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Editorial

Letter from the Editors-in-Chief of AIMS Animal Science

  • Citation: Peipei Ma, He Meng, Guosheng Su. 2025: Letter from the Editors-in-Chief of AIMS Animal Science, AIMS Animal Science, 1(1): 1-2. doi: 10.3934/aas.2025001

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  • In this paper, we consider the multiplicity of positive solutions for the following singular tempered fractional equation with lower order tempered fractional derivative

    { R0Dtα,λu(t)=p(t)h(eλtu(t),R0Dtβ,λu(t)),  t(0,1),R0Dtβ,λu(0)=0,  R0Dtβ,λu(1)=0, (1.1)

    where α(1,2],β(0,1) with αβ>1, hC([0,+)×[0,+),[0,+)), pL1((0,1),(0,+)), which implies that the weight function can have infinite many singular points in [0,1].

    The equation (1.1) contains a tempered fractional derivative R0Dtα,λ, which is actually obtained by multiplying an exponential factor in the Riemann-Liouville fractional derivative R0DtDtα, i.e., the following relationship exists between tempered fractional derivative and Riemann-Liouville fractional derivative

    R0Dtα,λu(t)=eλtR0DtDtα(eλtu(t)). (1.2)

    For the definition of the standard Riemann-Liouville fractional derivative and integral, we refer the reader to [1,2,3,4,5].

    As the optimization of the Riemann-Liouville fractional derivative, the tempered fractional derivative has many advantages, which not only overcomes the defect of using the power law of the classical fractional derivative in the mathematical sense, such as the Riemann-Liouville fractional derivatives [6,7], the Caputo fractional derivatives [8], Hadamard fractional derivatives [9,10,11] etc, but also brings many practical applications. It especially describes the anomalous diffusion phenomena in Brownian motion with the semi-heavy tails or semi-long range dependence, such as the limits of random walk with an exponentially tempered jump distribution [12,13], transient super-diffusion [14], anomalous diffusions in heterogeneous systems [15], exponentially tempering Lévy flights with both the α-stable and Gaussian trends [16], and pure jump Lévy process with the fractional derivatives in the risk management of financial derivatives traded over the counter [17].

    In recent work [18], an upper and lower solutions technique has been employed to establish the existence of positive solutions for a singular tempered fractional turbulent flow model in a porous medium

    { R0Dtα,λ(φp(R0Dtβ,λu(t)))=f(t,u(t)),  t(0,1),u(0)=0,  R0Dtβ,λu(0)=0,  u(1)=10eλ(1t)u(t)dt, (1.3)

    with α(0,1], β(1,2] and the nonlinearity f is decreasing in the second variable. In [19], in view of the monotone iterative method, the iterative properties of positive solutions for a tempered fractional equation

    { R0Dtϑ,μu(t)=f(t,eμtu(t),R0Dtδ,μu(t)),  t(0,1),R0Dtδ,μu(0)=0,  R0Dtδ,μu(1)=10eμ(1t)R0Dtδ,μu(t)dt, (1.4)

    were established, where ϑ(1,2], δ(0,1) with ϑδ>1, μ is a positive constant, and f:(0,1)×[0,+)×[0,+)[0,+) is a continuous and nondecreasing function with respect to the two space variables. Recently, by using spaces theories [20,21,22,23,24], smooth theories [25,26,27], operator method [28,29], the method of moving sphere [30], critical point theories [31,32,33,34] and so on, some other types of fractional equations was also studied [35,36,37,38,39,40,41,42,43,44,45,46].

    However, when the nonlinearity contains lower order tempered fractional derivative, the results of multiplicity of positive solutions have not yet been obtained. In order to fill this gap, by applying the reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the above equation are established in this paper. The interesting points are that the nonlinearity contains the lower order tempered fractional derivative and the weight function can have infinitely many singular points in [0,1].

    In this section, we first recall definitions and some useful properties of the Riemann-Liouville fractional derivative and integral.

    Definition 2.1 ((2.1.1) on page 69 in [1]). The Riemann-Liouville fractional integral of order α>0 of a function u:(0,+)R is given by

    Iαu(t)=1Γ(α)t0(ts)α1u(s)ds

    provided that the righthand side is pointwise and defined on (0,+).

    Definition 2.2 ((2.1.5) on page 70 in [1]). The Riemann-Liouville fractional derivative of order α>0 of a function u:(0,+)R is given by

    DDαu(t)=1Γ(nα)(ddt)nt0(ts)nα1u(s)ds,

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

    The following properties of the Riemann-Liouville fractional derivative and integral can be found on pages 73–75 (Lemmas 2.3–2.5) in [1].

    Lemma 2.1. Suppose u(t)C[0,1]L1[0,1] and α>β>0. Let n=[α]+1, then

    (i)

    IαR0DtDtαu(t)=u(t)+c1tα1+c2tα2++cntαn, (2.1)

    where ciR,i=1,2,3,...,n.

    (ii)

    IαIβu(t)=Iα+βu(t),  R0DtDtβIαu(t)=Iαβu(t),  R0DtDtβIβu(t)=u(t).

    Lemma 2.2. Let pL1([0,1],(0,+)), then the singular linear tempered fractional equation

    { R0Dtαβ,λu(t)=p(t),u(0)=0,  u(1)=0, (2.2)

    has a unique positive solution u(t) provided that 1<αβ2, which can be expressed by

    u(t)=10H(t,s)p(s)ds, (2.3)

    where

    H(t,s)={ tαβ1(1s)αβ1(ts)αβ1Γ(αβ)eλteλs,0st1;tαβ1(1s)αβ1Γ(αβ)eλteλs,  0ts1. (2.4)

    is the Green function of (2.2).

    Proof. In fact, it follows from 1<αβ2, (1.2) and (2.1) that

    eλtu(t)=1Γ(αβ)t0(ts)αβ1eλsp(s)ds+b1tαβ1+b2tαβ2,t[0,1].

    Since u(0)=0 and u(1)=0, we have b2=0 and

    b1=1Γ(αβ)10(1s)αβ1eλsp(s)ds.

    Thus

    u(t)=1Γ(αβ)[10(1s)αβ1eλteλsp(s)dstαβ1t0(ts)αβ1eλteλsp(s)ds]=1Γ(αβ)t0[tαβ1(1s)αβ1(ts)αβ1]eλteλsp(s)ds+1Γ(αβ)1ttαβ1(1s)αβ1eλteλsp(s)ds=10H(t,s)p(s)ds,  t[0,1].

    The following Lemma has been proven (see Lemma 2.3 of [6] or Lemma 3 of [4]).

    Lemma 2.3. Suppose α(1,2],β(0,1) with αβ>1, then H(t,s) is a nonnegative continuous function in [0,1]×[0,1] satisfying, for any (t,s)[0,1]×[0,1],

    tαβ1(1t)eλtΓ(αβ)(1s)αβ1seλsH(t,s)tαβ1(1t)eλtΓ(αβ) or ((1s)αβ1seλsΓ(αβ)). (2.5)

    Suppose α(1,2],β(0,1) with αβ>1. In order to use the reducing order technique, we introduce the following integral transformation

    u(t)=eλtIβ(eλty(t)),  t[0,1] (2.6)

    and then consider the following reducing order problem

    { R0Dtαβ,λy(t)=p(t)h(Iβ(eλty(t)),y(t)),  t(0,1),y(0)=0,  y(1)=0. (2.7)

    Lemma 2.4. Suppose α(1,2],β(0,1) with αβ>1. The reducing order problem (2.7) is equivalent to the singular tempered fractional equation (1.1). In particular, if y is a positive solution of the problem (2.7), then u(t)=eλtIβ(eλty(t)) is a positive solution of the singular tempered fractional equation (1.1).

    Proof. We first suppose that u is a positive solution of the singular tempered fractional equation (1.1). Let

    u(t)=eλtIβ(eλty(t)).

    Noticing that α(1,2],β(0,1) with αβ>1, let n=[α]+1, i.e., n is the smallest integer greater than or equal to α, then it follows from (1.2) and Lemma 2.1 that

    R0Dtα,λu(t))=eλtR0DtDtα(eλtu(t))=eλtdndtnInα(eλtu(t))=eλtdndtnInα(Iβ(eλty(t)))=eλtdndtnInα+β(eλty(t)))=eλtR0DtDtαβ(eλty(t)))=R0Dtαβ,λy(t), (2.8)

    and

    R0Dtβ,λu(t))=eλtR0DtDtβ(eλtu(t))=eλtR0DtDtβ(Iβ(eλty(t)))=y(t). (2.9)

    By (2.8) and (2.9), we have

    { R0Dtαβ,λy(t)=R0Dtα,λu(t))=p(t)h(eλtu(t),R0Dtβ,λu(t))=p(t)h(Iβ(eλty(t)),y(t)),t(0,1),y(0)=R0Dtβ,λu(0)=0,   y(1)=R0Dtβ,λu(1)=0,

    thus, y solves the equation (2.7).

    Conversely, suppose that y is any solution of the reducing order problem (2.7), then we have

    { R0Dtαβ,λy(t)=p(t)h(Iβ(eλty(t)),y(t)),  t(0,1),y(0)=0,  y(1)=0.

    Make integral transformation (2.6), similar to (2.8) and (2.9), we have

    R0Dtα,λu(t))=R0Dtαβ,λy(t),  R0Dtβ,λu(t))=y(t).

    Substituting the above formulas into (2.7), we get

    R0Dtα,λu(t))=p(t)h(Iβ(eλty(t)),y(t))=p(t)h(eλtu(t),R0Dtβ,λu(t)),   0<t<1,

    and

    R0Dtβ,λu(0)=0,  R0Dtβ,λu(1)=0,

    which implies that u(t)=eλtIβ(eλty(t)) solves the singular tempered fractional equation (1.1).

    Suppose E=C([0,1];R) with the norm

    ||y||=maxt[0,1]|y(t)|.

    Define a cone of E and an operator T, respectively:

    P={yE:y(t)tαβ1(1t)eλt||y||},

    and

    (Ty)(t)=10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds. (2.10)

    In order to obtain the existence of a positive solution of the Eq (1.1), Lemma 2.4 indicates that we only consider the fixed points of the operator T.

    Now, we list the following hypotheses, which are used in the rest of this paper.

    (C1)hC([0,+)×[0,+),[0,+)) and pL1((0,1),(0,+)).

    (C2) There is a constant n>0 such that for any 0u+v(1+eλΓ(β+1))n,

    h(u,v)<μn,

    where

    μ=[neλΓ(αβ)10p(s)ds]1.

    (C3) There is a constant ρ>0 such that for any

    (Γ(αβ)(14)α1Γ(α)+(14)αβe34λ)ρu+v(1+eλΓ(β+1))ρ

    it implies

    h(u,v)ϱρ,

    where

    ϱ=8αβe14λΓ(αβ)[3414p(s)ds]1.

    Lemma 2.5. Assume α(1,2],β(0,1) with αβ>1 and (C1) holds, then the operator T:PP is completely continuous.

    Proof. First by (C1), T is continuous on [0,1]. For any yP, there exists a constant M>0 such that ||y||M, then

    0Iβ(eλsy(s))=t0(ts)β1eλsy(s)Γ(β)dsMeλΓ(β). (2.11)

    Let

    =max(u,v)[0,MeλΓ(β)]×[0,M]h(u,v),

    then by (2.11), for any yP, we have

    ||Ty||=maxt[0,1]10H(t,s)p(s)h(Iβ(eλty(s)),y(s))ds10(1s)αβ1seλsΓ(αβ)p(s)h(Iβ(eλty(s)),y(s))ds10(1s)αβ1seλsΓ(αβ)p(s)dseλΓ(αβ)10p(s)ds<+, (2.12)

    which implies that T:PE is well-defined. In addition, by (2.5) and (2.12), we have

    (Ty)(t)10(1s)αβ1seλsΓ(αβ)p(s)h(Iβ(eλty(s)),y(s))ds×tαβ1(1t)eλt||Ty||tαβ1(1t)eλt,

    which implies that T(P)P.

    In the end, by using the standard arguments and combining the Ascoli-Arzela theorem, we know T(P)P is completely continuous.

    Our proof of main results depends on the fixed point theorem of cone expansion and compression (see Theorem 2.3.3 on page 93 of [47]).

    Lemma 2.6. [47] Suppose P is a cone of real Banach space E, the bounded open subsets Ω1,Ω2 of E satisfy θΩ1,¯Ω1Ω2. Let T:P(¯Ω2Ω1)P be a completely continuous operator such that either

    (1) Tzz,zPΩ1 and Tzz,zPΩ2, or

    (2) Tzz,zPΩ1 and Tzz,zPΩ2;

    then T has a fixed point in P(¯Ω2Ω1).

    For the convenience of the proof, we first introduce the following notations whenever the limits exist or not:

    h0=limu+v0h(u,v)u+v,   h=limu+v+h(u,v)u+v,

    and then state our main results as follows.

    Theorem 3.1. Assume that (C1) and (C2) hold, and

    h0=+,h=+,

    then the tempered fractional equation (1.1) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that

    A1tα1eλtu1(t)B1eλ(1t),  A2tα1eλtu2(t)B2eλ(1t).

    Proof. It follows from h0=+ that there exist 0<m<n and a sufficiently large constant

    N>32αβeλΓ(αβ)3414p(s)ds

    such that for any 0<u+v(1+eλΓ(β+1))m,

    h(u,v)N(u+v).

    Take Ωm={yE:||y||<m}, and Ωm={yE:||y||=m}, then for any yPΩm, one has

    Iβ(eλsy(s))=t0(ts)β1Γ(β)eλsy(s)dseλmtβΓ(β+1)eλmΓ(β+1), (3.1)

    then

    Iβ(eλsy(s))+y(s)(1+eλΓ(β+1))m. (3.2)

    Thus,

    ||Ty||(Ty)(12)=10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds2α+βe12λΓ(αβ)10(1s)αβ1seλsp(s)N(Iβ(eλsy(s))+y(s))ds2α+βe12λΓ(αβ)10(1s)αβ1seλsp(s)Ny(s)ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)Ny(s)dsN8αβe54λΓ(αβ)3414(1s)αβ1seλsp(s)ds||y||N32αβeλΓ(αβ)3414p(s)ds||y||||y||. (3.3)

    So, for any yPΩm, we have ||Ty||||y||.

    Next, let Ωn={yE:||y||<n} and Ωn={yE:||y||=n}, then it follows from (3.2) and (C2) that for any yPΩn, one has

    (Ty)(t)=10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds10(1s)αβ1seλsΓ(αβ)p(s)h(Iβ(eλty(s),y(s))dseλΓ(αβ)10p(s)h(Iβ(eλty(s),y(s))dsμneλΓ(αβ)10p(s)dsn. (3.4)

    Therefore for any yPΩn, we have ||Ty||||y||.

    On the other hand, it follows from h=+ that there exists M>0 and

    η32αβeλΓ(αβ)(3414p(s)ds)1

    such that for any u+v>M, we have

    h(u,v)η(u+v).

    Take R=n+4αβe34λM and let ΩR={yE:||y||<R} and ΩR={yE:||y||=R}, then for any yPΩR and [14,34], one has

    y(t)(14)αβe34λ||y||(14)αβe34λ(n+4αβe34λM)M.

    Thus, for any yPΩR, we have

    ||Ty||(Ty)(12)=10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)η(Iβ(eλsy(s))+y(s))ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)ηy(s)ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)ηy(s)dsη8αβe54λΓ(αβ)3414(1s)αβ1seλsp(s)ds||y||η32αβeλΓ(αβ)3414p(s)ds||y||||y||. (3.5)

    Therefore, for any yPΩR, we have ||Ty||||y||.

    According to Lemma 2.6, T has two fixed points y1P(¯ΩmΩn) and y2P(¯ΩRΩn) with m||y1||n||y2||R. Thus, it follows from Lemma 2.4 that the tempered fractional equation (1.1) has at least two positive solutions satisfying

    mΓ(αβ)tα1eλtΓ(α)u1(t)=eλtIβ(eλty1(t))neλ(1t)Γ(β+1),

    and

    nΓ(αβ)tα1eλtΓ(α)u2(t)=eλtIβ(eλty2(t))Reλ(1t)Γ(β+1).

    Theorem 3.2. Assume that (C1) and (C3) hold and

    h0=0,h=0,

    then the tempered fractional equation (1.1) has at least two positive solutions u3,u4; moreover, there exist four constants A3,B3,A4,B4>0 such that

    A3tα1eλtu3(t)B3eλ(1t),  A4tα1eλtu4(t)B4eλ(1t).

    Proof. First, notice that h0=0, for any ϵ>0. Let us select 0<κ<ρ such that for any 0<u+v<(1+eλΓ(β+1))κ,

    h(u,v)ϵ(u+v).

    Choose ϵ such that

    ϵeλΓ(αβ)(1+eλΓ(β+1))10p(s)ds1.

    Now, let Ωκ={yE:||y||<κ} and Ωκ={yE:||y||=κ}, then for any yPΩκ, the same as (3.2), we also have

    Iβ(eλsy(s))+y(s)(1+eλΓ(β+1))κ, (3.6)

    which implies that for any yPΩκ, the following estimation is valid:

    h(Iβ(eλty(s)),y(s)ϵ(Iβ(eλty(s))+y(s)).

    Consequently, for any yPΩκ, one gets

    (Ty)(t)=10H(t,s)p(s)h(Iβ(eλty(s)),y(s))ds10(1s)αβ1seλsΓ(αβ)p(s)h(Iβ(eλty(s)),y(s))dseλΓ(αβ)10p(s)h(Iβ(eλty(s)),y(s))dseλΓ(αβ)10p(s)ϵ(Iβ(eλty(s))+y(s))dseλΓ(αβ)10p(s)ϵ(1+eλΓ(β+1))κdsϵeλΓ(αβ)(1+eλΓ(β+1))10p(s)dsκκ. (3.7)

    (3.7) implies that ||Ty||||y||,  yPΩκ.

    Next, let Ωρ={yE:||y||<ρ}, and Ωρ={yE:||y||=ρ}, then for any yPΩρ, we have

    Iβ(eλssαβ1y(s))=t0(ts)β1Γ(β)sαβ1eλsy(s)dsΓ(αβ)tα1ρΓ(α). (3.8)

    Thus, it follows from (3.8) that for any yPΩρ and [14,34], one has

    (Γ(αβ)(14)α1Γ(α)+(14)αβe34λ)ρ=Γ(αβ)(14)α1Γ(α)ρ+(14)αβe34λρ=Iβ(eλssαβ1y(s))+y(t)(1+eλΓ(β+1))ρ. (3.9)

    So, for any yPΩρ, we have

    ||Ty||(Ty)(12)=10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds2α+βe12λΓ(αβ)10(1s)αβ1seλsp(s)h(Iβ(eλsy(s)),y(s))ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)h(Iβ(eλsy(s)),y(s))ds2α+βe12λΓ(αβ)3414(1s)αβ1seλsp(s)ϱρdsϱ8αβe14λΓ(αβ)3414p(s)dsρ||y||, (3.10)

    which implies that ||Ty||||y|| holds and yPΩρ.

    On the other hand, since h=0, for any ϵ>0, there exists M>0 such that for any u+v>M

    h(u,v)ϵ(u+v).

    For the above ϵ>0, choose a sufficiently small one such that

    ϵeλΓ(αβ)(1+eλΓ(β+1))10p(s)ds<1,

    and take

    R=max{eλΓ(αβ)max0u+vMh(u,v)10p(s)ds1ϵeλΓ(αβ)(1+eλΓ(β+1))10p(s)ds,ρ+M}.

    Next, let ΩR={yE:||y||<R}, and ΩR={yE:||y||=R}, then for any yPΩR, one has

    (Ty)(t)=10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds10(1s)αβ1seλsΓ(αβ)p(s)h(Iβ(eλty(s),y(s))dseλΓ(αβ){0Iβ(eλty(s)+y(s)Mp(s)h(Iβ(eλty(s),y(s))ds     +MIβ(eλty(s)+y(s)(1+eλΓ(β+1))Rp(s)h(Iβ(eλty(s),y(s))ds}eλΓ(αβ){max0u+vMh(u,v)10p(s)ds     +MIβ(eλty(s)+y(s)(1+eλΓ(β+1))Rp(s)ϵ(Iβ(eλty(s)+y(s))ds}eλΓ(αβ){max0u+vMh(u,v)+ϵ(1+eλΓ(β+1))R)10p(s)dsR. (3.11)

    Therefore, for any yPΩR, we have ||Ty||||y||.

    According to Lemma 2.6, T has two fixed points y3P(¯ΩρΩκ) and y4P(¯ΩRΩρ) with κ||y3||ρ||y4||R. Thus, it follows from Lemma 2.4 that the tempered fractional equation (1.1) has at least two positive solutions satisfying

    κΓ(αβ)tα1eλtΓ(α)u3(t)=eλtIβ(eλty3(t))ρeλ(1t)Γ(β+1),

    and

    ρΓ(αβ)tα1eλtΓ(α)u4(t)=eλtIβ(eλty4(t))Reλ(1t)Γ(β+1).

    The tempered fractional diffusion equation has many important applications, including the tempered fractional Langevin and Vasicek differential equations [48] and the space-time tempered fractional diffusion-wave equation [49]. The new theorems established in the present paper are very useful in the area of tempered fractional calculus. We can find the multiple positive solutions for the singular tempered fractional equations with the lower order tempered fractional derivatives using the proposed theorems.

    Let us apply the main results to solve two singular tempered fractional equations with lower order tempered fractional derivative.

    Example 4.1. Let α=53, β=13, λ=3, and

    h(u,v)={ (u+v)12,   0u+v47,(u+v)24732,   u+v>47.

    We consider the multiplicity of positive solutions for the following singular tempered fractional equation with the lower order tempered fractional derivative:

    { R0Dt53,3u(t)=|12t|12h(e3tu(t),R0Dt13,3u(t)),0t1,R0Dt13,3u(0)=0,  R0Dt13,3u(1)=0. (4.1)

    Conclusion. The singular tempered fractional equation (4.1) has at least two positive solutions u1,u2: moreover, there exist four constants A1,B1,A2,B2>0 such that

    A1t23e3tu1(t)B1e3(1t),  A2t23e3tu2(t)B2e3(1t).

    Proof. Here,

    p(t)=|12t|12,  h(u,v)={ (u+v)12600,   0u+v47,(u+v)2600×4732,   u+v>47.

    Clearly, (C1) holds and

    h0=limu+v0h(u,v)u+v=+,   h=limu+v+h(u,v)u+v=+.

    In the following, we verify the condition (C2). In fact, take n=2, then

    μ=[neλΓ(αβ)10p(s)ds]1=[2e3Γ(43)10|12s|12ds]1=0.0079,

    and for any 0u+v(1+eλΓ(β+1))n=2(1+e3Γ(43))=46.9853, we have

    h(u,v)<4712600=0.0114<μn=0.0158,

    which implies that (C2) holds.

    Consequently, it follows from Theorem 3.1 that the tempered fractional equation (4.1) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that

    A1t23e3tu1(t)B1e3(1t),  A2t23e3tu2(t)B2e3(1t).

    Example 4.2. Let α=53, β=13, λ=3, and

    h(u,v)={ 1002(u+v)2,   0u+v12,50(u+v)12,   u+v>12.

    We consider the multiplicity of positive solutions for the following singular tempered fractional equation with the lower order tempered fractional derivative

    { R0Dt53,3u(t)=|12t|12h(e3tu(t),R0Dt13,3u(t)),0t1,R0Dt13,3u(0)=0,  R0Dt13,3u(1)=0. (4.2)

    Conclusion. The tempered fractional equation (4.2) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that

    A1t23e3tu1(t)B1e3(1t),  A2t23e3tu2(t)B2e3(1t).

    Proof. Here

    p(t)=|12t|12,  h(u,v)={ 1002(u+v)2,   0u+v12,50(u+v)12,   u+v>12.

    Clearly, (C1) holds and

    h0=limu+v0h(u,v)u+v=0,   h=limu+v+h(u,v)u+v=0.

    Take ρ=2, and we have

    ϱ=8αβe14λΓ(αβ)[3414p(s)ds]1=843e34Γ(43)[3414=|12s|12ds]1=15.1238.

    For any

    0.8184=2(Γ(43)(14)23Γ(53)+(14)43e94)u+v2(1+e3Γ(43))=46.9844,

    one has

    h(u,v)50×0.818412=45.232>ϱρ=30.2476,

    which implies that (C3) holds.

    Consequently, it follows from Theorem 3.2 that the tempered fractional equation (4.2) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that

    A1t23e3tu1(t)B1e3(1t),  A2t23e3tu2(t)B2e3(1t).

    This work studies the multiplicity of positive solutions for a class of singular tempered fractional equations with the lower order tempered fractional derivative. By applying reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the equation are established. The interesting points are the nonlinearity contains the lower order tempered fractional derivative and the weight function can have infinitely many singular points in [0,1]. However, in this study, the conditions α(1,2],β(0,1) with αβ>1 are required; if 0<αβ<1 or h has singularity at space variables, these interesting problems are still worth future studying.

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

    This work is granted by the Natural Science Foundation of Shandong Province of China (ZR2022AM015) and an ARC Discovery Project Grant.

    Xinguang Zhang is a guest editor of the special issue for ERA and was not involved in the editorial review or the decision to publish this article. All authors declare that there are no competing interests.



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