# Existence of Solutions for Integrodifferential Equations of Fractional Order with Antiperiodic Boundary Conditions

Authors: Ahmed Alsaedi
Journal: International Journal of Differential Equations
 (2009)

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## Abstract

We discuss the existence of solutions for a nonlinear antiperiodic boundary value problem of integrodifferential equations of fractional
orderq∈(1,2]. The contraction mapping principle and Krasnoselskii's fixed point theorem are applied to establish the results.


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## Body

## 1. Introduction

Recently, the subject of fractional differential equations has emerged as an important area of investigation. Fractional differential equations arise in many engineering and scientific disciplines as the fractional derivatives describe numerous events and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, and so forth. For some recent development on the subject, see [1–15] and the references therein.Integrodifferential equations arise in many engineering and scientific disciplines, often as approximation to partial differential equations, which represent much of the continuum phenomena. Many forms of these equations are possible. For details, see [16–20] and the references therein.Antiperiodic boundary value problems have recently received considerable attention as antiperiodic boundary conditions appear in numerous situations, for instance, see [21–25].In this paper, we prove some existence and uniqueness results for the following antiperiodic fractional boundary value problem:(1.1)cDqx(t)=f(t,x(t),(χx)(t)),t∈[0,T],T>0,1<q≤2,x(0)=-x(T),x′(0)=-x′(T),
where cDq denotes the Caputo fractional derivative of order q, f:[0,T]×X×X→X, and for γ:[0,T]×[0,T]→[0,∞),
(1.2)(χx)(t)=∫0tγ(t,s)x(s)ds,withγ0=max{∫0tγ(t,s)ds:(t,s)∈[0,T]×[0,T]}. Here, (X,∥·∥) is a Banach space and C=C([0,T],X) denotes the Banach space of all continuous functions from [0,T]→X endowed with a topology of uniform convergence with the norm denoted by ∥·∥.


## 2. Preliminaries

First of all, we recall some basic definitions [26–28].Definition 2.1.
For a functiong:[0,∞)→ℝ, the Caputo derivative of fractional order q is defined as
(2.1)cDqg(t)=1Γ(n-q)∫0t(t-s)n-q-1g(n)(s)ds,n-1<q<n,n=[q]+1,where [q] denotes the integer part of the real number q.Definition 2.2.
The Riemann-Liouville fractional integral of orderq is defined as
(2.2)Iqg(t)=1Γ(q)∫0tg(s)(t-s)1-qds,q>0,provided that the integral exists.Definition 2.3.
The Riemann-Liouville fractional derivative of orderq for a function g(t) is defined by
(2.3)Dqg(t)=1Γ(n-q)(ddt)n∫0tg(s)(t-s)q-n+1ds,n=[q]+1,provided that the right-hand side is pointwise defined on (0,∞).Lemma 2.4 (see [8]).
Forq>0, the general solution of the fractional differential equation cDqx(t)=0 is given by
(2.4)x(t)=c0+c1t+c2t2+⋯+cn-1tn-1,where ci∈ℝ, i=0,1,2,…,n-1 (n=[q]+1).In view of Lemma2.4, it follows that
(2.5)IqcDqx(t)=x(t)+c0+c1t+c2t2+⋯+cn-1tn-1,
for some ci∈ℝ, i=0,1,2,…,n-1 (n=[q]+1).Now, we state a known result due to Krasnoselskii [29] which is needed to prove the existence of at least one solution of (1.1).Theorem 2.5.
LetM be a closed convex and nonempty subset of a Banach space X. Let A,B be the operators such that (i)Ax+By∈M whenever x,y∈M; (ii)A is compact and continuous; (iii)B is a contraction mapping. Then there exists z∈M such that z=Az+Bz.Lemma 2.6.
For anyω∈C[0,T], the unique solution of the boundary value problem
(2.6)cDqx(t)=ω(t),0<t<T,1<q≤2,x(0)=-x(T),x′(0)=-x′(T)
is given by
(2.7)x(t)=∫0TG(t,s)ω(s)ds,where G(t,s) is the Green's function given by
(2.8)G(t,s)={(t-s)q-1-(1/2)(T-s)q-1Γ(q)+(T-2t)(T-s)q-24Γ(q-1),s≤t,-(T-s)q-12Γ(q)+(T-2t)(T-s)q-24Γ(q-1),t≤s.Proof.
Using (2.5), for some constants b1,b2∈ℝ, we have
(2.9)x(t)=Iqω(t)-b1-b2t=∫0t(t-s)q-1Γ(q)ω(s)ds-b1-b2t.In view of the relations cDqIqx(t)=x(t) and IqIpx(t)=Iq+px(t) for q,p>0,x∈L(0,1), we obtain
(2.10)x′(t)=∫0t(t-s)q-2Γ(q-1)ω(s)ds-b2.Applying the boundary conditions x(0)=-x(T),x′(0)=-x′(T), we find that
(2.11)b1=12Γ(q)∫0T(T-s)q-1ω(s)ds-T4Γ(q-1)∫0T(T-s)q-2ω(s)ds,b2=12Γ(q-1)∫0T(T-s)q-2ω(s)ds.
Thus, the unique solution of (2.6) is
(2.12)x(t)=∫0t(t-s)q-1Γ(q)ω(s)ds-12∫0T(T-s)q-1Γ(q)ω(s)ds+14(T-2t)∫0T(T-s)q-2Γ(q-1)ω(s)ds=∫0TG(t,s)ω(s)ds,where G(t,s) is given by (2.8). This completes the proof.


## 3. Main Results

To prove the main results, we need the following assumptions:(A1)
∥f(t,x(t),(χx)(t))-f(t,y(t),(χy)(t))∥≤L1∥x-y∥+L¯1∥χx-χy∥, for all t∈[0,T], x,y∈X;(A2)
∥f(t,x(t),(χx)(t))∥≤μ(t), for all (t,x,χx)∈[0,T]×X×X and μ∈L1([0,T],R+).Theorem 3.1.
Letf:[0,T]×X×X→X be a jointly continuous function satisfying the assumption (A1) with (L1+γ0L¯1)≤Γ(q+1)/Tq(3+q/2). Then the antiperiodic boundary value problem (1.1) has a unique solution.Proof.
DefineΘ:C→C by
(3.1)(Θx)(t)=∫0t(t-s)q-1Γ(q)f(s,x(s),(χx)(s))ds-12∫0T(T-s)q-1Γ(q)f(s,x(s),(χx)(s))ds+14(T-2t)∫0T(T-s)q-2Γ(q-1)f(s,x(s),(χx)(s))ds,t∈[0,T].
Setting supt∈[0,T]∥f(t,0,0)∥=M and choosing r≥(MTq/Γ(q+1))(3+q/2), we show that ΘBr⊂Br, where Br={x∈C:∥x∥≤r}. For x∈Br, we have(3.2)∥(Θx)(t)∥≤∫0t(t-s)q-1Γ(q)∥f(s,x(s),(χx)(s))∥ds+12∫0T(T-s)q-1Γ(q)∥f(s,x(s),(χx)(s))∥ds+14|T-2t|∫0T(T-s)q-2Γ(q-1)∥f(s,x(s),(χx)(s))∥ds≤∫0t(t-s)q-1Γ(q)[∥f(s,x(s),(χx)(s))-f(s,0,0)∥+∥f(s,0,0)∥]ds+12∫0T(T-s)q-1Γ(q)[∥f(s,x(s),(χx)(s))-f(s,0,0)∥+∥f(s,0,0)∥]ds+14|T-2t|∫0T(T-s)q-2Γ(q-1)[∥f(s,x(s),(χx)(s))-f(s,0,0)∥+∥f(s,0,0)∥]ds≤((L1+γ0L¯1)r+M)[1Γ(q)∫0t(t-s)q-1ds+12Γ(q)∫0T(T-s)q-1ds+14Γ(q-1)|T-2t|∫0T(T-s)q-2ds]≤((L1+γ0L¯1)r+M)[Tq2Γ(q+1)(3+q2)]≤r.Now, for x,y∈C and for each t∈[0,T], we obtain
(3.3)∥(Θx)(t)-(Θy)(t)∥≤∫0t(t-s)q-1Γ(q)∥f(s,x(s),(χx)(s))-f(s,y(s),(χy)(s))∥ds+12∫0T(T-s)q-1Γ(q)∥f(s,x(s),(χx)(s))-f(s,y(s),(χy)(s))∥ds+14|T-2t|∫0T(T-s)q-2Γ(q-1)∥f(s,x(s),(χx)(s))-f(s,y(s),(χy)(s))∥ds≤(L1+γ0L¯1)∥x-y∥[1Γ(q)∫0t(t-s)q-1ds+12Γ(q)∫0T(T-s)q-1ds+14Γ(q-1)|T-2t|∫0T(T-s)q-2ds]≤(L1+γ0L¯1)Tq2Γ(q+1)(3+q2)∥x-y∥≤ΛL1,L¯1,γ0,T,q∥x-y∥,where ΛL1,L¯1,γ0,T,q=((L1+γ0L¯1)Tq/2Γ(q+1))(3+q/2), which depends only on the parameters involved in the problem. As ΛL1,L¯1,γ0,T,q<1, therefore Θ is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem).Theorem 3.2.
Letf:[0,T]×X×X→X be a jointly continuous function mapping bounded subsets of [0,1]×X×X into relatively compact subsets of X, and the assumptions (A1)-(A2) hold with ((L1+γ0L¯1)Tq/2Γ(q+1))(1+q/2)<1. Then the antiperiodic boundary value problem (1.1) has at least one solution on [0,T].Proof.
Let us fix(3.4)r≥∥μ∥L1Tq2Γ(q+1)(3+q2),and consider Br={x∈C:∥x∥≤r}. We define the operators Φ and Ψ on Br as
(3.5)(Φx)(t)=1Γ(q)∫0t(t-s)q-1f(s,x(s),(χx)(s))ds,(Ψx)(t)=-12∫0T(T-s)q-1Γ(q)f(s,x(s),(χx)(s))ds+14(T-2t)∫0T(T-s)q-2Γ(q-1)f(s,x(s),(χx)(s))ds.For x,y∈Br, we find that
(3.6)∥Φx+Ψy∥≤∥μ∥L1Tq2Γ(q+1)(3+q2)≤r.Thus, Φx+Ψy∈Br. It follows from the assumption (A1) that Ψ is a contraction mapping for ((L1+γ0L¯1)Tq/2Γ(q+1))(1+q/2)<1. Continuity of f implies that the operator Φ is continuous. Also, Φ is uniformly bounded on Br as
(3.7)∥Φx∥≤∥μ∥L1TqΓ(q+1).Now we prove the compactness of the operator Φ. In view of (A1), we define sup(t,x,χx)∈Ω∥f(t,x,(χx))∥=fmax, Ω=[0,T]×Br×Br, and consequently we have
(3.8)∥(Φx)(t1)-(Φx)(t2)∥=∥1Γ(q)∫0t1[(t2-s)q-1-(t1-s)q-1]f(s,x(s),(χx)(s))ds+∫t1t2(t2-s)q-1f(s,x(s),(χx)(s))ds∥≤fmaxΓ(q+1)|2(t2-t1)q+t1q-t2q|,which is independent of x. So Φ is relatively compact on Br. Hence, by Arzela Ascoli theorem, Φ is compact on Br. Thus all the assumptions of Theorem 2.5 are satisfied and the conclusion of Theorem 2.5 implies that the antiperiodic boundary value problem (1.1) has at least one solution on [0,T].Example 3.3.
Consider the following antiperiodic boundary value problem:(3.9)cDqx(t)=1(t+4)2∥x∥1+∥x∥+∫0te-(s-t)16x(s)ds,t∈[0,1],1<q≤2,x(0)=-x(1),x′(0)=-x′(1).
Here, f(t,x)=(1/(t+4)2)(∥x∥/(1+∥x∥)),γ(t,s)=e-(s-t)/16, T=1. Clearly,
(3.10)∥f(t,x,χx)-f(t,y,χy)∥≤116(∥x-y∥+∥χx-χy∥).So (A1) is satisfied with L1=L¯1=1/16. Further
(3.11)(L1+γ0L¯1)TqΓ(q+1)(3+q2)<1⇔(3+q/2)Γ(q+1)<16e.Thus, by Theorem 3.1, the boundary value problem (3.9) has a unique solution on [0,1].



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*Source: 417606-2009-12-23.xml*