# Existence of the Unique Nontrivial Solution for Mixed Fractional Differential Equations

Authors: Yujing Liu; Chenguang Yan; Weihua Jiang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we consider the differential equations with right-sided Caputo and left-sided Riemann-Liouville fractional derivatives. Furthermore, the expression of Green’s function is derived, and its properties are investigated. By the fixed-point theorem for bothφ−h,e-concave operators and mixed monotone operators, we get the existence and uniqueness of the solution, respectively. As applications, some examples are provided to illustrate our main results.


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## Body

## 1. Introduction

Fractional differential equations are generalization of the ordinary differential equations to a nonintegral order, and they have been widely used in other fields of mathematics (as discussed in [1–10]). In recent decades, many authors devoted themselves to fractional equations. More fractional boundary value problems have been applied to physics, biology, medicine, software engineering, neural network, and other related sciences (see [11–16]). With the publication of works discussed in [17–22], the theory of fractional boundary value problems are gradually enriched and systematized.In [23], Song and Cui concerned the existence of solutions of nonlinear mixed fractional differential equation with the integral boundary value problem under resonance:
(1)CD1α_D0+βut=ft,ut,D0+β+1ut,D0+βut,0<t<1,u0=u′0=0,u1=∫01utdAt,where CD1α_ is the left Caputo fractional derivative of order α∈1,2, and D0+β is the right Riemann-Liouville fractional derivate of order β∈0,1. The coincidence degree theory is the main theoretical basis to prove the existence of solutions of such problems. In recent years, there have been some studies on the existence of solutions for mixed fractional differential equations (see [24, 25]).By using the fixed-point theorem for mixed monotone operators, Jong et al. [26] dealt with the existence of positive solutions of the following multipoint boundary value problems for nonlinear fractional differential equations
(2)D0+βφpD0+αxt=ft,xt,0<t<1,x0=0,D0+γx1=∑i=1m−2ξiD0+γxηi,D0+αx0=0,φpD0+αx1=∑i=1m−2ζiφpD0+αxηi,where D0+α, D0+β, and D0+γ are the standard Riemann-Liouville derivatives with 1<α,β≤2,0<γ≤1,0<ξi,ηi,ζi<1,i=1,2,⋯m−2. The fixed-point theorem for mixed monotone operators was also used to prove the existence of solutions of boundary value problems [27, 28].Moreover, many researchers focused on theφ−h,e-concave operators in [29–31] and its applications in [32].Based on above works, this paper investigates the existence of solutions for the fractional differential equations(3)CD1−αD0+βxt+ft,xt=b,0<t<1,x0=0,x′1=D0+βx1=0,where DC1−α is the right-sided Caputo fractional derivative with 0<α≤1,D0+β is the left-sided Riemann-Liouville fractional derivative with 1<β≤2 and α+β>2. Here, f:0,1×−∞,+∞⟶−∞,+∞ is continuous, and b>0 is a constant real number.In Section2, there are some definitions, properties, and lemmas related to this article. Then, we obtain the Green’s function and prove some lemmas of Green’s function. In Section 3, two important theorems are obtained. In Theorem 13, set Ph,e is defined. According to the fixed-point theorem of increasing concave operator, the existence of the unique solution of boundary value problem (3) is obtained. In Theorem 14, set Ph is defined, the existence of solutions is also obtained by the fixed-point theorem of mixed monotone operators. In the last section, some examples are given to illustrate the validity of the theorems.


## 2. Preliminaries

In this part, we present some basic definitions, properties, and lemmas.Definition 1 (see [20]).
The left-sided and right-sided Riemann-Liouville fractional integrals of orderαα>0 of a function g:0,∞⟶R are given by
(4)I0+αgt=∫0tt−sα−1Γαgsds,I1−αgt=∫t1s−tα−1Γαgsds,where Γα is the Gamma function.Definition 2 (see [20]).
The left-sided Riemann-Liouville fractional derivative and right-sided Caputo fractional derivative of orderαα>0 of a function g∈Cn0,∞,R are given by
(5)D0+αgt=dndtnI0+n−αgt,CD1−αgt=−1nI1−n−αgnt,where n−1<α<n.Property 3 (see [20]).
Letα>0 and n=α+1. If ft∈Cn0,1, then
(6)I0+αD0+αft=ft+∑j=1nCjtα−j,I1−αCD1−αft=ft+∑k=0n−1Ck′1−tk,where Cj,Ck′∈R is arbitrary constant.Lemma 4.
Letα ∈ (0,1], β ∈ (1,2]. For y ∈ C[0,1], then the unique solution of the fractional differential equation
(7)CD1−αD0+βxt=−yt,0<t<1,x0=0,x′1=D0+βx1=0,is xt=∫01Gt,sysds where
(8)Gt,s=1ΓαΓβ−∫0st−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ,0≤s≤t≤1,1ΓαΓβ−∫0tt−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ,0≤t≤s≤1.Proof.
Applying the right-sided fractional integralI1−α to both sides of the Equation (7) and by Property 3, we can obtain that
(9)D0+βxt=−I1−αyt+C0,where C0∈R is arbitrary constant. Applying the left-sided fractional integral I0+β to both sides Equation (9) above and by Property 3, we can obtain that
(10)xt=−I0+βI1−αyt+C0Γ1+βtβ+C1tβ−1+C2tβ−2,where Ci∈Ri=0,1,2. By x0=0 and D0+βx1=0, we get C2=C0=0, then
(11)xt=−I0+βI1−αyt+C1tβ−1.
Finding the derivative of (11), we have
(12)x′t=−I0+β−1I1−αyt+β−1C1tβ−2,
Sincex′1=0, it follows
(13)x′1=−1ΓαΓβ−1∫011−τβ−2dτ∫τ1s−τα−1ysds+β−1C1=0.
Exchanging the order of the above double integral, we have(14)C1=1ΓαΓβ∫01ysds∫0s1−τβ−2s−τα−1dτ.
SubstituteC1 into (11), we know that
(15)xt=−1ΓαΓβ∫0tt−τβ−1dτ∫τ1s−τα−1ysds+tβ−1ΓαΓβ∫01ysds∫0s1−τβ−2s−τα−1dτ.
Exchange the order of the first double integral in the above formula, we get that(16)xt=−1ΓαΓβ∫0tysds∫0st−τβ−1s−τα−1dτ+∫t1ysds∫0tt−τβ−1s−τα−1dτ+tβ−1ΓαΓβ∫01ysds∫0s1−τβ−2s−τα−1dτ.
Calculated that the Green’s function of the fractional differential Equation (7) is
(17)Gt,s=1ΓαΓβ−∫0st−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ,0≤s≤t≤1,1ΓαΓβ−∫0tt−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ,0≤t≤s≤1.
The proof is completed.Consequently, the boundary value problem (3) has a unique solution if and only if xt satisfies the integral equation:
(18)xt=∫01Gt,sfs,xs−bds=∫01Gt,sfs,xsds−b∫01Gt,sds.Lemma 5.
The Green’s functionsGt,s defined by Lemma 4 satisfy the following properties:
(1)
Gt,s≥0, for all s,t∈0,1(2)
s/2Jstβ−1≤Gt,s≤2Jstβ−1≤J0tβ−1, for all s,t,∈0,1, where Js=1/ΓαΓβ∫s/2s1−τβ−2s−τα−1dτ,J0=2/ΓαΓβα+β−2.Proof.
First, we prove that the functionGt,s is nonnegative.
For any0≤s≤t≤1,
(19)Gt,s=1ΓαΓβ∫0stβ−11−τβ−2−t−τβ−1s−τα−1dτ≥1ΓαΓβ∫0st−tτβ−1−t−τβ−1s−τα−1dτ≥0.
For any0≤t≤s≤1,
(20)Gt,s=1ΓαΓβ−∫0tt−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ≥1ΓαΓβ−∫0st−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ≥0.
In a word, for anyt,s∈0,1, Gt,s≥0.
Then, we proveGt,s≤2Jstβ−1≤J0tβ−1.
For anyt,s∈0,1,
(21)Gt,s≤tβ−1ΓαΓβ∫0s1−τβ−2s−τα−1dτ=tβ−1ΓαΓβ∫0s21−τβ−2s−τα−1dτ+∫s2s1−τβ−2s−τα−1dτ≤2tβ−1ΓαΓβ∫s2s1−τβ−2s−τα−1dτ=2Jstβ−1≤2tβ−1ΓαΓβ∫s2ss−τα+β−3dτ=2tβ−1sα+β−2ΓαΓβα+β−22α+β−2≤J0tβ−1.
Finally, we prove thatGt,s≥s/2Jstβ−1.
For anyt,s∈0,1,
(22)Gt,s≥1ΓαΓβ−∫0st−τβ−1s−τα−1dτ+tβ−1∫0s1−τβ−2s−τα−1dτ=tβ−1ΓαΓβ∫0s1−τβ−2−1−τtβ−1s−τα−1dτ≥∫0s1−τβ−2−1−τβ−1s−τα−1dτΓαΓβtβ−1=∫0sτ1−τβ−2s−τα−1dτΓαΓβtβ−1≥∫s/2s1−τβ−2s−τα−1dτΓαΓβtβ−1≥s2∫s/2s1−τβ−2s−τα−1dτΓαΓβtβ−1=s2Jstβ−1.
The proof is completed.Next, we summarize two fixed-point lemmas and some basic concepts in ordered Banach space.LetE,· be a real Banach space which is partially ordered by a cone P⊂E, i.e., x≤y if and only if y−x∈P. If x≤y and x≠y, then we denote x<y or y>x. θ denotes the zero element of E. P is called normal if there exists M>0 such that, for all x,y∈E, θ≤x≤y implies x≤My; in this case, M is called the normality constant of P. We say that an operator A:E⟶E is increasing if x≥y implies Ax≥Ay [29].Givenh∈E and h>θ, we define the set
(23)Ph=x∈E∃0<λ1≤λ2:λ1h≤x≤λ2h.Remark 6.
Ifx∈Ph, let λ0=max1/λ1,λ2, we have 1/λ0h≤x≤λ0h.
Lete∈P and θ≤e≤h, we define the set
(24)Ph,e=x∈E∃0<µ1≤µ2:µ1h≤x+e≤µ2h.Definition 7 (see [29, 31]).
LetA:P⟶P be a given operator. For any x∈P and r∈0,1, there exists φr∈r,1 such that Arx≤φrAx. Then, A is called a generalized concave operator.Definition 8 (see [29]).
LetA:Ph,e⟶E be a given operator. For any x∈Ph,e and λ∈0,1, there exists φλ>λ such that
(25)Aλx+λ−1e≥φλAx+φλ−1e.
Then,A is called a φ−h,e-concave operator.Lemma 9 (see [29]).
LetP be normal and A be an increasing φ−h,e-concave operator with Ah∈Ph,e. Moreover, for any ω0∈Ph,e, making the sequence ωn=Aωn−1,n=1,2,⋯, then we obtain ωn−x∗⟶0 as n⟶∞.Definition 10 (see [28]).
LetD⊂E. Operator A:D×D⟶E is said to be mixed monotone if Ax,y is nondecreasing in x and nonincreasing in y, i.e., x1≤x2,y2≤y1,xi,yi∈Di=1,2 implies Ax1,y1≤Ax2,y2.Definition 11 (see [28]).
LetA:P×P⟶E a mixed monotone operator. Assume that for all 0<t<1, there exists 0<σ=σt<1 such that
(26)Atx,1ty≥tσtAx,yholds for all x,y∈P; then, A is called a t−σt mixed monotone model operator.Lemma 12 (see [28]).
Leth>θ. A:Ph×Ph⟶Ph is a t−σt mixed monotone operator. Then, A has exactly one fixed-point x∗ in Ph. Moreover, constructing successively the sequences
(27)xn=Axn−1,yn−1,yn=Ayn−1,xn−1,n=1,2,⋯,for any initial point x0,y0∈Ph, we have xn−x∗⟶0 and yn−x∗⟶0 as n⟶∞.


## 3. Main Results

We consider the spaceE=C0,1 with the usual maximum norm u=maxt∈0,1ut. Clearly, E is a Banach space. Set P=x∈Ext≥0,t∈0,1. Obviously, P is a normal cone and P⊂E, the normality constant is 1.Theorem 13.
Leth and e are defined by ht=Htβ−1,et=b∫01Gt,sds, where t∈0,1 and H=2b/ΓαΓβα+β−2, Gt,s is the Green’s functions of (3) defined by Lemma 4. Suppose the following conditions hold:
(H1). f:0,1×−e∗,+∞⟶−∞,+∞ is continuous and increasing with respect to the second variable, that is, for any 0≤x≤y, we get ft,xt≤ft,yt, where e∗=max0≤t≤1ex
(H2). For any λ∈0,1 there is φλ>λ such that ft,λx+λ−1y≥φλft,x, where t∈0,1,x∈0,+∞,y∈0,e∗
(H3). ft,0≥0 with ft,0≢0 for t∈0,1.
Then, the fractional differential equations (3) have a unique nontrivial solution x∗ in Ph,e. Moreover, for any given initial value x0∈Ph,e, making the sequence xn=Axn−1,n=1,2,⋯, then we obtain xn−x∗⟶0 as n⟶∞.Proof.
For any∈0,1, it is easy to see et≥0, that is, e∈P.
Further,(28)et≤b∫01J0tβ−1ds=b2ΓαΓβα+β−2tβ−1=ht.
Hence,0≤et≤ht.
Let(29)Axt=∫01Gt,sfs,xsds−et,t∈0,1.
The boundary (3) has an integral formulation given by
(30)xt=∫01Gt,sfs,xsds−b∫01Gt,sds=∫01Gt,sfs,xsds−et=Axt.
So,xt is the solution of the problem (3) if and only if xt is the fixed point of the operator of A.
Firstly, it is apparent from the definition ofA that A is Ph,e⟶E.
From (H2), we know that(31)Aλx+λ−1e=∫01Gt,sfs,λxs+λ−1esds−et≥∫01Gt,sφλfs,xsds−et=φλ∫01Gt,sfs,xsds−φλet+φλet−et=φλ∫01Gt,sfs,xsds−et+φλ−1et=φλAxt+φλ−1et.
By Definition8, A is φ−h,e-concave operator.
Secondly, forx,y∈Ph,e and x≤y, we get
(32)Axt=∫01Gt,sxsds−et≤∫01Gt,sysds−et=Ayt.
So,A is increasing.
Thirdly, we would prove thatAh∈Ph,e.
Letl1=∫01s/21−s/2α+β−2−1−sα+β−2fs,0ds/HΓαΓβα+β−2,l2=2∫01fs,Hds/HΓαΓβα+β−2.(33)Aht+et=∫01Gt,sfs,hsds=∫01Gt,sfs,Hsβ−1ds≤∫01J0tβ−1fs,Hds≤2HΓαΓβα+β−2∫01fs,HdsHtβ−1=l2ht,Aht+et=∫01Gt,sfs,Hsβ−1ds≥∫01s2Jstβ−1fs,0ds≥∫01s/21−s/2α+β−2−1−sα+β−2fs,0dsHΓαΓβα+β−2Htβ−1=l1ht.
It is not difficult to verify that(34)0<∫01s/21−s/2α+β−2−1−sα+β−2fs,0dsHΓαΓβα+β−2≤2∫01fs,HdsHΓαΓβα+β−2.
That is to say,l2≥l1>0. Then, we have l1h≤Ah+e≤l2h. So, Ah∈Ph,e.
Consequently, by using Lemma9, the operator A has a unique fixed-point x∗ in Ph,e, i.e., x∗t=∫01Gt,sfs,x∗sds−et,t∈0,1. And for any x0∈Ph,e, there exists the sequence xn=Axn−1,n=1,2,⋯, satisfies xn⟶x∗ as n⟶∞,
(35)xn=∫01Gt,sfs,xn−1sds−b∫01Gt,sds.
The proof is completed.Theorem 14.
Letht=Htβ−1, where t∈0,1 and H is defined by Theorem 13. Suppose the following conditions hold:
(H4). ft,x−b=ϕt,x+ψt,x≥0, where ϕ,ψ:0,1×0,+∞⟶0,+∞ are continuous functions and for any fixed t∈0,1, ϕt,x is nondecreasing in x>0, and ψt,x is nonincreasing in x>0
(H5). For 0<η<1, there exists σ∈0,β such that ϕt,ηx≥ησϕt,x,ψt,η−1x≥ησψt,x. For η>1, from above inequality, we can get that ϕt,ηx≤ησϕt,x,ψt,η−1x≤ησψt,x, where t∈0,1 and x>0;
(H6) ∫01t−σβ−1ψt,Hdt<+∞.
Then, the boundary value problem (3) has a unique positive solution in Ph.Proof.
According to definition ofPh in Theorem 13, Ph is normal cone. Let
(36)Tx,yt=∫01Gt,sfs,xs−bds=∫01Gt,sϕs,xs+ψs,ysds.
From Remark6, for any x,y∈Ph, there exist two positive constants M1 and M2 such that 1/M1ht≤xt≤M1ht,1/M2ht≤yt≤M2ht. Let M=maxM1,M2, it is easy to know that M>1 and
(37)1Mht≤xt≤Mht,1Mht≤yt≤Mht.
By (H5), we have
(38)ϕt,xt≤ϕt,Mht≤ϕt,MH≤Mσϕt,H,ϕt,xt≥ϕt,1Mht=ϕt,1MHtβ−1≥M−σtσβ−1ϕt,H,ψt,yt≤ψt,1Mht=ψt,1MHtβ−1≤Mσt−σβ−1ψt,H,ψt,yt≥ψt,Mht≥ψt,MH≥M−σψt,H.
Hence, for anyt∈0,1, we have the followings:
(39)Tx,yt=∫01Gt,sϕs,xs+ψs,ysds≤∫01J0tβ−1Mσϕs,H+Mσs−σβ−1ψs,Hds=J0Mσ∫01ϕs,H+s−σβ−1ψs,Hdstβ−1=H−1J0Mσ∫01ϕs,H+s−σβ−1ψs,Hdsht.
It follows from the above (H6) that Tx,yt<+∞.
(40)Tx,yt≥∫01s2Jstβ−1ϕs,xs+ψs,ysds≥∫01s2Jstβ−1M−σsσβ−1ϕs,H+M−σψs,Hds=M−σ∫01s2Jstβ−1sσβ−1ϕs,H+ψs,Hdstβ−1=H−1M−σ∫01s2Jstβ−1sσβ−1ϕs,H+ψs,Hdsht.
Supposem1−1=H−1∫01s/2Jstβ−1sσβ−1ϕs,H+ψs,Hds,
(41)m2=H−1J0∫01ϕs,H+s−σβ−1ψs,Hds,then
(42)1m1Mσht≤Tx,yt≤m2Mσht.
So,T:Ph×Ph⟶Ph.
By (H4), we obtain that for any x1≤x2x1,x2∈Ph and y∈Ph,
(43)Tx1,yt=∫01Gt,sϕs,x1s+ψs,ysds≤∫01Gt,sϕs,x2s+ψs,ysds=Tx2,yt.
For anyy1≤y2y1,y2∈Ph and x∈Ph,
(44)Tx,y1t=∫01Gt,sϕs,xs+ψs,y1sds≥∫01Gt,sϕs,xs+ψs,y2sds=Tx,y2t.
Consequently,T is a mixed monotone operator.
For anyη∈0,1 and σ∈0,β in (H5),
(45)ϕt,ηxt≥ησϕt,xt,ψt,η−1xt≥ησψt,xt.
So, we get(46)Tηx,η−1yt=∫01Gt,sϕs,ηxs+ψs,η−1ysds≥ησ∫01Gt,sϕs,xs+ψs,ysds=ησTx,yt.
From Definition11, T is a t−σt mixed monotone model operator. By Lemma 12, T has exactly one fixed-point x∗ in Ph. Constructing successively the sequences
(47)xn=Txn−1,yn−1,yn=Tyn−1,xn−1,n=1,2,.⋯
For any initial pointx0,y0∈Ph, we have xn−x∗⟶0 and yn−x∗⟶0 as n⟶∞.
The proof is completed.


## 4. Application

Now, we give two concrete examples to illustrate our main theorems.Example 1.
Consider the mixed fractional differential equation(48)DC1−3/4D0+3/2xt+ete∗x+et1/3=1,0<t<1,x0=0,x′1=D0+3/2x1=0.
Here,α=3/4,β=3/2,b=1>0; therefore, α+β=9/4>2,et=1×∫01Gt,sds=∫01Gt,sds>0,e∗=max0≤t≤1ex, H=2b/ΓαΓβα+β−2=8/Γ3/4Γ3/2=7.3665.
(1)
It is obvious thatft,x=et/e∗x+et1/3>0 is continuous and increasing with respect to x(2)
Takingφλ=λ1/3>λ for any λ∈0,1,t∈0,1,x∈0,+∞ and y∈0,e∗, we have
(49)ft,λx+λ−1y=ete∗λx+λ−1y+et1/3=λ1/3ete∗x+1−1λy+1λet1/3=λ1/3ete∗x+1−1λete∗y+1λet1/3≥λ1/3ete∗x+1−1λete∗e∗+1λet1/3=λ1/3ete∗x+et1/3=φλft,x.(3)
Obviously,ft,0=et1/3≥0, and ft,0≢0, for all t∈0,1.
Therefore, it follows from Theorem13 that the problem (48) has a unique nontrivial solution x∗∈Ph,e, where et=∫01Gt,sds and ht=7.3665tβ−1.Example 2.
Consider the mixed fractional differential equation(50)DC1−2/3D0+3/2xt+tx+1x+12+tx+1x+12=1,0<t<1,x0=0,x′1=D0+3/2x1=0.
Here,α=2/3,β=3/2,b=1>0, therefore, α+β=13/6>2,H=2b/ΓαΓβα+β−2=12/Γ2/3Γ3/2=9.9995.
(1)
It is obvious thatft,x−1=tx+1/x+tx+1/x≥0 is continuous. Letting ϕt,x=tx+tx≥0,ψt,x=t/x+t/x≥0, it is easy to know that for any t∈0,1, ϕt,x is nondecreasing in x>0, and ψt,x is nonincreasing in x>0(2)
Forη∈0,1, there exists σ=1∈0,3/2 such that ϕt,ηx=tηx+tηx≥ηtx+tx=ηϕt,x, ψt,η−1x=ηt/x+ηt/x≥ηt/x+t/x=ηψt,x, where x>0 and t∈0,1(3)
Obviously, forH=9.9995, ψt,H=t/9.9995+t/9.9995, ∫01t−σβ−1ψt,Hdt=∫01t−1/2t/9.9995+t/9.9995dt≈0.3829<+∞.
Therefore, it follows from Theorem14 that the problem (50) has a unique nontrivial solution x∗∈Ph, where ht=9.9995tβ−1.


## 5. Conclusion

In this paper, we discuss the boundary value problem of mixed fractional differential equation with Caputo fractional derivatives and Riemann-Liouville fractional derivatives. Firstly, integrating Equation (7) and applying the boundary value conditions, we construct the Green’s function of the boundary value problem. Unfortunately, the expression of the Green’s function is very complex. Fortunately, we still prove the fundamental property of the Green’s function.Secondly, by using the fixed-point theorems of increasing convex operators which defined in setPh,e, we obtain the existence of solution of the mixed fractional differential Equation (3). In addition, by using the fixed-point theorems of monotone mixed operators which defined in set Ph, we also obtain the existence of solution of the mixed fractional differential Equation (3). The two theorems restrict different assumptions on the nonlinear term f, respectively, and state the existence of solutions for the boundary value problem (3). These two theorems are not contradictory, but two cases. That gives the existence of solution proved in this paper more sufficiency.At last, the validity of main theorems in this paper is obtained from the examples.



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