# Existence Results for Nonlinear Multiorder Fractional Differential Equations with Integral and Antiperiodic Boundary Conditions

Authors: HuiChol Choi; YongSim Sin; KumSong Jong
Journal: Journal of Applied Mathematics
 (2020)

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## Abstract

In this paper, we study the solvability of a class of nonlinear multiorder Caputo fractional differential equations with integral and antiperiodic boundary conditions. By using some fixed point theorems including the Banach contraction mapping principle and Schaefer’s fixed point theorem, we obtain new existence and uniqueness results for our given problem. Also, we give some examples to illustrate our main results.


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## Body

## 1. Introduction

Fractional calculus has a history of several hundred years, and many valuable results that have contributed to the development of mathematical theories and their application to practice have been created during its historical process (see [1]). Also, fractional differential equations are one of the powerful means to model and solve scientific and technological problems that have been arisen in physics, chemistry, biology, mechanics, and many other fields, and it has developed more and more in-depth (see [2]). In particular, boundary value problems of fractional differential equations are often used as mathematical models for many phenomena in a variety of physical, biological, mechanical, and chemical studies such as analysis of turbulent flow, simulation of chemical reaction, and image processing technique (see [3–6]).In recent years, antiperiodic boundary value problems have been put forward in various practical phenomena and have attracted the attention of a large number of researchers because of the specific properties of their solutions (see [7]). Based on many works about the solvability of antiperiodic boundary value problems for integer-order differential equations (see [8–10]), a lot of attempts have been made to extend the existence results for them to the case of fractional differential equations (see [11–20]). For instance, Agarwal and Ahmad [11] established the existence of solutions of the following single-term Caputo fractional differential equations with antiperiodic boundary conditions by using the nonlinear alternative of Leray-Schauder type and Leray-Schauder degree theory:
(1)Dcqxt=ft,xt,t∈0,T,T>0,3<q≤4,x0=−xT,x′0=−x′T,x″0=−x″T,x‴0=−x‴T.In [17], Choudhary and Daftardar-Gejji considered the antiperiodic boundary value problem of the nonlinear multiorder Caputo fractional differential equation
(2)∑i=0nλiDcαiut=ft,ut,t∈0,T,u0=−uT,where λi∈R,i=0,1,⋯,n,λn≠0,0≤α0<α1<⋯<αn<1. They proved the existence and uniqueness of solutions to their problem in terms of the two-parametric functions of Mittag-Leffler type. Their equation in problem (2) is a generalization of the classical relaxation equation and governs some fractional relaxation processes.Analyzing the higher-order fractional differential equations like that in problem (1), some new research papers considered not only antiperiodic boundary conditions but also mixed-type boundary conditions which are composed of both integral and antiperiodic boundary conditions (see [21–25]). Xu [24] obtained new existence and uniqueness results for the following single-term fractional differential equations with integral and antiperiodic boundary conditions by means of the Krasnosel’skii fixed point theorem, contraction mapping principle, and Leray-Schauder degree theory:
(3)Dcqxt=ft,xt,t∈0,1,1<q<2,x1=μ∫01xsds,x′0+x′1=0.Taking the previous results together, we can know that very little has been done on the multiterm fractional differential equations with integral and antiperiodic boundary conditions. In particular, as far as we know, the research on the mixed-type boundary value problems of nonlinear multiorder fractional differential equations like that in problem (2) which is of great significance in practice has not been carried out at all.Motivated by above analysis, in this paper, we investigate the existence and uniqueness of solutions for the following mixed-type fractional boundary value problems of combining the nonlinear multiorder Caputo fractional differential equations, which are similar to the equation in problem (2) and have higher fractional orders, with the integral and antiperiodic boundary conditions in problem (3):
(4)Dc0+αut+∑i=1nλitDc0+αiut+∑i=1mμitDc0+βiut+σtut=ft,ut,t∈0,1,(5)u1=μ∫01usds,u′0+u′1=0,where 1<α1<⋯<αn<α≤2,0<β1<⋯<βm<1,0<μ<1,λi∈C0,1i=1,2,⋯,n,μi∈C0,1i=1,2,⋯,m,σ∈C0,1,f∈C0,1×R,R.


## 2. Derivation of the Integral Equation

The Riemann-Liouville fractional integral and the Caputo fractional derivative of orderα>0 of a function f:0,∞⟶R are given by
(6)I0+αft≔1Γα∫0tt−sα−1fsds,Dc0+αft≔1Γn−α∫0tt−sn−α−1fnsds,where n=α+1, provided that the right-hand sides are pointwise defined on 0,∞ (see [1]).Definition 1.
A functionu∈C0,1 with a Caputo fractional derivative of order α that belongs to C0,1 (i.e., Dc0+αu∈C0,1) is said to be a solution of problem (4)–(5) if it satisfies equation (4) and the boundary conditions (5).Lemma 2.
If a functionu is a solution of problem (4)–(5), then yt≔Dc0+αut is a solution of the integral equation
(7)yt=ft,∫01Gt,sysds−∑i=1nλitI0+α−αiyt−∑i=1mμitI0+α−βiyt+12∑i=1mμit⋅t1−βiΓ2−βiI0+α−1ytt=1−σt∫01Gt,sysds,in C0,1, and conversely, if y∈C0,1 is a solution of the integral equation (7), then a function u which is given by
(8)ut=∫01Gt,sysds,
is a solution of problem (4)–(5), where
(9)Gt,s=t−sα−1Γα+μ41−μΓα+141−sα−4α1−sα−1+αα−11−sα−2+12Γα1−tα−11−sα−2−21−sα−1,0≤s≤t≤1,μ41−μΓα+141−sα−4α1−sα−1+αα−11−sα−2+12Γα1−tα−11−sα−2−21−sα−1,0≤t≤s≤1.Proof.
Let a functionu be a solution of problem (4)–(5). Applying I0+α on both sides of the expression yt=Dc0+αut, it is obvious that
(10)ut=∫01Gt,sysds,(see [24]).
We can rewrite (10) as
(11)ut=μ1−μI0+α+1ytt=1−μ1−μI0+αytt=1+μ41−μI0+α−1ytt=1+I0+αyt−I0+αytt=1+12I0+α−1ytt=1−t2I0+α−1ytt=1=I0+αyt+μ1−μI0+α+1ytt=1−11−μI0+αytt=1+μ41−μ+1−t2I0+α−1ytt=1.
From (11), we have some equalities as follows:
(12)Dc0+αiut=I0+α−αiyt,i=1,⋯,n,Dc0+βiut=I0+α−βiyt−12Γ2−βi⋅t1−βiI0+α−1ytt=1,i=1,⋯,m.
Substituting the aboveDc0+αiut and Dc0+βiut into (4), it can be easily seen that
(13)yt+∑i=1nλitI0+α−αiyt+∑i=1mμitI0+α−βiyt−12Γ2−βi⋅t1−βiI0+α−1ytt=1+σt∫01Gt,sysds=ft,∫01Gt,sysds.
This yields the integral equation (7).
Conversely, let a functiony∈C0,1 be a solution of the integral equation (7). Substituting the expression
(14)ut=∫01Gt,sysds,into (7), we can get that
(15)yt+∑i=1nλitI0+α−αiyt+∑i=1mμitI0+α−βiyt−12∑i=1mμit⋅t1−βiΓ2−βiI0+α−1ytt=1+σtut=ft,ut.
On the other hand, using the expression of Green’s functionGt,s, we can see that
(16)ut=I0+αyt+μ1−μI0+α+1ytt=1−11−μI0+αytt=1+μ41−μ+1−t2I0+α−1ytt=1.
So, we know thatDc0+αut, Dc0+αiut, and Dc0+βiut exist for any t∈0,1 and Dc0+αu∈C0,1.
Considering the relations(17)Dc0+αiut=I0+α−αiyt,i=1,⋯,n,Dc0+βiut=I0+α−βiyt−12Γ2−βi⋅t1−βiI0+α−1ytt=1,i=1,⋯,m,we can rewrite (15) as
(18)Dc0+αut+∑i=1nλitDc0+αiut+∑i=1mμitDc0+βiut+σtut=ft,ut.
That is,u satisfies equation (4).
Now, we prove thatu satisfies the boundary conditions (5). By simple calculation, we have
(19)u1=μ1−μI0+α+1ytt=1−μ1−μI0+αytt=1+μ41−μI0+α−1ytt=1+I0+αytt=1−I0+αytt=1+12I0+α−1ytt=1−12I0+α−1ytt=1=μ1−μI0+α+1ytt=1−μ1−μI0+αytt=1+μ41−μI0+α−1ytt=1,∫01usds=μ1−μI0+α+1ytt=1−μ1−μI0+αytt=1+μ41−μI0+α−1ytt=1+I0+α+1ytt=1−I0+αytt=1+12I0+α−1ytt=1−14I0+α−1ytt=1=11−μI0+α+1ytt=1−11−μI0+αytt=1+141−μI0+α−1ytt=1.
This means thatu1=μ∫01usds. Also, since
(20)u′0=I0+α−1ytt=0−12I0+α−1ytt=1=−12I0+α−1ytt=1,u′1=I0+α−1ytt=1−12I0+α−1ytt=1=12I0+α−1ytt=1,we can see that u′0+u′1=0. Therefore, it is proved that u satisfies the boundary conditions (5). The proof is completed.Remark 3.
By Lemma2, the existence of solutions for problem (4)–(5) refers to the solvability of the integral equation (7) in C0,1.Remark 4.
As we can see from the expression(21)∫01Gt,sysds=I0+αyt+μ1−μI0+α+1ytt=1−11−μI0+αytt=1+μ41−μ+1−t2I0+α−1ytt=1,the function ∫01Gt,sysds is continuous in 0,1 for any y∈C0,1. Also, for any y∈C0,1, the following inequality holds:
(22)∫01G⋅,sysds=maxt∈0,1I0+αyt+μ1−μI0+α+1ytt=1−11−μI0+αytt=1+μ41−μ+1−t2I0+α−1ytt=1≤1Γα+1+μ1−μ⋅1Γα+2+11−μ⋅1Γα+1+μ41−μ+12⋅1Γαy=wα,μy,where wα,μ is denoted as
(23)wα,μ≔11−μΓα+12−μα+44+μα+1.


## 3. Main Results

Define the operatorsP and Q on C0,1 as follows:
(24)Pyt≔ft,∫01Gt,sysds,Qyt≔−∑i=1nλitI0+α−αiyt−∑i=1mμitI0+α−βiyt+12∑i=1mμit⋅t1−βiΓ2−βiI0+α−1ytt=1−σt∫01Gt,sysds.Then, the integral equation (7) can be regarded as the operator equation
(25)yt=Pyt+Qyt.Lemma 5.
The following hold:(i)
For anyα∈R+, the fractional integral operator I0+α:C0,1⟶C0,1 is compact.
(ii)
For anyd∈C0,1, the operator A:C0,1⟶C0,1 which is defined by
(26)Axt≔dt⋅xt,is a bounded linear operator.Proof.
(i)
It is sufficient to prove that for any bounded setΩ≔u∈C0,1u≤r, I0+αΩ is relatively compact. Obviously, we can see
(27)∀u∈Ω,I0+αut≤1Γα+1u.
So, we can know thatI0+αΩ is uniformly bounded. Also, we have that for any t1,t2∈0,1t1<t2,
(28)I0+αut1−I0+αut2=1Γα∫0t1t1−sα−1usds−1Γα∫0t2t2−sα−1usds≤1Γα∫0t1t1−sα−1usds−∫0t1t2−sα−1usds+∫t1t2t2−sα−1usds≤1Γα∫0t1t2−sα−1−t1−sα−1⋅usds+∫t1t2t2−sα−1⋅usds≤uΓα+12t2−t1α+t1α−t2α.
This yields thatI0+αΩ is equicontinuous. Therefore, using the Ascoli-Arzelà theorem, it can be easily seen that I0+αΩ is relatively compact.
(ii)
It is obvious that for anyd∈C0,1,
(29)∀x∈C0,1,Axt≤maxt∈0,1dt⋅x.
This completes the proof of (ii).In this article, the following hypothesis will be used:H1.
There existL,l∈0,+∞ and σ1,σ2∈0,1 such that for any t1,t2∈0,1 and for any x1,x2∈R,
(30)ft1,x1−ft2,x2≤lt1−t2σ1+Lx1−x2σ2.Lemma 6.
Assume that hypothesisH1 holds. Then, the operator P:C0,1⟶C0,1 is compact.Proof.
As in the proof of Lemma5, put Ω≔u∈C0,1u≤r. Then, we have that for any u∈Ω,
(31)Put=ft,∫01Gt,susds≤ft,∫01Gt,susds−ft,0+ft,0≤L∫01Gt,susdsσ2+ft,0≤Lrσ2⋅∫01G⋅,sdsσ2+f⋅,0.
This implies thatPΩ is uniformly bounded.
On the other hand, we can get that for anyt1,t2∈0,1t1<t2,
(32)Put1−Put2=ft1,∫01Gt1,susds−ft2,∫01Gt2,susds≤lt1−t2σ1+L∫01Gt1,susds−∫01Gt2,susdsσ2.
Since(33)∫01Gt1,susds−∫01Gt2,susds=I0+αutt=t1−t12I0+α−1utt=1−I0+αutt=t2+t22I0+α−1utt=1≤I0+αutt=t1−I0+αutt=t2+I0+α−1utt=1⋅t1−t22≤uΓα+1⋅2t2−t1α+t1α−t2α+u2Γα⋅t1−t2,the following inequality holds:
(34)Put1−Put2≤lt1−t2σ1+Lrσ2Γασ2⋅2t2−t1α+t1α−t2αα+t1−t22σ2.
This yields thatPΩ is equicontinuous. The conclusion then follows from the Ascoli-Arzelà theorem.Lemma 7.
The operatorQ:C0,1⟶C0,1 is compact.Proof.
From Lemma5 and the fact that the composition of the bounded linear operator and compact operator is also compact, every term of the operator Q is compact. Since the sum of two compact operators is also compact, the proof is completed.Lemma 8 (see [26]).
LetX be a Banach space. Assume that Ω is an open bounded subset of X with θ∈Ω, and let T:Ω¯⟶X be a compact operator such that
(35)Tu≤u,∀u∈∂Ω.
Then,T has a fixed point in Ω¯.Denoteω∗ as follows:
(36)ω∗≔∑i=1nλiΓα−αi+1+∑i=1mμiΓα−βi+1+12∑i=1mμiΓ2−βiΓα+σ⋅wα,μ.Here, we list more hypotheses to be used throughout this paper.H2.
ω∗<1.Theorem 9.
Assume that hypothesesH1 and H2 hold. If the nonlinear function f satisfies that
(37)limx→0maxt∈0,1ft,xx=0,then problem (4)–(5) has at least one solution.Proof.
Sincelimu→0∫01Gt,susds=0 and limx→0maxt∈0,1ft,x/x=0, we obtain
(38)∀d1>0,∃r1>0,∀u∈C0,1u≤r1,f⋅,∫01G⋅,susds≤d1⋅u.
Putd≔1−ω∗. Then, it holds that
(39)∃r>0,∀u∈C0,1u≤r,f⋅,∫01G⋅,susds≤d⋅u.
Now, consider the ballBr≔u∈C0,1∣u≤r, and for any u0∈∂Br, estimate the norm P+Qu0. Since u0=r, we can see that
(40)Pu0+Qu0≤f⋅,∫01G⋅,su0sds+∑i=1nλi⋅u0Γα−αi+1+∑i=1mμi⋅u0Γα−βi+1+12∑i=1mμi⋅u0Γ2−βiΓα+σ⋅∫01G⋅,su0sds≤f⋅,∫01G⋅,su0sds+∑i=1nλi⋅u0Γα−αi+1+∑i=1mμi∣⋅u0Γα−βi+1+12∑i=1mμi∣⋅u0Γ2−βiΓα+σ⋅wα,μu0≤d+ω∗⋅r=r=u0.
Therefore, the operatorP+Q has a fixed point in terms of Lemma 8. This yields the conclusion.Example 1.
Consider the following fractional boundary value problem:(41)Dc0+1.8ut+0.3t⋅Dc0+1.2ut+0.01t2⋅Dc0+0.6ut+0.2⋅ut=1+u2t−1,u1=0.1∫01usds,u′0+u′1=0.
Check that the conditions of Theorem9 are satisfied. Putting ft,x=1+x2−1, it can be easily seen that
(42)ft1,x1−ft2,x2=1+x12−1+x22≤x1−x2,limx→0maxt∈0,1ft,xx=limx→01+x2−1x=0.
By simple calculation, we also getω∗≈0.72<1. So, problem (41) has at least one solution.Lemma 10 (Schaefer’s fixed point theorem) (see [27]).
LetX be a Banach space and A:X⟶X a compact operator. Then, either
(i)
the equationx=λAx has a solution for λ=1 or
(ii)
the set of all such solutionsx, for 0<λ<1, is unbounded.Theorem 11.
Assume that hypothesesH1 and H2 hold. And suppose that
(i)
∃a>0,∃ψ∈C0,1×R,R, and
(43)limx→+∞maxt∈0,1ψt,xx=0∧∀t∈0,1,∀x∈R,ft,x≤a+ψt,x.
(ii)
∀t∈0,1,x1≤x2⟹ft,x1≤ft,x2
Then, problem (4)–(5) has at least one solution.Proof.
From Lemmas6 and 7 and hypothesis H1, the operator P+Q is compact. Using condition (i), we can get
(44)∀d1>0,∃r1>0,∀x≥r1,ft,x≤a+d1⋅x.
Putd≔1−ω∗/2wα,μ. Then, it follows that
(45)∃r>0,∀x≥r,f⋅,x≤a+d⋅x.
Now consider the setS≔u∈C0,1∣u=λPu+λQu,0<λ<1. From Remark 4, it is obvious that for any u∈S,
(46)∫01G⋅,susds∈C0,1.
There are two cases∫01G⋅,susds≤r and ∫01G⋅,susds>r.
If∫01G⋅,susds≤r, then because of condition (ii), we have
(47)ft,∫01Gt,susds≤ft,∫01G⋅,susds≤ft,r≤a+d⋅r.
Thus, it follows that(48)Pu+Qu≤a+d⋅r+∑i=1nλiΓα−αi+1+∑i=1mμiΓα−βi+1+12∑i=1mμiΓ2−βiΓα+σ⋅wα,μ⋅u=a+d⋅r+ω∗u.
This yields that(49)u≤λa+dr1−λω∗≤a+dr1−ω∗.
If∫01G⋅,susds>r, then as in the first case, using condition (ii) and Remark 4, we can see that
(50)ft,∫01Gt,susds≤ft,∫01G⋅,susds≤a+d⋅∫01G⋅,susds≤a+d⋅wα,μu.
And we obtain(51)Pu+Qu≤a+d⋅wα,μu+∑i=1nλiΓα−αi+1+∑i=1mμiΓα−βi+1+12∑i=1mμiΓ2−βiΓα+σ⋅wα,μ⋅u=a+d⋅wα,μ+ω∗u.
Therefore, it holds that(52)u≤λa1−λd⋅wα,μ+ω∗≤a1−d⋅wα,μ−ω∗=2a1−ω∗.
These inequalities (49) and (52) imply the boundedness of the set S. By using Lemma 10., the operator P+Q has a fixed point. This completes the proof.Example 2.
Consider the boundary value problem(53)Dc0+1.8ut+0.3t⋅Dc0+1.2ut+0.01t2⋅Dc0+0.6ut−0.8ut=sint−1+u2t+1,u1=0.1∫01usds,u′0+u′1=0.
Problem (53) is equal to the following problem:
(54)Dc0+1.8ut+0.3t⋅Dc0+1.2ut+0.01t2⋅Dc0+0.6ut+0.2ut=sint−1+u2t+1+ut,u1=0.1∫01usds,u′0+u′1=0.
For problem (54), check that the conditions of Theorem 11 are satisfied. Putting ft,x=sint−1+x2+1+x, we can see that
(55)ft1,x1−ft2,x2≤t1−t2+2x1−x2,a≔1,ψt,x≔1+x−1+x2,limx→+∞maxt∈0,1ψt,xx=0,∂f∂xt,x=1−x1+x2>0.
Also, sinceω∗≈0.72<1, all the conditions of Theorem 11 are satisfied. So, problem (53) has at least one solution.Theorem 12.
Suppose that the following hold:(i)
There existsL≥0 such that for any t∈0,1 and for any x1,x2∈R,
(56)ft,x1−ft,x2≤Lx1−x2.
(ii)
q≔L+σ⋅wα,μ+∑i=1nλi/Γα−αi+1+∑i=1m1/Γα−βi+1+1/2Γ2−βiΓαμi<1
Then, problem (4)–(5) has a unique solution.Proof.
Define the operatorT:C0,1⟶C0,1 as
(57)Tyt≔Pyt+Qyt.
Obviously, we can obtain that for anyy1,y2∈C0,1 and for any t∈0,1,
(58)Ty1t−Ty2t≤ft,∫01Gt,sy1sds−ft,∫01Gt,sy2sds+∑i=1nλitI0+α−αiy1t−y2t+∑i=1mμitI0+α−βiy1t−y2t+12∑i=1mμitΓ2−βiI0+α−1y1t−y2tt=1+σt⋅∫01Gt,s⋅y1s−y2sds≤L+σ⋅wα,μ+∑i=1nλiΓα−αi+1+∑i=1m1Γα−βi+1+12Γ2−βiΓαμi⋅y1−y2=qy1−y2.
This means thatT:C0,1⟶C0,1 is a contraction operator. Therefore, the operator T has a unique fixed point in terms of the Banach contraction mapping principle, and problem (4)-(5) has a unique solution.Example 3.
Consider the following boundary value problem:(59)Dc0+1.8ut+0.3t⋅Dc0+1.2ut+0.01t2⋅Dc0+0.6ut+0.2ut=sint+0.031+u2t−1,u1=0.1∫01usds,u′0+u′1=0.
Puttingft,x=sint+0.031+x2−1, we can see easily that
(60)ft,x1−ft,x2≤0.03x1−x2.
This shows that condition (i) of Theorem12 is satisfied. On the other hand, we can get
(61)q≈0.78<1.
Thus, problem (59) has a unique solution.


## 4. Conclusion

In this paper, we consider the solvability of nonlinear multiorder fractional differential equations with integral and antiperiodic boundary conditions. At first, we derive an integral equation, transforming our fractional boundary value problems. Then, using some fixed point theorems such as the Banach contraction mapping principle and Schaefer’s fixed point theorem, we prove the existence and uniqueness of solutions. Since the problem considered in [24] is a special case of our problem, the results of this work can be regarded as generalizations of those results.



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