# Positive Solutions of Fractional Differential Equations withp-Laplacian

Authors: Yuansheng Tian; Sujing Sun; Zhanbing Bai
Journal: Journal of Function Spaces
 (2017)

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## Abstract

The multiplicity of positive solution for a new class of four-point boundary value problem of fractional differential equations withp-Laplacian operator is investigated. By the use of the Leggett-Williams fixed-point theorem, the multiplicity results of positive solution are obtained. An example is given to illustrate the main results.


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## Body

## 1. Introduction

In recent years, boundary value problems of nonlinear fractional differential equations have been studied extensively (see [1–23] and the references therein). By the use of some fixed-point theorem, the existence results of positive solutions are obtained for singular factional problem [2, 4, 24, 25], impulsive fractional problem [3, 20], nonlocal problem [1, 12, 14, 16, 22, 23], numerical solution problem [17, 18], initial value problem [19, 20], Dirichlet value problem [6, 9, 15], iterative solution problem [5], and so on. Cui [7] considered the following boundary values problems: (1)D0+αxt+ptft,xt+qxt=0,0<t<1,u0=u′0=u1=0, where 2<α≤3 is a real number and D0+α is the standard Riemann-Liouville differentiation. Under the assumption that f(t,x) is a Lipschitz continuous function, by the use of u0-positive operator, they studied the uniqueness results for the fractional differential equation.On the other hand, because of the wide mathematical and physical background, the existence of positive solutions for nonlinear integer-order boundary values problems withp-Laplacian operator has received wide attention (see [8, 12, 13, 21, 24–29]). For example, Su et al. [28] considered the following four-point boundary values problems with p-Laplacian operator: (2)ϕpu′t′+atfut=0,0<t<1,αϕpu0-βϕpu′ξ=0,γϕpu1-δϕpu′η=0,where ϕps=sp-2s,p>1. Liu et al. [13], Dong et al. [8], and Zhang et al. [25] studied p-Laplacian boundary value problems with fractional derivative. By using the fixed-point index theory, they obtained the existence of positive solutions.In this paper, we investigate the multiplicity of positive solution for a new class of four-point boundary value problem of fractional differential equations withp-Laplacian operator: (3)D0+γϕpD0+αut=ft,ut,0<t<1,u0=D0+αu0=0,D0+βu1=λuξ,D0+αu1=μD0+αuη, where α,β,γ∈R,1<α,γ≤2,β>0,1+β≤α and ξ,η∈(0,1),λ,μ∈[0,+∞),(1-β)Γ(α)≥λΓ(α-β)ξα-2,1-μp-1ηγ-2≥0 and ϕps=sp-2s,p>1,D0+α is the standard Riemann-Liouville differentiation and f∈C([0,1]×[0,+∞),[0,+∞)). By using the Leggett-Williams fixed-point theorem on a cone, the multiplicity results of positive solution are obtained.


## 2. Preliminaries

Lemma 1 (see [10]).
Assume thatx∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then (4)I0+αD0+αxt=xt+c1tα-1+c2tα-2+⋯+cNtα-N,ci∈R,i=1,2,…,N,where N is the smallest integer greater than or equal to α.Lemma 2.
Ifh∈C[0,1],ϕp(s)=sp-2s,p>1,ϕq=(ϕp)-1,1/p+1/q=1,α,β,γ∈R,1<α,γ≤2,β>0,1+β≤α and ξ,η∈(0,1),λ,μ∈[0,+∞),M≔λΓ(α-β)ξα-1,N≔μp-1ηγ-1, then the problem (5)D0+γϕpD0+αut=ht,0<t<1,u0=D0+αu0=0,D0+βu1=λuξ,D0+αu1=μD0+αuη has a unique solution (6)ut=∫01Gt,sϕq∫01Ks,τhτdτds, where (7)Gt,s=Γαtα-11-sα-β-1-λΓα-βtα-1ξ-sα-1ΓαΓα-M-t-sα-1Γα,0≤s≤t≤1,s≤ξ,Γαtα-11-sα-β-1-Γα-Mt-sα-1ΓαΓα-M,0<ξ≤s≤t≤1,Γαtα-11-sα-β-1-λΓα-βtα-1ξ-sα-1ΓαΓα-M,0≤t≤s≤ξ<1,tα-11-sα-β-1Γα-M,0≤t≤s≤1,ξ≤s.Ks,τ=s1-τγ-1-μp-1sη-τγ-1-1-Ns-τγ-11-NΓγ,0≤τ≤s≤1,τ≤η,s1-τγ-1-1-Ns-τγ-11-NΓγ,0<η≤τ≤s≤1,s1-τγ-1-μp-1sη-τγ-11-NΓγ,0≤s≤τ≤η<1,s1-τγ-11-NΓγ,0≤s≤τ≤1,η≤τ.Proof.
Supposeu is a solution of the problem (5). By Lemma 1, there is (8)ϕpD0+αut=Iγht+d1tγ-1+d2tγ-2, for some d1,d2∈R. Taking into account the fact that D0+αu(0)=0, we have d2=0 and (9)ϕpD0+αut=1Γγ∫0tt-τγ-1hτdτ+d1tγ-1. Thus, (10)ϕpD0+αu1=1Γγ∫011-τγ-1hτdτ+d1,ϕpD0+αuη=1Γγ∫0ηη-τγ-1hτdτ+d1ηγ-1. By D0+αu(1)=μD0+αu(η), (10), it holds that(11)d1=-∫011-τγ-1Γγ1-Nhτdτ+∫0ημp-1η-τγ-1Γγ1-Nhτdτ. So, (12)ϕpD0+αut=∫0tt-τγ-1Γγhτdτ-∫01tγ-11-τγ-1Γγ1-Nhτdτ+∫0ημp-1tγ-1η-τγ-1Γγ1-Nhτdτ=-∫01Kt,τhτdτ,(13)D0+αut+ϕq∫01Kt,τhτdτ=0. Applying Lemma 1, we can reduce (13) to an equivalent integral equation (14)ut=-I0+αϕq∫01Kt,τhτdτ+c1tα-1+c2tα-2, for some c1,c2∈R. With the condition u(0)=0, there is c2=0. Consequently, (15)ut=-I0+αϕq∫01Kt,τhτdτ+c1tα-1. By (15), one has (16)D0+βut=-D0+βI0+αϕq∫01Kt,τhτdτ+c1D0+βtα-1=-I0+α-βϕq∫01Kt,τhτdτ+c1ΓαΓα-βtα-β-1. So, (17)Dβu1=-∫011-sα-β-1Γα-βϕq∫01Ks,τhτdτds+c1ΓαΓα-β,uξ=-∫0ξξ-sα-1Γαϕq∫01Ks,τhτdτds+c1ξα-1. By Dβu(1)=λu(ξ), (17), we have (18)c1=Γα-βΓα-M∫011-sα-β-1Γα-βϕq∫01Ks,τhτdτds-λ∫0ξξ-sα-1Γαϕq∫01Ks,τhτdτds. So, the unique solution of problem (5) is (19)ut=-∫0tt-sα-1Γαϕq∫01Ks,τhτdτds+Γα-βtα-1Γα-M∫011-sα-β-1Γα-βϕq∫01Ks,τhτdτds-λ∫0ξξ-sα-1Γαϕq∫01Ks,τhτdτds=∫01Gt,sϕq∫01Ks,τhτdτds. The proof is completed.Lemma 3.
Suppose that(1-β)Γ(α)≥λΓ(α-β)ξα-2,1-μp-1ηγ-2≥0. The functions G(t,s) and K(t,s) satisfy the following:(1)
G(t,s),K(t,s)∈C([0,1]×[0,1]);G(t,s)>0,K(t,s)>0 for t,s∈(0,1),(2)
G(t,s)≤G(s,s),K(t,s)≤K(s,s) for t,s∈[0,1],(3)
There exists a positive functionψ(s)∈C(0,1) such that (20)minξ≤t≤1Gt,s≥ψsmax0≤t≤1Gt,s=ψsGs,s,s∈0,1.Proof.
The Proof of the Statement (1). From definitions, it is clear that G(t,s),K(t,s)∈C([0,1]×[0,1]). By (1-β)Γ(α)≥λΓ(α-β)ξα-2,1-μp-1ηγ-2≥0, we have (21)Γα≥1-βΓα≥λΓα-βξα-2>λΓα-βξα-1=M,1-N=1-μp-1ηγ-1>1-μp-1ηγ-2≥0.
If0<s≤t<1,s≤ξ, let(22)gt,s=tα-11-sα-β-1-t-sα-1Γα. It is obvious that g(t,s)>0 for 0<s≤t<1. Hence, we have (23)Gt,s=Γαtα-11-sα-β-1-Γα-Mt-sα-1-λΓα-βtα-1ξ-sα-1ΓαΓα-M=1Γα1+MΓα-Mtα-11-sα-β-1-t-sα-1Γα-λΓα-βtα-1ξ-sα-1ΓαΓα-M=tα-11-sα-β-1-t-sα-1Γα+λΓα-βtα-1ξα-11-sα-β-1-ξ-sα-1ΓαΓα-M=gt,s+λΓα-βtα-1Γα-Mgξ,s>0.
By using the analogous argument, it holds thatG(t,s)>0 for other situations. Hence, G(t,s)>0 for t,s∈(0,1).
If0<s≤t<1,s≤η, let (24)kt,s=t1-sγ-1-t-sγ-1Γγ. It is obvious that k(t,s)>0 for 0<s≤t<1. Hence, we have (25)Kt,s=t1-sγ-1-μp-1tη-sγ-1-1-Nt-sγ-11-NΓγ=1+N1-Nt1-sγ-1Γγ-t-sγ-1Γγ-μp-1tη-sγ-11-NΓγ=t1-sγ-1-t-sγ-1Γγ+μp-1tγ-1ηγ-11-sγ-1-η-sγ-11-NΓγ=kt,s+μp-1tγ-11-Nkη,s>0. Similarly, it holds thatK(t,s)>0 for 0<η≤s≤t<1 or 0<t≤s≤η<1 or 0<t≤s<1,η≤s.
Hence,K(t,s)>0 for t,s∈(0,1).
The Proof of the Statement (2). Let (26)g1t,s=Γαtα-11-sα-β-1-λΓα-βtα-1ξ-sα-1ΓαΓα-M-t-sα-1Γα,0≤s≤t≤1,s≤ξ,g2t,s=Γαtα-11-sα-β-1-Γα-Mt-sα-1ΓαΓα-M,0<ξ≤s≤t≤1,g3t,s=Γαtα-11-sα-β-1-λΓα-βtα-1ξ-sα-1ΓαΓα-M,0≤t≤s≤ξ<1,g4t,s=tα-11-sα-β-1Γα-M,0≤t≤s≤1,ξ≤s.
It is easy to check thatg3(t,s) and g4(t,s) are increasing with respect to t on [0,s]. We will show that g1(t,s) and g2(t,s) are decreasing with respect to t on [s,1]. For 0≤s≤t≤1,s≤ξ, let (27)h1t,s=ΓαΓα-Mg1t,s; then we have (28)h1t,s=Γαtα-11-sα-β-1-Γα-Mt-sα-1-λΓα-βtα-1ξ-sα-1,∂h1t,s∂t=α-1tα-2Γα1-sα-β-1-Γα-M1-stα-2-λΓα-βξ-sα-1≤α-1tα-2Γα1-sα-β-1-Γα-M1-sα-2-λΓα-βξ-sα-1=α-1tα-21-sα-2Γα1-s1-β-Γα-M-λΓα-βξ-sα-11-s2-α≤α-1tα-21-sα-2Γα1-1-βs-Γα-M-λΓα-βξ-sα-11-s2-α=α-1tα-21-sα-2M-Γα1-βsλΓα-βξ-sα-11-s2-α.
For0≤s≤ξ, let (29)ωs=M-Γα1-βs-λΓα-βξ-sα-11-s2-α=λΓα-βξα-1-Γα1-βs-λΓα-βξ-sα-11-s2-α. We have (30)ω0=λΓα-βξα-1-λΓα-βξα-1=0,ωξ=λΓα-βξα-1-1-βξΓα=ξλΓα-βξα-2-1-βΓα≤0,ω′s=β-1Γα+λΓα-βξ-sα-11-s2-αα-11-s+ξ-s2-α1-sξ-s.
If there existss∗∈(0,ξ) such that ω′(s∗)=0, then (31)λΓα-βξ-s∗α-11-s∗2-α=1-βΓα1-s∗ξ-s∗α-11-s∗+ξ-s∗2-α. Therefore, we have (32)ωs∗=λΓα-βξα-1-1-βΓαs∗+1-s∗ξ-s∗α-11-s∗+ξ-s∗2-α=λΓα-βξα-1-1-βΓα×s∗α-11-s∗+ξ-s∗2-α+1-s∗ξ-s∗α-11-s∗+ξ-s∗2-α=λΓα-βξα-1-1-βΓαξα-11-s∗-ξ-s∗α-11-s∗α-11-s∗+ξ-s∗2-α+ξξ-s∗2-α-ξ-s∗22-α+1-s∗ξ-s∗α-11-s∗+ξ-s∗2-α=λΓα-βξα-1-1-βΓαξ+ξ-s∗2-α1-s∗-ξ-s∗22-αα-11-s∗+ξ-s∗2-α=λΓα-βξα-1-1-βΓαξ+ξ-s∗2-α1-ξα-11-s∗+ξ-s∗2-α=λΓα-βξα-1-1-βΓαξ-1-βΓαξ-s∗2-α1-ξα-11-s∗+ξ-s∗2-α≤λΓα-βξα-2-1-βΓαξ≤0. In fact, taking into account the fact that λΓ(α-β)ξα-2-(1-β)Γ(α)≤0 and 2-α≥0, one has (33)-1-βΓαξ-s∗2-α1-ξα-11-s∗+ξ-s∗2-α≤0. So, we have max0≤s≤ξω(s)≤0, which implies that ∂h1(t,s)/∂t≤0. Hence g1(t,s) is decreasing with respect to t on [s,1].
For0<ξ≤s≤t≤1, let h2(t,s)=Γ(α)(Γ(α)-M)g2(t,s); then we have (34)h2t,s=Γαtα-11-sα-β-1-Γα-Mt-sα-1,(35)∂h2t,s∂t=α-1tα-2Γα1-sα-β-1-Γα-M1-stα-2≤α-1tα-2Γα1-sα-β-1-Γα-M1-sα-2=α-1tα-21-sα-2Γα1-s1-β-Γα-M≤α-1tα-21-sα-2Γα1-1-βs-Γα-M≤α-1tα-21-sα-2λΓα-βξα-1-1-βξΓα=α-1tα-21-sα-2ξλΓα-βξα-2-1-βΓα≤0, which implies that g2(t,s) is decreasing with respect to t on [s,1].
We can conclude thatG(t,s) is increasing with respect to t for t≤s and G(t,s) is decreasing with respect to t for t≥s. Hence, G(t,s)≤G(s,s) for t,s∈[0,1].
By using the analogous method, we can conclude thatK(t,s) is increasing with respect to t for t≤s and K(t,s) is decreasing with respect to t for t≥s. Hence, K(t,s)≤K(s,s) for t,s∈[0,1].
The Proof of the Statement (3). Taking into account the definition of G(t,s), there is (36)minξ≤t≤1Gt,s=minξ≤t≤1g1t,s,0≤s≤ξ,minminξ≤t≤1g2t,s,minξ≤t≤1g4t,s,ξ≤s≤1,=g11,s,0≤s≤ξ,φs,ξ≤s≤1, where φ(s)=min⁡{g2(1,s),g4(ξ,s)}. It is easy to see that φ(s) is continuous on [ξ,1] and φ(s)>0 for all s∈[ξ,1). Let (37)ψs=g11,sGs,s,0<s≤ξ,φsGs,s,ξ≤s<1, where (38)Gs,s=Γαsα-11-sα-β-1-λΓα-βsα-1ξ-sα-1ΓαΓα-M,0≤s≤ξ,sα-11-sα-β-1Γα-M,ξ≤s≤1. Then (39)minξ≤t≤1Gt,s≥ψsmax0≤t≤1Gt,s=ψsGs,s,s∈0,1. The proof is completed.Lemma 4 (see [11]).
LetP be a cone in a real Banach space E,Pc={u∈P∣u≤c},θ be a nonnegative continuous concave functional on a cone P such that θ(u)≤u, for all u∈P¯c, and P(θ,b,d)={u∈P∣b≤θ(u),u≤d}. Suppose T:P¯c→P¯c is completely continuous and there exist constants 0<a<b<d≤c such that(A1)
{u∈P(θ,b,d)∣θ(u)>b}≠∅ and θ(Tu)>b for u∈P(θ,b,d);(A2)
Tu<a for u≤a;(A3)
θ(Tu)>b for u∈P(θ,b,c) with Tu>d.
ThenT has at least three fixed points u1,u2,u3 satisfyingu1<a,b<θ(u2) and u3>a with θ(u3)<b.


## 3. Main Result

LetE=C[0,1] be a Banach space with u=max0≤t≤1ut. Define the cone P⊂E by P={u∈E∣u(t)≥0,0≤t≤1}. Let the nonnegative continuous concave functional θ on the P be defined by (40)θu=minξ≤t≤1ut.Lemma 5.
LetT:P→E be an operator defined by (41)Tut=∫01Gt,sϕq∫01Ks,τfτ,uτdτds. Then T:P→P is completely continuous.Proof.
T
:
P
→
P is continuous in view of nonnegativity and continuity of G(t,s),K(t,s) and f(t,u). Furthermore it is easy to see that by the Arzela-Ascoli theorem and Lebesgue dominated convergence theorem T:P→P is completely continuous.For convenience, we introduce the following notations:(42)N=∫01Gs,sϕq∫01Ks,τdτds-1,L=∫ξ1ψsGs,sϕq∫01Ks,τdτds-1.Theorem 6.
Supposef∈C([0,1]×[0,+∞),[0,+∞)) and there exist constants 0<a<b<c such that(B1)
f(t,u)<ϕp(Na), for (t,u)∈[0,1]×[0,a];(B2)
f(t,u)≥ϕp(Lb), for (t,u)∈[ξ,1]×[b,c];(B3)
f(t,u)≤ϕp(Nc), for (t,u)∈[0,1]×[0,c].
Then problem (3) possesses at least three positive solutions u1,u2, and u3 with (43)max0≤t≤1u1t<a,b<minξ≤t≤1u2t<max0≤t≤1u2t≤c,a<max0≤t≤1u3t≤c,minξ≤t≤1u3t<b.Proof.
In order to apply Lemma4, we divide the proof into four steps.
Step  1. If u∈P¯c, then u≤c. Assumption (B3) implies f(t,u(t))≤ϕp(Nc) for 0≤t≤1. Consequently, (44)Tu=max0≤t≤1∫01Gt,sϕq∫01Ks,τfτ,uτdτds≤∫01Gs,sϕq∫01Ks,τϕpNcdτds≤Nc∫01Gs,sϕq∫01Ks,τdτds=c. Hence T:P¯c→P¯c.
Step  2. We claim that the condition (A1) of Lemma 4 is satisfied. Let u(t)=(b+c)/2,0≤t≤1. we can easily see that u(t)=(b+c)/2∈P(θ,b,c),θ(u)=θ((b+c)/2)>b; consequently, {u∈P(θ,b,c)∣θ(u)>b}≠∅. Hence, if u∈P(θ,b,c), then b≤u(t)≤c for ξ≤t≤1. From assumption (B2), we have (45)ft,ut≥ϕpLb,ξ≤t≤1. So (46)θTu=minξ≤t≤1Tut=minξ≤t≤1∫01Gt,sϕq∫01Ks,τfτ,uτdτds≥minξ≤t≤1∫01Gt,sϕq∫01Ks,τϕpLbdτds>∫ξ1ψsGs,sϕq∫01Ks,τϕpLbdτds=Lb∫ξ1ψsGs,sϕq∫01Ks,τdτds=b. Then, we have (47)θTu>b,∀u∈Pθ,b,c. This implies that condition (A1) of Lemma 4 is satisfied.
Step  3. We now prove the condition (A2) of Lemma 4 is satisfied. If u∈P¯a, then u≤a. Assumption (B1) implies f(t,u(t))<ϕp(Na) for 0≤t≤1. Thus (48)Tu=max0≤t≤1∫01Gt,sϕq∫01Ks,τfτ,uτdτds<∫01Gs,sϕq∫01Ks,τϕpNadτds=Na∫01Gs,sϕq∫01Ks,τdτds=a. Hence, the condition (A2) of Lemma 4 is also satisfied.
Step  4. Finally, we prove that the condition (A3) of Lemma 4 is satisfied. If u∈P(θ,b,c), byStep  2, we have θ(Tu)>b. Hence, the condition (A3) of Lemma 4 is also satisfied.
By Lemma4, problem (3) has at least three positive solutions u1,u2, and u3 with (49)max0≤t≤1u1t<a,b<minξ≤t≤1u2t<max0≤t≤1u2t≤c,a<max0≤t≤1u3t≤c,minξ≤t≤1u3t<b.The proof is complete.Example 7.
Consider the following boundary value problem:(50)D3/2ϕ3/2D3/2ut=ft,ut,0<t<1,u0=D3/2u0=0,D1/2u1=14u12,D3/2u1=18D3/2u14, where (51)ft,u=18t2+2u2,u<1,137+18t2+17u,u≥1. Choose p=α=γ=3/2,β=ξ=1/2,η=λ=1/4,μ=1/8. It is easy to check that (1-β)Γ(α)-λΓ(α-β)ξα-2>0 and 1-μp-1ηγ-2>0; by simple computation, we have (52)N≈1.3788,L≈3.9213.
Leta=1/4,b=1,c=7; one can check that the function f(t,u) satisfies(S1)
f(t,u)=1/8t2+2u2≤0.25<ϕ3/2(Na)≈0.5871, for (t,u)∈[0,1]×[0,1/4];(S2)
f(t,u)=13/7+1/8t2+1/7u>2.0312>ϕ3/2(Lb)≈1.9802, for (t,u)∈[1/2,1]×[1,7];(S3)
f(t,u)=13/7+1/8t2+1/7u<2.9822<ϕ3/2(Nc)≈3.1067, for (t,u)∈[0,1]×[0,7].
That is to say that all the conditions of Theorem6 hold. Thus Theorem 6 implies the problem (50) has at least three positive solutions u1,u2, and u3 such that (53)max0≤t≤1u1t<14,1<min1/2≤t≤1u2t<max0≤t≤1u2t≤7,14<max0≤t≤1u3t≤7,min1/2≤t≤1u3t<1.



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