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Research article

Existence results for a class of nonlinear singular p-Laplacian Hadamard fractional differential equations

  • Received: 17 October 2023 Revised: 18 December 2023 Accepted: 02 January 2024 Published: 16 January 2024
  • Based on properties of Green's function and the some conditions of f(t,u), we found a minimal and a maximal positive solution by the method of sequence approximation. Moreover, based on the properties of Green's function and fixed point index theorem, the existence of multiple positive solutions for a singular p-Laplacian fractional differential equation with infinite-point boundary conditions was obtained and, at last, an example was given to demonstrate the validity of our main results.

    Citation: Limin Guo, Weihua Wang, Cheng Li, Jingbo Zhao, Dandan Min. Existence results for a class of nonlinear singular p-Laplacian Hadamard fractional differential equations[J]. Electronic Research Archive, 2024, 32(2): 928-944. doi: 10.3934/era.2024045

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  • Based on properties of Green's function and the some conditions of f(t,u), we found a minimal and a maximal positive solution by the method of sequence approximation. Moreover, based on the properties of Green's function and fixed point index theorem, the existence of multiple positive solutions for a singular p-Laplacian fractional differential equation with infinite-point boundary conditions was obtained and, at last, an example was given to demonstrate the validity of our main results.



    Fractional-order differential operator has become one of the most important tools for mathematical modeling of complex mechanics and physical processes because it can describe mechanical and physical processes with historical memory and spatial global correlation succinctly and accurately. Additionally, and the fractional-order derivative modeling is simple, the physical meaning of parameters is clear and the description is accurate. In recent years, fractional derivative has become an important tool to describe all kinds of complex mechanical and physical behaviors, so the study of a positive solutions of fractional differential equations has attracted much attention. For an extensive collection of such literature, readers can refer to [1,2,2,3,4,5,6,7,8,9,10,11,12]. Alsaedi et al. [13] investigated the following equation

    ρDα1(φP(ρDβ0+u(t)))=v1f(t,u(t),ρDβ0+u(t)))+vρ2Iξ0+g(t,u(t),ρDβ0+u(t)))

    with boundary value condition

    u(0)=0,u(1)=λ1u(μ),ρDβ0+u(1)=0,φp(ρDβ0+u(0))=λ2φp(ρDβ0+u(η))

    where φp(t)=|t|p2t, 1p+1q=1,p,q>1,1<α,β2,ρ>0,ζ>0,0<μ,η<1, 0λ1<1μρ(β1),0λ2<1(1ηρ)α1, and ρDα0+u and ρDβ0+u denote the right and left fractional derivatives of orders α and β with respect to a power function, respectively. The authors proved the uniqueness of positive solutions for the given problem for the cases 1<p2 and p>2 by applying an efficient novel approach together with the Banach contraction mapping principle. Li and Liu [8] considered the fractional differential equation

    C0Dαgu(t)+f(t,u)=0,0<t<1

    with boundary value condition

    x(0)=0, C0D1gu(0)=0, C0D1gu(1)=10h(t)C0Dvgu(t)g(t)dt

    where 2<α<3,1<v<2,αv1>0, fC([0,1]×R+,R+), g>0, hC([0,1],R+), R+=[0,+), and the existence of multiple solutions for the following system by the fixed point theorem are on cone.

    In recent years, more and more scientists have devoted themselves to the study of Hadamard fractional differential systems. For the part of outstanding results of Hadamard's research on fractional differential systems, please refer to [14,15,16]. Ardjouni[14] studied the following Hadamard fractional differential equations

    HDα1+u(t)+ϕ(t,u(t))=HDβ1+φ(t,u(t)),1<t<e

    with integral boundary conditions

    u(1)=0,u(e)=1Γ(αβ)e1(loges)αβ1g(s,x(s))dss

    where 1<α2,0<βα1, g,f:[1,e]×[0,)[0,) are given continuous functions, ϕ is not required for any monotone assumption, and φ is nondecreasing on x. The authors get the existence and uniqueness of the positive solution by the method of upper and lower solutions and Schauder and Banach fixed point theorems. In [15], Berhail and Tabouche studied the following fractional differential equation

    HDα1+u(t)=f(t,u(t),u(t)),1<t<T,T<e

    with integral boundary conditions

    u(1)=0,u(3)(1)=0,u(1)+au(1)=T1g1(s)u(s)ds,u(T)bu(T)=T1g2(s)u(s)ds

    where 3<α4, g1,g2C([1,T],[0,+)), a,b>0, and HDα1+ denotes the Hadamard factional order of α. Based on the properties of Green's function, the authors get the existence of a positive solution for the equation in [15] by the Avery-Peterson fixed point theorem. In [16], the authors studied the existence of a positive solution and stability analysis of the following equation

    HDβ1+(ϕp(Dα1+x))(t)=f(t,x),1<t<e

    with integral boundary conditions

    x(1)=HDα1+x(1)=u(1)=u(e)=0,ϕp(HDα1+u(e))=μe1ϕp(HDα1+x(t))dtt

    where α,β, and μ are three positive real numbers with α(2,3],β(1,2], and μ[0,β), ϕp(s)=|s|p2s is the P-Laplacian for p>1,sR, f is a continuous function on [1,e]×R, and HDα1+,HDβ1+ is the Hadamard fractional differential equation.The authors get the positive solutions by using the fixed point methods.

    Motivated by the excellent results above, in this paper, we will devote to considering the following infinite-point singular p-Laplacian Hadamard fractional differential equation:

    HDα1+(φp(HDγ1+u))(t)+f(t,u(t))=0,1<t<e (1.1)

    with infinite-point boundary condition

    u(j)(1)=0,j=0,1,2,,n2;HDr11+u(e)=j=1ηHjDr21+u(ξj),HDγ1+u(1)=0;φp(HDγ1+u(e))=i=1ζjφp(HDγ1+u(ξj)) (1.2)

    where α,γR+=[0,+), 1<α2, n1<γn(n3), r1,r2[2,n2], r2r1, p-Laplacian operator φp is defined as φp(s)=|s|p2s, p, q>1, 1p+1q=1, and 0<ηi,ζi<1,1<ξi<e(i=1,2,,), fC([1,e]×R1+,R1+))(R1+=[0,+), and HDα1+u,HDγ1+u, HDri1+u(i=1,2) are the standard Hadamard fractional-order derivatives.

    In this paper, we investigate the existence of positive solutions for a singular infinite-point p-Laplacian boundary value problem. Compared with [15], the equation in this paper is a p-Laplacian fractional differential equation and the method in which we used is a fixed point index and sequence approximation. Compared with [16,17], value at infinite points are involved in the boundary conditions of the boundary value problem (1.1, 1.2) and the minimal positive solution and maximal positive solution are obtained in this paper.

    For some basic definitions and lemmas about the theory of Hadamard fractional calculus, the reader can refer to the recent literature such as[6,9,18].

    Definition 2.1([18,19]). The Hadamard fractional integral of α(α>0) order of a function :(0,)R1+ is given by

    HIα1+(t)=1Γ(α)t1(lnts)α1(s)sds

    Definition 2.2([18,19]). The Hadamard fractional derivative of α(α>0) order of a continuous function :(0,)R1+ is given by

    HDα1+(t)=1Γ(nα)(tddt)nt1(s)s(lnts)αn+1ds

    where n=[α]+1 and  [α] denotes the integer part of the number α, provided that the righthand side is pointwise defined on (0,).

    Lemma 2.1([18,19]). If α,β>0, then

    HIα1+(lnx)β1=Γ(β)Γ(β+α)(lnx)β+α1,HDα1+(lnx)β1=Γ(β)Γ(βα)(lnx)βα1

    Lemma 2.2([19]). Suppose that α>0 and C[0,)L1[0,), then the solution of Hadamard fractional differential equation HDα1+(t)=0 is

    (t)=c1(lnt)α1+c2(lnt)α2++cn(lnt)αn,ciR(i=0,1,,n),n=[α]+1

    Lemma 2.3([19]). Suppose that α>0,α is not a natural number and C[1,)L1[1,), then

    (t)=HIα1+HDα1+(t)+nk=1ck(lnt)αk

    for t(1,e] where ckR(k=1,2,,n), and n=[α]+1.

    Lemma 2.4. Let yL1(1,e)C(1,e), then the equation of the BVPs

    HDγ1+u(t)=y(t), 1<t<e (2.1)

    with boundary condition u(j)(1)=0(j=0,1,2,,n2),HDr11+u(e)=j=1ηHjDr21+u(ξj) has integral representation

    u(t)=e1Ψ(t,s)y(s)dss (2.2)

    where

    Ψ(t,s)=1ΔΓ(γ){Γ(γ)(lnt)γ1Ξ(s)(lnelns)γr11Δ(lntlns)γ1,1ste,Γ(γ)(lnt)γ1Ξ(s)(lnelns)γr11, 1tse (2.3)

    in which

    Ξ(s)=1Γ(γr1)1Γ(γr2)sξjηj(lnξjlnslnelns)γr21(lnelns)r1r2, 
    Δ=Γ(γ)Γ(γr1)Γ(γ)Γ(γr2)j=1ηjlnξγr21j0 (2.4)

    Proof.By means of the Lemma 2.3, we reduce (2.1) to an equivalent integral equation

    u(t)=HIγ1+y(t)+C1(lnt)γ1+C2(lnt)γ2++Cn(lnt)γn

    for Ci(i=1,2,,n)R. From u(0)=0, we have Cn=0, then taking the first derivative, we have

    u(t)=HIγ11+y(t)+C1(γ1)(lnt)γ11t+C2(γ2)(lnt)γ31t++

    By u(1)=0, we have Cn1=0, and taking the derivative step by step and combining u(i)(1)=0,(i=2,,n2), we have Ci=0(i=2,3,,n2). Consequently, we get

    u(t)=C1(lnt)γ1HIγ1+y(t)

    By some properties of the fractional integrals and fractional derivatives, we have

    HDr11+u(t)=C1Γ(γ)Γ(γr1)(lnt)γr11HIγr11+y(t),HDr21+u(t)=C1Γ(γ)Γ(γr2)(lnt)γr21HIγr21+y(t) (2.5)

    On the other hand, HDr11+u(1)=j=1ηHjDr21+u(ξj), and combining with (2.5), we get

    C1=e1(lnelns)γr11Γ(γr1)Δy(s)dssj=1ηjξj1(lnξjlns)γr21Γ(βr2)Δy(s)dss=e1(lnelns)γr11Ξ(s)Δy(s)dss

    where Ξ(s),Δ are as (2.3), then,

    u(t)=C1(lnt)γ1HIγ1+y(t)=t1Δ(lntlns)γ1Γ(γ)Δy(s)dss+e1(lnelns)γr11(lnt)γ1Ξ(s)Δy(s)dss=e1Ψ(t,s)y(s)dss

    Therefore, Ψ(t,s) is as (2.3).

    Lemma 2.5. The Green functions (2.3) have the following properties:

    (i)Ψ(t,s)>0,tΨ(t,s)>0,1<t,s<e;

    (ii)maxt[1,e]Ψ(t,s)=Ψ(e,s)=1ΔΓ(γ)[Γ(γ)Ξ(s)(lnelns)γr11Δ(lnelns)γ1];

    (iii)Ψ(t,s)(lnt)α1Ψ(e,s),1t,se

    Proof. (i) By simple calculation, we have Ξ(s)>0. For s[1,e], we have Ξ(s)Ξ(1), then by the expression of Ξ(s) and Δ, we have

    Γ(γ)Ξ(s)Δ=Γ(γ)Ξ(1)

    For 1<st<e by the preceding formula, we have

    Ψ(t,s)=1ΔΓ(γ)[Γ(γ)(lnt)γ1Ξ(s)(lnelns)γr11Δ(lntlns)γ1]=1ΔΓ(γ)(lnt)γ1[Γ(γ)Ξ(s)(lnelns)γr11Δ(1lnslnt)γ1]1Γ(γ)(lnt)γ1[(lnelns)γ1(1lnslnt)γ1]0

    For 1<ts<e, obviously, Ψ(t,s)>0. Furthermore, by direct calculation, we get

    tΨ(t,s)=1ΔΓ(γ){(γ1)Γ(γ)(lnt)γ2Ξ(s)(lnelns)γr11Δ(lntlns)γ1t,1ste,(γ1)Γ(γ)(lnt)γ2Ξ(s)(lnelns)γr11t, 1tse (2.6)

    Clearly, tΨ(t,s) is continuous on [1,e]×[1,e]. By the similar method, for 1<st<e, we get

    tΨ(t,s)=1ΔΓ(γ)t[(γ1)Γ(γ)(lnt)γ2Ξ(s)(lnelns)γr11Δ(γ1)(lntlns)γ2]=γ1ΔΓ(γ)(lnt)γ2[Γ(γ)Ξ(s)(lnelns)γr11Δ(1lnslnt)γ2]γ1Γ(γ)t(lnt)γ2[(lnelns)γr11(1lnslnt)γ2]γ1Γ(γ)t(lnt)γ2[(lnelns)γ2(1lnslnt)γ2]0

    (ii) By (i), we can easily get Ψ(t,s) is increasing on t; hence, we get

    maxt[1,e]Ψ(t,s)=Ψ(e,s)=1ΔΓ(γ)[Γ(γ)Ξ(s)(lnelns)γr11Δ(lnelns)γ1],1se

    (iii) For 1ste, we have

    Ψ(t,s)=1ΔΓ(γ)[Γ(γ)(lnt)γ1Ξ(s)(lnelns)γr11Δ(lntlns)γ1]=1ΔΓ(γ)(lnt)γ1[Γ(γ)Ξ(s)(lnelns)γr11Δ(1lnslnt)γ1]1ΔΓ(γ)(lnt)γ1[Γ(γ)Ξ(s)(lnelns)γr11Δ(1lns)γ1]=(lnt)γ1Ψ(e,s)

    For 1tse, we have

    Ψ(t,s)=1Δ(lnt)γ1Ξ(s)(lnelns)γr11=(lnt)γ11ΔΞ(s)(lnelns)γr11(lnt)γ1Ψ(e,s)

    Lemma 2.6. Let fC([1,e]×(0,+),[0,+)), then the BVP (1.1,1.2) has a solution

    u(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dss (2.7)

    where

    H(t,s)=1¯ΔΓ(α){(lnt)α1Γ(α)¯Ξ(s)(lnelns)α1¯Δ(lntlns)α1,1ste,(lnt)α1Γ(α)¯Ξ(s)(lnelns)α1, 1tse, (2.8)

    in which

    ¯Ξ(s)=1Γ(α)1Γ(α)sξjζj(lnξjlnslnelns)α1,
    ¯Δ=1j=1ζjlnξα1j0

    Proof. Let v=φp(HDγ1+u),¯h(t)C[1,e], then the Eqs (1.1),(1.2) can be changed into the following equation

    HDα1+v(t)+¯h(t)=0,1<t<e,v(1)=0,v(e)=j=1ζjv(ξj)

    then by the similar method with Lemma 2.4, we get v(t)=e1H(t,s)dss, and H(t,s) is as (2.8). Let u(t) be the solution of BVP (1.1,1.2) and let κ(t)=HDγ1+u(t). By Lemma 2.4, we have

    u(t)=e1Ψ(t,s)κ(s)dss (2.9)

    Putting v(t)=φp(κ(t)), we have

    v(t)=e1H(t,s)f(s,u(s))dss (2.10)

    Combining (2.9) and (2.10), we have (2.7). The proof of Lemma 2.6 is completed.

    Lemma 2.7. The Green functions (2.8) have the following properties:

    (i)H(t,s)>0,tH(t,s)>0,1<t,s<e;

    (ii)maxt[1,e]H(t,s)=H(e,s)=1¯ΔΓ(α)[Γ(α)¯Ξ(s)(lnelns)α1¯Δ(lnelns)α1];

    (iii)H(t,s)(lnt)α1H(e,s),1t,se

    Proof. The proof is similar to Lemma 2.5 of this paper and we omit it here.

    Lemma 2.8[20]. Let E Be a real Banach space, PE be a cone, and Ωr=uP:ur. Let the operator T:PΩrP be completely continuous and satisfy Txx,xΩr, then

    (i) If Txx,xΩr, then i(T,Ωr,P)=1;

    (ii) If Txx,xΩr, then i(T,Ωr,P)=0

    Now, we define the Banach space E=C([1,e],R), which is assigned a maximum norm, that is, u=sup1te|u(t)|. Let P={uE|u(t)0}, then P is a cone in E. Define an operator T:PP by

    (Tu)(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dss (3.1)

    then equation (1.1,1.2) has a solution if, and only if, the operator T has a fixed point.

    Lemma 3.1 If fC([1,e]×[0,+),[0,+)), then the operator T:PP is completely continuous.

    Proof. From the continuity and nonnegativeness of Ψ(t,s) and f(t,u(t)), we know that T:PP is continuous. Let ΩP be bounded, then for all t[1,e] and uΩ, there exists a positive constant M such that |f(t,u(t))|M. Thus,

    |(Tu)(t)|=|e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dss||e1Ψ(e,s)φq(e1H(e,τ)dττ)dss×Mq1|=ωMq1L

    where

    L=e1Ψ(e,s)dss,  ω=φq(e1H(e,τ)dττ)

    which means that T(Ω) is uniformly bounded.

    On the other hand, from the continuity of Ψ(t,s) on [1,e]×[1,e], we have Ψ(t,s) is uniformly continuous on [1,e]×[1,e]. Hence, for fixed s[1,e] and for any ε>0, there exists a constant δ>0 such that t1,t2[1,e] and |t1t2|<δ, |Ψ(t1,s)Ψ(t2,s)|<1ωMq1ε, then for all uΩ, we have

    |(Tu)(t2)(Tu)(t1)|e1|Ψ(t2,s)Ψ(t1,s)|φq(e1H(s,τ)f(τ,u(τ))dττ)dssωMq1e1|Ψ(t2,s)Ψ(t1,s)|dss<ε

    which implies that T(Ω) is equicontinuous. By the Arzela-Ascoli theorem, we have that T:PP is completely continuous. The proof is complete.

    Theorem 3.1 If fC([1,e]×[0,+),[0,+)), f(t,u) is nondecreasing in u, and λ(0,+), then BVP(1.1, 1.2) has a minimal positive solution ¯v in Br and a maximal positive solution ¯ϱ in Br. Moreover, vm(t)¯v(t), ϱm(t)¯ϱ(t) as m uniformly on [1,e], where

    vm(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,vm1(τ))dττ)dss (3.2)

    and

    ϱm(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,ϱm1(τ))dττ)dss (3.3)

    Proof. Let

    Br={uP:ur},
    rωMq11e1Ψ(e,s)dss

    For uBr, there exists a positive constant M1 such that |f(t,u(t))|M1,

    |(Tu)(t)|=|e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dss||e1Ψ(e,s)φq(e1H(e,τ)dττ)dss×Mq11|=ωMq11e1G(e,s)dssr

    hence,

    T:BrBr

    By Lemma 3.1, we get that T:BrBr is completely continuous. Therefore, by the Schauder fixed point theorem, the operator T has at least one fixed point, and so BVP(1.1, 1.2) has at least one solution in Br. Next, we show that BVP(1.1, 1.2) has a positive solution in Br, which is a minimal positive solution.

    From (3.1) and (3.2), we have that

    vm(t)=(Tvm1)(t),t[1,e],m=1,2,3, (3.4)

    This, together with f(t,u) being nondecreasing in u, yields that

    0=v0(t)v1(t)vm(t),t[1,e]

    Since T is compact, we have that {vm} is a sequentially compact set. Hence, there exists ¯vBr such that vm¯v(m).

    Let u(t) be any positive solution of BVP(1.1, 1.2) in Br. Obviously, 0=v0(t)u(t)=(Tu)(t); thus,

    vm(t)u(t),m=0,1,2,. (3.5)

    Taking limits as m in (3.5), we have ¯v(t)u(t) for t[1,e]. Thus, ¯v is a minimal positive solution, then, we show BVP(1.1, 1.2)has a positive solution in Br, which is a maximal positive solution.

    Let ϱ0(t)=r,t[1,e] and ϱ1(t)=Tϱ0(t). From T:BrBr, we get ϱ1Br. Therefore,

    0ϱ1(t)r=ϱ0

    This, together with f(t,u) being nondecreasing in u, yields that

    ϱm(t)ϱ1(t)ϱ0(t),t[1,e]

    Using a proof similar to that of the above, we can show that

    ϱm(t)¯ϱ(t)(m)

    and

    ¯ϱ(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,¯ϱ(τ))dττ)dss

    Let u(t) be any positive solution of BVP(1.1, 1.2) in Br. It is obvious that

    u(t)ϱ0(t)

    so

    u(t)ϱm(t) (3.6)

    Taking limits as m in (3.6), we have u(t)¯ϱ(t) for t[1,e]. The proof is complete.

    Define

    f0=limu0+supt[1,e]f(t,u)φp(l1u),f0=limu0+supt[1,e]f(t,u)φp(l2u),
    f=limu+supt[1,e]f(t,u)φp(l3u),f=limu+supt[1,e]f(t,u)φp(l4u)

    Let

    B1=e1G(e,s)(lns)(α1)(q1)dss

    Theorem 3.2 If fC([1,e]×[0,+),[0,+)) and the following conditions hold:

    (H1) f0=f=+.

    (H2) There exists constants λ,>0 such that f(t,u)λq1φp(l5u) for t[1,e],u[0,], then BVP(1.1, 1.2)has at least two positive solutions u1 and u2 such that

    0<u1<<u2

    for

    λ(1(l2B1)1q1κ,1(l5L)1q1κ)(1(l4B1)1q1κ,1(l5l)1q1κ) (3.7)

    where

    l2B1>l5L  and  l4B1>l5L

    Proof. Since

    f0=limu0+supt[1,e]f(t,u)φp(l2u)=+

    there is 0(0,) such that

    f(t,u)φp(l2u) for t[1,e],u[0,0]

    Let

    Ω0={uP:u0}

    then, for any uΩ0, it follows from Lemma 2.4 that

    (Tu)(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dssλq1e1(lnt)γ1Ψ(e,s)φq(e1H(s,τ)φp(l2u)dττ)dssλq1l2e1(lnt)γ1Ψ(e,s)φq(e1(lns)α1H(e,τ)dττ)dssu=λq1l2φq(κ)e1(lnt)γ1Ψ(e,s)(lns)(α1)(q1)dssu=λq1l2B1φq(κ)u (3.8)

    where e is as (3.8), κ=e1H(e,τ)dττ,B1=e1Ψ(e,s)(lns)(α1)(q1)dss. Thus,

    Tuλq1l2B1φq(κ)u

    This, together with (3.7), yields that

    Tuu,uΩ0

    By Lemma 2.6, we get

    i(T,Ω0,P)=0 (3.9)

    In view of

    f=limu+supt[1,e]f(t,u)φp(l4u)=+

    there is 0,0>, such that

    f(t,u)λq1φp(l4u),for  t[1,e],u[0,+)

    Let

    Ω0={uP:u0}

    then, for any uΩ0, it follows from Lemma 2.4 that

    (Tu)(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dssλq1e1(lnt)γ1Ψ(e,s)φq(e1H(s,τ)φp(l4u)dττ)dssλq1l4e1(lnt)γ1Ψ(e,s)φq(e1(lns)α1H(e,τ)dττ)dss=λq1l4φq(κ)e1(lnt)γ1Ψ(e,s)(lns)(α1)(q1)dssu=λq1l4B1φq(κ)u

    Thus, by (3.7), we have

    Tuλq1L4B1φq(κ)u

    This, together with (3.7), yields that

    Tuu,uΩ0

    By Lemma 2.6, we get

    i(T,Ω0,P)=0 (3.10)

    Finally, let Ω={uP:u}. For any uΩ, it follows from Lemma 2.3 and (H2) that

    (Tu)(t)=e1Ψ(t,s)φq(e1H(s,τ)f(τ,u(τ))dττ)dssλq1e1Ψ(e,s)φq(e1H(e,τ)φp(l5u)dττ)dssλq1l5e1Ψ(e,s)φq(e1H(e,τ)dττ)dssu=λq1l5φq(κ)e1Ψ(e,s)dssu=λq1l5Lφq(κ)u

    where e is as (3.8), L=e1Ψ(e,s)dss. Thus,

    Tuλq1l5Lφq(κ)u

    This, together with (3.7), yields that

    Tuu,uΩ

    By Lemma 2.6, we get

    i(T,Ω,P)=1 (3.11)

    From (3.9)–(3.11) and 0<<0, we get

    i(T,Ω0¯Ω,P)=1,i(T,Ω¯Ω0,P)=1

    Hence, T has a fixed point u1Ω¯Ω0 and a fixed point u2Ω0¯Ω. Obviously, u1,u2 are both positive solutions of BVP (1.1, 1.2) and 0<u1<<u2. The proof of Theorem 3.2 is completed. In a similar way, we get the following result.

    Corollary 3.1 If fC([1,e]×[0,+),[0,+)), and the following conditions hold:

    (H3) f0=f=0.

    (H4) There exists constants λ,2>0 such that f(t,u)λq1φp(l6u) for t[1,e],u[0,2], then BVP(1.1, 1.2)has at least two positive solutions u1 and u2 such that

    0<u1<2<u2

    for

    λ(1(l6B1)1q1κ,1(l3L)1q1κ)(1(l6B1)1q1κ,1(l1L)1q1κ) (3.12)

    where

    l6B1>l3L  and   l6B1>l1L

    Example. Consider the following infinite-point pLaplacian fractional differential equations

    {HD321+(φ3(HD521+u))(t)+f(t,u(t))=0, 1<t<e,u(1)=u(1)=0;HDr11+u(1)=j=1ηHjDr21+u(ξj),HDα1+u(1)=0;φp(HD521+u(e))=j=1ζiφp(HD521+u(ηi)) (4.1)

    where γ=52,α=32,r1=32,r2=12,p=3,q=32, ηj=12j2,ξj=e1j4, ζj=12j2,

    f(t,u)=(lnt+2)|u(t)|3+|u(t)|

    Clearly, for fC([1,e]×[0,),[0,)), one can have

    |f(t,u)|=|(lnt+2)|u(t)|3+|u(t)||3

    By simple calculation, we have

    Δ=Γ(γ)Γ(γr1)Γ(γ)Γ(γr2)j=1ηjlnξγr21j=Γ(52)Γ(5232)Γ(52)Γ(5212)j=112j2(1j4)238.18,
    ¯Δ=1i=1ζilnξα1i=1i=1ζilnξ12i=112i=11j410.5411=0.4589,
    P(s)=1Γ(γr1)1Γ(γr2)sξjηj(lnξjlnslnelns)γr21(lnelns)r1r2=1Γ(5232)1Γ(5212)s1j412j2(1j4lnslnelns)52121(lnelns)3212=112sξj(1j4lnslnelns)(1s),
    L=e1Ψ(e,s)dss=1Δe1P(s)(lnelns)γr11ds1Γ(γ)e1(lnelns)γ1dss138.18e1(112sξj(1j4lnslnelns)(lnelns))(lnelns)dss1Γ(52)e1(lnelns)32dss0.5600,
    κ=e1H(e,s)dss=1¯Δe1¯P(s)(lnelns)α1dss1Γ(α)e1(lnelns)α1dss10.25e1(1Γ(32)1Γ(32)sξjηj(lnξjlnslnelns)12)(lnelns)12dss1Γ(32)e1(lnelns)12dss5.08

    and

    B1=e1Ψ(e,s)(lns)(α1)(q1)dss=e1Ψ(e,s)(lns)14dss138.18e1(112sξj(1j4lnslnelns)(lnelns))(lnelns)(lns)14dss1Γ(52)e1(lnelns)32(lns)14dss3.0235

    Hence, take l2=10,l5=8,L=0.56,l4=6, and we have

    l2B1=10×3.0235>l5L=8×0.56,l4B1=6×3.0235>l5L=8×0.56,

    then there exists λ=0.005 such that

    λ(1(l2B1)1q1κ,1(l5L)1q1κ)(1(l4B1)1q1κ,1(l5l)1q1κ)=(0.00022,0.01)(0.0006,0.01),

    Hence, all the conditions of Theorem 3.2 hold and boundary value problem (4.1) has two positive solutions u1,u2.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to thank the referee for his/her valuable comments and suggestions. This research was supported by the National Natural Science Foundation of China (12101086, 12272064, 62273061), the major project of Basic science (Natural science) research in colleges and universities of Jiangsu Province(22KJA460001, 23KJA580001), Changzhou Science and Technology Plan Project (CE20235049), and Open Project of Applied Mechanics and Structure Safety Key Laboratory of Sichuan Province (SZDKF-202102).

    The authors declare there is no conflicts of interest.



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