The multiplicity of positive solution for a new class of four-point boundary value problem of fractional differential equations with p-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.

National Natural Science Foundation of China11571207Education Department of Hunan Province16A198Natural Science Foundation of Hunan Province2015JJ6101Construct Program of the Key Discipline in Hunan Province1. 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 with p-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 with p-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. PreliminariesLemma 1 (see [<xref ref-type="bibr" rid="B12">10</xref>]).

Assume that x∈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.

If h∈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.

Suppose u 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:

G(t,s),K(t,s)∈C([0,1]×[0,1]);G(t,s)>0,K(t,s)>0 for t,s∈(0,1),

G(t,s)≤G(s,s),K(t,s)≤K(s,s) for t,s∈[0,1],

There exists a positive function ψ(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.

If 0<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 that G(t,s)>0 for other situations. Hence, G(t,s)>0 for t,s∈(0,1).

If 0<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 that

K(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 that g3(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.

For 0≤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 exists s∗∈(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].

For 0<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 that G(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 that K(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 [<xref ref-type="bibr" rid="B13">11</xref>]).

Let P 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

{u∈P(θ,b,d)∣θ(u)>b}≠∅ and θ(Tu)>b for u∈P(θ,b,d);

Tu<a for u≤a;

θ(Tu)>b for u∈P(θ,b,c) with Tu>d.

Then T has at least three fixed points u1,u2,u3 satisfying

u1<a,b<θ(u2) and u3>a with θ(u3)<b.

3. Main Result

Let E=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.

Let T: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.

Suppose f∈C([0,1]×[0,+∞),[0,+∞)) and there exist constants 0<a<b<c such that

f(t,u)<ϕp(Na), for (t,u)∈[0,1]×[0,a];

f(t,u)≥ϕp(Lb), for (t,u)∈[ξ,1]×[b,c];

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 Lemma 4, 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), by Step 2, we have θ(Tu)>b. Hence, the condition (A3) of Lemma 4 is also satisfied.

By Lemma 4, 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.

Let a=1/4,b=1,c=7; one can check that the function f(t,u) satisfies

f(t,u)=1/8t2+2u2≤0.25<ϕ3/2(Na)≈0.5871, for (t,u)∈[0,1]×[0,1/4];

f(t,u)=13/7+1/8t2+1/7u>2.0312>ϕ3/2(Lb)≈1.9802, for (t,u)∈[1/2,1]×[1,7];

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 Theorem 6 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.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The Project is supported by NSFC (11571207), the Scientific Research Foundation of Hunan Provincial Education Department (16A198), the Hunan Provincial Natural Science Foundation of China (2015JJ6101), and the Construct Program of the Key Discipline in Hunan Province.

AhmadB.SivasundaramS.On four-point nonlocal boundary value problems of nonlinear integro-differential equations of fractional orderBaiZ.ChenY.LianH.SunS.On the existence of blow up solutions for a class of fractional differential equationsBaiZ.DongX.YinC.Existence results for impulsive nonlinear fractional differential equation with mixed boundary conditionsBaiZ.QiuT.Existence of positive solution for singular fractional differential equationBaiZ.ZhangS.SunS.YinC.Monotone iterative method for fractional differential equationsZhouY.ChenF.LuoX.Existence results for nonlinear fractional difference equationCuiY.Uniqueness of solution for boundary value problems for fractional differential equationsDongX.BaiZ.ZhangS.Positive solutions to boundary value problems of p-Laplacian with fractional derivativeJiaoF.ZhouY.Existence results for fractional boundary value problem via critical point theoryKilbasA. A.SrivastavaH. M.TrujilloJ. J.LeggettR. W.WilliamsL. R.Multiple positive fixed points of nonlinear operators on ordered Banach spacesLiuX.JiaM.GeW.The method of lower and upper solutions for mixed fractional four-point boundary value problem with p-Laplacian operatorLiuX.JiaM.GeW.Multiple solutions of a p-Laplacian model involving a fractional derivativeQiT.LiuY.CuiY.Existence of solutions for a class of coupled fractional differential systems with nonlocal boundary conditionsSongQ.DongX.BaiZ.ChenB.Existence for fractional Dirichlet boundary value problem under barrier strip conditionsTianY.Positive solutions to m-point boundary value problem of fractional differential equationWangZ.A numerical method for delayed fractional-order differential equationsWangZ.HuangX.ZhouJ.A numerical method for delayed fractional-order differential equations: based on G-L definitionWeiZ.LiQ.CheJ.Initial value problems for fractional differential equations involving Riemann-Liouville sequential fractional derivativeWangJ.FeckanM.ZhouY.Fractional order differential switched systems with coupled nonlocal initial and impulsive conditionsZhangX.GeW.Impulsive boundary value problems involving the one-dimensional p-LaplacianZouY.LiuL.CuiY.The existence of solutions for four-point coupled boundary value problems of fractional differential equations at resonanceZouY.CuiY.Existence results for a functional boundary value problem of fractional differential equationsTianY.ChenA.GeW.Multiple positive solutions to multipoint one-dimensional p-Laplacian boundary value problem with impulsive effectsZhangX.LiuL.WuY.CuiY.Entire blow-up solutions for a quasilinear p-Laplacian Schrödinger equation with a non-square diffusion termChenC.SongH.YangH.Liouville type theorems for stable solutions of p-Laplace equation in RNCuiY.SunJ.A generalization of Mahadevan's version of the Krein-Rutman theorem and applications to p-Laplacian boundary value problemsSuH.WeiZ.WangB.The existence of positive solutions for a nonlinear four-point singular boundary value problem with a p-Laplacian operatorWangY.HouC.Existence of multiple positive solutions for one-dimensional p-Laplacian