# Existence and Uniqueness Results for Fractional Differential Equations with Riemann-Liouville Fractional Integral Boundary Conditions

Authors: Mohamed I. Abbas
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We prove the existence and uniqueness of solution for fractional differential equations with Riemann-Liouville fractional integral boundary conditions. The first existence and uniqueness result is based on Banach’s contraction principle. Moreover, other existence results are also obtained by using the Krasnoselskii fixed point theorem. An example is given to illustrate the main results.


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## Body

## 1. Introduction

Very recently, fractional differential equations have gained much attention due to extensive applications of these equations in the mathematical modelling of physical, engineering, biological phenomena and viscoelasticity (see, e.g., [1, 2]). There has been a significant development in fractional differential equations. One can see the monographs of Kilbas et al. [3], Miller and Ross [4], Lakshmikantham et al. [5], and Podlubny [6]. In particular, Agarwal et al. [7] establish sufficient conditions for the existence and uniqueness of solutions for various classes of initial and boundary value problem for fractional differential equations and inclusions involving the Caputo fractional derivative in finite dimensional spaces.Moreover, the theory of boundary value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, population dynamics [8], and cellular systems [9] can be reduced to the nonlocal problems with integral boundary conditions. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers [10–13] and so forth.In this paper, we study the existence and uniqueness of solution for fractional differential equations with Riemann-Liouville fractional integral boundary conditions of the following form:(1)DqCxt=ft,xt,t∈J≔0,1,1<q≤2,α1x0+β1Ipxtt=0=γ1,α2x1+β2Ipxtt=1=γ2,0<p<1,whereCDq is the Caputo fractional derivative of order q, Ip is the Riemann-Liouville fractional integral of order p, and αi,βi,γi(i=1,2) are real constants.Moreover,f:J×E→E is a continuous function satisfying some assumptions that will be specified later and E is a Banach space with norm ·.


## 2. Preliminaries

We now gather some definitions and preliminary facts which will be used throughout this paper. Denote byC(J,E) the Banach space of continuous functions x:J→E, with the usual supremum norm (2)x∞=supt∈J⁡xt.We need some basic definitions and properties [6, 14] of fractional calculus which are used in this paper.Definition 1.
The Riemann-Liouville fractional integral of orderq>0 of the function h∈L1([a,b]) is defined as(3)Iqht=1Γq∫0thst-s1-qds.Definition 2.
For a functionh defined on the interval [a,b], the Caputo fractional order derivative of h is defined by(4)DqCht=1Γn-q∫0thnst-sq-n+1ds, where n=[q]+1 and [q] denotes the integer part of q.From the definition of the Caputo derivative, the following auxiliary results have been established in [3].Lemma 3.
Letq>0; then the differential equationCDtqh(t)=0 has solutions (5)ht=c0+c1t+c2t2+⋯+cm-1tm-1,for some ci∈R, i=0,1,…,m-1.Lemma 4.
Letq>0; then(6)IqDtqCht=ht+c0+c1t+c2t2+⋯+cm-1tm-1,for some ci∈R, i=0,1,…,m-1, m=-[-q].To study the boundary value problem (1), we first consider the associated linear problem and obtain its solution.Lemma 5.
For a givenσ∈C(J,E), the unique mild solution of the fractional boundary value problem,(7)DtqCxt=σt,0<q<1,t∈J,α1x0+β1Ipxtt=0=γ1,α2x1+β2Ipxtt=1=γ2,0<p<1,satisfies the following integral equation:(8)xt=γ1α1+ν1tγ2-γ1ν2α1-∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1σsds+1Γq∫0tt-sq-1σsds,where(9)ν1=Γp+2α2Γp+2+β2,ν2=Γp+1α2Γp+1+β2.Proof.
By Lemma4, we reduce (7) to an equivalent integral equation(10)xt=Iqσt-c0-c1t=1Γq∫0tt-sq-1σsds-c0-c1t.Then we have(11)Ipxt=1Γp∫0tt-sp-1xsds=1Γp∫0tt-sp-11Γq∫0ss-τq-1στdτ-c0-c1sds=1ΓpΓq∫0t∫τtt-sp-1s-τq-1dsστdτ-c0tpΓp+1-c1Γp∫0tt-sp-1sds=Bp,qΓpΓq∫0tt-τp+q-1στdτ-c0tpΓp+1-c1Γp∫0tt-sp-1sds=1Γp+q∫0tt-τp+q-1στdτ-c0tpΓp+1-c1Γp∫0tt-sp-1sds,where the integral (12)∫τtt-sp-1s-τq-1ds=t-τp+q-1∫011-zp-1zq-1dz=t-τp+q-1Bp,qis evaluated with the help of the substitution s=τ+z(t-τ) and the definition of the beta function B(p,q)=∫01‍(t-s)p-1sq-1ds=Γ(p)Γ(q)/Γ(p+q).
Applying the boundary conditions in (7), we have(13)-c0α1+β1×0=γ1,whichimpliesthatc0=-γ1α1,(14)α2∫011-sq-1Γqσsds-c0-c1+β2∫011-sp+q-1Γp+qσsds-c0∫011-sp-1Γpds-c1∫011-sp-1sΓpds=γ2.Using (13) in (14) together with (9), we obtain (15)c1=ν1∫01α21-sq-1Γq+β21-sp+q-1Γp+qσsds-ν1γ2-γ1ν2α1.Substituting the values of c0 and c1 in (10), we get (8). This completes the proof.Lemma 6 (Arzelà-Ascoli, [15]).
If a sequence{xn} in a compact subset of X is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.Lemma 7 (Krasnoselskii, [15]).
LetΩ be a closed convex and nonempty subset of a Banach space X. Let A and B be two operators such that(i)
Ax+By∈Ω, wherever x,y∈Ω;(ii)
A is compact and continuous;(iii)
B is a contraction mapping.Then there exists z∈Ω such that z=Az+Bz.


## 3. Main Results

In view of Lemma5, we define the operator T:C(J,E)→C(J,E) as follows: (16)Txt=γ1α1+ν1tγ2-γ1ν2α1-∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1fs,xsds+1Γq∫0tt-sq-1fs,xsds,t∈J.For the forthcoming analysis we impose suitable conditions on the functions involved in the boundary value problem (1). Namely, we assume the following:(H1)
The functionf:J×E→E is continuous and satisfies the following Lipschitz condition with constant k>0: (17)ft,x-ft,y≤kx-y,x,y∈E,t∈J.(H2)
Letd and r be two nonnegative real numbers such that 0<d<1 and (18)k1+ν1α2Γq+1+ν1β2Γp+q+1<d,M1+ν1α2Γq+1+ν1β2Γp+q+1+ν1γ2+γ1ν2α1+γ1α1≤1-dr,whereM≔supt∈Jft,0.Our first result is based on Banach’s contraction principle.Theorem 8.
Assume that conditions (H1) and (H2) are satisfied. Then the boundary value problem (1) has a unique solution in C(J,E).Proof.
We need to prove that the operatorT has a fixed point on the set Br={x∈E:x≤r}. For x∈Br, we have (19)Txt≤γ1α1+ν1γ2+γ1ν2α1+∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1fs,xs-fs,0ds+∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1fs,0ds+1Γq∫0tt-sq-1fs,xs-fs,0ds+1Γq∫0tt-sq-1fs,0ds≤γ1α1+ν1γ2+γ1ν2α1+kx∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1ds+M∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1ds+kxΓq∫0tt-sq-1ds+MΓq∫0tt-sq-1ds≤γ1α1+ν1γ2+γ1ν2α1+kα2Γq+1+β2Γp+q+1r+Mα2Γq+1+β2Γp+q+1+krΓq+1+MΓp+q+1=k1+ν1α2Γq+1+ν1β2Γp+q+1r+M1+ν1α2Γq+1+ν1β2Γp+q+1+ν1γ2+γ1ν2α1+γ1α1.Therefore, (20)Tx≤dr+1-dr,which implies that TBr⊂Br. Hence, T maps Br into itself.
Now, we will prove thatT is a contraction mapping on Br. For x,y∈Br, we have(21)Txt-Tyt≤ν1∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1fs,xs-fs,ysds+1Γq∫0tt-sq-1fs,xs-fs,ysds≤k1+ν1α2Γq+1+ν1β2Γp+q+1x-y.Therefore, (22)Tx-Ty≤dx-y.Hence, the operator T is a contraction. Then T has a unique fixed point which is a solution of the boundary value problem (1).
The following result is based on the Krasnoselskii fixed point theorem. To apply this theorem, we need the following hypothesis:(H3)
There existϕ(t)∈L1(J,R+) such that (23)ft,x≤ϕt,∀t,x∈J×E.Theorem 9.
Assume that conditions (H1) and (H3) are satisfied. Then the boundary value problem (1) has at least a solution in C(J,E), provided that(24)l≔kν1α2Γq+1+β2Γp+q+1<1.Proof.
Lettingϕ∗≔supt∈Jϕ(t), we fix (25)ρ≥γ1α1+ν1γ2+γ1ν2α1+1+ν1α2Γq+1+ν1β2Γp+q+1ϕ∗. On Bρ={x∈E:x≤ρ}, we define the two operators R and S as follows: (26)Rxt=γ1α1+ν1tγ2-γ1ν2α1-∫01α21-sq-1Γq+β21-sp+q-1Γp+qfs,xsds,t∈J,Sxt=1Γq∫0tt-sq-1fs,xsds,t∈J.For x,y∈Bρ, we have(27)Rx+Sy≤γ1α1+ν1γ2+γ1ν2α1+∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1fs,xsds+1Γq∫0tt-sq-1fs,ysds≤γ1α1+ν1γ2+γ1ν2α1+∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1ϕsds+1Γq∫0tt-sq-1ϕsds≤γ1α1+ν1γ2+γ1ν2α1+ϕ∗∫01α2Γq1-sq-1+β2Γp+q1-sp+q-1ds+ϕ∗1Γq∫0tt-sq-1ds≤γ1α1+ν1γ2+γ1ν2α1+ϕ∗α2Γq+1+β2Γp+q+1+ϕ∗Γq+1=γ1α1+ν1γ2+γ1ν2α1+1+ν1α2Γq+1+ν1β2Γp+q+1ϕ∗≤ρ.Hence, Rx+Sy∈Bρ.
On the other hand, from(H1) together with (24), it is easy to see that (28)Rx-Ry≤lx-y,and since l<1, then R is a contraction mapping.
Moreover, continuity off implies that the operator S is continuous. Also, S is uniformly bounded on Bρ and (29)Sx≤ϕ∗Γq+1.
Now we prove the compactness of the operatorS.
In view of(H1) we define L≔sup(t,x)∈J×Bρf(t,x) and consequently we have for t1,t2∈J, t1<t2, and x∈Bρ(30)Sxt2-Sxt1=supt,x∈J×Bρ⁡1Γq∫0t2t2-sq-1-t1-sq-1fs,xsds+1Γq∫t1t2t1-sq-1fs,xsds≤LΓqt2q-t1q,which is independent of x and tends to zero as t2-t1→0. Thus, S is equicontinuous. Hence, by the Arzelà-Ascoli Theorem (Lemma 6), S is compact on Bρ. Thus, all the assumptions of Lemma 7 are satisfied. So the conclusion of the Krasnoselskii fixed point theorem implies that the boundary value problem (1) has at least one solution on J. This completes the proof of Theorem 9.


## 4. An Example

Consider the following fractional boundary value problem:(31)Dt4/3Cxt=1et+2xtxt+1,t∈0,1,x0+I1/3xtt=0=12,12x1+13I1/3xtt=1=2,where q=4/3, p=1/3, α1=1, α2=1/2, β1=1, β2=1/3, γ1=1/2, γ2=2, and ν1≃1.282114, ν2≃1.145105.Here,f:[0,1]×R→R is given by (32)ft,x=1et+2xx+1,∀t,x∈0,1×R.Then we have (33)ft,x-ft,y≤1e2x-y. Therefore, (H1) is satisfied with k=1/e2≃0.135335.Moreover,(34)ft,x=1et+2xx+1≤1et+2xx+1≤1et+2. Therefore, ϕt=1/et+2, ∀t∈[0,1], and ϕ∗=1/e2≃0.135335.Also, we getM=supt∈J|f(t,0)|=1/e2≃0.135335.Finally, simple calculations give(35)k1+ν1α2Γq+1+ν1β2Γp+q+1≃0.224974<1,l=kν1α2Γq+1+β2Γp+q+1≃0.130704<1.Clearly, all the assumptions of Theorems 8 and 9 are satisfied. So there exists at least one solution of the boundary value problem (31) on [0,1].


## 5. Conclusion

In this paper, we study the existence and uniqueness for fractional order differential equations with Riemann-Liouville integral boundary conditions (1) in Banach spaces. Existence and uniqueness results of solutions are established by virtue of fractional calculus, Banach’s contraction principle, and the Krasnoselskii fixed point theorem. As applications, an example is presented to illustrate the main results.



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