# Boundary Value Problems of Nonlinear Mixed-Type Fractional Differential Equations

Authors: Ping Yu; Hongju Li; Jian Ding; Yanli Ma
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, by means of a fixed point theorem for monotone decreasing operators on a cone, we discuss the existence of positive solutions for boundary value problems of nonlinear fractional singular differential equation. The proof of the main result is based on Gatica–Oliker–Waltman fixed-point theorem. At last, an example is given to illustrate our main conclusion.


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## Body

## 1. Introduction

Fractional differential equations with boundary conditions have been studied by many researchers because of their wide applications in many fields of mechanics, engineering, robotics, electrical networks, and so on [1–4]. Many authors consider the applications of fractional differential equations in different fields and obtain some basic results of fractional differential equations (see [5–9] and its references for details). In particular, the singularity of the nonlinear fractional boundary value problem of spatial variables and its references are discussed in literature [10–15].In the paper [10], by Guo-krasnosl’skil fixed point theorem, the authors proved the existence and multiplicity of positive solutions for a system of Riemann–Liouville fractional differential equations with singular nonnegative nonlinearities and p-Laplacian operators:(1)D0+α1ϕr1D0+β1ut+ft,ut,vt=0,t∈0,1,D0+α2ϕr1D0+β2ut+gt,ut,vt=0,t∈0,1,uj0=0,j=0,…,n−2,D0+β1u0=0,D0+γ0u1=∑i=1p∫01D0+γiutdHit,vj0=0,j=0,…,m−2,D0+β2v0=0,D0+δ0v1=∑i=1q∫01D0+δivtdKit,where α1,α2∈0,1,β1∈n−1,n,β2∈m−1,m, the functions f and g are nonnegative, and they may be singular at t=0 and/or t=1.In [14], the authors studied the existence of positive solutions for a nonlinear Riemann–Liouville fractional differential equation with a sign-changing nonlinearity:(2)D0+αut+λft,ut=0,t∈0,1,u0=u′0=⋯=un−20=0,D0+put|t=1=∑i=1maiD0+qut|t=ξi,where λ is a positive parameter, α∈ℝ,α∈n−1,n,n∈ℕ,n≥3,ξi∈ℝ for all i=1,…,mm∈ℕ,0<ξ1<⋯<ξm<1,p∈1,n−2,q∈0,p,D0+α denotes the Riemann–Liouville derivative of order α, and the nonlinearity f may change sign and may be singular at t=0 or t=1. The authors also proved relevant conclusion by Guo-krasnosl’skil fixed point theorem.On the basis of the above literatures, the singular fractional boundary value problem (BVP) in this text,(3)D0+αut+ft,ut=0,(4)u0=0,u1=a1∫01q1θuθdA1θ+a2∫0ηq2θuθdA2θ+a3∑i=1∞μiuζi,is studied, where 1<α≤2; aj≥0j=1,2,3; μi≥0,0<η<ζ1<ζ2<⋯<ζi<⋯<1i=1,2,…; 1−a3∑i=1∞μiζiα−1>0; q1,q2:0,1⟶ℝ+ are continuous with q1,q2∈L10,1; and ∫01q1θuθdA1θ and ∫01q2θuθdA2θ denote the Riemann–Stieltjes integrals, in which A1,A2:0,1⟶ℝ is the function of bounded variation. And, D0+α is the fractional derivative of Riemann–Liouville. At x=0, ft,x is singular. Let us make the following hypothesis:(B1)ft,x:0,1×0,∞⟶0,∞ is a continuous function(B2)ft,x decreases monotonically with respect to x for each fixed t(B3)limx⟶0+ft,x=∞ and limx⟶∞ft,x=0, uniformly on compact subsets of (0,1)(B4)0<∫01ft,hλtdt<∞ for all λ>0 and hλ=λhtThe layout of this paper is as follows: Section1 is a brief introduction of the research. Section 2 introduces some definitions and basic properties of fractional order derivative and integral. And, the Green function is proposed for the BVP (3) (4). Section 3 is the main results. Section 4 presents an example to illustrate the correctness of our main conclusion.


## 2. Preliminaries

For convenience, in this section, we present some definitions and lemmas which will be used in the following conclusions.Definition 1 (see [6]).
The Riemann–Liouville type integral of orderα>0 of a function f:0,∞⟶ℝ is defined by(5)I0+αft=1Γα∫0tt−θα−1fθdθ,where n−1<α≤n and Γθ=∫0∞tθ−1e−tdt is the Gamma function.Definition 2 (see [6]).
On the basis of Riemann–Liouville, the Riemann–Liouville fractional derivative of orderα>0 of a function f:0,∞⟶ℝ that is considered as antifractional integral of a function f is given as(6)D0+αft=1Γn−αddtn∫0tt−θα−n+1fθdθ.Lemma 1 (see [6]).
Letα>0. If we assume u∈C0,1∩L10,1, then the fractional differential equation D0+αu=0 has ut=c1tα−1+c2tα−2+⋯+cntα−n,ci∈ℝi=1,2,…,n, as the unique solution, where n=α is the smallest integer greater than or equal to α.Lemma 2 (see [6]).
We assume thatu∈C0,1∩L10,1 with a fractional derivative of order α>0 of belonging to u∈C0,1∩L10,1. Then(7)I0+αD0+αut=ut+c1tα−1+c2tα−2+⋯+cntα−n,for some ci∈ℝi=1,2,…,n, where n=α is the smallest integer greater than or equal to α.Lemma 3 (see [16]).
Givenf∈C0,1, then for t∈0,1,(8)ut=∫01Gt,θfθdθ,is the unique solution of the following equation:(9)D0+αut+ft=0,satisfying the boundary condition (4), where α∈1,2 and(10)Gt,θ=G1t,θ+k1tα−1Q1θ+k2tα−1Q2θ,(11)G1t,θ=t1−θα−1lθ−σt−θα−1σΓα,0≤θ≤t≤1,t1−θα−1lθσΓα,0≤t≤θ≤1,Q1θ=∫01G1t,θq1tdA1t,Q2θ=∫0ηG1t,θq2tdA2t,θ∈0,1,σ=1−a3∑i=1∞μiζiα−1>0,lθ=1−a3bθ1−θα−1,θ∈0,1,bθ=∑i=1∞μiζi−θα−1,0≤θ≤ζ1,∑i=2∞μiζi−θα−1,ζ1≤θ≤ζ2,…∑i=i0∞μiζi−θα−1,ζi0−1≤θ≤ζi0,…0,limi⟶∞ζi≤θ≤1.Apparently,l0=σ,(12)k1=a1σρ¯11+a1ρ1σρ¯1+1a2ρ2σρ¯2,k2=a1ρ1σρ¯1+1a2σρ¯2,ρ1=∫01tα−1q1tdA1t>0,ρ¯1=1−a1ρ1σ>0,ρ2=∫0ηtα−1q2tdA2t>0,ρ¯2=1−a1a2ρ1σ2ρ¯1+a2σρ2>0.Obviously,Gt,θ is continuous on 0,1×0,1.Lemma 4.
Suppose thata3∑i=1∞μiζiα−1<1, then the positive function lθ=1−a3bθ/1−θα−1 defined in Lemma 3 is nondecreasing on 0,1.


## 3. Main Conclusions

To simplify the proof of our main results, in this section, we firstly present and prove some lemmas.Lemma 5.
Leta3∑i=1∞μiζiα−1<1 and Ait is increasing on t∈0,1i=1,2. So, the Green function Gt,θ of equation (10) satisfies the following:(i)
Gt,θ≥0, when 0<t,θ<1(ii)
htG1,θ≤Gt,θ≤Hθ, for 0≤t,θ≤1, where ht=tα−1 and(13)Hθ=1−θα−1σΓαlθ+k1Q1θ+k2Q2θ.Proof.
(i)
For0≤θ≤t≤1,(14)Gt,θ=1σΓαt1−θα−1lθ−σt−θα−1+k1tα−1Q1θ+k2tα−1Q2θ≥1σΓαt1−θα−1lθ−σt1−θα−1+k1tα−1Q1θ+k2tα−1Q2θ=t1−θα−1σΓαlθ−l0+k1tα−1Q1θ+k2tα−1Q2θ≥0,For0≤t≤θ≤1,(15)Gt,θ=1σΓαt1−θα−1lθ+k1tα−1Q1θ+k2tα−1Q2θ≥0.(ii)
If0≤θ≤t≤1, we have(16)Gt,θ=1σΓαt1−θα−1lθ−σt−θα−1+k1tα−1Q1θ+k2tα−1Q2θ≥1σΓαt1−θα−1lθ−σt1−θα−1+k1tα−1Q1θ+k2tα−1Q2θ=tα−1G1,θ=htG1,θ,Gt,θ=1σΓαt1−θα−1lθ−σt−θα−1+k1tα−1Q1θ+k2tα−1Q2θ≤1σΓαt1−θα−1lθ−l0+l0α−1σΓα∫t−θt1−θxα−2dx+k1tα−1Q1θ+k2tα−1Q2θ≤1σΓαt1−θα−1lθ−l0+l0α−1σΓαt1−θα−2θ1−t+k1tα−1Q1θ+k2tα−1Q2θ≤1σΓαt1−θα−1lθ−2−αl0+k1tα−1Q1θ+k2tα−1Q2θ≤tα−11σΓα1−θα−1lθ+k1Q1θ+k2Q2θ≤Hθ.If0≤t≤θ≤1, we have(17)Gt,θ=1σΓαt1−θα−1lθ+k1tα−1Q1θ+k2tα−1Q2θ=tα−11σΓα1−θα−1lθ+k1Q1θ+k2Q2θ≥tα−11σΓα1−θα−1lθ−σ1−θα−1+k1Q1θ+k2Q2θ=tα−1G1,θ=htG1,θ,Gt,θ=1σΓαt1−θα−1lθ+k1tα−1Q1θ+k2tα−1Q2θ≤tα−11σΓα1−θα−1lθ+k1Q1θ+k2Q2θ≤Hθ.Assume thatE=C0,1 is a Banach space that the supremum norm is u=maxt∈0,1ut. Let P=u∈E|ut≥0,t∈0,1. It is easy to see that P is a cone in E. For λ>0, let hλt=λht, where ht is given in Lemma 5. In what follows, a set D⊂P and an integral operator T:D⟶P are given, respectively, as(18)D=u∈P|∃λu>0such thatut≥hλt,t∈0,1,Tut=∫01Gt,θfθ,uθdθ.Sinceft,x is singular at x=0, it suffices to define D as above. Moreover, note that, for u∈D, there exists λu>0 such that ut≥hλt for all t∈0,1. We get ft,ut≤ft,hλt for t∈0,1, because ft,x decreases with respect to x. Therefore,(19)0≤∫01Gt,θfθ,uθdθ≤∫01G1,θfθ,hλθ<∞.Likewise,T is decreasing with respect to D.Lemma 6 (see [17]).
Assume thatE is a Banach space and P is a normal cone on E. And, D is a subset in P when x,y∈D with x≤y; therefore, x,y⊂D. Assume that T:D⟶P is a continuous and decreasing mapping which is compact on every closed order interval contained in D, and suppose there is an x0∈D such that T2x0=TTx0 is defined and Tx0andT2x0 are orders comparable to x0; Therefore, T has a fixed point in D in case that either(1)
Tx0≤x0, T2x0≤x0(2)
x0≤Tx0, x0≤T2x0 or(3)
Define a complete sequence of iteratesTnx0n=0∞ and there is y0∈D so as to Ty0∈D with y0≤Tnx0 for each n∈N.Lemma 7.
u∈D is a solution of BVP (3) (4) if and only if Tu=u.Proof.
First, we find that one direction of the lemma is clearly correct. Now let us prove the other side. Letu∈D. Then, Tut=∫01Gt,θfθ,uθdθ, and Tu satisfies BVP (3) (4). Furthermore, by Lemma 5, for each t∈0,1, we have(20)Tut=∫01Gt,θfθ,uθdθ≥ht∫01G1,θfθ,uθdθ=htu.
Therefore, there exists someλTu such that Tut≥hλt, which implies that Tu∈D. That is, T:D⟶D.Now, we present another lemma which establishes a priori upper bound and a priori lower bound on solutions to prepare for Lemma6.Lemma 8.
Suppose thatf satisfies B1−B4, then there are constants mandM such that m≤u≤M for any solution u∈D of BVP (3) (4).Proof.
Prove the lemma by the inverse method.
Firstly, let us proveu≤M. Suppose a solution sequence unn=1∞ of BVP ((3) (4)) such that un≤un+1 with limn⟶∞un=∞. For any solution un∈D of BVP ((3) (4)), from equation (20), for t∈0,1,n≥1, we have(21)unt=Tunt≥htun.
Hence, assumptionsB2 and B4 yield, for t∈0,1 and all n>1:(22)unt=Tunt=∫01Gt,θfθ,unθdθ≤∫011−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,qu1θdθ=N1,for some 0<N1<+∞. Particularly, un≤N1, for each n≥1, which contradicts limn⟶∞un=∞.
Secondly, let us proveu≥m. And, let us suppose the conclusion is not true. Therefore, there is a solution sequence unn=1∞ of BVP (3) (4) such that unt⟶0 uniformly on 0,1 as n⟶∞. Let N2=infGt,θ:t,θ∈1/5,4/5×1/5,4/5>0. From B3, there exists some ξ>0 such that for t∈1/5,4/5 and 0<x<ξ, we have ft,x≥5/3N2. On the other hand, there is an n0∈ℕ such that n≥n0 implies 0<unt<ξ/2, for t∈0,1. Then, we have, for t∈1/5,4/5 and n≥n0,(23)unt=Tunt=∫01Gt,θfθ,unθdθ≥∫1/54/5Gt,θfθ,unθdθ≥N2∫1/54/5fθ,ξ2dθ≥N2∫1/54/553N2dθ=1.
It is easy to see that this conclusion contradicts the hypothesis thatunt⟶0 uniformly on 0,1 as n⟶∞. Then, there is an m>0 such that m≤u.Next let us prove the main conclusion of the article.Theorem 1.
Assume thatf satisfies B1−B4, then BVP ((3) (4)) has at least one positive solution.Proof.
Defineϕn:0,1⟶0,+∞ by(24)ϕnt=∫01Gt,θfθ,ndθ,n≥1.
By conditionsB1−B4, for all n≥1,(25)0<ϕn+1t≤ϕnt,on0,1,(26)limn⟶∞ϕnt=0uniformly on0,1.
Next define a sequence of functionsfn:0,1×0,+∞,n≥1, with fnt,x=ft,maxx,ϕnt.
So,fn is continuous and satisfies B2, for n≥1. Moreover,(27)fnt,x≤ft,xon0,1×0,+∞,fnt,x≤ft,ϕnton0,1×0,+∞.
It is easy to see thatfn effectively eliminates the singularity in ft,x at x=0. Next, we defined a sequence of operators Tn:P⟶P,n≥1, by Tnut=∫01Gt,θfnθ,uθdθ, u∈P. By the Arzela–Ascoli Theorem, Tn is a compact mapping on P. In addition, Tn0≥0 and Tn20≥0. From Lemma 6, for all n≥1,t∈0,1, there is un∈P such that Tnunt=unt. Then, un conforms to the conditions of BVP (3) (4). Moreover, for all un,(28)Tnunt=∫01Gt,θfnθ,unθdθ=∫01Gt,θfnθ,maxunθ,ϕnθdθ≤∫01Gt,θfnθ,ϕnθdθ≤Tϕnt,which implies unt=Tnunt≤Tϕnt, t∈0,1,n∈ℕ. By Lemma 8, there exists constants m,M>0 such that m≤un≤M for all n∈ℕ. Then, we can follow the argument of equation (20) to show that unt≥htm, on 0,1, for n≥1. Because T:D⟶D is a compact mapping, there exists a subsequence of Tun which converges to some u∗∈D. We relabel the subsequence as the original sequence such that Tun⟶u∗ as n⟶∞.
And then, Let us prove that(29)limn⟶∞Tun−un=0.
Letλ=m and ϵ>0. From B4, there exists 0<δ<1 such that(30)∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,hλθdθ<ε2.
Furthermore, combining equation (26), there exists an n1 such that, for n≥n1, vnt≤hλt on δ,1, so that ϕnt≤hλt≤unt on δ,1. Thus, for θ∈δ,1 and n≥n0(31)fnθ,unθ=fθ,maxunθ,vnθ=fθ,unθ,and for t∈0,1,(32)Tunt−unt=Tunt−Tnunt=∫01Gt,θfθ,unθ−fnθ,unθdθ.
Then, fort∈0,1,(33)Tunt−unt≤∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,unθdθ+∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,maxunθ,ϕnθdθ≤∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,unθdθ+∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,unθdθ≤2∫0δ1−θα−1σΓαlθ+k1Q1θ+k2Q2θfθ,hλθdθ<ϵ.
Then,∀t∈0,1, we conclude that Tun−un<ϵ for all n≥n1. Therefore, u∗∈hm,M and for t∈0,1(34)Tu∗t=Tlimn⟶∞Tunt=Tlimn⟶∞unt=limn⟶∞Tun=u∗t.


## 4. Example

Example 1.
We consider the following fractional equations:(35)D0+2/3ut+t+u−3=0,0<t<1,u0=0,u1=13∫0144θ2+1uθdA1θ+5∫01/1044θ2+5uθdA2θ+3∑i=1∞2i2u1−1i+15,where ft,u=t+u−3.α=3/2,a1=1/3,a2=5,a3=3,η=1/10,μi=2/i2,ζi=1−1/i+15,andq1t=4/4t2+1,q2t=4/4t2+5 and(36)A1t=3,t∈0,12,4,t∈12,1,A2t=4,t∈0,12,5,t∈12,1.
It is obvious that the hypothesesB1−B4 are satisfied, that is to say, the conditions of Theorem 1 are satisfied. Then, the BVP (35) has at least one positive solution.



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