# On Multiple Positive Solutions for Singular Fractional Boundary Value Problems with Riemann-Stieltjes Integrals

Authors: Lufeng Gu; Qiuyan Zhong; Zhuyan Shao
Journal: Journal of Function Spaces
 (2023)

---

## Abstract

In this paper, the existence result of at least two positive solutions is obtained for a nonlinear Riemann-Liouville fractional differential equation subject to nonlocal boundary conditions, where fractional derivatives and Riemann-Stieltjes integrals are involved. The nonlinearity possesses singularities on both its time and space variables. The discussion is based on the fixed point index theory on cones.


---

## Body

## 1. Introduction

We consider the existence of at least two positive solutions for the following nonlinear fractional differential equation with integral boundary value conditions(1)D0+αut+ft,ut=0,0<t<1,u0=u′0=⋯=un−20=0,D0+β0u1=∑i=1m∫01D0+βiutdHit,whereα∈ℝ,α∈n−1,n,n,m∈ℕ,n≥3,βi∈ℝ for all i=0,1,⋯,m,0≤β1<β2<⋯<βm≤β0<α−1,β0≥1,f∈C0,1×0,+∞,0,+∞.D0+μ represents the Riemann-Liouville derivative of order μ (for μ=α,β0,β1,⋯,βm). The integrals involved in the boundary conditions are Riemann-Stieltjes integrals; here, Hii=1,2,⋯,m are functions of bounded variation. For more general integral conditions, see [1] and the references therein. The nonlinearity f permits singularities at both t=0,1 and u=0.In recent years, there has been a gradual increase in the investigation of fractional differential equations and systems of fractional differential equations with nonlocal boundary value conditions owing to their better descriptions in very important phenomena in science and technology than in integers. For fractional calculus and its applications in nonlocal problems, see monographs [2–6] and papers [7–25] and the references therein. Very recently, by means of the fixed point theory, principal characteristic value, and fixed point theorems together with height functions, Tudorache et al. [7, 9] investigated the existence of one, two, or three positive solutions for BVP (1). Existence results can be found in [15–19] for the system of fractional differential equations with boundary conditions related to BVP (1).As is well known, Riemann-Stieltjes integral boundary conditions are more general, and they include many special cases such as two-point, three-point, and other classical integral conditions or a combination of them. As a consequence, boundary conditions in BVP (1) are generalizations of those adopted in the literature [10–14], which can be listed below:
(2)u0=u′0=⋯=un−20=0,D0+pu1=∑i=1maiD0+qu1,u0=u′0=⋯=un−20=0,D0+βu1=λ∫0ηD0+βhtutdt,u0=u′0=⋯=un−20=0,D0+pu1=∫01D0+qutdHt,u0=u′0=⋯=un−20=0,D0+βu1=λ∫0ηD0+γutdHt.We make an effort in this paper to investigate the existence of multiple positive solutions for BVP (1). By a positive solution of BVP (1), we mean a function u∈C0,1 satisfying BVP (1) with ut>0 for all t∈0,1. This paper admits the following features. Firstly, compared with [16–18], the nonlinearity f in this paper possesses singularities not only on the time but also on the space variables. Secondly, compared with [7–9], super linear conditions on the nonlinearity at 0 and ∞ are imposed to obtain the existence of at least two positive solutions. Thirdly, conditions given in this paper are shown to be easy to verify by an example. The main tools employed in this paper are the cone theory and fixed point index theorems on cones.


## 2. Preliminaries and Several Lemmas

First, we introduce some useful lemmas from [7, 9] which will be used in the latter. For notational convenience, denote
(3)Δ=ΓαΓα−β0−∑i=1mΓαΓα−βi∫01sα−βi−1dHis.Consider the fractional differential equation(4)D0+αut+xt=0,0<t<1,u0=u′0=⋯=un−20=0,D0+β0u1=∑i=1m∫01D0+βiutdHit,where x∈C0,1∩L10,1.Lemma 1 (see [7, 9]).
IfΔ≠0, then the unique solution u∈C0,1 of problems (4) is given by
(5)ut=∫01Gt,sxsds,t∈0,1,where
(6)Gt,s=g1t,s+tα−1Δ∑i=1m∫01g2iτ,sdHiτ,(7)g1t,s=1Γαtα−11−sα−β0−1−t−sα−1,0≤s≤t≤1,tα−11−sα−β0−1,0≤t≤s≤1,(8)g2it,s=1Γα−βitα−βi−11−sα−β0−1−t−sα−βi−1,0≤s≤t≤1,tα−βi−11−sα−β0−1,0≤t≤s≤1,for all t,s∈0,1×0,1,i=1,2,⋯,m.Lemma 2 (see [7, 9]).
We suppose thatΔ>0. Then, the Green function G given by (6) is a continuous function on 0,1×0,1 and satisfies the following inequalities:
(i)
Gt,s≤Js for all t,s∈0,1, where(9)Js=h1s+1Δ∑i=1m∫01g2iτ,sdHiτ,s∈0,1,h1s=1Γα1−sα−β0−11−1−sβ0,s∈0,1(ii)
Gt,s≥tα−1Js for all t,s∈0,1(iii)
Gt,s≤σtα−1 for all t,s∈0,1, where(10)σ=1Γα+1Δ∑i=1m1Γα−βi∫01τα−βi−1dHiτ.
We make the following assumptions:
(H1) α∈ℝ,α∈n−1,n,n,m∈ℕ,n≥3,βi∈ℝ for all i=0,1,⋯,m,0≤β1<β2<⋯<βm≤β0<α−1,β0≥1,Hi:0,1⟶ℝi=1,2,⋯,m are nondecreasing functions and Δ>0.
(H2) f∈C0,1×0,+∞,0,+∞.
(H3) There exist a,b∈C0,1,0,+∞∩L10,1,g∈C0,+∞,0,+∞ such that
(11)ft,u≤atgu+bt,∀t∈0,1,u∈0,+∞,apq~∗=∫01atgpqtdt<+∞,b∗~=∫01btdt<+∞,for any q≥p>0, where
(12)gpqt=maxgu:tα−1p≤u≤q.
(H4) There exists c∈C0,1,0,+∞ such that
(13)ft,uctu⟶+∞asu⟶+∞,uniformly for t∈0,1, and
(14)c∗=∫01ctdt<+∞.
(H5) There exists d∈C0,1,0,+∞ such that
(15)ft,udt⟶+∞asu⟶0+,uniformly for t∈0,1, and
(16)d∗=∫01dtdt<+∞.
In addition, considering the boundedness ofJt,t∈0,1, it is easy to know that
(17)apq∗=∫01Jtatgpqtdt<+∞,b∗=∫01Jtbtdt<+∞,ap∗=∫01Jtatgpptdt<+∞,where
(18)gppt=maxgu:tα−1p≤u≤p,forallp>0.
LetE=C0,1 be the traditional Banach space of all continuous functions defined on 0,1 with the maximum norm u=max0≤t≤1ut and P the cone
(19)P=u∈E:ut≥tα−1u,t∈0,1.
DenotePpq=u∈P:p≤u≤q,Pr=u∈P:u≤r,∂Pr=u∈P:u=r for q>p>0,r>0.
Define the operatorT as follows:
(20)Tut=∫01Gt,sfs,usds,0≤t≤1.Clearly, T:P\0⟶C0,1.Lemma 3.
Suppose that (H1)–(H3) hold; then, for any q>p>0, T:Ppq⟶P is completely continuous.Proof.
For anyu∈Ppq, we have p≤u≤q. It follows from the definition of cone P that
(21)tα−1p≤ut≤q,∀t∈0,1.
By (H2), (H3), and Lemma 2, we get that
(22)ft,ut≤atgpqt+bt,∀t∈0,1,u∈Ppq,(23)Tut=∫01Gt,sfs,usds≤∫01Jsfs,usds≤∫01Jsasgpqs+bsds=apq∗+b∗,∀t∈0,1,which means that T is well defined. For any t∈0,1, we have from (23) that
(24)Tut=∫01Gt,sfs,usds≤∫01Jsfs,usds.
Therefore,(25)Tu≤∫01Jsfs,usds.
On the other hand, it follows from Lemma2 and (25) that
(26)Tut=∫01Gt,sfs,usds≥tα−1∫01Jsfs,usds≥tα−1Tu,∀t∈0,1.
Thus, we have proven thatT maps Ppq into P.
In the following, we are in the position to show thatT is completely continuous. First, we prove that T is continuous. For un,u¯∈Ppq with un−u¯⟶0n⟶∞, we have limn⟶∞unt=u¯t,t∈0,1. By (H1), we know
(27)limn⟶∞ft,unt=ft,u¯t,0<t<1.
Similar to (22), for un,u¯∈Ppq, we have
(28)ft,unt≤atgpqt+bt,ft,u¯t≤atgpqt+bt,∀t∈0,1.
Thus,(29)Gt,sfs,uns−Gt,sfs,u¯s≤2Jsasgpqs+bs=σs∈L10,1.
It follows from (27), (29), (H3), and the Lebesgue-dominated convergence theorem that
(30)limn⟶∞Tun−Tu¯≤limn⟶∞maxt∈0,1∫01Gt,sfs,uns−Gt,sfs,u¯sds=0,which means that T is continuous.
Next, we will show thatT is a compact operator. Let V be a bounded set in Ppq. For any u∈V, we have p≤u≤q. Similar to (23), we know
(31)Tut≤apq∗+b∗,∀t∈0,1,u∈Ppq,which means that TV is bounded uniformly. In the following, we shall prove that TV is equicontinuous. To this end, we estimate Tu′ for u∈V.
(32)Tu′t=∫01∂Gt,s∂tfs,usds=∫01∂g1t,s∂t+α−1tα−2Δ∑i=1m∫01g2iτ,sdHiτfs,usds≤α−1∫011Γαtα−21−sα−β0−1+t−sα−2+tα−2Δ∑i=1m1Γα−βiτα−βi−11−sα−β0−1dHiτfs,usds≤α−1∫012Γα+1Δ∑i=1m1Γα−βiHi1−Hi0asgpqs+bsds≤α−12Γα+1Δ∑i=1m1Γα−βiHi1−Hi0apq~∗+b∗~=ΔF,∀t∈0,1.
Thus, for any0≤t1≤t2≤1 and u∈V, one has
(33)Tut1−Tut2=∫t1t2Tnu′sds≤Ft1−t2.
Thus,TV is equicontinuous. It follows from the Arzelà-Ascoli theorem that TV is relatively compact, and then, T is a compact operator. Hence, T:Ppq⟶P is completely continuous.Lemma 4 (see [26]).
LetE be a Banach space and P⊂E a cone in E. Assume that T:Pr⟶P is a compact map such that Tu≠u for u∈∂Pr,
(i)
Ifu≤Tu,∀u∈∂Pr, then(34)iT,Pr,P=0(ii)
Ifu≥Tu,∀u∈∂Pr, then(35)iT,Pr,P=1.


## 3. Main Result

Theorem 5.
Assume that (H1)–(H5) hold. In addition, there exists R0>0 such that
(36)aR0∗+b∗<R0.
Then, the BVP (1) has at least two positive solutions u∗ and u∗∗ with 0<u∗<R0<u∗∗.Proof.
It follows from Lemma3 that for any q>p>0, the operator T:Ppq⟶P is completely continuous. In the following, we shall prove that T has two different fixed points u∗ and u∗∗ in P satisfying 0<u∗<R0<u∗∗.
Chooseθ∈0,1/2. We know from (H4) that there exists r1>0 such that
(37)ft,u≥θ1−α∫θ1−θG1/2,scsds−1ctu,∀t∈0,1,u≥r1.
Let(38)R1>maxθ1−αr1,R0.
Foru∈∂PR1, we have, by the construction of cone P, that
(39)ut≥tα−1R1≥θα−1R1>r1,∀t∈θ,1−θ.
It follows from (37) to (39) that
(40)Tu12=∫01G12,sfs,usds>θ1−α∫θ1−θG12,scsds−1∫θ1−θG12,scsusds≥θ1−α∫θ1−θG12,scsds−1∫θ1−θG12,scsds·θα−1R1=R1.
Thus,(41)Tu=maxt∈0,1Tut≥Tu12>R1=u,∀u∈∂PR1.
Hence, by Lemma4,
(42)iT,PR1,P=0.
By condition (H4), there exists r2>0 such that
(43)ft,u≥∫θ1−θG12,sdsds−1dtR0,∀t∈0,1,0<u<r2.
Choose(44)0<R2<minr2,R0.
Foru∈PR2, we have
(45)0<R2tα−1≤ut≤u=R2<r2,∀t∈0,1.
Consequently, we have from (43) to (45) and (H5) that
(46)Tu12=∫01G12,sfs,usds≥∫θ1−θG12,sdsds−1∫θ1−θG12,sdsR0ds≥∫θ1−θG12,sdsds−1∫θ1−θG12,sdsds·R0=R0>R2.
Thus,(47)Tu=maxt∈0,1Tut≥Tu12>R2=u,∀u∈PR2.
As a consequence, we get(48)iT,PR_2,P=0.
On the other hand, foru∈∂PR0, by (H3), Lemma 2, and (36), we get
(49)Tut=∫01Gt,sfs,usds≤∫01Jsfs,usds≤∫01JsasgR0R0s+bsds≤aR0∗+b∗<R0,∀t∈0,1,
i.e.,(50)Tu≤u,u∈∂PR0.
Then, Lemma4 guarantees that
(51)iT,PR0,P=1.
It follows from (42), (48), (51) and the additivity of the fixed point index that
(52)iT,PR1\PR0∘,P=−1,iT,PR0\PR2∘,P=1.
Hence,T has two distinct fixed points u∗ and u∗∗ belonging to PR0\PR2∘ and PR1\PR0∘, respectively, with 0<R2<u∗<R0<u∗∗≤R1.


## 4. An Example

Example 1.
Consider the following fractional differential equations with nonlocal boundary value problems(53)D0+11/3ut+175t21−t45u3+u−1/7+125t1−t34=0,0<t<1,u0=u′0=u″0=0,D0+13/6u1=∫01D0+2/3utdt+D0+5/3u12.
Conclusion: BVP (53) has at least two positive solutions u∗ and u∗∗ with 0<u∗<5<u∗∗.Proof.
In this problem,α=11/3,n=4,m=2,β0=13/6,β1=2/3,β2=5/3,H1t=t for all t∈0,1, H2t=0fort∈0,1/2;1fort∈1/2,1. By simple computation, we have Δ=1.852483495372207>0. It is clear that (H1) and (H2) are satisfied. Furthermore,
(54)h1s=1Γ11/31−s1/21−1−s13/6,s∈0,1,(55)g21t,s=1Γ3t21−s1/2−t−s2,0≤s≤t≤1,t21−s1/2,0≤t≤s≤1,(56)g22t,s=1Γ2t1−s1/2−t−s,0≤s≤t≤1,t1−s1/2,0≤t≤s≤1,(57)Js=h1s+1Δ1Γ41−s1/2−1Γ41−s3+1Γ212·1−s1/2−12−s,0≤s≤12,h1s+1Δ1Γ41−s1/2−1Γ41−s3+1Γ2·12·1−s1/2,12<s≤1.
For anyr>0, (H3) holds for at=1/75t21−t45,gu=u3+u−1/7,bt=1/25t1−t34, and
(58)apq~∗=∫01atgpqtdt=∫01175t21−t45q3+1t8/3p7dt=175∫01t−2/51−t−4/5dt·q3+∫01t−82/1051−t−4/5dt·p−1/7=175B35,15q3+B23105,15p−1/7<+∞.
b~∗=∫011/25t1−t34dt=1/25B3/4,1/4≈0.177715317526335. Thus, (H3) is verified. Obviously, (H4) and (H5) are valid for ct=dt=1/75t21−t45,c∗=d∗≈0.078296677374705.
Next, we focus on checking (36). Take R0=5. By (57), we know that
(59)b∗=∫01Jsbsds≤∫01h1s+1Δ1Γ41−s1/2+1Γ31−s1/2·125s1−s34ds≤∫011Γα+1Δ1Γ41−s1/2+1Γ31−s1/2·125s1−s34ds≤125·1Γ11/3·∫01s−1/41−s−3/4ds+1Δ1Γ4+1Γ3·∫01s−1/41−s−1/4ds=125·1Γ11/3B34,14+11.8524834953722071Γ4+1Γ3·B34,34≈1250.249239737672306·4.442882938158366+0.539815875552012·16+12·1.694426169587958≈0.068685136355131,aR0∗=∫01JsasgR0ds≤175∫011Γα+1Δ1Γ4+1Γ31−s1/2·1s21−s45R03+1s8/3R07ds≤175·1Γ11/3·B35,15·R03+B23105,15·R0−1/7+11.852483495372207·1Γ4+1Γ3·B35,710·R03+B23105,710·R0−1/7≈175·0.249239737672306·5.872250803102905·53+9.049460301355431·5−1/7+0.539815875552012·16+12·2.153890871161322·53+5.136337054837259·5−1/7≈3.774703981617563.
Hence,(60)aR0∗+b∗=3.774703981617563+0.068685136355131=3.843389117972694<5=R0,which implies that (36) holds. Consequently, our conclusion follows from Theorem 5.


## 5. Conclusions

In this paper, we focus on the existence and multiplicity of positive solutions for a class of a higher-order Riemann-Liouville fractional differential equation with Riemann-Stieltjes integrals. The nonlinearity possesses singularities on both its time and space variables. By means of the fixed point index theory on cones, the existence result of at least two positive solutions is obtained. Conditions imposed on the nonlinearity are shown to be easy to verify by an example.



---

*Source: 6154626-2023-02-03.xml*