# On Antiperiodic Nonlocal Three-Point Boundary Value Problems for Nonlinear Fractional Differential Equations

Authors: Bashir Ahmad; Jorge Losada; Juan J. Nieto
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

We introduce boundary value conditions involving antiperiodic and nonlocal three-point boundary conditions. We solve a nonlinear fractional differential equation supplemented with those conditions. We obtain some existence results for the given problem by applying some standard tools of fixed point theory. These results are well illustrated with the aid of examples.


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## Body

## 1. Introduction

In recent years, several kinds of boundary value problems of nonlinear fractional differential equations, supplemented with a variety of boundary conditions (including Dirichlet, Neumann, mixed, periodic, antiperiodic, multipoint, integral type, and nonlocal), have been investigated by several researchers. This investigation includes a wide collection of results ranging from theoretical to analytic and numerical methods. For details and examples, see [1–9] and the references therein. This surge has been mainly due to the extensive applications of fractional operators in basic and technical sciences and engineering. One can easily witness from literature (special issues and books) on the topic that the tools of fractional calculus have helped in improving the mathematical modeling of several phenomena of practical nature; for instance, see [10–17].In this paper, we study a new class of problems of fractional differential equations supplemented with antiperiodic and three-point nonlocal boundary conditions. Precisely, we consider the following fractional problem:(1)Dcqxt+ft,xt=0,t∈0,1,1<q≤2,x0=-x1,αx′0+βx′1=-γxθ,0<θ<1,where cDq is the usual Caputo fractional derivative, f:[0,1]×R→R is a continuous function, α,β are positive real constants, and γ is a nonnegative constant. Further, it is assumed that γ(1-2θ)-2(α+β)≠0.We emphasize that the second boundary condition in (1) can be interpreted as the sum of scalar multiples of the values of the derivative of the unknown function at x=0 and x=1 is proportional to the value of the unknown function at an arbitrary value θ∈(0,1). In case of α=β=1,  γ=0, problem (1) reduces to the one with antiperiodic boundary conditions. Thus the proposed problem generalizes antiperiodic problems to semiantiperiodic three-point nonlocal problems.


## 2. Auxiliary Lemma and Notations

In order to define the solutions for the given problem, we consider the following lemma.Lemma 1.
Letg∈C[0,1]. Then the following linear semiantiperiodic three-point fractional boundary value problem (2)Dcqxt+gt=0,t∈0,1,1<q≤2,x0=-x1,αx′0+βx′1=-γxθ,0<θ<1has a unique solution given by (3)xt=-∫0tt-sq-1Γqgsds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqgsds+1-2t·γ∫0θθ-sq-1Γqgsds+β∫011-sq-2Γq-1gsds.Proof.
It is well known that the solution of equationcDqx(t)+g(t)=0,  1<q≤2 can be written as(4)xt=-∫0tt-sq-1Γqgsds+b1+b2t,where b1,b2∈R are unknown arbitrary constants. Since x(0)=-x(1), (4) gives(5)2b1+b2=∫011-sq-1Γqgsds.Using the boundary conditions αx′(0)+βx′(1)=-γx(θ) in (4), we get (6)γb1+α+β+γθb2=γ∫0θθ-sq-1Γqgsds+β∫011-sq-2Γq-1gsds.Solving system of (5) and (6) for b1 and b2, it is found that (7)b1=1γ1-2θ-2α+β-θγ+α+β·∫011-sq-1Γqgsds+β∫011-sq-2Γq-1gsds+γ∫0θθ-sq-1Γqgsds,b2=1γ1-2θ-2α+βγ∫011-sq-1Γqgsds-2β∫011-sq-2Γq-1gsds-2γ∫0θθ-sq-1Γqgsds.Substituting these values of b1 and b2 in (4) completes solution (3).
LetD=C([0,1],R) denote the Banach space of all continuous functions from [0,1] into R endowed with the usual supremum norm defined by x=sup{|x(t)|,  t∈[0,1]}.
In view of Lemma1, we consider a fixed point problem equivalent to the nonlinear antisymmetric problem (1) as follows: (8)Hx=x,where the operator H:D→D is defined as (9)Hxt=-∫0tt-sq-1Γqfs,xsds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqfs,xsds+1-2t·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds.Next we set (10)σ=1Γq+11+γ1-θ-α-β+βq+γθqγ1-2θ-2α+β.


## 3. Main Results

In this section, we present our main results. The first result relies on classical Banach’s contraction mapping principle.Theorem 2.
Letf:[0,1]×R→R be a continuous function satisfying the Lipschitz condition; that is, it exists l>0 such that (H1)
|f(t,x)-f(t,y)|≤l|x-y|,∀t∈[0,1],x,y∈R.
Then problem (1) has a unique solution if lσ<1, where σ is given by (10).Proof.
In the first step, it will be shown thatHBr⊂Br, where Br={x∈D:x≤r} with r>σμ(1-σl)-1, and supt∈[0,1]|f(t,0)|=μ. For x∈Br,t∈[0,1], using |f(s,x(s))|=|f(s,x(s))-f(s,0)+f(s,0)|≤lr+μ, we get (11)Hx≤supt∈0,1⁡∫0tt-sq-1Γqfs,xs-fs,0+fs,0ds+1γ1-2θ-2α+βγt-θ-α-β∫011-sq-1Γqfs,xs-fs,0+fs,0ds+1-2tγ∫0θθ-sq-1Γqfs,xs-fs,0+fs,0ds+β∫011-sq-2Γq-1fs,xs-fs,0+fs,0ds≤lr+μsupt∈0,1⁡∫0tt-sq-1Γqds+1γ1-2θ-2α+βγt-θ-α-β∫011-sq-1Γqds+1-2tγ∫0θθ-sq-1Γqds+β∫011-sq-2Γq-1ds≤lr+μσ≤r,which implies that HBr⊂Br, where we have used (10). Now, for x,y∈D, and for each t∈[0,1], we obtain (12)Hx-Hy≤supt∈0,1⁡∫0tt-sq-1Γqfs,xs-fs,ysds+1γ1-2θ-2α+βγt-θ-α-β∫011-sq-1Γqfs,xs-fs,ysds+1-2tγ∫0θθ-sq-1Γqfs,xs-fs,ysds+β∫011-sq-2Γq-1fs,xs-fs,ysds≤lσx-y.Since lσ<1 by the given assumption, the operator H is a contraction. Thus, by Banach’s contraction mapping principle, there exists a unique solution for problem (1). This completes the proof.The next existence result is based on the following Schaefer’s fixed point theorem [18], Th. 4.3.2.Theorem 3.
LetX be a Banach space. Assume that T:X→X is completely continuous operator and the set (13)V=u∈X∣u=νTu,0<ν<1is bounded. Then T has a fixed point in X.Theorem 4.
Assume that there exists a positive constantM such that |f(t,x)|≤M for t∈[0,1],x∈R. Then problem (1) has at least one solution.Proof.
We first show that the operatorH is completely continuous. Obviously continuity of the operator H follows from continuity of f. Let B⊂D be a bounded set. By the assumption |f(t,x)|≤M, for x∈B, we have(14)Hxt≤∫0tt-sq-1Γqfs,xsds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqfs,xsds+1-2t·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds≤Mσ=M1,which implies that (Hx)≤M1. Further, we find that(15)Hx′t≤∫0tt-sq-2Γq-1fs,xsds+1γ1-2θ-2α+βγ∫011-sq-1Γqfs,xsds+2γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds≤MΓq+1q+γ+2βq+γθqγ1-2θ-2α+β=M2.Hence, for t1,t2∈[0,T], we have(16)Hxt2-Hxt1≤∫t1t2Hx′sds≤M2t2-t1.This implies that H is equicontinuous on [0,1]. Thus, by the Arzelà-Ascoli theorem, the operator H:D→D is completely continuous.
Next, we consider the set(17)A=x∈D∣x=νHx,0<ν<1,and show that the set A is bounded. Let x∈A,  and then x=νHx,0<ν<1. For any t∈[0,1], we have (18)xt=-∫0tt-sq-1Γqfs,xsds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqfs,xsds+1-2t·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsdsand |x(t)|=ν|(Hx)(t)|≤|(Hx)(t)|≤Mσ=M1 for any t∈[0,1]. So the set A is bounded. Thus, the conclusion of Theorem 3 applies and the operator H has at least one fixed point. This, in turn, implies that problem (1) has at least one solution on [0,1].Now we show the existence of solutions for problem (1) by means of Leray-Schauder degree theory.Theorem 5.
Suppose that there exist constants0≤ϱ<1/σ and N>0 such that |f(t,x)|≤ϱ|x|+N for all t∈[0,1],x∈D. Then problem (1) has at least one solution.Proof.
Define a ballBR⊂D with radius R>0 by (19)BR=x∈D:maxt∈0,1⁡xt<R,where R will be fixed later. Then, it is enough to show that the operator H:B¯R→D (given by (10)) is such that (20)x≠λHx,∀x∈∂BR,∀λ∈0,1.Now we set (21)Kλ,x=λHx,λ∈0,1.Then, by Arzelà-Ascoli theorem, ωλ(x)=x-K(λ,x)=x-λHx is completely continuous. If condition (20) holds true, then the following Leray-Schauder degrees are well defined and, by the homotopy invariance of topological degree, we have that (22)deg⁡ωλ,BR,0=deg⁡I-λH,BR,0=deg⁡ω1,BR,0=deg⁡ω0,BR,0=deg⁡I,BR,0=1≠0,0∈BR,where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, ω1(t)=x-λHx=0 for at least one x∈BR. In order to justify condition (20), it is assumed that x=λHx for some λ∈[0,1] and for all t∈[0,1] so that (23)xt=λHxt≤∫0tt-sq-1Γqfs,xsds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqfs,xsds+1-2t·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds≤ϱx+N·∫0tt-sq-1Γqds+1γ1-2θ-2α+βγt-θ-α-β·∫011-sq-1Γqds+1-2t·γ∫0θθ-sq-1Γqds+β∫011-sq-2Γq-1ds,which, on taking norm over the interval [0,1], yields (24)x≤ϱx+Nσ,where σ is given by (10). In consequence we have (25)x≤σN1-ϱσ.Letting R=σN/(1-ϱσ)+1, (20) holds. This completes the proof.The next result is based on Krasnoselskii’s fixed point theorem [18], Th. 4.4.1.Theorem 6.
Letf:[0,1]×R→R be a continuous function satisfying (H1) and (H2)
|f(t,x)|≤μ(t),∀(t,x)∈[0,1]×R, and μ∈C([0,1],R+).Then problem (1) has at least one solution on [0,1] if (26)γ1-θ-α-β+βq+γθqΓq+1γ1-2θ-2α+β<1.Proof.
Forr≥μσ, let us define a closed set (ball) Br={x∈D:x≤r} and introduce the operators H1 and H2 on Br as (27)H1xt=-∫0tt-sq-1Γqfs,xsds,H2xt=+1γ1-2θ-2α+βγt-θ-α-β∫011-sq-1Γqfs,xsds+1-2t·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds.For x,y∈Br, it is easy to show that (H1x)+(H2y)≤μσ≤r, where σ is given by (10). This implies that H1x+H2y∈Br.
In view of condition (26), the operator H2 is a contraction. Continuity of the operator H1 follows from that of f. Also, H1x≤μ/Γ(q+1) implies that H1 is uniformly bounded on Br. Furthermore, with sup(t,x)∈[0,1]×Br|f(t,x)|=fb<∞ and t1,t2∈(0,1], we have (28)H1xt2-H1xt1≤fbΓq+1t2q-t1q+2t2-t1q⟶0,independent of x as t1→t2. This shows that H1 is relatively compact on Br. Hence, we infer by the Arzelà-Ascoli theorem that H1 is compact on Br. Thus all the conditions of Krasnoselskii’s fixed point theorem hold true. Hence, problem (1) has at least one solution on [0,1]. This completes the proof.Finally, we make use of Leray-Schauder nonlinear alternative to show the existence of solutions for problem (1).Lemma 7 (nonlinear alternative for single valued maps [19]).
LetE1 be a closed, convex subset of a Banach space E and V be an open subset of E1 with 0∈V. Suppose that U:V¯→E1 is continuous, compact (i.e., U(V¯) is a relatively compact subset of E1) map. Then either (i)
U has a fixed point in V¯ or(ii)
there isx∈∂V (the boundary of  V in E1) and κ∈(0,1) with x=κU(x).Theorem 8.
Letf:[0,1]×R→R be a continuous function. Assume that (H3)
there exist a functionp∈C([0,1],R+) and a nondecreasing function ψ:R+→R+ such that |f(t,x)|≤p(t)ψ(x),∀(t,x)∈[0,1]×R;(H4)
there exists a constantM>0 such that (29)MψMpσ>1,whereσ is given by (10).
Then problem (1) has at least one solution on [0,1].Proof.
As a first step, we show that the operatorH:D→D defined by (10) maps bounded sets into bounded sets in D. For a positive number r, let Br={x∈D:x≤r} be a bounded set in D. Then, for x∈Br together with (H3) and (H4), we obtain (as before)(30)Hxt≤pσψM<M.
Next, it will be shown thatH maps bounded sets into equicontinuous sets of D. Let t1,t2∈[0,1] and x∈Br. Then (31)Hxt2-Hxt1=-∫0t2t2-sq-1Γqfs,xsds+∫0t1t1-sq-1Γqfs,xsds+1γ1-2θ-2α+βγt2-t1·∫011-sq-1Γqfs,xsds-t2-t1·γ∫0θθ-sq-1Γqfs,xsds+β∫011-sq-2Γq-1fs,xsds≤ψrpΓq+1t2q-t1q+2t2-t1q+δt2-t1,where (32)δ=γ1+θq+βqγ1-2θ-2α+β.Clearly, the right-hand side of the above inequality tends to zero independent of x∈Br as t2→t1. Thus, by the Arzelà-Ascoli theorem, the operator H is completely continuous.
Letx be a solution for the given problem. Then, for λ∈(0,1), as before, we obtain (33)x=supt∈0,1⁡λHxt≤ψxpσ.In view of (H4), there exists M such that x≠M. Let us choose P={x∈D:x<M+1}.
Notice that the operatorH:P¯→D is continuous and completely continuous. From the choice of P, there is no x∈∂P such that x=λH(x) for some λ∈(0,1). Consequently, the conclusion of Lemma 7 applies and hence the operator H has a fixed point x∈P¯ which is a solution of problem (1). This completes the proof.


### 3.1. Examples

Consider a three-point boundary value problem of nonlinear fractional differential equations given by(34)Dc3/2xt+ft,xt=0,t∈0,1,(35)x0=-x1,x′0+x′1=-x12.Here q=3/2,  α=β=γ=1, and θ=1/2. With the given values, it is found that(36)σ=1Γq+11+γ1-θ-α-β+βq+γθqγ1-2θ-2α+β=142+162π⋍1.3829.(a)
In (34), let us choose a continuous function (37)ft,xt=e-t49+t2π2+x+tan-1⁡x,t∈0,1.Clearly|f(t,x)-f(t,y)|≤2/7|x-y|,∀t∈[0,1],  x,y∈R, and l/σ<1, where l=2/7 and σ is given by (36). Thus problem (34)-(35) with f(t,x) given by (37) has a unique solution on [0,1] by Theorem 2.(b)
Taking(38)ft,x=t+1216e-2t+1sin⁡x+3cos2⁡tin (34), one can notice that |f(t,x)|≤1/2|x|+3/4. Thus, with ϱ=1/2,  N=3/4, the assumptions of Theorem 5 are satisfied and consequently its conclusion applies to problem (34)-(35) with f(t,x) given by (38).(c)
Choosing(39)ft,x=cos⁡t2+14+tx31+x3+x+12in (34), we find that |f(t,x)|≤p(t)ψ(x),∀(t,x)∈[0,1]×R, where (40)pt=cos⁡t2+14+t,ψx=x+32.Using condition (H4), that is, M/ψ(M)pσ>1, we get M>M1,M1≃1.1205. Clearly all the conditions of Theorem 8 are satisfied. Hence there exists a solution of problem (34)-(35) with f(t,x) given by (39).


## 3.1. Examples

Consider a three-point boundary value problem of nonlinear fractional differential equations given by(34)Dc3/2xt+ft,xt=0,t∈0,1,(35)x0=-x1,x′0+x′1=-x12.Here q=3/2,  α=β=γ=1, and θ=1/2. With the given values, it is found that(36)σ=1Γq+11+γ1-θ-α-β+βq+γθqγ1-2θ-2α+β=142+162π⋍1.3829.(a)
In (34), let us choose a continuous function (37)ft,xt=e-t49+t2π2+x+tan-1⁡x,t∈0,1.Clearly|f(t,x)-f(t,y)|≤2/7|x-y|,∀t∈[0,1],  x,y∈R, and l/σ<1, where l=2/7 and σ is given by (36). Thus problem (34)-(35) with f(t,x) given by (37) has a unique solution on [0,1] by Theorem 2.(b)
Taking(38)ft,x=t+1216e-2t+1sin⁡x+3cos2⁡tin (34), one can notice that |f(t,x)|≤1/2|x|+3/4. Thus, with ϱ=1/2,  N=3/4, the assumptions of Theorem 5 are satisfied and consequently its conclusion applies to problem (34)-(35) with f(t,x) given by (38).(c)
Choosing(39)ft,x=cos⁡t2+14+tx31+x3+x+12in (34), we find that |f(t,x)|≤p(t)ψ(x),∀(t,x)∈[0,1]×R, where (40)pt=cos⁡t2+14+t,ψx=x+32.Using condition (H4), that is, M/ψ(M)pσ>1, we get M>M1,M1≃1.1205. Clearly all the conditions of Theorem 8 are satisfied. Hence there exists a solution of problem (34)-(35) with f(t,x) given by (39).



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