# Existence and Uniqueness of Positive Solutions for a Class of Nonlinear Fractional Differential Equations with Singular Boundary Value Conditions

Authors: Yan Debao
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

This paper focuses on a singular boundary value (SBV) problem of nonlinear fractional differential (NFD) equation defined as follows:D0+βυτ+fτ,υτ=0,τ∈0,1, υ0=υ′0=υ″0=υ″1=0, where 3<β≤4,D0+β is the standard Riemann–Liouville fractional (RLF) derivative. The nonlinear function fτ,υτ might be singular on the spatial and temporal variables. This paper proves that a positive solution to the SBV problem exists and is unique, taking advantage of Green’s function through a fixed-point (FP) theory on cones and mixed monotone operators.


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## Body

## 1. Introduction

Last few decades, significant progress has been achieved in solving NFD equations led by both the advance of fractional calculus and its applications to physics, mechanics, chemistry, economics, engineering, and other fields [1–8].The initial value problem and the boundary problem were studied in [9–21], respectively. The SBV problem was recently studied for NFD [22–33], mostly focused on investigating the positive solution. They were mainly based on nonlinear analysis techniques such as Leray–Schuader theory, FP, topological theories, and mixed monotone method [34–39].In [22], the presence of positive solutions for the SVB problem for NFD equation was investigated, which is defined as(1)CD0+αυτ+fτ,υτ=0,τ∈0,1,υ0=υ′0=υ″0=0,where 2<α≤3, DC0+α is the standard Caputo’s fractional derivative, and f:0,1×0,+∞⟶0,+∞ with limτ⟶0+fτ,⋅=+∞. Assuming some hypotheses on f, they gained positive solutions through the nonlinear Leray–Schauder-type alternative in a cone and the Guo–Krasnoselskii FP theory.Xu [28] investigated the following SBV problem for NFD:(2)D0+αυτ=fτ,υτ,τ∈0,1,υ0=υ1=υ′0=υ′1=0,where 3<α≤4, D0+α represents the standard RLF derivative, and f:0,1×0,+∞⟶0,+∞ with limυ⟶0+f⋅,υ=+∞. They obtained the presence of multiple solutions under the condition of fτ,υ=gυ+hυ and the uniqueness of solution for fτ,υ=qτgυ+hυ using the Guo–Krasnoselskii FP theory, mixed monotone scheme, and Leray–Schauder’s nonlinear alternative.Zhang and Zhong [31] investigated the boundary value problem of singular NFD written as(3)D0+αυτ+fτ,υτ=0,τ∈0,1,υ0=0,D0+βυ0=0,D0+βυ1=∑i=1nξiD0+βυη,where the function f permits singularities at τ=0,τ=1, and υ=0. The presence of multiple positive solutions for fτ,υ≤aτgυ+hυ was proved through the Guo–Krasnoselskii FP theory.In this paper, motivated by those works, the following SBV problem for NFD is investigated:(4)D0+βυτ+fτ,υτ=0,τ∈0,1,(5)υ0=υ′0=υ″0=υ″1=0,where 3<β≤4, D0+α is the standard RLF derivative, and fτ,υτ becomes singular when τ=0,τ=1, and υ=0. In analogy with other works, the corresponding Green’s NFD and its positive characteristics are inferred. As application, the presence of multiple positive solutions of (4) for fτ,υ≤a1τf1υ+a2τf2υ and the uniqueness of positive solution of (4) for fτ,υ=a1τf1υ+a2τf2υ are investigated, respectively.Unlike the existing studies, two novel features are induced in this paper. (1) To the best of our knowledge, boundary condition (5) is firstly considered for NFD. (2) Conditions (H2) and (H5) imposed on f are different from those in [26–34].The remainder of this paper is structured as follows. Preliminaries are given in Section2, including definitions, lemmas, the deduction of Green’s function for problem (4), and new positive properties. Section 3 proves the presence of positive solutions of (4) by the Guo–Krasnoselskii FP theory and demonstrates an example. Section 4 discusses the uniqueness of the positive solution of (4) by a mixed monotone operator and demonstrates another example.


## 2. Preliminaries

The lemmas and definitions from [3] are given for the convenience of the reader as follows:Definition 1 (see [3]).
The RLF integral of the orderβ>0 of a function fx:0,+∞⟶R is formulated:(6)I0+βfx=1Γβ∫0xx−sβ−1fsds,provided the right side is pointwise defined on 0,+∞.Definition 2 (see [3]).
The RLF derivative of the orderβ>0 of a continuous function fx:0,+∞⟶R is formulated as(7)D0+βfx=1Γn−βddxn∫0xfsx−sβ−n+1ds,where n=β+1 and ⋅ represents the integer part of number.Lemma 1 (see [3]).
The solution of the NFD equation defined as(8)D0+βμτ=0is μτ=C1τβ−1+C2τβ−2+⋯+CNτβ−N, Ci∈ℜ, i=1,2,…,N, where N is the smallest integer greater than or equal to β and μ∈C0,1∩L0,1.Lemma 2 (see [3]).
Under the assumption thatμ∈C0,1∩L0,1 with a NFD of order β, then(9)I0+βD0+βμτ=μτC1τβ−1+C2τβ−2+⋯+CMτα−M,where Ci∈ℜ, i=1,2,…,M, where M is the smallest integer greater than or equal to β.Lemma 3.
Provided thatκτ∈C0,1 and 3<β≤4,(10)D0+βμτ+κτ=0,0<τ<1,(11)μ0=μ′0=μ″0=μ″1=0.
The solution of (10) is unique and as follows:(12)μτ=∫01Gτ,sκsds,where Green’s function Gτ,s is denoted as(13)Gτ,s=1Γβτβ−11−sβ−3−τ−sβ−1,0≤s≤τ≤1,τβ−11−sβ−3,0≤τ≤s≤1.Proof.
(10) is rewritten as follows through Lemma 2:(14)μτ=−1Γβ∫0ττ−sβ−1κsds+C1τβ−1+C2τβ−2+C3τβ−3+C4τβ−4,where Ci∈ℜ, i=1,2,3,4. From the boundary conditions μ0=μ′0=μ″0=0, we have C2=C3=C4=0. Then,(15)μτ=−1Γβ∫0ττ−sβ−1κsds+C1τβ−1,μ′τ=−β−1Γβ∫0ττ−sβ−2κsds+C1β−1τβ−2,μ″τ=−β−1β−2Γβ∫0ττ−sβ−3κsds+C1β−1β−2τβ−3.
By the conditionu″1=0, we have(16)C1=1Γβ∫011−sβ−3κsds.
Accordingly, the unique solution of problem (10) and (11) is given as(17)μτ=−1Γβ∫0ττ−sβ−1κsds+τβ−1Γβ∫011−sβ−3κsds=∫01Gτ,sκsds.
Lemma3 is proved with this.Lemma 4.
The properties ofGτ,s defined by (13) are as follows:(1)
τβ−1s2−s1−sβ−3≤ΓβGτ,s≤s2−s1−sβ−3,0≤τ,s≤1(2)
τβ−1s1−sβ−2≤ΓβGτ,s≤τβ−11−sβ−3,0≤τ,s≤1(3)
Gτ,s>0,0<τ,s<1(4)
Gτ,s∈C0,1×0,1Proof.
Property (4) is obvious and (3) holds from (1). Thus, here (1) and (2) will be proved.(1)
Whens≤τ,(18)ΓβGτ,s=τβ−11−sβ−3−τ−sβ−1=τ2τβ−31−sβ−3−τ−sβ−1=τ2−τ−τs2τ−τsβ−3=τ2s2−sτ−τsβ−3=τβ−1s2−s1−sβ−3,ΓβGττ,s=β−1τβ−21−sβ−3−β−1τ−sβ−2,ΓβGτττ,s=β−1β−2τβ−31−sβ−3−β−1β−2τ−sβ−3=β−1β−2τ−τsβ−3−τ−sβ−3≥0.Obviously,ΓβGτττ,s is nondecreasing in τ, and thus, for 0≤s≤τ≤1, we have(19)ΓβGτ,s≤max0≤s≤τ≤1ΓβGτ,s=ΓβG1,s=1−sβ−3−1−sβ−1=1−1−s21−sβ−3=s1−s2−sβ−3.When0≤τ≤s≤1, note that 0≤s2−s≤1, we have(20)ΓβGτ,s=τβ−11−sβ−3≥τβ−1s2−s1−sβ−3,ΓβGτ,s=τβ−11−sβ−3≤sβ−11−sβ−3=s2⋅sβ−3⋅1−sβ−3≤s21−sβ−3.Note thats2−s−s2=2s−2s2=2s1−s≥0; then,(21)ΓβGτ,s≤s2−s1−sβ−3.From (18)–(21), we have the first conclusion in Lemma 4 which holds.(2)
Whens≤τ,(22)ΓβGτ,s=τβ−11−sβ−3−τ−sβ−1=τ2τβ−31−sβ−3−τ−sβ−1=τ2τ−τsβ−3−τ−sβ−1≥τ2τ−τsβ−3−τ−τsβ−1=τ2−τ−τs2τ−τsβ−3=τ2s2−sτ−τsβ−3=τβ−1s2−s1−sβ−3,(23)ΓβGτ,s=τβ−11−sβ−3−τ−sβ−1≤τβ−11−sβ−3.
Whenτ≤s,(24)ΓβGτ,s=τα−11−sβ−3.
Since0≤s2−s≤1 for 0≤s2−s≤1, so(25)ΓβGτ,s=τβ−11−sβ−3≥τβ−1s2−s1−sβ−3.
Also,(26)ΓβGτ,s=τβ−11−sβ−3.
From (22)–(26), we have the second conclusion in Lemma 4 which holds.
Now, we give the following definitions and lemmas (see [34–39]), which are essential in proving the results.Lemma 5.
For a Banach spaceΨ, let Λ⊂Ψ denote a normal cone in Ψ and Φ1andΦ2 denote open subsets of Ψ with θ∈Φ1⊂Φ2¯. Then, let a completely continuous operator F:Λ⟶Λ satisfy either Fρ≤ρ,ρ∈Λ∩∂Φ1,Fρ≥ρ,ρ∈Λ∩∂Φ2 or Fρ≥ρ,ρ∈Λ∩∂Φ1,Fρ≤ρ,ρ∈Λ∩∂Φ2.
Then,F has an FP in Λ∩Φ¯2/Φ1.
Letς∈Λ with ς≤1,ς≠θ and Qς=x∈Λ|x≠θ,there exists constantsm,M>0,such thatmς≤x≤Mς, where θ is the zero element in Λ.Definition 3.
A:Qς×Qς⟶Qς is a mixed monotone operator when satisfying the monotone condition of Ax,y in x,y∈Qς such that Ax1,y≤Ax2,y when x1≤x2 and Ax,y1≤Ax,y2 when y1≥y2. x∗∈Qς is an FP of A when Ax∗,x∗=x∗.Lemma 6.
The unique fixed pointx∗ofA exists when a constant γ∈0,1 satisfies(27)Aτx,1τy≥τγAx,y,∀x,y∈Qς,0<τ<1,where A is a mixed monotone operator.


## 3. Presence of Positive Solutions of SVB

The presence and multiplicity of positive solutions of (4) and (5) is investigated here. The nonlinear function fτ,x∈C0,1×0,+∞,0,+∞, f may be singular when τ=0,τ=1, and x=0.For a Banach spaceΨ=C0,1 with the maximum norm max0≤τ≤1μτ, let K∈Ψ denote a nonnegative cone defined as(28)K=μ∈Ψ|μτ≥τβ−1μ,τ∈0,1.The operatorT is defined as follows:(29)Tμτ=∫01Gτ,sfs,μsds,0≤τ≤1.Clearly,T:K∖θ⟶C0,1. Denote Br=μτ∈Ψ|μτ<r and(30)Kr=Λ∩Br=μ∈Ψ|τβ−1μ≤μτ<r.The following are assumed for later use:H1f∈C0,1×0,+∞,0,+∞.H2 There exist a1,a2∈C0,1,0,+∞ and f1,f2∈C0,+∞,0,+∞ satisfying(31)fτ,μ≤a1τf1μ+a2τf2μ,∀τ∈0,1,μ∈0,+∞,and for anyr>0,(32)∫01a1τf1rτ+a2τf2rτdt<+∞,wherefirτ=maxfiμ:rτβ−1≤μ≤r,i=1,2.H3 There exist R1>0,0<c1<1/2 and c1,d1⊂0,1 and a nonnegative function b1τ∈L10,1 with 0<2β−1<∫c1d1G1/2,sb1sds satisfying(33)fx,μ≥b1τμ,∀τ,μ∈c1,d1×R1,+∞.H4 There exist c2,d2⊂0,1 and a nonnegative function b2τ∈L10,1 with 0<∫01b2sds<+∞ satisfying(34)limμ⟶0+infminτ∈c2,d2fτ,μb2τ=+∞.Lemma 7.
For anyr>0,T:K∖Br⟶K is completely continuous.Proof.
For anyμ∈K/Br, we have τβ−1μ≤μτ≤μ. From H2 and (1) of Lemma 4(35)Tμτ=∫01Gτ,sfs,μsds≤∫01s2−s1−sβ−3fs,μsds≤∫01s2−s1−sβ−3a1sf1rs+a2sf2rsds<+∞,meaning that T is well defined.
And, by (1) of Lemma4,(36)Tμτ=∫01Gτ,sfs,μsds≥τβ−1∫01s2−s1−sβ−3fs,μsds≥τβ−1Tμτ.
So,T maps K,Br into K.
For a bounded setD∈K,Br, a number R exists such that R>r and r≤μ≤R for any μ∈D. So, we have(37)Tμτ≤∫01s2−s1−sβ−3a1sf1rs+a2sf2rsds<+∞,which means TD is uniformly bounded. Gτ,s is uniformly continuous on 0,1×0,1. Accordingly, δ>0 exists for any ε>0, such thatτ′−τ″<δ,s∈0,1, for τ′,τ″∈0,1,(38)Gτ′,s−Gτ″,s<ε1+∫01a1sf1rs+a2sf2rsds.
Consequently,(39)Tμτ′−Tμμ″≤∫01Gτ′s−Gτ″,sfs,μsds≤∫01Gτ′,s−Gτ″,sa1sf1rs+a2sf2rsds≤∫01ε1+∫01a1sf1rs+a2sf2rsdsa1sf1rs+a2sf2rsds<ε,implying TD is equicontinuous. According to the Arzela–Ascoli theorem, T:K∖Br⟶K is compact.
In the following, the continuity ofT will be proved. Given μn⊂K∖Br and μn−μ0⟶0n⟶∞, R>r satisfies(40)r≤μn<R,n=1,2,….
Forε>0, δ∈0,1/2 satisfies through the absolute continuity of integral,(41)∫0δs2−s1−sβ−3a1sf1rs+a2sf2rsds<ε3,∫1−δ1s2−s1−sβ−3a1sf1rs+a2sf2rsds<ε3.
Due to the uniform continuity offτ,μ on δ,1−δ×rτβ−1,r and μn−μ0⟶0n⟶∞,n>N exists. Then,(42)fτ,μnτ−fτ,μ0τ<ε3∫01s2−s1−sβ−3ds,τ∈δ,1−δ.
Then,(43)Tμn−Tμ0≤max0≤t≤1∫01Gτ,sfs,μns−fs,μ0sds≤∫01s2−s1−sβ−3fs,μns−fs,μ0sds≤2∫0δs2−s1−sβ−3a1sf1rs+a2sf2rsds+∫δ1−δs2−s1−sβ−3fs,μns−fs,μ0sds+2∫1−δ1s2−s1−sβ−3a1sf1rs+a2sf2rsds≤ε3+ε3+ε3=ε.
So,T is continuous. The proof is finished.Theorem 1.
Under the conditionsH1−H4 and the existence of a positive constant r>0 and λ>β>0 such that(44)∫01a1sf1rs+a2sf2rsds<2λr,at least two positive solutions μ1 and μ2 of (4) exists with 0<μ1<2λr<μ2.Proof.
According to Lemma7, the complete continuity of T is valid from K∖Br into K for any r>0. Then, the existence of two fixed points μ1 and μ2 with 0<μ1<2λr<μ2 is proved here. For any r>0 and condition H3, we choose(45)r1>max2λ−1R1,2λr.
Whenμ∈K and μ=r1, we have(46)μτ≥τβ−1r1≥12β−1r1>R1,∀τ∈12,1.
Thus, we get fromH3 and (45) and (46):(47)Tμ12=∫01G12,sfs,μsds≥∫c1d1G12,sfs,μsds≥∫c1d1G12,sb1sμsds≥∫c1d1G12,sb1sds⋅12β−1r1>2β−112β−1r1=r1.
Therefore,(48)Tμ=maxτ∈0,1Tμτ≥Tμ12≥r1=μ,∀μ∈K,μ=r1.
The conditionH4 guarantees that, for M=r∫c2d2G1/2,sb2sds−1, there exists R2∈0,1 satisfying(49)fτ,μ>Mb2τ,∀τ,μ∈c2,d2×0,R2.
Choose(50)0<r2<minr,R2.
Whenμ∈K and μ=r2, we have(51)Tμ12=∫01G12,sfs,μsds≥r∫c2d2G12,sb2sds−1∫c2d2G12,sb2sds=r>r2.
Therefore,(52)Tμ=maxt∈0,1Tμτ≥Tμ12≥r2=μ,∀μ∈K,μ=r2.
Forμ∈K, where μ=2λr, H2, (1) of Lemma 4, and 0<s2−s1−sβ−3<1, it is similar to (35) and (44), one can get(53)Tμτ=∫01Gτ,sfs,μsds≤∫01s2−s1−sβ−3fs,μsds≤∫01s2−s1−sβ−3a1sf1rs+a2sf2rsds≤∫01a1sf1rs+a2sf2rsds<2λr.
Thus, we have(54)Tμ≤μ,∀μ∈K,μ=2λ.
From Lemma5, (48), (52), and (54), two fixed points μ1,μ2 of T satisfy(55)0<r2≤μ1<2λr<μ2≤r1.
The proof is completed.Example 1.
Consider the following SBV problem:(56)D0+7/2μτ+τ22μ−1+33Γ7/23−2τ−1/21−τ−1/2μ=0,0<τ<1,μ0=μ′0=μ″0=μ″1=0.β=7/2 and fτ,μ=τ2/2μ−1+33Γ7/2/3−2τ−1/21−τ−1/2μ. We set f1μ=μ−1, a1τ=τ2/2, f2μ=μ, and a2τ=33Γ7/2/3−2τ−1/21−τ−1/2. By simple computation, we have f1r=r−1τ−5/2,f2r=r, and(57)∫01a1sf1rs+a2sf2rsds=r−1+33Γ7/23−2rπ<+∞.So, the conditionsH1andH2 hold.Next, we setb1τ=a2τ,b2τ=a1τ,c1,d1=c2,d2=1/4,3/4. Then, it is obviously fτ,μ≥b1τμ,∀τ,μ∈c1,d1×R1,+∞ for any R1>0. By simple computation, we have ∫c1d1G1/2,sb1sds=33⋅2−5/2>27/2−1. And limμ⟶0+infminτ∈1/4,3/4fτ,μ/b2τ=+∞. So, the conditions H3andH4 also hold. Taking r=1,λ=10.25, we have by (57)(58)∫01a1sf1rs+a2sf2rsds≈1084.4<210.25⋅1.Consequently, condition (44) holds. Then, from Theorem 1 in Example 1 at least two positive solutions μ1 and μ2 exist with 0<μ1<2λr<μ2.


## 4. Uniqueness of Singular Problem Solution

By property (2) of Green’s function,(59)ςτps≤ΓβGτ,s≤ςτqs,τ,s∈0,1,where ςτ=τβ−1,ps=s1−sβ−2,andqs=1−sβ−3.From Lemma3 and (59), a solution of (4) μτ is(60)μτ=∫01Gτ,sfs,μsds,τ∈0,1,ςτ∫01psΓβfs,μsds≤μτ≤ςτ∫01qsΓβfs,μsds.Then,μτ∈Qς.For a Banach spaceΨ,Λ is a normal cone in Ψ when Λ=ρτ∈C0,1:ρτ≥0,∀τ∈0,1.Throughout this section, the following is assumed:(61)H5fτ,μ=a1tf1μ+a2tf2μ,τ∈0,1,where f1:0,+∞⟶0,+∞ is nondecreasing and continuous, f2:0,+∞⟶0,+∞ is nonincreasing and continuous, and a1τ,a2τ∈C0,1,0,+∞.Theorem 2.
A unique positive solutionμ∗ of (4) exists with ∀γ∈0,1 satisfying(62)f1τμ≥τγf1μ,(63)f2τ−1μ≥tγf2μ,where τ∈0,1 and μ>0, and ∀a1τ,a2τ satisfy(64)∫01qsς−γsa1sf11+a2sf21ds<+∞.Proof.
Taking the similar process in [34], for τ∈0,1,μ>0,(65)f2τμ≤τ−γf2μ,(66)f2τ≤τ−γf21,(67)f2τ−1≥τγf21,(68)f1τ≥τγf11,(69)f1μ≤μγf11,μ>1.
Define(70)Qς=μ∈Λ|1Mςτ≤μτ≤Mςτ,τ∈0,1,where ςτ=τβ−1, and M>1 is defined as(71)M>maxξ∫011Γβqsς−γsa1sf11+a2sf21ds1/1−γ,ξ∫011Γβpsςγsa1sf11+a2sf21ds−1/1−γ.
Forμ,ν∈Qς, we define(72)Tξμ,ντ=ξ∫01Gτ,sa1sf1μs+a2sf2νsds,∀τ∈0,1.
We firstly show thatTξ:Qς×Qς⟶Qς. Let μ,ν∈Qς, then from (69),(73)f1μτ≤f1Mςτ≤f1M≤Mγf11,∀τ∈0,1,and from (65) and (66), we have(74)f2ντ≤f21Mςτ≤f21Mς−γτ≤Mγf11ς−γτf21,∀t∈0,1.
So, we have(75)Tξμ,ντ≤ξ∫011Γβςτqsa1sf1μs+a2sf2νsds≤Mγςτξ∫011Γβqsa1sf11+a2sς−γsf21ds≤Mγςτξ∫011Γβqsς−γsa1sf11+a2sf21ds≤MγςτM1−γ=Mςτ,∀τ∈0,1.
And, from (62), (67), and (68),(76)f1μt≥f11Mςτ≥f11Mςγτ≥M−γf11ςγτf11,∀τ∈0,1,f2νt≥f2Mςτ≥f2M≥M−γf21,∀τ∈0,1,for all μ,ν∈Qς. So,(77)Tξμ,ντ≥ξ∫011Γβςτpsa1sf1μs+a2sf2νsds≥M−γςτξ∫011Γβpsa1sςγsf11+a2sf21ds≥M−γςτξ∫011Γβpsςγsa1sf11+a2sf21ds≥M−γςτM−1−γ=M−1ςτ,∀τ∈0,1.
It is easy to check thatTξμ,ν is nondecreasing in μ and nonincreasing in ν.
Next, for anyσ∈0,1 and μ,ν∈Qς, we have(78)Tξσμ,σ−1ντ=ξ∫01Gτ,sa1sf1σμs+a2sf2σ−1νsds≥ξ∫01Gτ,sa1sσγf1μs+a2sσγf2νsds=σγTξμ,ντ,∀τ∈0,1.
Lemma6 accordingly holds. Also, a unique μ∗∈Qς satisfying Tξμ∗,μ∗=μ∗ exists. Theorem 2 is proved.Example 2.
The following example for SBV is considered here:(79)D0+7/2μτ+1τ9μ1/3+τμ−1/4=0,0<τ<1,μ0=μ′0=μ″0=μ″1=0.
We letf1μτ=μ1/3τ,a1τ=1/τ9,f2μτ=μ−1/4τ,a2τ=τ,andγ=1/3. Accordingly,(80)f1τμ=τ1/3μ1/3≥τγf1μ,f2τ−1μ=τ1/4μ−1/4≥τγf2μ,∫01qsς−γsa1sf11+a2sf21ds=∫011−s7/2−3s−7/2−1⋅1/3s−1/9+sds=18+67<+∞.
So, Theorem2 is validated, indicating the presence of a unique positive solution μ∗.



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