# Existence and Nonexistence of Positive Solutions for Mixed Fractional Boundary Value Problem with Parameter andp-Laplacian Operator

Authors: Ying Wang
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

This paper mainly studies a class of mixed fractional boundary value problem with parameter andp-Laplacian operator. Based on the Guo-Krasnosel’skii fixed point theorem, results on the existence and nonexistence of positive solutions for the fractional boundary value problem are established. An example is then presented to illustrate the effectiveness of the results.


---

## Body

## 1. Introduction

In this paper, we consider the existence and nonexistence of positive solution for the following mixed fractional boundary value problem (BVP):(1)D0+αφpDc0+βut+λft,ut=0,0<t<1,u1=u′0=u′′0=⋯=un-10=0,Dc0+βu0=0,Dc0+βu1=aDc0+βuη,where λ>0 is a parameter, 1<α≤2, n-1<β≤n, n≥2, D0+α is the Riemann-Liouville derivative operator, Dc0+β is the Caputo fractional derivative operator, φp is the p-Laplacian operator defined by φp(s)=|s|p-2s, (φp)-1=φq, 1/p+1/q=1, p>1, η∈(0,1), and a>0 and satisfies 1-ap-1ηα-1>0, and f:[0,1]×[0,+∞)→[0,+∞) is a continuous function.The theory of fractional differential equation has gained interesting by many researchers due to its deep real world background and, in recent years, more and more papers concern the boundary value problems for fractional-order differential equations; see [1–27]. In [28], Wang, Xiang and Liu use Krasnoselskii’s fixed point theorem and Leggett-Williams theorem to obtain the existence results of BVP, which is given in the following:(2)D0+γφpD0+αut=ft,ut,0<t<1,D0+αu0=u0=0,u′1=auξ,D0+αu1=bD0+αuη,where 1<α,γ≤2, 0≤a,b≤1, 0<ξ,η<1, D0+α and D0+γ are Riemann-Liouville derivative operators, φp is the p-Laplacian operator defined by φp(s)=|s|p-2s, and f:[0,1]×[0,+∞)→[0,+∞) is a continuous function.Lu et al. in [29] investigated the BVP(3)D0+αφpD0+βut=ft,ut,0<t<1,u0=u′0=u′1=0,D0+βu0=D0+βu1=0,where 1<α≤2, 2<β≤3, D0+α and D0+β are Riemann-Liouville derivative operators, φp is the p-Laplacian operator defined by φp(s)=|s|p-2s, and f:[0,1]×[0,+∞)→[0,+∞) is a continuous function. By the use of Guo-Krasnosel’skii fixed point theorem, the existence and multiplicity results of BVP (3) are obtained.In this paper, we study the existence and nonexistence of positive solutions for the mixed fractional boundary value problem as BVP (1), which leads to lots of difference and new features. On one hand, compared to the papers mentioned above, which only involve one derivative, our study involves both the Riemann-Liouville fractional derivative and the Caputo fractional derivative, which making the studied problems difficult. On the other hand, under different combinations of superlinearity and sublinearity of the function f, results on the existence and nonexistence of positive solutions are received and the impact of the parameter on the existence and nonexistence of positive solutions is also obtained.


## 2. Preliminaries and Lemmas

Definition 1 (see [30, 31]).
The Caputo fractional-order derivative of ordersα>0 and n-1<α<n, n∈N, is defined as (4)Dc0+αut=1Γn-α∫0tt-sn-α-1unsds,where u∈Cn(J,R), R=(-∞,+∞), N denotes the natural number set, n=[α]+1, and [α] denotes the integer part of α.Definition 2 (see [30, 31]).
Letα>0 and let u be piecewise continuous on (0,+∞) and integrable on any finite subinterval of J. Then for t>0, we call (5)I0+αut=1Γα∫0tt-sα-1usds,the Riemann-Liouville fractional integral of u of order α.Lemma 3 (see [30, 31]).
Letn-1<α≤n, u∈Cn[0,1]. Then (6)I0+αDc0+αut=ut+c0+c1t+c2t2+⋯+cn-1tn-1,where ci∈R(i=1,2,…,n-1) and n is the smallest integer greater than or equal to α.Letφp(Dc0+βu(t))=v(t); then v(0)=0,v(1)=ap-1v(η), and we now consider the following BVP:(7)D0+αvt+yt=0,0<t<1,v0=0,v1=ap-1vη,Lemma 4 (see [32]).
Ify∈C[0,1], then BVP (7) has a unique solution(8)vt=∫01Ht,sysds,where(9)
H
t
,
s
=
h
t
,
s
+
a
p
-
1
t
α
-
1
1
-
a
p
-
1
η
α
-
1
,
(10)
h
t
,
s
=
t
1
-
s
α
-
1
-
t
-
s
α
-
1
Γ
α
,
0
≤
s
≤
t
≤
1
,
t
1
-
s
α
-
1
Γ
α
,
0
≤
t
≤
s
≤
1
.For any giveny∈C[0,1], consider the following BVP:(11)D0+αφpDc0+βut+yt=0,0<t<1,u1=u′0=u′′0=···=un-10=0,Dc0+βu0=0,Dc0+βu1=aDc0+βuη.By analysis, we know that (11) can be decomposed into the BVP (7) and the BVP(12)Dc0+βut+φq∫01Ht,sysds=0,0<t<1,u1=u′0=u′′0=···=un-10=0.Lemma 5.
Ify∈C[0,1], then BVP(13)Dc0+βut+yt=0,0<t<1,u1=u′0=u′′0=···=un-10=0has a unique solution(14)ut=∫01Gt,sysds,where (15)
G
t
,
s
=
1
-
s
β
-
1
-
t
-
s
β
-
1
Γ
β
,
0
≤
s
≤
t
≤
1
,
1
-
s
β
-
1
Γ
β
,
0
≤
t
≤
s
≤
1
.Proof.
By Lemma3, BVP (13) is equivalent to the following integral equation: (16)ut=-∫0tt-sβ-1Γβysds+c1+c2t+c3t2⋯+cntn-1.Conditions u′(0)=u′′(0)=⋯=u(n-1)(0)=0 imply that c2=c3=⋯=cn=0. That is, (17)ut=-∫0tt-sβ-1Γβysds+c1.By u(1)=0, we get (18)c1=-∫011-sβ-1Γα1ysds.Combining (16) and (18), we can obtain (14). The proof is completed.Lemma 6.
The Green functionsH(t,s) and G(t,s) defined separately by (9) and (15) have the following properties:
(1)H(t,s) and G(t,s):[0,1]×[0,1]→[0,+∞) are continuous.
(2)(1-s)β-1(1-tβ-1)/Γ(β)≤G(t,s)≤(1-s)β-1/Γ(β).Proof.
Obviously, (1) holds, in the following, and we proof (3). From the definition of G(t,s), for 0≤t≤s≤1, we know that (3) holds.
For0≤s≤t≤1, we have t-ts≥t-s and then (19)1-sβ-1-t-sβ-1≥1-sβ-1-t-tsβ-1≥1-sβ-1-tβ-11-sβ-1=1-sβ-11-tβ-1,so we know that (1-s)β-1(1-tβ-1)/Γ(β)≤G(t,s). It is also defined by G(t,s), and we obtain that G(t,s)≤(1-s)β-1/Γ(β). Thus, we get that (3) holds. The proof is completed.LetX=C[0,1]; then X is a Banach space with the norm u=supt∈[0,1]|u(t)|. For any γ1,γ2∈(0,1), γ1<γ2, denote (20)K=u∈X:mint∈γ1,γ2ut≥ωu,where ω=mint∈[γ1,γ2](1-tβ-1) and then K is a positive cone in X. Define an integral operator T:K→X by(21)Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds,t∈0,1.We know that u is a positive solutions of BVP (1) if and only if u is a fixed point of T in K.Lemma 7.
T : K → K is a completely continuous operator.Proof.
By the routine discussion,T:K→X is well defined and we only prove T(K)⊆K. For any u∈K, t∈[0,1], by Lemma 6, we have(22)Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds≤φqλΓβ∫011-sβ-1φq∫01Hs,τfτ,uτdτds,and then(23)Tu≤φqλΓβ∫011-sβ-1φq∫01Hs,τfτ,uτdτds.On the other hand, for t∈[γ1,γ2], by Lemma 6, we get(24)Tut≥φqλΓβ∫011-sβ-11-tβ-1φq∫01Hs,τfτ,uτdτds=1-tβ-1φqλΓβ∫011-sβ-1φq∫01Hs,τfτ,uτdτds≥ωφqλΓβ∫011-sβ-1φq∫01Hs,τfτ,uτdτds.By (23) and (24), we can prove that Tu(t)≥ωTu. Therefore, we have TK⊆K.
According to the Ascoli-Arzela theorem and the continuity off, we get that T:K→K is completely continuous. The proof is completed.Lemma 8 (see [33]).
LetP be a positive cone in a Banach space; E, Ω1, and Ω2 are bounded open sets in E, θ∈Ω1, and Ω¯1⊂Ω2, A:P∩(Ω¯2∖Ω1)→P is a completely continuous operator. If the conditionsAx≤x, ∀x∈P∩∂Ω1, Ax≥x, and ∀x∈P∩∂Ω2,
orAx≥x, ∀x∈P∩∂Ω1, Ax≤x, and ∀x∈P∩∂Ω2,
are satisfied, thenA has at least one fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

Denote(25)f0=liminfx→0+inft∈γ1,γ2ft,xφpx,f0=limsupx→0+supt∈0,1ft,xφpx,f∞=liminfx→+∞inft∈γ1,γ2ft,xφpx,f∞=limsupx→+∞supt∈0,1ft,xφpx,L1=∫011-sβ-1Γβφq∫01Hs,τdτds,L2=ω2∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τdτds.


### 3.1. Existence of BVP (1)

Theorem 9.
Assumef∞φp(L1-1)>f0φp(L2-1); then BVP (1) has at least one positive solution for(26)λ∈φpL2-1f∞,φpL1-1f0,where we impose 1/f∞=0, if f∞=+∞, and 1/f0=+∞, if f0=0.Proof.
For anyλ satisfying (26), there exists ε0>0 such that(27)φpL2-1f∞-ε0≤λ≤φpL1-1f0+ε0.By the definition of f0, there exists R1>0 such that(28)ft,x≤f0+ε0φpx,0≤x≤R1,t∈0,1.Let KR1={u∈K:u<R1}. For any u∈∂KR1, t∈[0,1], by the definition of ·, we know that (29)ut≤ut≤u≤R1,t∈0,1.Thus, for any u∈∂KR1, by (28) and (29), we have(30)ft,ut≤f0+ε0φput,t∈0,1.Hence, for any u∈∂KR1, by Lemmas 6 and (30), we conclude that(31)Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds≤φqλ∫011-sβ-1Γβφq∫01Hs,τf0+ε0φpuτdτds≤φqλf0+ε0u∫011-sβ-1Γβφq∫01Hs,τdτds≤u.Then, we have (32)Tu≤u,u∈∂KR1.
On the other hand, by the definition off∞, there exists R′>0 such that(33)ft,x≥f∞-ε0φpx,x≥R′,t∈γ1,γ2.Choose R2=maxR′/ω,2R1. Let KR2={u∈K:u<R2}. For any u∈∂KR2, by the definition of ·, we have (34)ut≥ωu≥ωR2≥R′,t∈γ1,γ2.For any u∈∂KR2, by (33) and (34), we have(35)ft,ut≥f∞-ε0φput≥f∞-ε0φpωR2,t∈γ1,γ2.Then, for any u∈∂KR2, by Lemma 6 and (35), we have(36)Tut≥φqλ∫011-sβ-1Γβ1-tβ-1φq∫01Hs,τfτ,uτdτds≥φqλmint∈γ1,γ21-tβ-1∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τfτ,uτdτds≥φqλω∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τf∞-ε0φpuτdτds≥φqλf∞-ε0uω2∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τdτds≥u.
It follows from the above discussion, (32), (36), and Lemmas 7 and 8, for any λ∈φp(L2-1)/f∞,φp(L1-1)/f0, that T has a fixed point u∈K¯R2∖KR1, so BVP (1) has at least one positive solution u; moreover u satisfies R1≤u≤R2. The proof is completed.By the similar proof as Theorem9, the following Theorem 10 holds.Theorem 10.
Assume thatf0φp(L1-1)>f∞φp(L2-1); then BVP (1) has at least one positive solution for λ∈φp(L2-1)/f0,φp(L1-1)/f∞, where we impose 1/f0=0, if f0=+∞, and 1/f∞=+∞, if f∞=0.


### 3.2. Nonexistence of BVP (1)

Theorem 11.
Assume thatf∞<+∞ and f0<+∞; then there exists λ0>0, such that for, λ∈(0,λ0), BVP (1) has no positive solution.Proof.
Byf∞<+∞ and f0<+∞, there exist positive constants M1 and M2 and r1 and r2(r1<r2), such that (37)ft,x≤M1φpx,0≤x≤r1,t∈0,1,ft,x≤M2φpx,x≥r2,t∈0,1.Setting (38)M0=max⁡M1,M2,maxt∈0,1r1≤x≤r2ft,xφpx,we get (39)ft,x≤M0φpx,x≥0,t∈0,1.Assume that u is a positive solution of BVP (1); we will show that it leads to a contradiction. Define λ0=(M0)-1φp(L1-1), and since λ∈(0,λ0), we conclude that(40)ut=Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds≤φqλ∫011-sβ-1Γβφq∫01Hs,τM0φpuτdτds≤φqλM0u∫011-sβ-1Γβφq∫01Hs,τdτds<φqλ0M0u∫011-sβ-1Γβφq∫01Hs,τdτds<u.Then, we have u<u, which is a contradiction. Therefore, BVP (1) has no positive solution for λ∈φp(L2-1)/f0,φp(L1-1)/f∞. The proof is completed.By the similar proof as Theorem11, the following Theorem 12 holds.Theorem 12.
Assume thatf∞>0, f0>0, f(t,x)>0 for t∈[γ1,γ2], and x>0; then there exists λ∗>0, such that, for λ∈(λ∗,+∞), BVP (1) has no positive solution.


## 3.1. Existence of BVP (1)

Theorem 9.
Assumef∞φp(L1-1)>f0φp(L2-1); then BVP (1) has at least one positive solution for(26)λ∈φpL2-1f∞,φpL1-1f0,where we impose 1/f∞=0, if f∞=+∞, and 1/f0=+∞, if f0=0.Proof.
For anyλ satisfying (26), there exists ε0>0 such that(27)φpL2-1f∞-ε0≤λ≤φpL1-1f0+ε0.By the definition of f0, there exists R1>0 such that(28)ft,x≤f0+ε0φpx,0≤x≤R1,t∈0,1.Let KR1={u∈K:u<R1}. For any u∈∂KR1, t∈[0,1], by the definition of ·, we know that (29)ut≤ut≤u≤R1,t∈0,1.Thus, for any u∈∂KR1, by (28) and (29), we have(30)ft,ut≤f0+ε0φput,t∈0,1.Hence, for any u∈∂KR1, by Lemmas 6 and (30), we conclude that(31)Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds≤φqλ∫011-sβ-1Γβφq∫01Hs,τf0+ε0φpuτdτds≤φqλf0+ε0u∫011-sβ-1Γβφq∫01Hs,τdτds≤u.Then, we have (32)Tu≤u,u∈∂KR1.
On the other hand, by the definition off∞, there exists R′>0 such that(33)ft,x≥f∞-ε0φpx,x≥R′,t∈γ1,γ2.Choose R2=maxR′/ω,2R1. Let KR2={u∈K:u<R2}. For any u∈∂KR2, by the definition of ·, we have (34)ut≥ωu≥ωR2≥R′,t∈γ1,γ2.For any u∈∂KR2, by (33) and (34), we have(35)ft,ut≥f∞-ε0φput≥f∞-ε0φpωR2,t∈γ1,γ2.Then, for any u∈∂KR2, by Lemma 6 and (35), we have(36)Tut≥φqλ∫011-sβ-1Γβ1-tβ-1φq∫01Hs,τfτ,uτdτds≥φqλmint∈γ1,γ21-tβ-1∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τfτ,uτdτds≥φqλω∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τf∞-ε0φpuτdτds≥φqλf∞-ε0uω2∫γ1γ21-sβ-1Γβφq∫γ1γ2Hs,τdτds≥u.
It follows from the above discussion, (32), (36), and Lemmas 7 and 8, for any λ∈φp(L2-1)/f∞,φp(L1-1)/f0, that T has a fixed point u∈K¯R2∖KR1, so BVP (1) has at least one positive solution u; moreover u satisfies R1≤u≤R2. The proof is completed.By the similar proof as Theorem9, the following Theorem 10 holds.Theorem 10.
Assume thatf0φp(L1-1)>f∞φp(L2-1); then BVP (1) has at least one positive solution for λ∈φp(L2-1)/f0,φp(L1-1)/f∞, where we impose 1/f0=0, if f0=+∞, and 1/f∞=+∞, if f∞=0.


## 3.2. Nonexistence of BVP (1)

Theorem 11.
Assume thatf∞<+∞ and f0<+∞; then there exists λ0>0, such that for, λ∈(0,λ0), BVP (1) has no positive solution.Proof.
Byf∞<+∞ and f0<+∞, there exist positive constants M1 and M2 and r1 and r2(r1<r2), such that (37)ft,x≤M1φpx,0≤x≤r1,t∈0,1,ft,x≤M2φpx,x≥r2,t∈0,1.Setting (38)M0=max⁡M1,M2,maxt∈0,1r1≤x≤r2ft,xφpx,we get (39)ft,x≤M0φpx,x≥0,t∈0,1.Assume that u is a positive solution of BVP (1); we will show that it leads to a contradiction. Define λ0=(M0)-1φp(L1-1), and since λ∈(0,λ0), we conclude that(40)ut=Tut=φqλ∫01Gt,sφq∫01Hs,τfτ,uτdτds≤φqλ∫011-sβ-1Γβφq∫01Hs,τM0φpuτdτds≤φqλM0u∫011-sβ-1Γβφq∫01Hs,τdτds<φqλ0M0u∫011-sβ-1Γβφq∫01Hs,τdτds<u.Then, we have u<u, which is a contradiction. Therefore, BVP (1) has no positive solution for λ∈φp(L2-1)/f0,φp(L1-1)/f∞. The proof is completed.By the similar proof as Theorem11, the following Theorem 12 holds.Theorem 12.
Assume thatf∞>0, f0>0, f(t,x)>0 for t∈[γ1,γ2], and x>0; then there exists λ∗>0, such that, for λ∈(λ∗,+∞), BVP (1) has no positive solution.


## 4. Examples

Consider the BVP(41)D0+3/2φ3/2Dc0+3/2ut+λft,ut=0,0<t<1,u1=u′0=0,Dc0+3/2u0=0,Dc0+3/2u1=aDc0+3/2u12,where λ>0 is a parameter, α=β=3/2, and η=1/2. Let f(t,x)=x3/1+t; choose [1/4,1/3]⊂(0,1), andthen f∞=+∞, f0=0. By Theorem 9, BVP (41) has at least one positive solution for λ∈(0,+∞).



---

*Source: 1462825-2018-09-03.xml*