# Existence and Nonexistence of Positive Solutions for Fractional Three-Point Boundary Value Problems with a Parameter

Authors: Yunhong Li; Weihua Jiang
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this work, we investigate the existence and nonexistence of positive solutions for p-Laplacian fractional differential equation with a parameter. On the basis of the properties of Green’s function and Guo-Krasnosel’skii fixed point theorem on cones, the existence and nonexistence of positive solutions are obtained for the boundary value problems. We also give some examples to illustrate the effectiveness of our main results.


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## Body

## 1. Introduction

Fractional calculus has played an important role in the modeling of different physical and natural phenomena, such as fluid mechanics, control system, and many other branches of engineering. In recent years, there are many papers concerning the existence of positive solutions for nonlinear fractional differential equations; see [1–18] and the references cited therein.In [1], Han et al. studied the existence of positive solutions for the following problems with the generalized p-Laplacian operator: (1)D0+βϕD0+αut=λfut,0<t<1,u0=u′0=u′1=0,ϕD0+αu0=ϕD0+αu1′=0. On the basis of the properties of Green’s function and Guo-Krasnosel’skii fixed point theorem on cones, some new existence results of at least one or two positive solutions are obtained by different eigenvalue interval for the aforementioned boundary value problems.Xu et al. [2] showed the existence of multiple positive solutions to singular positone and semipositone m-point boundary value problems of nonlinear fractional differential equations (2)D0+αut+ft,ut=0,0<t<1,u0=0,u1=∑i=1m-2βiuηi, where 1<α<2 and D0+α is the standard Riemann-Liouville fractional derivative. By means of the Leray-Schauder nonlinear alternative and a fixed point theorem on cones, they deduce multiple positive solutions to singular positive and semipositive m-point boundary value problems.Shen et al. [3] investigate the following fractional thermostat model with a parameter: (3)​cDαut+λft,ut=0,t∈0,1,u′0=0,βcDα-1u′1+uη=0, where 1<α≤2,0≤η≤1,β>0, and ​cDα is the Caputo’s fractional derivatives. Using the fixed point theorem on cones, the existence and nonexistence results for positive solutions are discussed for the boundary value problems.Motivated by the aforementioned works, this paper is concerned with the positive solutions for p-Laplacian fractional differential equation with a parameter(4)D0+βϕpD0+αut=λft,ut,t∈0,1,(5)ϕpD0+αu0i=0,i=0,1,2,⋯,l-2,(6)ϕpD0+αut′t=1=bϕpD0+αut′t=ξ,(7)uj0=0,j=0,1,2,⋯,n-2,(8)u′1=au′ξ, where λ>0,ϕp(u)=up-2u,p>1. D0+β and D0+α are the Riemann-Liouville fractional derivatives, 3<n-1<α≤n,3<l-1<β≤l, and l+n-1<α+β≤l+n. Based on the properties of Green’s function and Guo-Krasnosel’skii fixed point theorem on cones, the existence of positive solutions are obtained for problems (4)-(8).We will always suppose the following conditions are satisfied:( H 1 )   a > 0 , b > 0,0 < ξ < 1 and 0<aξα-2<1,0<bξβ-2<1;( H 2 )   f ( t , u ) : [ 0,1 ] × ( 0 , ∞ ) → ( 0 , ∞ ) is continuous.


## 2. Background and Definitions

For the convenience of the reader, we state some basic definitions and lemmas about fractional calculus theory, which can be found in [19, 20].Definition 1.
The fractional integral of orderα>0 of a function y:(0,+∞)→R is given by(9)I0+αyt=1Γα∫0tt-sα-1ysds, provided that the right side is pointwise defined on (0,+∞), where (10)Γα=∫0+∞e-xxα-1dx.Definition 2.
For a continuous functiony:(0,+∞)→R, the Riemann-Liouville derivative of fractional order α>0 is defined as(11)D0+αyt=1Γn-αddtn∫0tt-sn-α-1ysds, where n=[α]+1, provided that the right side is pointwise defined on (0,+∞).Lemma 3.
Letα>0. Assume that u,D0+αu∈L1(0,1). Then (12)I0+αD0+αut=ut+c1tα-1+c2tα-2+⋯+cNtα-N,ci∈R,i=1,2,⋯,Nas the unique solution, where N is the smallest integer greater than or equal to α.In order to obtain our main results, we need the following Guo-Krasnoselskii fixed point theorem in [21].Theorem 4.
Let(E,·) be a Banach space, and P⊂E be a cone in E. Assume that Ω1 and Ω2 are open subsets of E such that 0∈Ω1,Ω¯1⊂Ω2. If (13)T:P∩Ω¯2∖Ω1→P is a completely continuous operator such that either
(i) ifTu≤u,u∈P∩∂Ω1, and Tu≥u,u∈P∩∂Ω2, or
(ii) ifTu≥u,u∈P∩∂Ω1, and Tu≤u,u∈P∩∂Ω2,
thenT has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Preliminary Lemmas

Lemma 5.
The boundary value problems (4)-(8) are equivalent to the following equation:(14)ut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds,where(15)Gt,s=1Atα-11-sα-2-atα-1ξ-sα-2-1-aξα-2t-sα-1,0≤s≤mint,ξ≤1,tα-11-sα-2-atα-1ξ-sα-2,0≤t≤s≤ξ≤1,tα-11-sα-2-1-aξα-2t-sα-1,0≤ξ≤s≤t≤1,tα-11-sα-2,0≤max⁡t,ξ≤s≤1,(16)Hs,τ=1Bsβ-11-τβ-2-bsβ-1ξ-τβ-2-1-bξβ-2s-τβ-1,0≤τ≤mins,ξ≤1,sβ-11-τβ-2-bsβ-1ξ-τβ-2,0≤s≤τ≤ξ≤1,sβ-11-τβ-2-1-bξβ-2s-τβ-1,0≤ξ≤τ≤s≤1,sβ-11-τβ-2,0≤maxs,ξ≤τ≤1,ϕq(s) is the inverse function of ϕp(s), a.e., ϕq(s)=sq-2s,1/p+1/q=1 and A=(1-aξα-2)Γ(α),B=(1-bξβ-2)Γ(β).Proof.
According to Lemma3, (4) is equivalent to the following integral equation: (17)ϕpD0+αut=λΓβ∫0tt-τβ-1fτ,uτdτ+c1tβ-1+c2tβ-2+⋯+cltβ-l. Conditions (5) imply that c2=⋯=cl=0. i.e.,(18)ϕpD0+αut=λΓβ∫0tt-τβ-1fτ,uτdτ+c1tβ-1. So, (19)ϕpD0+αut′=λΓβ-1∫0tt-τβ-2fτ,uτdτ+c1β-1tβ-2. From (6), we have(20)c1=-λB∫011-τβ-2fτ,uτdτ+λbB∫0ξξ-τβ-2fτ,uτdτ. By use of (18) and (20), we get (21)ϕpD0+αut=λΓβ∫0tt-τβ-1fτ,uτdτ-λtβ-1B∫011-τβ-2fτ,uτdτ+λbtβ-1B∫0ξξ-τβ-2fτ,uτdτ=-λ∫01Ht,τfτ,uτdτ. Therefore, (22)D0+αut=-ϕqλ∫01Ht,τfτ,uτdτ.
In view of Lemma3, we have (23)ut=-1Γα∫0tt-sα-1ϕqλ∫01Hs,τfτ,uτdτds+d1tα-1+d2tα-2+⋯+dntα-n. Conditions (7) imply that d2=⋯=dn=0. i.e.,(24)ut=-1Γα∫0tt-sα-1ϕqλ∫01Hs,τfτ,uτdτds+d1tα-1. So, (25)u′t=-1Γα-1∫0tt-sα-2ϕqλ∫01Hs,τfτ,uτdτds+d1α-1tα-2. From (8), we get(26)d1=1A∫011-sα-2ϕqλ∫01Hs,τfτ,uτdτds-aA∫0ξξ-sα-2ϕqλ∫01Hs,τfτ,uτdτds. By use of (24) and (26), we can obtain (27)ut=-1Γα∫0tt-sα-1ϕqλ∫01Hs,τfτ,uτdτds+tα-1A∫011-sα-2ϕqλ∫01Hs,τfτ,uτdτds-atα-1A∫0ξξ-sα-2ϕqλ∫01Hs,τfτ,uτdτds=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds.
The proof is complete.Lemma 6.
FunctionsG(t,s) and H(s,τ) defined by (15) and (16), respectively, are continuous on [0,1]×[0,1] and satisfy
(i)G(t,s)≥0,H(s,τ)≥0, for t,s,τ∈[0,1];
(ii)G(t,s)≥(1/Γα)tα-1s(1-s)α-2,H(s,τ)≥(1/Γβ)sβ-1τ(1-τ)β-2, for t,s,τ∈[0,1];
(iii)G(t,s)≤k1s(1-s)α-2,H(s,τ)≤k2τ(1-τ)β-2, for t,s,τ∈[0,1], where(28)k1=1Aα-1+α-2aξα-3,k2=1Bβ-1+β-2bξβ-3.Proof.
If0≤s≤min{t,ξ}≤1, then (29)AGt,s=tα-11-sα-2-atα-1ξ-sα-2-1-aξα-2t-sα-1≥tα-11-sα-2-atα-1ξ-ξsα-2-1-aξα-2t-tsα-1=tα-11-sα-2-aξα-2tα-11-sα-2-1-aξα-2tα-11-sα-1=1-aξα-2tα-11-sα-2-1-aξα-2tα-11-sα-1=1-aξα-2tα-11-sα-21-1-s=1-aξα-2tα-1s1-sα-2≥0,AGt,s=tα-11-sα-2-atα-1ξ-sα-2-1-aξα-2t-sα-1=tα-11-sα-2-atα-1ξ-sα-2-t-sα-1+aξα-2t-sα-1=tα-11-sα-2-t-sα-1-atα-1ξ-sα-2+aξα-2t-sα-1=tt-tsα-2-tt-sα-2+st-sα-2-attξ-tsα-2+atξ-sξα-2t-s≤α-2t-tsα-3s-st+st-sα-2-attξ-tsα-2+attξ-sξα-2-astξ-sξα-2≤α-2s1-sα-2+s1-sα-2-attξ-tsα-2+attξ-sξα-2=α-1s1-sα-2+attξ-sξα-2-tξ-tsα-2.
If0≤s≤t≤ξ≤1, then at[(tξ-sξ)α-2-(tξ-ts)α-2]≤0.
If0≤s≤ξ≤t≤1, then (30)attξ-sξα-2-tξ-tsα-2≤atα-2tξ-sξα-3st-sξ≤α-2aξα-3t-sα-3t-ξs≤α-2aξα-3s1-sα-2. So we get if 0≤s≤min{t,ξ}≤1, then (31)1Γαtα-1s1-sα-2≤Gt,s≤k1s1-sα-2, where k1=(1/A)[(α-1)+(α-2)aξα-3]. Obviously, Ak1>1.
If0≤t≤s≤ξ≤1, then (32)AGt,s=tα-11-sα-2-atα-1ξ-sα-2≥tα-11-sα-2-atα-1ξ-ξsα-2=tα-11-sα-2-aξα-2tα-11-sα-2=1-aξα-2tα-11-sα-2≥1-aξα-2tα-1s1-sα-2≥0,(33)AGt,s=tα-11-sα-2-atα-1ξ-sα-2≤tα-11-sα-2≤s1-sα-2≤Ak1s1-sα-2.If 0≤ξ≤s≤t≤1, then (34)AGt,s=tα-11-sα-2-1-aξα-2t-sα-1=tα-11-sα-2-t-sα-1+aξα-2t-sα-1≥tα-11-sα-2-t-sα-1≥tα-11-sα-2-t-tsα-1=tα-11-sα-2-tα-11-sα-1=tα-1s1-sα-2≥1-aξα-2tα-1s1-sα-2≥0,AGt,s=tα-11-sα-2-1-aξα-2t-sα-1=tt-tsα-2-t-sα-1+aξα-2t-sα-1=tt-tsα-2-tt-sα-2+st-sα-2+aξα-2t-sα-1≤α-2t-tsα-3s-st+s1-sα-2+aξα-3s1-sα-2≤α-2s1-sα-2+s1-sα-2+aξα-3s1-sα-2=α-1+aξα-3s1-sα-2≤Ak1s1-sα-2.
If0≤max{t,ξ}≤s≤1, we get (35)AGt,s=tα-11-sα-2≥1-aξα-2tα-1s1-sα-2,AGt,s=tα-11-sα-2≤Ak1s1-sα-2. Consequently, we get(36)1Γαtα-1s1-sα-2≤Gt,s≤k1s1-sα-2fort,s∈0,1.
Similarly, we can get(37)1Γβsβ-1τ1-τβ-2≤Hs,τ≤k2τ1-τβ-2fors,τ∈0,1, where k2=(1/B)[(β-1)+(β-2)bξβ-3].
The proof is complete.LetE be the real Banach space C[0,1] with the maximum norm and define the cone P⊂E by (38)P=u∣u∈Eandut≥ktα-1u,t∈0,1.where (39)k=Bβ-1q-1+2,α-1k1ΓαB2,α-1k2Γβq-1,Bρ,η=∫01xρ-11-xη-1dx.Define the operatorT on P by(40)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds.Lemma 7.
T : P → P is completely continuous.Proof.
For anyu∈P, in view of (26) and Lemma 6, we get (41)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≤∫01k1s1-sα-2λ∫01k2τ1-τβ-2fτ,uτdτq-1ds=k1B2,α-1λk2∫01τ1-τβ-2fτ,uτdτq-1=k1B2,α-1k2q-1λ∫01τ1-τβ-2fτ,uτdτq-1. So we obtain (42)Tut≤k1B2,α-1k2q-1λ∫01τ1-τβ-2fτ,uτdτq-1,Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≥∫011Γαtα-1s1-sα-2ϕqλ∫011Γβsβ-1τ1-τβ-2fτ,uτdτds=tα-1Γα∫01s1-sα-21Γβsβ-1q-1λ∫01τ1-τβ-2fτ,uτdτq-1ds=tα-1Γα1Γβq-1∫01s1-sα-2sβ-1q-1dsλ∫01τ1-τβ-2fτ,uτdτq-1=Bβ-1q-1+2,α-1ΓαΓβq-1tα-1λ∫01τ1-τβ-2fτ,uτdτq-1≥ktα-1Tu. Therefore, T:P→P. In view of continuity of G,H and f, we have T:P→P is continuous.
LetΩ⊂P be bounded, i.e., there exists a positive constant M>0 such that u≤M, for all u∈Ω, let L=max0≤t≤1,0≤u≤M,ft,u+1, then, for t∈[0,1] and u∈Ω, we get (43)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≤∫01k1s1-sα-2ϕqλ∫01k2τ1-τβ-2Ldτds=k1B2,α-1λk2B2,β-1Lq-1<+∞. Hence, T(Ω) is uniformly bounded.
On the other hand, sinceG(t,s) is continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1]. Thus, for any ε>0, there exists a constant δ>0, such that t1,t2∈[0,1] with |t1-t2|<δ imply(44)Gt1,s-Gt2,s<ελk2B2,β-1Lq-1for  anys∈0,1.
Then, for allu∈Ω, (45)Tut1-Tut2≤∫01Gt1,s-Gt2,sϕqλ∫01Hs,τfτ,uτdτds<ελk2B2,β-1Lq-1λk2B2,β-1Lq-1=ε. Hence, T(Ω) is equicontinuous. By Arzela-Ascoli theorem, we have T:P→P is completely continuous.


## 4. Main Results

For convenience we introduce the following notations. Let(46)A1limsupu→0+maxt∈0,1ft,uϕpu=f0,A2liminfu→+∞mint∈0,1ft,uϕpu=f∞,A3liminfu→0+mint∈0,1ft,uϕpu=f0,A4limsupu→+∞maxt∈0,1ft,uϕpu=f∞.The following theorems are the main results in this paper.Theorem 8.
If(A1) and (A2) hold, and (47)k1B2,α-1ΓαkBβ-1q-1+2,α-1p-1<Bα-1p-1+2,β-1f∞k2B2,β-1f0Γβ, then for each (48)λ∈ΓαkBβ-1q-1+2,α-1p-1ΓβBα-1p-1+2,β-1f∞,1k1B2,α-1p-11k2B2,β-1f0, the boundary value problems (4)-(8) have at least one positive solution. Here we impose 1/f∞=0 if f∞=+∞ and 1/f0=+∞ if f0=0.Proof.
Suppose(A1) holds, we may choose H1>0, so that, for each (t,u)∈[0,1]×[0,H1], there are f(t,u)≤(f0+ε)ϕp(u). Thus, if u∈P and u=H1, then by (40) and Lemma 6, we have (49)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≤∫01k1s1-sα-2ϕqλ∫01k2τ1-τβ-2f0+εϕpudτds≤k1B2,α-1λk2B2,β-1f0+εq-1H1≤H1. Let Ω1={u∈E:u<H1}, then the previous inequality show that Tu≤u for u∈P∩∂Ω1.
On the other hand, suppose (A2) holds, there ish2>0 such that f(t,u)≥(f∞-ε)ϕp(u), for (t,u)∈[0,1]×[h2,+∞). Thus, if u∈P and u=H2=max{2H1,h2}, by (40) and Lemma 6, we have (50)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≥∫011Γαtα-1s1-sα-2ϕqλ∫011Γβsβ-1τ1-τβ-2f∞-εϕpudτds=1Γα∫01sβ-1q-1+11-sα-2dsλf∞-εΓβ∫01τ1-τβ-2up-1dτq-1≥Bβ-1q-1+2,α-1Γαλf∞-εΓβ∫01τ1-τβ-2kτα-1H2p-1dτq-1≥Bβ-1q-1+2,α-1ΓαλBα-1p-1+2,β-1f∞-εΓβq-1kH2≥H2.Let Ω2={u∈E:u<H2}, then the previous inequality show that Tu≥u for u∈P∩∂Ω2.
Thus, from Theorem4, we know that the operator T has a fixed point in P∩(Ω¯2∖Ω1).Theorem 9.
If(A3) and (A4) hold, and (51)k1B2,α-1ΓαkBβ-1q-1+2,α-1p-1<Bα-1p-1+2,β-1f0k2B2,β-1f∞Γβ, then for each (52)λ∈ΓαkBβ-1q-1+2,α-1p-1ΓβBα-1p-1+2,β-1f0,1k1B2,α-1p-11k2B2,β-1f∞, the boundary value problems (4)-(8) have at least one positive solution. Here we impose 1/f∞=+∞ if f∞=0 and 1/f0=0 if f0=+∞.Proof.
Suppose(A3) holds, we may choose H3>0 so that, for each (t,u)∈[0,1]×[0,H3] there are f(t,u)≥(f0-ε)ϕp(u). Thus, if u∈P and u=H3, then by (40) and Lemma 6, we have (53)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≥∫011Γαtα-1s1-sα-2ϕqλ∫011Γβsβ-1τ1-τβ-2f0-εϕpudτds=1Γα∫01sβ-1q-1+11-sα-2dsλΓβ∫01τ1-τβ-2f0-εϕpudτq-1≥Bβ-1q-1+2,α-1ΓαλΓβ∫01τ1-τβ-2f0-εϕpkτα-1H3dτq-1≥Bβ-1q-1+2,α-1ΓαλBα-1p-1+2,β-1f0-εΓβq-1kH3≥H3.Let Ω3={u∈E:u<H3}, then the previous inequality shows that Tu≥u for u∈P∩∂Ω3.
Suppose (A4) holds, we consider two cases.
Case 1. Suppose f is bounded, then there exists some l1>0, such that f(t,u)≤l1, for u∈[0,1],u∈(0,+∞). Thus, if u∈P and if(54)u=H41=max2H3,k1B2,α-1λk2B2,β-1lq-1, by (40) and Lemma 6, we have (55)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≤∫01k1s1-sα-2ϕqλ∫01k2τ1-τβ-2l1dτds≤k1B2,α-1λk2B2,β-1l1q-1≤H41.
Case 2. Suppose f is unbounded; there is h4>0 such that f(t,u)≤(f∞+ε)ϕp(u), for (t,u)∈[0,1]×[h4,+∞). Then there exists some H42>max{2H3,h4}, such that f(t,u)≤maxt∈[0,1]|f(t,H42)|, for u∈(0,H42).
Thus, ifu∈P and u=H42, by (40) and Lemma 6, we have (56)Tut=∫01Gt,sϕqλ∫01Hs,τfτ,uτdτds≤∫01k1s1-sα-2ϕqλ∫01k2τ1-τβ-2f∞+εϕpudτds≤k1B2,α-1λk2B2,β-1f∞+εq-1H42≤H42.
LetΩ4={u∈E:u<H4},H4=max{H41,H42}, then the previous inequality shows that Tu≤u for u∈P∩∂Ω4. Thus, from Theorem 4, we know that the operator T has a fixed point in P∩(Ω¯4∖Ω3).Theorem 10.
Iff0<+∞ and f∞<+∞, then there exists λ1>0 such that for all 0<λ<λ1, the boundary value problems (4)-(8) have no positive solution.Proof.
Sincef0<+∞ and f∞<+∞, there exist positive constants M1,M2,r1,r2, such that r1<r2 and
f ( t , u ) ≤ M 1 ϕ p ( u ) for t∈[0,1],u∈[0,r1];
f ( t , u ) ≤ M 2 ϕ p ( u ) for t∈[0,1],u∈[r2,+∞).
Let(57)M0=maxM1,M2,supu∈r1,r2maxt∈0,1⁡ft,uϕpu>0, then we have (58)ft,u≤M0ϕpufort∈0,1,u∈0,+∞.
Assumew(t) is a positive solution of the boundary value problems (4)-(8); let (59)λ1=1k1B2,α-1p-11k2B2,β-1M0, then for all t∈[0,1], we get (60)wt=Twt=∫01Gt,sϕqλ∫01Hs,τfτ,wτdτds≤∫01k1s1-sα-2ϕqλ∫01k2τ1-τβ-2M0ϕpwdτds<k1B2,α-1λ1k2B2,β-1M0q-1wt=wt,which is a contraction. Therefore, the boundary value problems (4)-(8) have no positive solution.Theorem 11.
Iff0>0 and f∞>0, then there exists λ2>0 such that, for all λ>λ2, the boundary value problems (4)-(8) have no positive solution.Proof.
Sincef0>0 and f∞>0, there exist positive constants M3,M4,r3,r4, such that r3<r4 and
f ( t , u ) ≥ M 3 ϕ p ( u ) for t∈[0,1],u∈[0,r3],
f ( t , u ) ≥ M 4 ϕ p ( u ) for t∈[0,1],u∈[r4,+∞).
Let(61)M5=minM3,M4,infu∈r3,r4mint∈0,1⁡ft,uϕpu>0, then we have (62)ft,u≥M5ϕpufort∈0,1,u∈0,+∞.
Supposew(t) is a positive solution of the problems (4)-(8); let (63)λ2=ΓαkBβ-1q-1+2,α-1p-1ΓβBα-1p-1+2,β-1M5, then for all t∈[0,1], we get (64)wt=Twt=∫01Gt,sϕqλ∫01Hs,τfτ,wτdτds≥∫011Γαtα-1s1-sα-2ϕqλ∫011Γβsβ-1τ1-τβ-2M5ϕpwdτds>Bβ-1q-1+2,α-1Γαλ2Bα-1p-1+2,β-1M5Γβq-1kwt=wt,which is a contraction. Therefore, the boundary value problems (4)-(8) have no positive solution.


## 5. Examples

In this section, we give some simple examples to explain our results.Example 1.
For the problems (4)-(8), let α=3.2,β=3.1,a=0.8,b=0.9,ξ=0.2p=1.5, then we get q=3,k=0.030716. Let (65)ft,u=u0.5100u+3u+3, then f0=1,f∞=100. From a direct calculation, we get(66)ΓαkBβ-1q-1+2,α-1p-1ΓβBα-1p-1+2,β-1f∞≈21.618258,1k1B2,α-1p-11k2B2,β-1f0≈30.516030. In view of Theorem 8, we get that problems (4)-(8) have at least one positive solution when λ∈(21.618258,30.516030).Example 2.
For the problems (4)-(8), let α=3.2,β=3.1,a=0.8,b=0.9,ξ=0.2p=1.5, then we get q=3,k=0.030716. Let (67)ft,u=u0.50.2u+20u+1, then f0=20,f∞=0.2. From a direct calculation, we get(68)ΓαkBβ-1q-1+2,α-1p-1ΓβBα-1p-1+2,β-1f0≈108.091288,1k1B2,α-1p-11k2B2,β-1f∞≈152.580152. In view of Theorem 8, we get that problems (4)-(8) have at least one positive solution when λ∈(108.091288,152.580152).


## 6. Conclusions

The fixed point theorem on cones is used to solve the three-point boundary value problems of a kind of nonlinear fractional differential equation with a parameter. Under certain conditions of the nonlinearity, the existence and nonexistence of positive solutions are obtained for the boundary value problems when the parameter belongs to different intervals.



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*Source: 9237856-2019-01-03.xml*