# Application of Fixed-Point Index Theory for a Nonlinear Fractional Boundary Value Problem with an Advanced Argument

Authors: Li Wu; Chuanzhi Bai
Journal: Discrete Dynamics in Nature and Society
 (2021)

---

## Abstract

In this paper, we investigate the existence of positive solutions of a class of fractional three-point boundary value problem with an advanced argument by using fixed-point index theory. Our results improve and extend some known results in the literature. Two examples are given to demonstrate the effectiveness of our results.


---

## Body

## 1. Introduction

Three-point boundary value problems arise in a variety of different applied mathematics and physics. In [1], Gupta studied the solvability of three-point BVPs for nonlinear second-order ordinary differential equations. Since then, many authors have investigated the existence and multiplicity of solutions for three-point boundary value problems for nonlinear integer-order ordinary differential equations and nonlinear fractional-order differential equations, see [2–19] and the references therein.In [2], Ma considered the following three-point boundary value problem:(1)u″t+atfu=0,0<t<1,u0=0,u1=αuη,where α>0, η∈0,1, αη<1, a∈C0,1,0,∞, and f∈C0,∞,0,∞ is suplinear or sublinear. The existence of at least one positive solution for BVP (1) is obtained by virtue of the Krasnosel’skii fixed-point theorem.In [3], due to the well-known Guo–Krasnoselskii fixed-point theorem [20], Wang et al. proved the existence of a positive solution to the following three-point BVPs of nonlinear fractional-order differential equation with an advanced argument:(2)CDαut+atfuθt=0,0<t<1,u0=u″0=0,u1=βuη,where 2<α≤3, η∈0,1, 0<β<1/η, DαC is the Caputo fractional derivative, and f:0,∞⟶0,∞ is a continuous function.Recently, in [4], Pourhadi et al. considered the following three-point boundary value problem:(3)CDαxt=ft,xt,x′t,a<t<b,1<α<2,xa=0,xb=μxη,a<η<b,μ>λ,where λ=b−a/η−a, DαaC stands for the Caputo fractional derivative, and f:a,b×ℝ×ℝ⟶ℝ is a continuous function. The existence of positive solutions for BVPs (3) is given by using the inequalities related with Green’s function and two well-known fixed-point theorems.Motivated by the mentioned works, in this paper, we consider the following fractional-order three-point boundary value problem:(4)CDαut+atfuθt=0,0<t<1,u0−βu′0=u″0=0,u1=γuη,where 2<α≤3, 0<η<1, β≥0, 0<γ<1+β/β+η, DαC is the Caputo fractional derivative, a∈C0,1,0,∞, and f∈C10,∞,0,∞.Based on the fixed-point index theory, the existence of at least one positive solution to BVP (4) has been obtained. The main results of the paper by Wang et al. [3] have been extended and improved.Remark 1.
Ifβ=0, then the fractional differential equations under boundary value conditions (4) reduces to the BVPs (2).
The paper is organized in this order. In Section2, we present basic definitions on fractional calculus theory and some lemmas which play an important role in the proof of our main results. In Section 3, under an adequate condition imposed on f, we prove the existence of positive solutions of BVPs (4). In Section 4, two examples are given to demonstrate our results.


## 2. Preliminaries

For convenience, some fundamental definitions on fractional calculus are given. For more details, we refer to [21].Definition 1.
Letα≥0 and f be a real function defined on a,b. The Riemann–Liouville fractional integral of order α is defined by I0f≡f and(5)Iαft=1Γα∫0tt−sα−1fsds,α>0,t∈0,1.Definition 2.
The Caputo derivative of fractional orderα≥0 is defined by D0Cf≡f and(6)DαCft=1Γn−α∫0tt−sn−α−1fnsds,α>0,t∈0,1,where n is the smallest integer greater or equal to α. In the following, we state some properties of fractional operators that will be used in proving our main results.Lemma 1 (see [21]).
Letα,β∈ℝ+ and u∈L10,1; then, IαIβux=Iα+βux holds almost everywhere on 0,1.Lemma 2 (see [21]).
Letα∈ℝ+ and u∈C0,1; then, DαCIαux=ux.Lemma 3 (see [21]).
Ifα>0, n is the smallest integer greater than or equal to α, and u∈ACn0,1, then(7)IαDαCux=ux+c0+c1t+⋯+cn−1tn−1,for some ci∈ℝ, i=0,1,…,n−1.Lemma 4.
Let2<α≤3 and g∈C10,1. A function u is a solution of the following boundary value problem:(8)DαCut+gt=0,0<t<1,(9)u0−βu′0=u″0=0,u1=γuη,if and only if u satisfies the integral equation(10)ut=∫01Gt,sgsds,where(11)Gt,s=t+β1−sα−1−t+βγη−sα−1−t−sα−11+β−β+γηΓα1+β−β+ηγ,0≤s≤t≤1,s≤η,t+β1−sα−1−t−sα−11+β−β+γηΓα1+β−β+ηγ,0<η≤s≤t≤1,t+β1−sα−1−t+βγη−sα−1Γα1+β−β+ηγ,0≤t≤s≤η<1,t+β1−sα−1Γα1+β−β+ηγ,0≤t≤s≤1,η≤s,is the Green function of BVP (8)-(9).Proof.
In view of Lemma3, we may reduce (8) to an equivalent integral equation:(12)ut=c1+c2t+c3t2−∫0tt−sα−1Γαgsds.
According to Lemmas1 and 2, we obtain(13)u′t=c2+2c3t−∫0tt−sα−2Γα−1gsds,u″t=2c3−∫0tt−sα−3Γα−2gsds.
Fromu″0=0, we know that c3=0. By u0−βu′0=0, it follows that c1=βc2. Using the boundary condition u1=γuη, we have(14)c2=−11+β−β+ηγ∫011−sα−1Γαgsds+γ1+β−β+ηγ∫0ηη−sα−1Γαgsds.
Therefore, the solution of BVP (8)-(9) is(15)ut=−∫0tt−sα−1Γαgsds+t+β1+β−β+ηγ∫011−sα−1Γαgsds−γt+β1+β−β+ηγ∫0ηη−sα−1Γαgsds=∫01Gt,sgsds,where the Green function Gt,s is defined by (11).Remark 2.
For Caputo-type fractional problems, Cichon and Salem in [22] studied the problem of the lack of equivalent between differential and integral forms and pointed out that many papers contain an error in the proof of the equivalent of the fractional-type differential problems and the corresponding integral forms. If g∈C10,1, then g is absolutely continuous on 0,1, and we know from [22] that (8) and (12) are equivalent.Lemma 5.
The Green functionGt,s, defined as in Lemma 4, has the following properties:(i)
Gt,s is a continuous function.(ii)(16)Gt,s≥Ht,s=t+βsγη2−αα−11−ηη−sα−2Γα1+β−β+ηγ,0≤s≤t≤1,s≤η,t+βγη1−sα−1Γα1+β−β+ηγ,0<η≤s≤t≤1,t+β1−γη1−sα−1Γα1+β−β+ηγ,0≤t≤s≤η<1,t+β1−sα−1Γα1+β−β+ηγ,0≤t≤s≤1,η≤s.Proof.
It is obvious that (i) holds by the expression ofGt,s. Now, we prove that (ii) holds. When t=0, G0,s=0. When t≠0, we show three different cases.Case 1.
For0≤s≤t≤1, s<η, let(17)g1t,s=t+β1−sα−1−t+βγη−sα−1−t−sα−11+β−β+ηγ.
Then, by Remark3,(18)g1t,s=t+β1−sα−1−t+βγηα−11−sηα−1−tα−11−stα−11+β−β+ηγ≥t+β1−sα−1−t+βγηα−11−sηα−1−tα−11−sα−11+β−β+ηγ≥t+β1−sα−1−t+βγη1−sηα−1−t1−sα−11+β−β+ηγ=1−sα−1t+β−t1+β−β+ηγ−t+βγη1−sηα−1=1−sα−1tηγ+β1−t+tγ−t+βγη1−sηα−1≥1−sα−1tηγ+βγη−t+βγη1−sηα−1=t+βγη1−sα−1−1−sηα−1=t+βγηα−1∫1−s/η1−sxα−2dx≥t+βγηα−11−sηα−21−s−1−sη=t+βsγη2−αα−11−ηη−sα−2.Case 2.
For0<η≤s≤t≤1, let(19)g2t,s=t+β1−sα−1−t−sα−11+β−β+ηγ.
Then, by Remark3,(20)g2t,s=t+β1−sα−1−tα−11−stα−11+β−β+ηγ≥t+β1−sα−1−t1−sα−11+β−β+ηγ=1−sα−1t+β−t1+β−β+ηγ≥t+βγη1−sα−1.Case 3.
For0≤t≤s≤η<1, let(21)g3t,s=t+β1−sα−1−t+βγη−sα−1.
Then,(22)g3t,s=t+β1−sα−1−γηα−11−sηα−1≥t+β1−sα−1−γηα−11−sα−1≥t+β1−γη1−sα−1.
The proof is complete.Remark 3.
We have inequality (18) by using 1−t+tγ≥γη. In fact, if γ≥1, since 0<γ<1+β/β+η that 1−γη>βγ−1≥0≥t1−γ, which implies that 1−t+tγ≥γη. If 0<γ<1, then 1−γη>1−γ≥t1−γ, which also yields that 1−t+tγ≥γη.Remark 4.
From Lemma5, we know that Gt,s≥Ht,s≥0, for all t,s∈0,1.Lemma 6.
Let2<α≤3 and 0<γ<1+β/β+η. Assume g∈C10,1,0,∞; then, the unique solution u of (8)-(9) satisfies ut≥0 for any t∈0,1. Moreover, the unique solution u of (8)-(9) satisfies(23)inft∈η,1ut≥δu,where δ=minγη,γ1−η/1−γη,η.Proof.
By Lemmas4 and 5 and Remark 4, we know that ut≥0, ∀t∈0,1. From Lemma 4, one has(24)u″t=−∫0tt−sα−3Γα−2gsds≤0.
By applying the concavity ofu, similar to the proof of Lemma 4 in [2], we can obtain that (23) holds.Lemma 7 (see [23]).
LetP be a reproducing cone in a real Banach space E and L:E⟶E be a compact linear operator with LP⊂P and spectral radius rL. If rL>0, then there exists φ∈P\0 such that Lφ=rLφ.Theorem 1 (see [23]).
Assume thatE is a Banach space, P⊂E is a cone, and ΩP is a bounded open subset in P. Moreover, assume that T:ΩP¯⟶P is a completely continuous operator. Then, the following conclusions hold:(i)
If there existsu0∈P\0 such that u≠Tu+λu0, for all u∈∂ΩP and λ≥0, then the fixed-point index iT,ΩP,P=0(ii)
If0∈ΩP and Tu≠λu, for all u∈∂ΩP and λ≥1, then the fixed-point index iT,ΩP,P=1


## 3. Main Results

LetE=C0,1 be the Banach space endowed with the sup norm and define the cone:(25)P=x∈E:x≥0,inft∈η,1xt≥δx,where δ is defined in Lemma 6. Define the linear operator L:P⟶P as follows:(26)Lut=∫01Gt,sasuθsds.Lemma 8.
Assume that(H1) a∈C10,1,0,∞ and a does not vanish identically on any subinterval(H2) The advanced argument θ∈C0,1,0,1 and t≤θt≤1, ∀t∈0,1hold. Then, the spectral radius of the operator L is positive, that is, rL>0.Proof.
Takingut=tα−1, then u=1. By (H1), (H2), and Lemma 5, we have(27)Lut=∫01Gt,sasuθsds≥∫01Ht,sasθsα−1ds≥∫01Ht,sassα−1ds≥∫0ηt+βsγη2−αα−11−ηη−sα−2Γα1+β−β+ηγassα−1ds≥tα−1γη2−αα−11−ηΓα1+β−β+ηγ∫0ηassαη−sα−2ds≔ωtα−1>0.
SinceL:P⟶P, in view of the monotonicity of L, we obtain(28)L2ut=LLut>Lωtα−1>ωLtα−1>ω2tα−1.
Repeating the process givesLnut>ωntα−1. Hence, we get Ln>ωn. So,(29)Ln1/n>ω,rL=limn⟶∞Ln1/n>ω>0.
The proof is completed.
Define the operatorT:P⟶E by(30)Tut=∫01Gt,sasfuθsds,0≤t≤1.
For convenience, set(31)f0=limu⟶0+fuu,f∞=limu⟶+∞fuu,Pc=u∈P:u<c.Lemma 9.
Letf∈C10,∞,0,∞. Assume that (H1) and (H2) hold and 0≤f0<1/rL. Then, there exists ρ0>0 such that, for ρ∈0,ρ0, iT,Pρ,P=1.Proof.
From Lemma6, we know that TP⊂P. It is easy to check that the operator T:P⟶P is completely continuous. It follows from f0<1/rL that there exist ε>0 and ρ0>0 such that, for 0≤u≤ρ0,(32)fu≤1rL−εu.
For0<ρ<ρ0, we will show that λu≠Tu, for all u∈∂Pρ and λ≥1. Otherwise, there exist u0∈∂Pρ and λ0≥1 such that(33)Tu0=λ0u0.
By (32), one has(34)λ0u0t=Tu0t=∫01Gt,sasfu0θsds≤1rL−εLu0t,that is,(35)u0t≤1/rL−ελ0Lu0t≤1/rL−ελ02L2u0t≤⋯≤1/rL−ελ0nLnu0t≤1/rL−ελ0nLnu0.
Thus,(36)u0≤1/rL−ελ0nLnu0,which implies that(37)λ01/rL−ε≤Ln1/n⟶rL,n⟶∞.
It means that(38)1≤λ0≤1−εrL.
This is a contradiction. By Theorem1 (ii), we obtain that iT,Pρ,P=1, for each ρ∈0,ρ0. The proof is completed.Lemma 10.
Letf∈C10,∞,0,∞. Assume that (H1) and (H2) hold and 1/rL<f∞≤∞. Then, there exists τ0>0 such that, for τ>τ0, iT,Pτ,P=0.Proof.
Sincef∞>1/rL, there exist ε>0 and τ0>0 such that, for u>τ0,(39)fu≥1rL+εu.
Forτ>τ0, assume that u≠Tu and u∈∂Pτ. By Lemma 7 and Theorem 1 (i), we need only to prove that(40)u≠Tu+λφ,λ>0,where φ∈P\0 with Lφ=rLφ. Otherwise, there exist u0∈∂Pτ and λ0>0 such that(41)u0=Tu0+λ0φ,which implies that u0≥Tu0 and u0≥λ0φ. So, by (39), we obtain that(42)Tu0t=∫01Gt,sasfu0θsds≥1rL+εLu0t≥1rL+ελ0Lφt=1rL+ελ0rLφt>λ0φt,by 1/rL+εrL>1. Together with (41), one has u0≥2λ0φ. From (42), we have that Tu0≥2λ0φ. Thus, u0≥3λ0φ. Repeating this process, we get that u0≥nλ0φ; this means that u0≥nλ0φ⟶∞, as n⟶∞. This is a contradiction. It follows from Theorem 1 (i) that iT,Pτ,P=0, for each τ>τ0. The proof is completed. By using Lemmas 9 and 10 and the properties of index, we have the following.Theorem 2.
Let2<α≤3, 0<η<1,β≥0, and γ>0 with γ<1+β/η+β. Suppose f∈C10,∞,0,∞. Assume that (H1) and (H2) hold. If 0≤f0<1/rL and 1/rL<f∞≤∞, then problem (4) has at least one positive solution on 0,1.Remark 5.
Letβ=0, f0=0, and f∞=∞; then, Theorem 2 reduces to Theorem 1.2 in [3].
By using Theorem1, similar to the proof of Lemmas 8 and 9 as in [10], we can prove the following lemmas.Lemma 11.
Letf∈C10,∞,0,∞. Assume that (H1) and (H2) hold and 1/rL<f0≤∞. Then, there exists ρ0>0 such that, for ρ∈0,ρ0, iT,Pρ,P=0.Lemma 12.
Letf∈C10,∞,0,∞. Assume that (H1) and (H2) hold and 0≤f∞<1/rL. Then, there exists τ0>0 such that, for τ>τ0, iT,Pτ,P=1.By using Lemmas11 and 12 and the properties of index, we have the following.Theorem 3.
Let2<α≤3, 0<η<1,β≥0, and γ>0 with γ<1+β/η+β. Suppose f∈C10,∞,0,∞. Assume that (H1) and (H2) hold. If 1/rL<f0≤∞ and 0≤f∞<1/rL, then problem (4) has at least one positive solution on 0,1.Remark 6.
Letβ=0, f0=∞, and f∞=0; then, Theorem 3 reduces to Theorem 1.1 in [3].


## 4. Examples

In this section, two examples are given to illustrate the existence of positive solutions of BVP (4).Example 1.
Consider the following boundary value problem:(43)CD5/2ut+t5/2fuθt=0,0<t<1,u0−23u′0=u″0=0,u1=65u13.where α=5/2, β=2/3, γ=6/5, η=1/3, at=t5/2, θt=t, and(44)fu=3320u3,0≤u≤1,33202u−u,u>1.
Clearly,f∈C10,∞,0,∞. A direct computation shows(45)ω=γη2−αα−11−ηΓα1+β−β+ηγ∫0ηassαη−sα−2ds=1837Γ5/2∫01/3s513−s1/2ds=1837Γ5/2Γ6Γ3/2Γ15/21313/2≈16631.
Note that conditions (H1) and (H2) hold. Through a simple calculation, we can obtain f0=0 and f∞=6640>6631=1/ω>1/rL (see Lemma 8). Thus, by Theorem 2, we see that problem (43) has at least one positive solution.Example 2.
Consider the following boundary value problem:(46)CD8/3ut+3t4fuθt=0,0<t<1,u0−13u′0=u″0=0,u1=35u23.
Here,α=8/3, β=1/3, γ=3/5, η=2/3, at=3t4, θt=3t, and(47)fu=932π2u2+112sinu,0≤u≤π2,1862πu+2sin2u,u>π2.
Obviously,f∈C10,∞,0,∞. A direct computation shows(48)ω=γη2−αα−11−ηΓα1+β−β+ηγ∫0ηassαη−sα−2ds=53/22/311·Γ8/3∫02/3s423−s2/3ds=53/22/311·Γ8/3Γ5Γ5/3Γ20/32317/3≈0.0022.
Note that conditions (H1) and (H2) hold. Obviously, we can get f0=511.5>454.5455=1/ω>1/rL and f∞=0. Thus, by Theorem 3, we see that problem (46) has at least one positive solution.



---

*Source: 6781541-2021-09-24.xml*