# Eigenvalue Criteria for Existence of Positive Solutions to Fractional Boundary Value Problem

Authors: Wen Lian; Zhanbing Bai
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The existence and multiplicity of positive solutions for the nonlinear fractional differential equation boundary value problem (BVP)DC0+αyx+fx,yx=0,0<x<1, y0=y′1=y″0=0 is established, where 2<α≤3, CD0+α is the Caputo fractional derivative, and f:0,1×0,∞⟶0,∞ is a continuous function. The conclusion relies on the fixed-point index theory and the Leray-Schauder degree theory. The growth conditions of the nonlinearity with respect to the first eigenvalue of the related linear operator is given to guarantee the existence and multiplicity.


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## Body

## 1. Introduction

In this paper, we concentrate on the existence and multiplicity of positive solutions for the following problem:(1)DC0+αyx+fx,yx=0,0<x<1,(2)y0=y′1=y″0=0,where 2<α≤3,DC0+α is the Caputo fractional derivative, and f:0,1×0,+∞⟶0,+∞ is a continuous function.In the past twenty years, the fractional differential equation has aroused great consideration [1–21] not only in its application in mathematics but also in other applications in science and engineering, for example, fluid mechanics, viscoelastic mechanics, electroanalytical chemistry, and biological engineering. Bai and Qiu [22, 23] have investigated the existence and multiplicity of positive solutions of (1) and (2) by using the nonlinear alternative of the Leray-Schauder type and Krasnoselskii’s fixed-point theorem in a cone, but they did not consider its eigenvalue criteria.The rest of the paper is organized as follows. In Section2, we recall some concepts relative to fractional calculus and give some lemmas with respect to the corresponding Green function. In Section 3, with the use of the fixed-point theory, some existence and multiplicity results of positive solutions are obtained. At last, two examples are given.


## 2. Background Materials

For the convenience of the reader, we give some definitions and lemmas.Definition 1 (see [23]).
The Caputo’s fractional derivative of orderα>0 of a continuous function y:0,+∞⟶ℝ is given by
(3)DC0+αyx=1Γn−α∫0xyntx−tα−n+1dt,where n−1≤α<n and n∈ℕ, provided that the right-hand side is pointwise defined on 0,+∞.Lemma 2 (see [15]).
Givenh∈C0,1, the unique solution of
(4)DC0+αyx+hx=0,0<x<1,y0=y′1=y″0=0,is given by
(5)yx=∫01Gx,thtdt,where
(6)Gx,t=α−1x1−tα−2−x−tα−1Γα,for0≤t≤x≤1,x1−tα−2Γα−1,for0≤x≤t≤1,Lemma 3 (see [23]).
The Green functionGx,t defined by (6) satisfies the following properties:
(i)
Gx,t>0, for all x,t∈0,1
(ii)
min1/4≤x≤3/4Gx,t≥1/4max0≤x≤1Gx,t=1/4G1,t,fort∈0,1Lemma 4 (see [24]).
LetK be a cone in a Banach space X, and Ω be a bounded open set in K. Suppose that T:Ω¯⟶K is a completely continuous operator. If there exists y0∈Kθ such that
(7)y−Ty≠μy0,forally∈∂Ω,μ≥0,then the fixed-point index iT,Ω,K=0.Lemma 5 (see [24]).
LetK be a cone in a Banach space X. Suppose that T:K⟶K is a completely continuous operator. If there exists a bounded open set Ω such that each solution of
(8)y=σTy,y∈K,σ∈0,1,satisfies y∈Ω, then the fixed-point index iT,Ω,K=1.Lemma 6 (see [25]).
Suppose thatA:C0,1⟶C0,1 is a completely continuous linear operator and AK⊂K. If there exist ψ∈C0,1\−K and a constant c>0 such that cAψ≥ψ, then the spectral radius rA≠0 and A has a positive eigenfunction φ∗ corresponding to its first eigenvalue λ1=rA−1.


## 3. Existence and Multiplicity

LetC0,1 be endowed with the maximum norm u=max0≤x≤1ux and the ordering u≤v if ux≤vx for all x∈0,1. Define
(9)K=u∈C0,1ux≥0,min1/4≤x≤3/4ux≥14u.Givenf∈C0,1×0,∞,0,∞. Let T,A:K⟶C0,1 be the operators defined by
(10)Tux≔∫01Gx,tft,utdt,u∈K,x∈0,1,and
(11)Aux≔∫01Gx,tutdt,u∈K,x∈0,1.It is well known thatT,A:K⟶K are all completely continuous [23].Denote(12)M=∫01max0≤x≤1Gx,tdt−1,N=∫1/43/4G12,tdt−1,where M,N are positive constants.Lemma 7.
SupposeA is defined by (11), then the spectral radius rA>0 and A has a positive eigenfunction φ1 corresponding to its first eigenvalue λ1=rA−1.Proof.
The operatorA:C0,1⟶C0,1 is a completely continuous linear operator and AK⊆K (see [23]). Choose ψ∈C0,1 and x1,x2⊂0,1 such that ψx≥0 for x∈0,1; ψx>0 for x∈x1,x2. By the use of Lemma 2, for x∈0,1, there holds
(13)Aψx=∫01Gx,tψtdt≥∫x1x2Gx,tψtdt>0.
So, we can choosec∈ℝ so large that
(14)cAψx≥ψx,forx∈0,1.
By Lemma6, we complete the proof.Theorem 8.
Suppose the following conditions hold:(15)I0liminfy→0+fx,yy>λ1,S∞limsupy→+∞fx,yy<λ1,where λ1 is the first eigenvalue of the operator A defined by (11). Then, BVP (1) and (2) have at least one positive solution.Proof.
By conditionI0, there exists r1>0 small enough such that
(16)fx,y≥λ1y,forall0≤x≤1,0≤y≤r1.
Letφ∗ be the positive eigenfunction of A corresponding to λ1, thus φ∗=λ1Aφ∗.
For everyφ∈Kr1¯, for x∈0,1
(17)Tφx=∫01Gx,tft,φtdt≥λ1∫01Gx,tφtdt=λ1Aφx.
Suppose without loss of generality thatT has no fixed point on ∂Kr1 (otherwise, the proof is completed). We claim that
(18)φ−Tφ≠μφ∗,forallφ∈∂Kr1,μ>0.
In fact, if there existφ1∈∂Kr1 and μ0>0 such that φ1−Tφ1=μ0φ∗, then
(19)φ1=Tφ1+μ0φ∗≥μ0φ∗.
Let(20)μ∗=supμφ1≥μφ∗.
It is easy to see that+∞>μ∗≥μ0>0 and φ1≥μ∗φ∗. Taking into account that A is a linear positive operator, we have
(21)λ1Aφ1≥μ∗λ1Aφ∗=μ∗φ∗.
Therefore, by (17),
(22)φ1=Tφ1+μ0φ∗≥λ1Aφ1+μ0φ∗≥μ∗+μ0φ∗,which contradicts the definition of μ∗. Hence (18) holds and we have from Lemma 3 that
(23)iT,Kr1,K=0.
On the other hand, byS∞, there exist 0<σ<1 and r2>r1 such that
(24)fx,y≤σλ1y,forally≥r2,x∈0,1.
DefineA1:C0,1⟶C0,1 as A1φ=σλ1Aφ,φ∈C0,1. Then A1 is a bounded linear operator and A1K⊂K. Denote
(25)B=max0≤x,t≤1Gx,tsupy∈B¯r2∫01ft,ytdt.
It is clear thatB<+∞. Let
(26)W=φ∈K∣φ=μTφ,0≤μ≤1.
In the following, we firstly prove that the setW is bounded.
For anyφ∈W, set φ¯x=minφx,r2 for x∈0,1 and denote Eφ=x∈0,1∣φx>r2, then for x∈0,1,
(27)φx=μTφx≤Tφx=∫01Gx,tft,φtdt=∫EφGx,tft,φtdt+∫0,1\EφGx,tftφ¯tdt≤∫01Gx,tσλ1φtdt+∫01Gx,tft,φ¯tdt≤A1φx+B.
ThusI−A1φx≤B,x∈0,1. Since λ1 is the first eigenvalue of A and 0<σ<1, the first eigenvalue of A1,rA1−1>1. Therefore, the inverse operator I−A1−1 exists and
(28)I−A1−1=I+A1+A12+⋯+A1n+⋯.
It follows fromA1K⊂K that I−A1−1K⊂K. So we have φx≤I−A1−1B,x∈0,1 and the set W is bounded.
Chooser3>maxr2,I−A1−1B. Then by Lemma 4, we have
(29)iT,Kr3,K=1.
By (23) and (29), one has
(30)iT,Kr3Kr1¯,K=iT,Kr3,K−iT,Kr1,K=1.
Then,T has at least one fixed point on Kr3\Kr1¯. This means that problem (1) and (2) have at least one positive solution. The proof is complete.Theorem 9.
Suppose the following conditions are met:(31)I∞liminfy→+∞fx,yy<λ1,S0limsupy→0+fx,yy<λ1,where λ1 is the first eigenvalue of the operator A defined by (11). Then BVP (1) and (2) have at least one positive solution.The proof is similar to Theorem 8.Theorem 10.
Suppose there exist two numbersb>a>0 such that the following conditions are met:
(32)C1fx,y≤Ma,for0≤x≤1and0≤y≤a,C2fx,y≥Nb,for14≤x≤34and14b≤y≤b.Then, BVP (1) and (2) have at least one positive solution.Proof.
If C1 and C2 hold, similar to Lemma 3 [6], we have
(33)iA,Ka,K=1,iA,Kb,K=0.
Consequently, the additivity of the fixed-point index implies(34)iA,KbKa¯,K=iA,Kb,K−iA,Ka,K=−1.
Consequently,A has a fixed point yx in Kb\Ka¯.Theorem 11.
The problem in (1) and (2) has at least two positive solutions if conditions I0, I∞, and C1 hold, where λ1 is the first eigenvalue of the operator A defined by (11).Proof.
BecauseI0 and I∞ hold, there exist 0<r<a<R such that
(35)iA,Kr,K=0,iA,KR,K=0.
On the other hand, C1 implies iA,Ka,K=1. So we have
(36)iA,KaKr¯,K=iA,Ka,K−iA,Kr,K=1,iA,KRKa¯,K=iA,KR,K−iA,Ka,K=−1,therefore, A has two fixed points, y1∈Ka\Kr¯,y2∈KR\Ka¯.Theorem 12.
The problem in (1) and (2) has at least two positive solutions if conditions S0, S∞, and C2 hold, where λ1 is the first eigenvalue of the operator A defined by (10).Proof.
BecauseS0 and S∞ hold, there exist 0<r<b<R such that
(37)iA,Kr,K=1,iA,KR,K=1.
On the other hand C2 implies iA,Kb,K=0. So we have
(38)iA,KbKr¯,K=iA,Kb,K−iA,Kr,K=−1,iA,KRKb¯,K=iA,KR,K−iA,Kb,K=1.Therefore, A has two fixed points, y1∈Kb\Kr¯,y2∈KR\Kb¯.


## 4. Example

To illustrate the main points, we give two examples.Example 13.
Let(39)fx,y=y1/2+2+sinx4.Consider the BVP(40)DC0+αyx+yx1/2+2+sinx4=0,0<x<1,(41)y0=y′1=y″0=0.It is not difficult to see that(42)fx,yy=y−1/2+2+sinx4y.Then(43)liminfy→0+fx,yy=+∞>λ1,limsupy→+∞fx,yy=0<λ1,where λ1 is the first eigenvalue of the operator A defined by (11). By Theorem 11, BVP (40) and (41) have at least one positive solution.Example 14.
Let(44)fx,y=y3+y2sinx.Consider the BVP(45)CD0+αyx+yx3+yx2sinx=0,0<x<1,(46)y0=y′1=y″0=0.It is not difficult to see that(47)fx,yy=y2+ysinx.Then(48)liminfy→+∞fx,yy=+∞>λ1,limsupy→+0+fx,yy=0<λ1,where λ1 is the first eigenvalue of the operator A defined by (11). By Theorem 12, BVP (45) and (46) have at least one positive solution.



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