# A Note on Fractional Equations of Volterra Type with Nonlocal Boundary Condition

Authors: Zhenhai Liu; Rui Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We deal with nonlocal boundary value problems of fractional equations of Volterra type involving Riemann-Liouville derivative. Firstly, by defining a weighted norm and using the Banach fixed point theorem, we show the existence and uniqueness of solutions. Then, we obtain the existence of extremal solutions by use of the monotone iterative technique. Finally, an example illustrates the results.


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## Body

## 1.  Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, and so forth. There has been a significant theoretical development in fractional differential equations in recent years (see [1–18]). Monotone iterative technique is a useful tool for analyzing fractional differential equations.In [3], Jankowski considered the existence of the solutions of the following problem:
(1)Dqx(t)=f(t,x(t),∫0t‍k(t,s)x(s)ds),0<q<1,000000000000000000000000000t∈(0,T],x~(0)=r,
where f∈C([0,T]×R2,R), x~(0)=t1-qx(t)|t=0 by using the Banach fixed point theorem and monotone iterative technique.Motivated by [3], in this paper we investigate the following nonlocal boundary value problem:
(2)Dαx(t)=f(t,x(t),∫0t‍k(t,s)x(s)ds)≡Fx(t),0<α<1,t∈(0,T],x~(0)=g(x),
where f∈C([0,T]×R2,R), g:C1-α([0,T])→R is a continuous functional, J=[0,T],x~(0)=t1-αx(t)|t=0, and k(t,s)∈C(Δ,R); here Δ={(t,s)∈J×J:0≤s≤t≤T}.Firstly, the nonlocal condition can be more useful than the standard initial condition to describe many physical and chemical phenomena. In contrast to the case for initial value problems, not much attention has been paid to the nonlocal fractional boundary value problems. Some recent results on the existence and uniqueness of nonlocal fractional boundary value problems can be found in [1, 2, 12, 14, 18]. However, discussion on nonlocal boundary value problems of fractional equations of Volterra type involving Riemann-Liouville derivative is rare. Secondly, in [3], in order to discuss the existence and uniqueness of problem (1), Jankowski divided q∈(0,1) into two situations to discuss; one is 0<q≤1/2 with an additional condition and the other is 1/2<q<1. In this paper, we unify the two situations without using the additional condition. Thirdly, for the study of differential equation, monotone iterative technique is a useful tool (see [9, 10, 16, 17]). We know that it is important to build a comparison result when we use the monotone iterative technique. We transform the differential equation into integral equation and use the integral equation to build the comparison result which is different from [3]. It makes the calculation easier and is suitable for the more complicated forms of equations.The paper is organized as follows. In Section2, we present some useful definitions and fundamental facts of fractional calculus theory. In Section 3, by applying Banach fixed point theorem, we prove the existence and uniqueness of solution for problem (2). In Section 4, by the utility of the monotone iterative technique, we prove that (2) has extremal solutions. At last, we give an example to illustrate our main results.


## 2. Preliminaries

LetC1-α(J,R)={x∈C((0,T],R):t1-αx(t)∈C(J,R)} with the norm ∥x∥C1-α=maxt∈J|t1-αe-λtx(t)|, where λ is a fixed positive constant which will be fixed in Section 3. Obviously, the space C1-α(J,R) is a Banach space. Now, let us recall the following definitions from fractional calculus. For more details, one can see [5, 11].Definition 1.
Forα>0, the integral
(3)Iαf(t)=1Γ(α)∫0t‍(t-s)α-1f(s)ds
is called the Riemann-Liouville fractional integral of order α.Definition 2.
The Riemann-Liouville derivative of orderα(n-1<α≤n) can be written as
(4)Dαf(t)=(ddt)n(In-αf(t))=1Γ(n-α)dndtn∫0t‍(t-s)n-α-1f(s)ds,t>0.Lemma 3 (see [5]).
Letn-1<α≤n. If f(t)∈L(0,T) and D0+α-nf(t)∈ACn[0,T], then one has the following equality:
(5)IαDαf(t)=f(t)-∑i=1n‍[Dα-if(t)]t=0tα-iΓ(α-i+1).


## 3. Existence and Uniqueness of Solutions

In what follows, to discuss the existence and uniqueness of solutions of nonlocal boundary value problems for fractional equations of Volterra type involving Riemann-Liouville derivative, we suppose the following.(H1)
There exist nonnegative constantsL1, L2, and W such that |k(t,s)|≤W, for all (t,s)∈Δ, and
(6)|f(t,v1,v2)-f(t,u1,u2)|≤L1|v1-u1|+L2|v2-u2|,0000000000000000000000000∀t∈J,∀v1,v2,u1,u2∈R.(H2)
There exists a nonnegative constantL3∈(0,1) such that
(7)|g(u1)-g(u2)|≤L3∥u1-u2∥C1-α,∀t∈J,000000∀u1,u2∈C1-α(J).Lemma 4.
Let (H1) hold.x∈C1-α(J) and x is a solution of the following problem:
(8)Dαx(t)=f(t,x(t),∫0t‍k(t,s)x(s)ds)≡Fx(t),x~(0)=g(x),
if and only if x(t) is a solution of the following integral equation:
(9)x(t)=g(x)tα-1+1Γ(α)∫0t‍(t-s)α-1Fx(s)ds.Proof.
Assume thatx(t) satisfies (8). From the first equation of (8) and Lemma 3, we have
(10)x(t)=I0+1-αx(t)|t=0tα-1Γ(α)+I0+αFx(t)=g(x)tα-1+1Γ(α)∫0t‍(t-s)α-1Fx(s)ds.
Conversely, assume that x(t) satisfies (9). Applying the operator Dα to both sides of (9), we have
(11)Dαx(t)=Fx(t).
In addition, by calculation, we can conclude x~(0)=t1-αx(t)|t=0=g(x). The proof is completed.Theorem 5.
Let (H1), (H2) hold,f∈C(J×R2,R),andk∈C(Δ,R). Then problem (2) has a unique solution.Proof.
Define the operatorN:C1-α(J)→C1-α(J) by
(12)Nx(t)=g(x)tα-1+1Γ(α)∫0t‍(t-s)α-1Fx(s)ds.
It is easy to check that the operator N is well defined on C1-α(J). Next we show that N is a contradiction operator on C1-α(J). For convenience, let
(13)ρ≡1q1/q(1-L3)×{L1Γ(α)[Tpα-p+1Γ(pα-p+1)2Γ(2pα-2p+2)]1/p0000+L2WαΓ(α)[Tpα+1Γ(pα-p+1)Γ(pα+1)Γ(2pα-p+2)]1/p[Tpα-p+1Γ(pα-p+1)2Γ(2pα-2p+2)]1/p},
and choose
(14)1<p<11-α,1p+1q=1,λ>ρq,
where λ is a positive constant defined in the norm of the space C1-α(J).
Then, for anyx,y∈C1-α(J), we have from (H1), (H2), and the Hölder inequality
(*)∥(Nx)(t)-(Ny)(t)∥C1-α00=maxt∈[0,T]|t1-αe-λt[(Nx)(t)-(Ny)(t)]|00≤maxt∈[0,T]e-λt|g(x)-g(y)|0000+maxt∈[0,T]1Γ(α)t1-αe-λt∫0t‍(t-s)α-1|Fx(s)-Fy(s)|ds00≤L3∥x-y∥C1-α0000+maxt∈[0,T]L1Γ(α)t1-αe-λt∫0t‍(t-s)α-1|x(s)-y(s)|ds0000+maxt∈[0,T]L2WΓ(α)t1-αe-λt0000×∫0t‍(t-s)α-1∫0s‍|x(τ)-y(τ)|dτds00≤L3∥x-y∥C1-α0000+maxt∈[0,T]L1Γ(α)t1-αe-λt∫0t‍(t-s)α-1sα-1eλsds∥x-y∥C1-α0000+maxt∈[0,T]L2WαΓ(α)t1-αe-λt∫0t‍(t-s)α-1sαeλsds∥x-y∥C1-α00≤L3∥x-y∥C1-α0000+maxt∈[0,T]L1Γ(α)t1-αe-λt(∫0t‍((t-s)α-1sα-1)pds)1/p0000×(∫0t‍eλsqds)1/q∥x-y∥C1-α0000+maxt∈[0,T]L2WαΓ(α)t1-αe-λt(∫0t‍((t-s)α-1sα)pds)1/p0000×(∫0t‍eλsqds)1/q∥x-y∥C1-α00≤L3∥x-y∥C1-α0000+maxt∈[0,T]L1Γ(α)e-λt(tpα-p+1∫01‍(1-η)pα-pηpα-pdη)1/p0000×eλt(λq)1/q∥x-y∥C1-α0000+maxt∈[0,T]L2WαΓ(α)e-λt(tpα+1∫01‍(1-η)pα-pηpαdη)1/p0000×eλt(λq)1/q∥x-y∥C1-α00≤L3∥x-y∥C1-α+L1Γ(α)0000×[Tpα-p+1Γ(pα-p+1)2Γ(2pα-2p+2)]1/p1(λq)1/q∥x-y∥C1-α0000+L2WαΓ(α)[Tpα+1Γ(pα-p+1)Γ(pα+1)Γ(2pα-p+2)]1/p0000×1(λq)1/q∥x-y∥C1-α00≤{L3+L1Γ(α)[Tpα-p+1Γ(pα-p+1)2Γ(2pα-2p+2)]1/p1(λq)1/q000000+L2WαΓ(α)[Tpα+1Γ(pα-p+1)Γ(pα+1)Γ(2pα-p+2)]1/p000000×1(λq)1/q[Tpα-p+1Γ(pα-p+1)2Γ(2pα-2p+2)]1/p}∥x-y∥C1-α.
According to λ>ρq and the Banach fixed point theorem, the problem (2) has a unique solution. The proof is completed.Remark 6.
Theorem5 is an essential improvement of [3, Theorem 1].


## 4. The Monotone Iterative Technique for Problem (2)

In this section, the monotone iterative technique is presented and constructed for problem (2). This method leads to a simple and yet efficient linear iterative algorithm. It yields two sequences of iterations that converge monotonically from above and below, respectively, to a solution of the problem.LetM,N∈C(J). We may assume |M(t)|≤M1, |N(t)|≤N1, for all t∈J, σ∈C1-α(J). Then, according to Lemma 4 and Theorem 5, the following linear problem
(15)Dαx(t)-M(t)x(t)-N(t)∫0t‍k(t,s)x(s)ds=σ(t),000000000000000000000000.00t∈(0,T],0<α<1,x~(0)=t1-αx(t)|t=0=g(x)
has a unique solution which satisfies
(16)x(t)=g(x)tα-1+1Γ(α)∫0t‍(t-s)α-1000000000×(∫0s‍M(s)x(s)000000000000+N(s)∫0s‍k(s,τ)x(τ)dτ+σ(s))ds.Lemma 7.
Let0<α<1, M,N∈C(J), |M(t)|≤M1, |N(t)|≤N1. Suppose that
(17)M1TαΓ(α)Γ(2α)+N1WTα+1Γ(α)Γ(2α+1)<1
and p∈C1-α(J) satisfies the problem
(18)p(t)≤p~(0)tα-1+1Γ(α)∫0t‍(t-s)α-1000000000×(∫0s‍M(s)p(s)000000000000+N(s)∫0s‍k(s,τ)p(τ)dτ)ds.p~(0)≤0,
Then p(t)≤0 for all t∈(0,T].Proof.
Suppose that the inequalityp(t)≤0, for all t∈(0,T], is not true. Therefore, there exists at least a t*∈(0,T] such that e-λt*t*1-αp(t*)>0. Without loss of generality, we assume e-λt*t*1-αp(t*)=max{e-λtt1-αp(t):t∈(0,T]}=ρ1>0.
We obtain that(19)e-λtt1-αp(t)00≤e-λtp~(0)+e-λtt1-αΓ(α)0000×∫0t‍(t-s)α-100000000×(M(s)p(s)+N(s)∫0s‍k(s,τ)p(τ)dτ)ds00≤e-λtt1-αΓ(α)∫0t‍(t-s)α-10000000000000×(∫0s‍M(s)p(s)0000000000000000+N(s)∫0s‍k(s,τ)p(τ)dτ)ds00≤e-λtM1t1-αΓ(α)∫0t‍(t-s)α-10000000000000000×sα-1eλse-λss1-α|p(s)|ds0000+e-λtN1Wt1-αΓ(α)∫0t‍(t-s)α-10000000000000000000×(∫0s‍τα-1eλτe-λττ1-α|p(τ)|dτ)ds.
Let t=t*; we have
(20)e-λt*t*1-αp(t*)00≤M1e-λt*t*1-αΓ(α)0000×∫0t*‍(t*-s)α-1sα-1eλse-λss1-α|p(s)|ds0000+N1We-λt*t*1-αΓ(α)0000×∫0t*‍(t*-s)α-100000000×(∫0s‍τα-1eλτe-λττ1-α|p(τ)|dτ)dsρ1≤M1t*1-αΓ(α)∫0t*‍(t*-s)α-1sα-1dsρ10000+N1Wt*1-ααΓ(α)∫0t*‍(t*-s)α-1sαdsρ1,ρ1≤(M1TαΓ(α)Γ(2α)+N1WTα+1Γ(α)Γ(2α+1))ρ1.
So
(21)M1TαΓ(α)Γ(2α)+N1WTα+1Γ(α)Γ(2α+1)≥1.
This is a contradiction. Hence p(t)≤0 for all t∈(0,T]. The proof is completed.Definition 8.
We say thatx0∈C1-α(J) is called a lower solution of problem (2) if
(22)x0(t)≤x~0(0)tα-1+1Γ(α)∫0t‍(t-s)α-1Fx0(s)ds,t∈(0,T],x~0(0)≤g(x0).
We say that y0∈C1-α(J) is called an upper solution of problem (2) if
(23)y0(t)≥y~0(0)tα-1+1Γ(α)∫0t‍(t-s)α-1Fy0(s)ds,t∈(0,T],y~0(0)≥g(y0).
In the following discussion, we need the following assumptions.(H3)
Assume thatg:C1-α(J)→R is a nondecreasing continuous function, f(t,β1,β2)∈C1-α(J) for all t∈J, x0≤β1≤y0, ∫0t‍k(t,s)x0(s)ds≤β2≤∫0t‍k(t,s)y0(s)ds. x0 and y0 are lower and upper solutions of problem (2), respectively, and x0≤y0.(H4)
Consider(24)f(t,v1,v2)-f(t,u1,u2)≥M(t)(v1-u1)+N(t)(v2-u2),wherex0≤u1≤v1≤y0, ∫0tk(t,s)x0(s)ds≤u2≤v2≤∫0t‍k(t,s)y0(s)ds. M,N∈C(J).
Let[x0,y0]={z∈C1-α(J):x0(t)≤z(t)≤y0(t),x~0(0)≤z~(0)≤y~0(0)}.Theorem 9.
Let inequality (17), (H2)–(H4) hold. Then there exist monotone sequences {xn},{yn}⊂[x0,y0] which converge uniformly to the extremal solutions of (2) in [x0,y0], respectively.Proof.
This proof consists of the following three steps.
Step
1. Construct the sequences {xn},{yn}.
For anyη∈[x0,y0], we consider the following linear problem:
(25)Dαx(t)-M(t)x(t)0000-N(t)∫0t‍k(t,s)x(s)ds00=f(t,η(t),∫0t‍k(t,s)η(s)ds)0000-M(t)η(t)0000-N(t)∫0t‍k(t,s)η(s)ds,t∈(0,T],x~(0)=g(η).
By Theorem 5, (25) has a unique solution which satisfies
(26)x(t)=x~(0)tα-1+1Γ(α)∫0t‍(t-s)α-100000000×[f(s,η(s),∫0s‍k(s,τ)η(τ)dτ)00000000000-M(s)(η(s)-x(s))00000000000-N(s)∫0s‍k(s,τ)0000000000000000000×(η(τ)-x(τ))dτ(s,η(s),∫0s‍k(s,τ)η(τ)dτ)]ds,x~(0)=g(η).
Define an operator A:[x0,y0]→[x0,y0] by x=Aη. It is easy to check that the operator A is well defined on [x0,y0]. Let η1,η2∈[x0,y0] with η1≤η2.
Settingp(t)=z1(t)-z2(t), z1(t)=Aη1(t), and z2(t)=Aη2(t), by (26), we obtain
(27)p(t)=p~(0)tα-1+1Γ(α)∫0t‍(t-s)α-1000000000×[f(s,η1(s),∫0s‍k(s,τ)η1(τ)dτ)00000000000-f(s,η2(s),∫0s‍k(s,τ)η2(τ)dτ)00000000000-M(s)(η1(s)-z1(s))00000000000+M(s)(η2(s)-z2(s))00000000000-N(s)∫0s‍k(s,τ)(η1(τ)-z1(τ))dτ00000000000+N(s)∫0s‍k(s,τ)(η2(τ)-z2(τ))dτ]ds≤p~(0)tα-1+1Γ(α)∫0t‍(t-s)α-1000000×(∫0s‍M(s)p(s)000000000+N(s)∫0s‍k(s,τ)p(τ)dτ)ds.
Besides,
(28)p~(0)=z~1(0)-z~2(0)=g(η1)-g(η2)≤0.
By Lemma 7, we know p(t)≤0, t∈(0,T]. It means that A is nondecreasing. Obviously, we can easily get that A is a continuous map. Let xn=Axn-1, yn=Ayn-1, n=1,2,….
Step
2. The sequences {t1-αxn}, {t1-αyn} converge uniformly to t1-αx*, t1-αy*, respectively.
In fact,xn, yn satisfy the following relation:
(29)x0≤x1≤⋯≤xn≤⋯≤yn≤⋯≤y1≤y0.
Setting p(t)=x0(t)-x1(t) and x0(t) is a lower solution of problem (2):
(30)p(t)≤x~0(0)tα-1+1Γ(α)∫0t‍(t-s)α-1000000000×f(s,x0(s),∫0s‍k(s,τ)x0(τ)dτ)ds-x~1(0)tα-1-1Γ(α)∫0t‍(t-s)α-1000000000×[f(s,x0(s),∫0s‍k(s,τ)x0(τ)dτ)000000000000-M(s)(x0(s)-x1(s))000000000000-N(s)∫0s‍k(s,τ)000000000000×(x0(τ)-x1(τ))dτ(s,x0(s),∫0s‍k(s,τ)x0(τ)dτ)]ds=p~(0)tα-1+1Γ(α)∫0t‍(t-s)α-100000000×(∫0s‍M(s)p(s)00000000000+N(s)∫0s‍k(s,τ)p(τ)dτ)ds.
Besides,
(31)p~(0)=x~0(0)-x~1(0)≤g(x0)-g(x0)=0.
By Lemma 7, we can obtain that x0≤x1 for all t∈(0,T]. Similarly, we can show that y1≤y0 for all t∈(0,T]. Applying the operator A to both sides of x0≤x1, y1≤y0, and x0≤y0, we can easily get (29). Obviously, the sequences {t1-αxn}, {t1-αyn} are uniformly bounded and equicontinuous. Then by using the Ascoli-Arzela criterion, we can conclude that the sequences {t1-αxn}, {t1-αyn} converge uniformly on (0,T] with limn→∞t1-αxn=t1-αx*, limn→∞t1-αyn=t1-αy* uniformly on (0,T].
Step
3. x*, y* are extremal solutions of (1).
x
*, y* are solutions of (1) on [x0,y0], because of the continuity of operator A. Let z∈[x0,y0] be any solution of (1). That is,
(32)z(t)=z~(0)tα-1+1Γ(α)∫0t‍(t-s)α-1Fz(s)ds,z~(0)=g(z).
Suppose that there exists a positive integer n such that xn(t)≤z(t)≤yn(t) on (0,T]. Let p(t)=xn+1(t)-z(t); we have
(33)p(t)=x~n+1(0)tα-1+1Γ(α)∫0t‍(t-s)α-100000000×[f(s,xn(s),∫0s‍k(s,τ)xn(τ)dτ)00000000000-M(s)(xn(s)-xn+1(s))00000000000-N(s)∫0s‍k(s,τ)000000000000000000×(xn(τ)-xn+1(τ))dτ(s,xn(s),∫0s‍k(s,τ)xn(τ)dτ)]ds-z~(0)tα-1-1Γ(α)∫0t‍(t-s)α-100000000×f(s,z(s),∫0s‍k(s,τ)z(τ)dτ)ds≤p~(0)tα-1+1Γ(α)∫0t‍(t-s)α-100000000×(∫0s‍M(s)p(s)00000000000+N(s)∫0s‍k(s,τ)p(τ)dτ)ds,p~(0)=x~n+1(0)-z~(0)=g(xn)-g(z)≤0.
By Lemma 7, we know that p(t)≤0 on (0,T], which implies xn+1(t)≤z(t) on (0,T]. Similarly, we obtain that z(t)≤yn+1(t) on (0,T]. Since x0(t)≤z(t)≤y0(t) on (0,T], by induction we get that xn(t)≤z(t)≤yn(t) on (0,T] for every n. Therefore, x*(t)≤z(t)≤y*(t) on (0,T] by taking n→∞. Thus, we completed this proof.


## 5. An Example

Example 1.
Consider the following problem:(34)D1/2x(t)=t+160x(t)+130∫0t‍tsx(s)ds,t∈(0,1],x~(0)=g(x)=η12x(η),0<η<1.
Obviously, T=1, α=1/2, k(t,s)=ts, and f(t,v1,v2)=t+(1/60)v1+(1/30)v2.
Letw=1, L1=1/60, L2=1/30, and L3=1/12.
It is easy to check that(35)|k(t,s)|≤1,|f(t,v1,v2)-f(t,u1,u2)|≤160|v1-u1|+130|v2-u2|,|g(x1)-g(x2)|≤112∥x1-x2∥C1-α.
So, (H1) and (H2) are satisfied. By the choice of p=3/2, q=3, we can get that λ>ρ3 and ρ≡(4/(11×31/3)){(1/20Γ(1/2))[Γ(1/4)2/Γ(1/2)]2/3+(1/15Γ(1/2))[Γ(1/4)Γ(7/4)]2/3}. According to Theorem 5, the problem (34) has a unique solution.
Consider the same equation as (34), taking x0(t)=0, y0(t)=t-1/2+6, and then we have y~0(0)=1.
Moreover,(36)y0(t)=t-1/2+6≥t-1/2+1Γ(α)∫0t‍(t-s)α-1×[∫0s‍s+160(s-1/2+6)000+130∫0s‍sτ(τ-1/2+6)dτ]ds,y~0(0)=1≥η1/212+η2,0<η<1.
On the other hand, it is easy to check that (H3) holds. And let M(t)=1/(t-1), N(t)=cost/30, and then we have
(37)f(t,v1,v2)-f(t,u1,u2)≥1t-1(v1-u1)+cost30(v2-u2),
where x0≤u1≤v1≤y0, ∫0tk(t,s)x0(s)ds≤u2≤v2≤∫0t‍k(t,s)y0(s)ds . So (H4) is satisfied. Obviously, M1=1/30, N1=1, and then we can get
(38)M1TαΓ(α)Γ(2α)+N1WTα+1Γ(α)Γ(2α+1)=31π1/260<1.
Inequality (17) holds. All conditions of Theorem 9 are satisfied, so problem (34) has extremal solutions.



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