] Received 4 January 2010; Accepted 8 February 2010
Academic Editor: Gaston Mandata N’Guerekata
Copyrightq2010 Zhi-Wei Lv et al. 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.
A new existence and uniqueness theorem is given for solutions to differential equations involving the Caputo fractional derivative with nonlocal initial condition in Banach spaces. An application is also given.
1. Introduction
Fractional differential equations have played a significant role in physics, mechanics, chemistry, engineering, and so forth. In recent years, there are many papers dealing with the existence of solutions to various fractional differential equations; see, for example,1–6.
In this paper, we discuss the existence of solutions to the nonlocal Cauchy problem for the following fractional differential equations in a Banach spaceE:
cDαxt ft, xt, 0≤t≤1, x0
1
0
gsxsds, 1.1
wherecDαis the standard Caputo’s derivative of order 0 < α <1,g ∈L10,1, R, gt ∈ 0,1,andfis a givenE-valued function.
2. Basic Lemmas
Let E be a real Banach space, and θ the zero element of E. Denote by C0,1, E the Banach space of all continuous functionsx : 0,1 → Ewith normxc supt∈0,1xt.
Let L10,1, E be the Banach space of measurable functions x : 0,1 → E which are Lebesgue integrable, equipped with the normxL1 1
0xsds. LetR 0,∞,R 0,∞,andμ1
0gsds. Afunctionx∈C0,1, Eis called a solution of1.1if it satisfies 1.1.
Recall the following defenition
Definition 2.1. LetB be a bounded subset of a Banach spaceX. The Kuratowski measure of noncompactness ofBis defined as
αB:inf
γ >0; Badmits a finite cover by sets of diameter≤γ
. 2.1
Clearly, 0≤αB<∞. For details on properties of the measure, the reader is referred to2.
Definition 2.2 see 7, 8. The fractional integral of order q with the lower limit t0 for a functionfis defined as
Iqft 1 Γ
q t
t0
t−sq−1fsds, t > t0, q >0, 2.2
whereΓis the gamma function.
Definition 2.3see7,8. Caputo’s derivative of orderqwith the lower limitt0for a function fcan be written as
cDqft 1
Γ n−q
t
t0
t−sn−q−1fnsds, t > t0, q >0, n
q 1. 2.3
Remark 2.4. Caputo’s derivative of a constant is equal toθ.
Lemma 2.5see7. Letα >0. Then we have
cDq Iqft
ft. 2.4
Lemma 2.6see7. Letα >0 andn α 1. Then
Iαc
Dαft
ft−n−1
k0
fk0
k! tk. 2.5
Lemma 2.7see9. IfH⊂C0,1, Eis bounded and equicontinuous, then aαCH αH0,1;
bαH0,1 maxt∈0,1αHt,whereH0,1 {xt:x∈H, t∈0,1}.
Lemma 2.8.
Qτ Γα < e,
t
0t−sα−1ds
Γα < e, 2.6
whereQτ 1
τgss−τα−1ds,t, τ ∈0,1.
Proof. A direct computation shows
Qτ Γα
1
τgss−τα−1ds ∞
0 sα−1e−sds
<
1
τs−τα−1ds ∞
0 sα−1e−sds
1−τ
0 sα−1ds ∞
0sα−1e−sds
≤ e1−τ
0 sα−1e−sds ∞
0 sα−1e−sds
< e
2.7
and
t
0t−sα−1ds Γα
t
0sα−1ds ∞
0sα−1e−sds ≤ et
0sα−1e−sds ∞
0sα−1e−sds < e. 2.8
3. Main Results
H1f ∈ 0,1× E, E, and there exist M > 0, pft ≤ M for t ∈ 0,1, pf ∈ L10,1, Rsuch thatft, x ≤pftxfort∈0,1and eachx∈E.
H2For anyt∈0,1andR >0,ft, BR {ft, x:x∈BR}is relatively compact inE, whereBR{x∈C0,1, E,xC ≤R}and
Λ1
2−μ e
1−μ M <1. 3.1
Lemma 3.1. IfH1holds, then the problem1.1is equivalent to the following equation:
xt 1
1−μ Γα
1
0
Qτfτ, xτdτ 1
Γα t
0
t−sα−1fs, xsds. 3.2
Proof. ByLemma 2.6and1.1, we have
xt x0 1
Γα t
0
t−sα−1fs, xsds. 3.3
Therefore,
x0 1
0
gsxsds
1
0
gs
x0 1
Γα s
0
s−τα−1fτ, xτdτ
ds
1
0
gsdsx0 1 Γα
1
0
gs s
0
s−τα−1fτ, xτdτds.
3.4
So,
x0 1
1−1
0gsds
Γα 1
0
gs s
0
s−τα−1fτ, xτdτds
1 1−μ
Γα 1
0
fτ, xτ 1
τ
s−τα−1gsds
dτ
1 1−μ
Γα 1
0
Qτfτ, xτdτ,
3.5
and then
xt 1
1−μ Γα
1
0
Qτfτ, xτdτ 1
Γα t
0
t−sα−1fs, xsds. 3.6
Conversely, ifxis a solution of3.2, then for everyt∈0,1, according toRemark 2.4 andLemma 2.5, we have
cDαxtcDα
1 1−μ
Γα 1
0
Qτfτ, xτdτ 1
Γα t
0
t−sα−1fs, xsds
cDα
1 1−μ
Γα 1
0
Qτfτ, xτdτ
cDα 1
Γα t
0
t−sα−1fs, xsds
θcDα
Iαft, xt ft, xt.
3.7
It is obvious thatx0 1
0gsxsds.This completes the proof.
Theorem 3.2. If (H1) and (H2) hold, then the initial value problem1.1has at least one solution.
Proof. Define operatorA:C0,1, E → C0,1, E, by
Axt 1 1−μ
Γα 1
0
Qτfτ, xτdτ 1
Γα t
0
t−sα−1fs, xsds. 3.8
Clearly, the fixed points of the operatorAare solutions of problem1.1.
It is obvious thatBRis closed, bounded, and convex.
Step 1. We prove thatAis continuous.
Let
xn, x∈C0,1, E, xn−xc−→0 n−→ ∞. 3.9
Thenr supnxnC <∞andxC ≤r.For eacht∈0,1,
Axnt−Axt ≤ e 1−μ
1
0
fτ, xnτ−fτ, xτdτ 1
Γα t
0
t−sα−1fs, xns−fs, xsds.
3.10
It is clear that
ft, xnt−→ft, xt, asn−→ ∞, t∈0,1,
ft, xnt−ft, xt≤2Mr. 3.11
It follows from3.11and the dominated convergence theorem that
Axn−AxC −→0, asn−→ ∞. 3.12
Step 2. We prove thatABR⊂BR.
Letx∈BR. Then for eacht∈0,1, we have
Axt ≤ 1 1−μ
1
0
Qτ
Γαfτ, xτdτ 1 Γα
t
0
t−sα−1fs, xsds
≤ 1 1−μ
1
0
Qτ
Γαpfτxτdτ 1 Γα
t
0
t−sα−1pfsxsds
≤ e
1−μMeM
xC
< R.
3.13
Step 3. We prove thatABRis equicontinuous.
Lett1, t2∈0,1,t1< t2,andx∈BR. We deduce that Axt2−Axt1
1 Γα
t2
0
t2−sα−1fs, xsds− t1
0
t1−sα−1fs, xsds
≤ 1 Γα
t1
0
t2−sα−1−t1−sα−1fs, xsds 1
Γα t2
t1
t2−sα−1fs, xsds
≤ t1
0
t2−sα−1−t1−sα−1ds t2
t1
t2−sα−1ds MR
Γα
≤
2t2−t1α
tα2−tα1 MR Γα1.
3.14
Ast1 → t2, the right-hand side of the above inequality tends to zero.
Step 4. We prove thatABRis relatively compact.
Let 5⊂BRbe arbitrarily given. Using the formula b
a
ytdt∈b−aco
yt:t∈0,1 3.15
fory∈Ca, b, EandH2, we obtain
αAVt≤α
co
Qs
1−uΓαfs, xs:s∈0,1, x∈V
α
co
t−sα−1
Γα fs, xs:s∈0, t, t∈0,1, x∈V
≤
Qs 1−uΓαα
fs, Vs
:s∈0,1
t−sα−1 Γα α
fs, Vs
:s∈0, t, t∈0,1
0.
3.16
It follows from 3.16thatαAVt 0 for t ∈ 0,1.This, together withLemma 2.7, yields that
αCAV 0. 3.17
From 3.17, we see thatABRis relatively compact. Hence, A : BR → BR is completely continuous. Finally, the Schauder fixed point theorem guarantees thatAhas a fixed point in BR.
Theorem 3.3. Besides the hypotheses ofTheorem 3.2, we suppose that there exists a constantLsuch that
0< L <Λ2, 3.18
ft, u−ft, w≤Lu−w, for everyu, w∈BR, 3.19
where
Λ2 1−μ 2−μ
e. 3.20
Then, the solutionxtof1.1is unique inBR.
Proof. From Theorem 3.2, we know that there exists at least one solution xt in BR. We suppose to the contrary that there exist two different solutionsutandwtinBR. It follows from3.8that
ut−wt ≤ e 1−μ
1
0
fτ, uτ−fτ, wτdτ 1
Γα t
0
t−sα−1fs, us−fs, wsds
≤ e 1−μ
1
0
Luτ−wτdτ
1 Γα
t
0
t−sα−1Lus−wsds.
3.21
Therefore, we get
u−wC≤ 2−μ
1−μeLu−wC. 3.22
By3.18, we obtainu−wC 0.So, the two solutions are identical inBR.
4. Example
Let
Ec0{x x1, . . . , xn, . . .:xn −→0} 4.1
with the norm x supn|xn|. Consider the following nonlocal Cauchy problem for the following fractional differential equation inE:
cDαxnt 1t
100n2xnt, t∈0,1, 0< α <1, xn0
1
0
1
2xnsds.
4.2
Conclusion. Problem4.2has only one solution on0,1.
Proof. Write
fnt, x 1t
100n2xn, f
f1, . . . , fn, . . . ,
gs 1
2, pft 1t 100n.
4.3
Then it is clear that
f ∈C0,1×E, E, pft≤ 1 50 M, pf ∈L0,1, R, ft, x≤pfx.
4.4
So,H1is satisfied.
In the same way as in Example 3.2.1 in 9, we can prove that ft, BRis relatively compact inc0.
By a direct computation, we get
Λ1
2−μ e 1−μ M≤
2−μ e 1−μ
1 50 3e
50 <1. 4.5
Hence, conditionH2is also satisfied.
Moreover, we have
fnt, u−fnt, w 1t
100n2un− 1t 100n2wn
≤ 1
50|un−wn|, 4.6 so
ft, u−ft, w≤ 1
50u−w. 4.7
Clearly,
Λ2 1−μ 2−μ
e 1−1/2 3e/2 1
3e. 4.8
Therefore,L1/50<1/3e. Thus, our conclusion follows fromTheorem 3.3.
Acknowledgments
This work was supported partially by the NSF of China 10771202, the Research Fund for Shanghai Key Laboratory for Contemporary Applied Mathematics08DZ2271900and the Specialized Research Fund for the Doctoral Program of Higher Education of China 2007035805.
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