doi:10.1155/2011/690653
Research Article
Existence of Solutions to Fractional Mixed Integrodifferential Equations with Nonlocal Initial Condition
A. Anguraj,
1P. Karthikeyan,
2and J. J. Trujillo
31Department of Mathematics, PSG College of Arts and Science, Coimbatore 641 014, India
2Department of Mathematics, KSR College of Arts and Science, Tiruchengode 637 215, India
3Departamento de An´alisis Matem´atico, Universidad de La Laguna, Tenerife 38271 La Laguna, Spain
Correspondence should be addressed to J. J. Trujillo,[email protected] Received 14 November 2010; Accepted 6 January 2011
Academic Editor: Dumitru Baleanu
Copyrightq2011 A. Anguraj 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.
We study the existence and uniqueness theorem for the nonlinear fractional mixed Volter- ra-Fredholm integrodifferential equation with nonlocal initial conditiondαxt/dtα ft, xt, t
0kt, s, xsds,1
0ht, s, xsds, x0 1
0gsxsds, wheret ∈ 0,1, 0 < α < 1, andf is a given function. We point out that such a kind of initial conditions or nonlocal restrictions could play an interesting role in the applications of the mentioned model. The results obtainded are applied to an example.
1. Introduction
Recently it have been proved that the differential models involving nonlocal derivatives of fractional order arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in many fields, for instance, physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, and so forthsee1–
6. In fact, such models can be considered as an efficient alternative to the classical nonlinear differential models to simulate many complex processes see7. For instance, fractional differential equations are an excellent tool to describe hereditary properties of viscoelastic materials and, in general, to simulate the dynamics of many processes on anomalous media.
Theory of fractional differential equations has been extensively studied by several authors as Delbosco and Rodino 8, Kilbas et al. 6, Lakshmikantham et al. 9–11, and also see 2,12–16.
Recently Mophou and N’Gu´er´ekata17, studied the Cauchy problem with nonlocal conditions
Dqxt Axt tnft, xt, Bxt, t∈0, T, n∈Z ,
x0 gx x0, 1.1
in general Banach space X with 0 < q < 1, and A is the infinitesimal generator of a C0- semigroup of bounded linear operators. By means of the Krasnoselskii’s Theorem, existence of solutions was also obtained.
Subsequently several authors have investigated the problem for different types of nonlinear differential equations and integrodifferential equations including functional differential equations in Banach spaces.
Very recently N’Gu´er´ekata 2, 18discussed the existence of solutions of fractional abstract differential equations with nonlocal initial condition. The nonlocal Cauchy problem is discussed by authors in 15 using the fixed-point concepts. Tidke 19 studied the nonfractional mixed Volterra-Fredholm integrodifferential equations with nonlocal conditions using Leray-Schauder Theorem.
Motivated by the above mentioned works in this manuscript we discuss the existence and the uniqueness of the solution for the following fractional integrodifferential equation with nonlocal integral initial condition in Banach Space:
dαxt dtα f
t, xt,
t
0
kt, s, xsds, 1
0
ht, s, xsds
,
x0 1
0
gsxsds,
1.2
wheret∈J 0,1, 0< α <1,gt∈0,1,g∈L10,1,R ,x∈Y CJ, Eis a continuous function onJ with values in the Banach spaceEandxY maxt∈JxtE, andf :J×E× E×E → E,k : D×E → E, andh: D0×E → Eare continuousE-valued functions. Here D {t, s ∈R2 : 0≤ s ≤ t ≤ 1}, andD0 J×J. The operatordα/dtαdenotes the Caputo fractional derivative of order α.
For the sake of the shortness let Kxt
t
0
kt, s, xsds, Hxt 1
0
ht, s, xsds. A
The paper is organized as follows. In Section 2, some definitions, lemmas and preliminary results are introduced to be used in the sequel.Section 3will involve the assumptions, main results and proofs of existence problem of 1.2, together with a nonlocal initial condition.
Finally an example is presented.
2. Preliminaries
LetEbe a real Banach space andθthe zero element ofE. LetL10,1, Ebe the Banach space of measurable functionsx : 0,1 → Ewhich are Lebesgue integrable, equipped with the
normxL11
0xsds. We will use the following notationR 0,∞,andR 0,∞. A functionx∈C0,1, Eis called a solution of1.2if it satisfies1.2.
Definition 2.1. A real functionftis said to be in the spaceCα, α ∈ Rif there exists a real numberp > α, such thatft tpgt, whereg ∈C0,∞, whileftis said to be in the space Cmα if and only iffm∈Cα, m∈N.
Definition 2.2. The fractionalarbitraryorder Riemann-Liouville integralon the right and on the leftof the functionf∈L1a, b, R of orderα, a, b∈R is defined by
Iaαft 1 Γα
t
a
t−sα−1
Γα fsds, Ib−α ft 1 Γα
b
t
s−tα−1
Γα fsds, 2.1
whereΓis the Gamma function of Euler.
Whena0, we writeIαft I0 αft f∗φαt, whereφαt tα−1/Γαfort >0, φαt 0 fort≤0, and∗represents the Convolution of Laplace. Then, it is well known that φαt → δtasα → 0, whereδis the Delta function.
Definition 2.3. The Riemann-Liouville fractional integral operator of orderα >0, of a function f∈Cμ, μ≥ −1 is defined as
Iαft φα∗ft, α >0, t >0,
I0fx fx. 2.2
Definition 2.4. The Caputo’s derivative of fractional orderα >0 for a suitable functionftis defined by
c Dαf
t 1 Γn−α
t
0
fns
t−sα−n 1ds, n−1< α < n, n α 1, 2.3 whereαdenotes the integer part of real number α.
It is obvious that the Caputo’s derivative of a constant is equal to 0.
Lemma 2.5. Letα >0 andn α 1. Then
Iαc Dαf
t ft−n−1
k0
fk0
k! tk. 2.4
Lemma 2.6. IfQτ, α ΓαI1−αgτ 1
τgss−τα−1dsforτ ∈0,1, and ifg ∈L10,1, R satisfies 0≤gs≤1 for 0≤s≤1 andα >0, then
Qτ, α Γα < e,
t
0t−sα−1ds
Γα < e. 2.5
Proof. A direct computation shows
Qτ, α Γα
1
τgss−τα−1ds ∞
0 sα−1e−sds 1
τs−τα−1ds ∞
0 sα−1e−sds
1−τ
0 sα−1ds ∞
0 sα−1e−sds ≤ e1−τ
0 sα−1e−sds ∞
0 sα−1e−sds < e, t
0t−sα−1ds Γα
t
0sα−1ds ∞
0 sα−1e−sds ≤ et
0sα−1e−sds ∞
0 sα−1e−sds < e.
2.6
Theorem 2.7Krasnoselkii. LetXbe a Banach space, letSbe a bounded closed convex subset ofX and letA,Bbe maps ofSintoXsuch thatAx By ∈Sfor every pairx, y ∈S. If A is completely continuous and B is a contraction then the equationAx Bxxhas a solution on S.
3. Main Results
We assume the following.
A1Iff ∈C0,1×E×E×E, Eand a nonnegative, boundedpf ∈L10,1, R , there existM >0,pft≤Mfort∈0,1such that
ft, x, Kx, Hx ≤pftx forx∈E. 3.1
A2There exist positive constantsL1,L2, andLsuch that f
t, x1, y1, z1
−f
t, x2, y2, z2 ≤L1
x1−x2 y1−y2 z1−z2
3.2 for allx1, y1, z1, x2, y2, z2∈Y,L2maxt∈Jft,0,0,0, andLmax{L1, L2}.
A3There exist positive constantsN1,N2, andNsuch that
kt, s, x1−kt, s, x2 ≤N1x1−x2 3.3 for allx1, x2∈Y,N2maxt,s∈Dkt, s,0, andNmax{N1, N2}.
A4There exist positive constantsC1,C2, andCsuch that
ht, s, x1−ht, s, x2< C1x1−x2 3.4 for allx1, x2∈Y,C2maxt,s∈D0ht, s,0, andCmax{C1, C2}.
A5p L/Γα 11 C N/α 1is such that 0≤p <1.
Firstly, we obtain the following lemmas to prove the main results on the existence of solutions to1.2.
Lemma 3.1. If (A1) holds withμ 1
0gsds, then the problem1.2is equivalent to the following equation:
xt 1
1−μ Γα
1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ φαt∗ft, xt, Kxt, Hxt.
3.5 Proof. ByLemma 2.5and1.2, we have
xt x0 1 Γα
t
0
t−sα−1fs, xs, Kxs, Hxsds. 3.6
Therefore,
x0 1
0
gsxsds
1
0
gsdsx0 1 Γα
1
0
gs s
0
s−τα−1fτ, xτ, Kxτ, Hxτdτ ds.
3.7
So,
x0 1
1−μ Γα
1
0
fτ, xτ, Kxτ, Hxτ 1
τ
s−τα−1gsds
dτ
1 1−μ
Γα 1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ,
3.8
and then
xt 1
1−μ Γα
1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ φαt∗ft, xt, Kxt, Hxt.
3.9 Conversely, ifxis a solution of3.5, then for everyt∈0,1, according toDefinition 2.4we have
cDαxt cDα
1 1−μ
Γα 1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ
φαt∗ft, xt, Kxt, Hxt θ cDα
Iαft, xt, Kxt, Hxt ft, xt, Kxt, Hxt.
3.10
It is obvious thatx0 1
0gsxsds. This completes the proof.
Lemma 3.2. If (A3) and (A4) are satisfied,K,Hare defined inA, then the conditions Kxt ≤tN1x N2,
Kx1t−Kx2t ≤N1tx1−x2, Hxt ≤C1x C2, Hx1t−Hx2t ≤C1x1−x2,
3.11
are satisfied for anyt∈J, andx, x1, x2∈Y. Proof. ByA3, we have
Kxt ≤ t
0
kt, s, xsds
t
0
kt, s, xs−kt, s,0 kt, s,0ds
≤ t
0
kt, s, xs−kt, s,0ds t
0
kt, s,0ds
≤N1tx N2t≤N1x N2.
3.12
On the other hand,
Kx1t−Kx2t ≤ t
0
kt, s, x1s−kt, s, x2sds
≤N1
t
0
x1s−x2sds
≤N1tx1−x2.
3.13
Similarly, for the other conditions, we use assumptionA4, to get Hxt ≤
1
0
ht, s, xsds≤C1x C2, Kx1t−Kx2t ≤C1x1−x2.
3.14
Theorem 3.3. If (A1)–(A5) are satisfied, then the fractional integrodifferential equation1.2has a unique solution continuous inJ.
Proof. We use the Banach contraction principle to prove the existence and uniqueness of the solution to1.2. LetBr {x∈Y :x ≤r} ⊆ Y, wherer ≥eM/1−μ L/Γα 11 C N/α 1and define the operatorΨon the Banach spaceY by
Ψxt 1 1−μ
Γα 1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ 1
Γα t
0
t−sα−1fs, xs, Kxs, Hxsds.
3.15
Firstly, we show that the operatorΨmapsBrinto itself. By usingA1and triangle inequality, we have
Ψxt ≤ eM
1−μx 1 Γα
t
0
t−sα−1fs, xs, Kxs, Hxsds
≤ eM
1−μx 1 Γα
t
0
t−sα−1 fs, xs, Kxs, Hxs ds
≤ eM 1−μx
1 Γα
t
0
t−sα−1fs, xs, Kxs, Hxs−fs,0,0,0 fs,0,0,0ds
≤ eM
1−μx 1 Γα
t
0
t−sα−1fs, xs, Kxs, Hxs−fs,0,0,0ds 1
Γα t
0
t−sα−1fs,0,0,0ds.
3.16 Now, ifA2is satisfied, then
Ψxt ≤ eM
1−μx L1 Γα
t
0
t−sα−1xs Kxs Hxsds L2
Γα t
0
t−sα−1ds
≤ eM
1−μx L1
Γα t
0
t−sα−1xsds L1
Γα t
0
t−sα−1Kxsds L1
Γα t
0
t−sα−1Hxsds L2 Γα
t
0
t−sα−1ds.
3.17
UsingLemma 3.2andA3, we have
Ψxt ≤ eM
1−μx L1
Γα 1tαx L1
ΓαN1x N2 t
0
t−sα−1sds L1
Γα 1C1x C2tα L2 Γα 1tα
≤ eM
1−μx L1
Γα 1tαx L1
Γα 2tα 1N1x N2 L1
Γα 1C1x C2tα L2 Γα 1tα eM
1−μx L1N2
Γα 2tα 1 L1C2
Γα 1tα L2 Γα 1tα L1
Γα 1tα
1 N1
α 1t C1
x,
3.18
ifx∈Br, we have
Ψxt ≤ eM
1−μx L
Γα 1
1 N
α 1 C
Lr
Γα 1
1 N
α 1 C
≤r.
3.19
Thus ΨBr ⊂ Br. Next, we prove that Ψis a contraction mapping. For this, let x1, x2 ∈ Y. ApplyingA2, we have
Ψx1t−Ψx2t
Qτ, α 1−μ
Γα 1
0
fs, x1τ, Kx1τ, Hx1τ−fs, x2τ, Kx2τ, Hx2τ dτ
1
Γα t
0
t−sα−1fs, x1s, Kx1s, Hx1sds
− 1 Γα
t
0
t−sα−1fs, x2s, Kx2s, Hx2sds
≤ e 1−μ
1
0
L1x1τ−x2τ Kx1τ−Kx2τ Hx1τ−Hx2τdτ L1
Γα t
0
t−sα−1x1s−x2sds t
0
t−sα−1Kx1s−Kx2sds t
0
t−sα−1Hx1s−Hx2sds
3.20 then usingA3,A4andLemma 3.2, one gets
Ψx1t−Ψx2t
≤ L1e
1−μx1−x2 1
0
dτ N1 1
0
τdτ C1 1
0
dτ
L1
Γαx1−x2 t
0
t−sα−1ds N1
t
0
t−sα−1sds C1
t
0
t−sα−1ds
≤ L1e 1−μ
1 N1
2 C1
x1−x2 L1 Γα
tα α
N1Γαtα 1 Γα 2
C1tα α
x1−x2
L1e
1−μ
1 N1
2 C1
L1
Γα 1
1 C1 N1
α 1t
tα
x1−x2
≤ Le
1−μ
1 N
2 C1
L Γα 1
1 C N
α 1
x1−x2
≤ Le
1−μ
1 N
2 C
p
x1−x2
≤ΩC,N,α,μ,px1−x2,
3.21
whereΩC,N,α,μ,p Le/1−μ1 N/2 C p < 1 depends on the parameter of the problem. ThereforeΨhas a unique fixed-pointx Ψx ∈ Br, which is a solution of3.5, and hence is a solution of1.2.
Theorem 3.4. Assume (A1)–(A4) holds. IfeM <1−μ, then1.2has at least one solution onI.
Proof. Chooser ≥ eM/1−μ pfL1/Γα 1and considerBr : {x ∈C :x ≤ r}. Now define onBr the operatorsA, Bby
Axt:φαt∗ft, xt, Kxt, Hxt, Bxt: 1
1−μ Γα
1
0
Qτ, αfτ, xτ, Kxτ, Hxτdτ. 3.22
Let us observe that ifx, y∈BrthenAx By∈Br. Indeed it is easy to check the inequality
Ax By ≤ eM
1−μ pf
L1
Γα 1 ≤r. 3.23
We can easily show that thatBis a contraction mapping. Letu, v∈Br. Then But−Bvt
: 1
1−μ Γα
1
0
Qτ, α fτ, uτ, Kuτ, Huτ−fτ, vτ, Kvτ, Hvτ dτ
≤ e 1−μ
1
0
L1uτ−vτ Kuτ−Kvτ Huτ−Hvτdτ
≤ L1e 1−μ
1 N1
2 C1
u−v
≤ΛN1,C1,L1u−v,
3.24 whereΛN1,C1,L1 L1e/1−μ1 N1/2 C1 <1 depends only on the parameter of the problem and henceBis contraction. Sincexis continuous, thenAxtis continuous in view ofA1. Let us now note thatAis uniformly bounded onBr. This follows from the inequality
Axt ≤ pf
L1
Γα 1. 3.25
Now let us prove thatAxtis equicontinuous.
Lett1, t2∈Jandx∈Br. Using the fact thatfis bounded on the compact setJ×Br thus supt,s∈J×B
rft, xs, Kxs, Hxs:c0<∞, we will get Axt1−Axt2 1
Γ q
t1
0
t1−sq−1 s
0
fs, xs, Kxs, Hxsds
− t2
0
t2−sq−1 s
0
fs, xs, Kxs, Hxsds
1 Γ
q t1
t2
t1−sq−1 s
0
fs, xs, Kxs, Hxsds
− t2
0
t2−sq−1−t1−sq−1 s
0
fs, xs, Kxs, Hxsds
≤ 1 Γ
q t1
t2
t1−sq−1 s
0
fs, xs, Kxs, Hxsds
1 Γ
q t2
0
t2−sq−1−t1−sq−1 s
0
fs, xs, Kxs, Hxsds
≤ c0
Γ
q 12t1−t2q tq2−tq1
≤ 2c0
Γ
q 1|t1−t2|q,
3.26 which does not depend onx. SoABris relatively compact. By the Arzela-Ascoli Theorem, Ais compact. We now conclude the result of the theorem based on the Krasnoselkii’s theorem above.
4. Example
Consider the following fractional integrodifferential equation:
cDαxt e−txt
9 et1 xt 1 10
t
0
e−1/2xsds 1 10
t
0
e−1/49xsds, t∈J 0,1,
x0 1
0
1
2xsds,
4.1
where 0 < α ≤ 1. Take E R . Set Kxt t
0e−1/2xsds, Hxt t
0e−1/49xsds, ft, x, Kx, Hx e−tx/9 et1 x Kx Hx,gs 1/2,pft e−txt/9 et. Then it is clear that
f∈C0,1×E×E×E, E, pft≤ 1 10 M, pf ∈L0,1, R , ft, x, Kx, Hx ≤pfx.
4.2
So,A1is satisfied. Letx, y∈Eandt∈J. Then we have Kx−Ky
t
0
e−1/2xsds− t
0
e−1/2ysds ≤ 1
2x−y, Hx−Hy
t
0
e−1/49xsds− t
0
e−1/49ysds ≤ 1
49x−y, ft, x, Kx, Hx−f
t, y, y, Hy
e−tx
9 et1 x− e−ty 9 et
1 y 1
10
Kx−Ky
Hx−Hy
≤ e−tx−y 9 et1 x
1 y 1
10
Kx−Ky Hx−Hy
≤ e−t
9 etx−y 1 10
Kx−Ky Hx−Hy
≤ 1
10x−y Kx−Ky Hx−Hy .
4.3
Hence the conditions A1–A4 hold withM 1/10,L1 1/10,N1 1/2, C1 1/49.
Chooser1 andμ1/2. Indeed L1e 1−μ
1 N1
2 C1
149e
490 <1, eM
1−μ e 5 <1.
4.4
FurtherA5is satisfied by a suitable choice of α. Then byLemma 3.2the problem1.2has a unique solution on0,1.
Acknowledgment
This paper has been partially supported by MICINNproject MTM2010-16499to which the authors are very thankful.
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