# The Existence and Uniqueness of Solutions for Variable-Order Fractional Differential Equations with Antiperiodic Fractional Boundary Conditions

Authors: Fang Wang; Lei Liu
Journal: Journal of Function Spaces
 (2022)

---

## Abstract

In this paper, we discuss the existence and uniqueness of solutions for nonlinear fractional differential equations of variable order with fractional antiperiodic boundary conditions. The main results are obtained by using fixed point theorem.


---

## Body

## 1. Introduction

Fractional calculus has become one of the important tools for the development of modern society; the fractional differential equation with variable order has gained lots of interest [1–4]. Some researchers have investigated the physical background and numerical analysis of fractional differential equations of variable order [5–8]. In [9], Bushnaq et al. used Bernstein polynomials with nonorthogonal basis to establish operational matrices for variable-order integration and differentiation which convert the considered problem to some algebraic type matrix equations and obtained numerical solution to variable-order fractional differential equations by numerical simulation. In [10], Shah et al. proposed a new algorithm for numerical solutions to variable-order partial differential equations, used properties of shifted Legendre polynomials to establish some operational matrices of variable-order differentiation and integration, and got the numerical solution by numerical experiments.In recent years, the antiperiodic boundary value problem of fractional differential equation has gradually become the focus of research, which have broad application in engineering and sciences such as physics, mechanics, chemistry, economics, and biology [11–17]. In [18], Ahmad and Nieto considered the following antiperiodic fractional boundary value problems:
(1)0CDtqxt=ft,xt,t∈0,T,x0=−xT,0CDtpx0=−0CDtpxT,where CDp denotes the Caputo fractional derivative of order q and f is a given continuous function.The problems related to the antiperiodic boundary value condition have been considered in [19–26], but the antiperiodic boundary value problem of fractional differential equation with variable order is almost not considered. In this paper, we investigate the existence of solutions for an antiperiodic fractional boundary value problem given by
(2)0CDtqtxt=ft,xt,t∈0,T,x0=−xT,0CDtpx0=−0CDtpxT,where CDp denotes the Caputo fractional derivative of order p, 0<p<1, CDqt denotes the Caputo fractional derivative of variable order qt, 1<qt≤2, T is a positive constant, and f:0,T×R⟶R is a given continuous function.


## 2. Preliminary Knowledge

In this section, we introduce some fundamental definitions and lemmas.Definition 1 (see [27]).
The Riemann-Liouville fractional integral of orderq for a continuous function f: 0,∞⟶R is defined as
(3)0Itqft=1Γq∫0tt−sq−1fsds,q>0,provided the integral exists.Definition 2 (see [27]).
Forn−1 times absolutely continuous function f:0,∞⟶R, the Caputo derivative of fractional order q is defined as
(4)0CDtqft=1Γn−q∫0tt−sn−q−1fnsds,n−1<q<n,n=q+1,where q denotes the integer part of the real number q.Definition 3 (see [3]).
The Riemann-Liouville fractional integral of variable orderqt for a continuous function f: 0,∞⟶R is defined as
(5)0Itqtft=∫0tt−sqt−1Γqtfsds,qt>0,t>0,provided that the right-hand side is pointwise defined.Definition 4 (see [3]).
Forn−1 times absolutely continuous function f: 0,∞⟶R, the Caputo fractional derivative of variable order qt is defined as
(6)0CDtqtft=∫0tt−sn−qt−1Γn−qtfnsds,n−1<qt<n,dnfdtn,n=qt.Definition 5 (see [25]).
LetI⊂R, I is called a generalized interval if it is either an interval, or a or ∅.
A finite setθ is called a partition of I if each x in I lies in exactly one of the generalized intervals ξ in θ.
A functionf: I⟶R is called piecewise constant with respect to partition θ of I if for any ξ∈θ, f is constant on ξ.Theorem 6 (see [27]).
LetE be a closed, convex, and nonempty subset of a Banach space X; let F: E⟶E be a continuous mapping such that FE is a relatively compact subset of X. Then, F has at least one fixed point in E.Lemma 7 (see [27]).
Letα>0, and let yt∈L∞a,b or yt∈Ca,b, then aCDtαaItαyt=yt.Lemma 8 (see [27]).
Letn=α+1 for α∉N0;n=α for α∈N0, if yt∈ACna,b or yt∈Cna,b, then
(7)aItαaCDtαyt=yt−∑k=0n−1ykak!t−ak.


## 3. Main Results

LetJ=0,T. Denote CJ,R be the Banach space of all continuous functions x:J⟶R with the norm x=supt∈Jxt and introduce the following assumption.H1 Let n∈N be an integer, θ=J1=0,T1,J2=T1,T2,⋯,Jn=Tn−1,Tn be a partition of the interval J, and qt:J⟶1,2 be a piecewise constant function with respect to θ with the following forms:
(8)qt=∑i=1nqiIit=q1,ift∈J1,q2,ift∈J2,⋮qn,ift∈Jn,where 1<qi≤2 are constants, and Ii is the indicator of the interval Ti−1,Ti, i=1,2,⋯,n (with T0=0, Tn=T), Iit=1 for t∈Ti−1,Ti, and Iit=0 for elsewhere.LetΩi=CJi,R be the Banach space of all continuous functions x:Ji⟶R with the norm xΩi=supt∈Jixt, i=1,2,⋯,n (with T0=0, Tn=T).The Caputo fractional derivative of variable orderqt for the function xt could be presented as a sum of Caputo fractional derivatives of constant orders qi by Definition 4, i=1,2,⋯,n:
(9)0CDtqtxt=∫0T1t−s1−q1Γ2−q1x2sds+⋯+∫Ti−1tt−s1−qiΓ2−qix2sds.Thus, according to (9), problem (2) can be written in the following form:
(10)0CDtqtxt=∫0T1t−s1−q1Γ2−q1x2sds+⋯+∫Ti−1tt−s1−qiΓ2−qix2sds=ft,xt.Definition 9.
The problem (2) has a solution, if there are functions xi, so that xi∈CJi,R, satisfy (10) and xi0=−xiT,0CDtpxi0=−0CDtpxiT.
Let the functionx∈CJ,R be such that xt≡xTi−1 on 0,Ti−1, then consider (2) as the following form:
(11)Ti−1CDtqixt=ft,xt,t∈Ti−1,Ti,xTi−1=−xTi,Ti−1CDtpxTi−1=−Ti−1CDtpxTi.Proposition 10.
For anyxt∈Ωi,ft,xt∈CJi×R,R, xt is a unique solution of problem (11) if and only if x satisfy the integral equation:
(12)xt=∫Ti−1TiGit,sfs,xsds,where Git,s is Green’s function given by
(13)Git,s=t−sqi−1Γqi+Ti−sqi−p−1Γ2−pTi−Ti−1−2t2Ti−Ti−11−pΓqi−p−Ti−sqi−12Γqi,s≤t,Ti−sqi−p−1Γ2−pTi−Ti−1−2t2Ti−Ti−11−pΓqi−p−Ti−sqi−12Γqi,t≤s.Proof.
Ifxt∈Ωi is a solution of problem (11), applying Ti−1Itqi on both sides of (11), according to Lemma 8, we get
(14)xt=Ti−1Itqift,xt+c1+c2t−Ti−1,
according to the facts thatTi−1CDtpc1=0,Ti−1CDtpt=t−Ti−11−p/Γ2−p, Ti−1CDtpTi−1Itqixt=Ti−1Itqi−pxt, and initial condition of problem (11), we get
(15)c2=−Γ2−pΓqi−pTi−Ti−11−p∫Ti−1TiTi−sqi−p−1fs,xsds,c1=Ti−Ti−1Γ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds−12Γqi∫Ti−1TiTi−sqi−1fs,xsds.
Thus, the solution of problem (11) is
(16)xt=1Γqi∫Ti−1tt−sqi−1fs,xsds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds−12Γqi∫Ti−1TiTi−sqi−1fs,xsds.
Green’s function can be written as(17)xt=∫Ti−1TiGit,sfs,xsds.
It implies thatxt is the solution to the integral equation (12). In turn, if xt∈Ωi is the solution to the integral equation (12), according to Lemma 7, we deduce that xt is the solution of the problem (11). Hence, we complete this proof.Theorem 11.
Assume thatH2ft,xt∈CJi×R,R and there exists a positive constant L such that ft,xt≤L, for any t∈Ti−1,Ti,xt∈R. Then, problem (11) has at least a solution.Proof.
According to Proposition10, problem (11) is equivalent to the following integral equation:
(18)xt=1Γqi∫Ti−1tt−sqi−1fs,xsds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds−12Γqi∫Ti−1TiTi−sqi−1fs,xsds.
Define operatorT:Ωi⟶Ωi by
(19)Txt=1Γqi∫Ti−1tt−sqi−1fs,xsds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds−12Γqi∫Ti−1TiTi−sqi−1fs,xsds,Bri=xt∈Ωi,xΩi≤ri,t∈Ji,where ri≥LTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1/2Γqi−p+1Γqi+1, observe that Bri is a closed, bounded, and convex subset of Banach space Ωi. For any xt∈Ωi, we have
(20)Txt=1Γqi∫Ti−1tt−sqi−1fs,xsds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds−12Γqi∫Ti−1TiTi−sqi−1fs,xsds≤L1Γqi∫Ti−1tt−sqi−1ds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1ds+12Γqi∫Ti−1TiTi−sqi−1ds≤Lt−Ti−1qiΓqi+1+Ti−Ti−1qi−pΓqi−p+1Ti−Ti−1−2tΓ2−p2Ti−Ti−11−p+Ti−Ti−1qi2Γqi+1≤LTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−12Γqi−p+1Γqi+1≤ri.
It impliesT:Ωi⟶Ωi is well defined.
Now, we consider the continuity of operatorT. Since ft,xt∈CJi×R,R, given an arbitrary ε>0, for any xt,yt∈Ωi,we can findδ>0such that ft,xt−ft,yt<2Γqi−p+1Γqi+1ε/Ti−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1. When xt−yt<δ for t∈Ji, for any t∈Ji, we have
(21)Txt−Tyt≤1Γqi∫Ti−1tt−sqi−1fs,xs−fs,ysds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xs−fs,ysds+12Γqi∫Ti−1TiTi−sqi−1fs,xs−fs,ysds≤1Γqi∫Ti−1tt−sqi−1ds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1ds+12Γqi∫Ti−1TiTi−sqi−1ds2Γqi−p+1Γqi+1εTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1≤t−Ti−1qiΓqi+1+Ti−Ti−1qi−pΓqi−p+1Ti−Ti−1−2tΓ2−p2Ti−Ti−11−p+Ti−Ti−1qi2Γqi+1×2Γqi−p+1Γqi+1εTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1<Ti−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−12Γqi−p+1Γqi+1×2Γqi−p+1Γqi+1εTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1=ε.
We get the operatorT is continuous.
For eachxt∈Ωi, we prove that if t1,t2∈Ji, and 0<t2−t1<δ, then Txt2−Txt1<ε:
(22)Txt2−Txt1=1Γqi∫Ti−1t2t2−sqi−1fs,xsds−1Γqi∫Ti−1t1t1−sqi−1fs,xsds+t1−t2Γ2−pTi−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds≤1Γqi∫Ti−1t1t2−sqi−1−t1−sqi−1fs,xsds+1Γqi∫t1t2t2−sqi−1fs,xsds+t2−t1Γ2−pTi−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds≤Lt2−Ti−1qi−t1−Ti−1qiΓqi+1+t2−t1Ti−Ti−1qi−1Γ2−pΓqi−p+1.
By the mean value theorem, we have(23)Txt2−Txt1≤Lt2−Ti−1qi−t1−Ti−1qiΓqi+1+t2−t1Ti−Ti−1qi−1Γ2−pΓqi−p+1≤LTi−Ti−1qi−1Γqi+LTi−Ti−1qi−1Γ2−pΓqi−p+1δ<ε.
Therefore,Txt2−Txt1<ε. According to the previous analysis, we know that is equicontinuous and uniformly bounded. We know by the Arzela-Ascoli theorem that T is compact on Bri, so the operator T is completely continuous. So, Theorem 11 implies that the antiperiodic boundary value problem of variable order (11) has at least a solution on Ji. This completes the proof.Theorem 12.
Assume thatH3ft,xt∈CJi×R,R such that ft,xt−ft,yt≤Kxt−yt, for any t∈Ji,xt,yt∈R, and if KTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1/Γqi−p+1Γqi+1<1, then problem (11) has a unique solution.Proof.
Letsupt∈Jift,0=M<∞, Bri=xt∈Ωi,xΩi≤ri,t∈Ji, where
(24)ri≥MTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−1Γqi−p+1Γqi+1.
For anyxt∈Bri, we have
(25)Txt≤1Γqi∫Ti−1tt−sqi−1fs,xsds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xsds+12Γqi∫Ti−1TiTi−sqi−1fs,xsds≤1Γqi∫Ti−1tt−sqi−1fs,xs−fs,0+fs,0ds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xs−fs,0+fs,0ds+12Γqi∫Ti−1TiTi−sqi−1fs,xs−fs,0+fs,0ds≤Kri+M1Γqi∫Ti−1tt−sqi−1ds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1ds+12Γqi∫Ti−1TiTi−sqi−1ds≤Kri+MTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−12Γqi−p+1Γqi+1≤ri.
It implies thatTBri⊆Bri.
For anyxt,yt∈Ωi,t∈Ji,
(26)Txt−Tyt≤1Γqi∫Ti−1tt−sqi−1fs,xs−fs,ysds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1fs,xs−fs,ysds+12Γqi∫Ti−1TiTi−sqi−1fs,xs−fs,ysds≤1Γqi∫Ti−1tt−sqi−1ds+Ti−Ti−1−2tΓ2−p2Ti−Ti−11−pΓqi−p∫Ti−1TiTi−sqi−p−1ds+12Γqi∫Ti−1TiTi−sqi−1dsKx−y≤KTi−Ti−1qi−13Γqi−p+1Ti−Ti−1+Γ2−pΓqi+1Ti+Ti−12Γqi−p+1Γqi+1x−y<12x−y.
It follows thatT is a contraction mapping. Thus, the Banach fixed point theorem yields that T has a unique fixed point which is the unique solution of the antiperiodic boundary value problem (11).Theorem 13.
Assume thatH1, H2 hold for all i =1, 2, ⋯, n, then the problem (2) has at least a solution in CJ,R.Proof.
According to Theorem11, problem (11) has at least a solution x∗∈Ωi. Define the function
(27)xi=xi∗Ti−1,t∈0,Ti−1,xi∗,t∈Ji.
Thus, the functionxi∈C0,Ti,R satify (10) and xi0=−xiT,0CDtpxi0=−0CDtpxiT. Then, the function
(28)xt=x1t,t∈J1,x2t=x2∗T1,t∈J1,x2∗,t∈J2,⋮xnt=xn∗Tn−1,t∈0,Tn−1,xn∗,t∈Jn,is a solution of problem (2) in CJ,R.Theorem 14.
Assume thatH1, H3 hold for all i=1,2,⋯,n, then problem (2) has a unique solution in CJ,R.The proof of Theorem14 is similar to Theorem 13.


## 4. Conclusion

This paper is devoted to considering the existence of solutions to the antiperiodic fractional boundary value problems for nonlinear differential equations of variable order, which is a piecewise constant function based on the essential difference about the variable order. Based on the fixed point theory, the results are obtained. It is also worth considering fractional differential equations of variable-order problems related to thermodynamics, fluid mechanics, resonance, etc.



---

*Source: 7663192-2022-09-05.xml*