Research article

Initial value problems for fractional p-Laplacian equations with singularity

  • Received: 19 February 2024 Revised: 23 March 2024 Accepted: 01 April 2024 Published: 03 April 2024
  • MSC : 26A33, 34A08, 34A12

  • We have studied initial value problems for Caputo fractional differential equations with singular nonlinearities involving the p-Laplacian operator. We have given a precise mathematical analysis of the equivalence of the fractional differential equations and Volterra integral equations studied in this paper. A theorem for the global existence of the solution was proven. In addition, an example was given at the end of the article as an application of the results found in this paper.

    Citation: Mahir Hasanov. Initial value problems for fractional p-Laplacian equations with singularity[J]. AIMS Mathematics, 2024, 9(5): 12800-12813. doi: 10.3934/math.2024625

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  • We have studied initial value problems for Caputo fractional differential equations with singular nonlinearities involving the p-Laplacian operator. We have given a precise mathematical analysis of the equivalence of the fractional differential equations and Volterra integral equations studied in this paper. A theorem for the global existence of the solution was proven. In addition, an example was given at the end of the article as an application of the results found in this paper.



    Interest in the subject of fractional differential equations has increased greatly over the past decades. Fractional differential equations appear in many fields such as physics, aerodynamics, electro-dynamics, and control theory (see [1,14,17,21,24,31,32,33]).

    We studied the following singular nonlinear initial value problem involving p-Laplacian

    {DβCϕp(DαCx(t))=tγf(t,x(t)),x(0)=x0,DαCx(0)=xα,x0,xαR,0γ<α,β1,xAC[0,1],DαCxAC[0,1]. (1.1)

    f:[0,1]×RR is a continuous function, ϕp(t)=|t|p2t,p>1, and DαCx(t) denotes the Caputo fractional derivative which is defined by DαCx(t):=I1αDx(t) for 0<α1, where Iα stands for the Riemann-Liouville fractional integral of order α>0 defined by

    Iαx(t):=1Γ(α)t0(ts)α1x(s)ds,xL1,

    where Lp:=Lp[0,1]={x(t):10|x(t)|p<,1p<} and Γ(α):=0sα1exp(s)ds,α>0 (see [14,31,32,33]). A literature review will be given at the end of this section. Since the problem studied in this paper contains a singular term, we believe that the obtained results are new.

    We denote the domain of an operator T by Dom(T). The domain of the Caputo derivative is defined by Dom(DαC)={x(t):xAC[0,1]}, where AC[0,1] is the set of absolutely continuous functions on the interval [0,1].

    In general, for any mN, and 0<α1, Dm+αCx(t)=I1αDm+1x(t) with Dom(Dm+αC)={x(t):DmxAC[0,1]}.

    Besides the Caputo derivative, we will also use the Riemann-Liouville and generalized Caputo derivatives.

    For 0<α1, the Riemann-Liouville derivative is defined by Dαx(t):=DI1αx(t),withDom(Dα)={x(t):I1αxAC[0,1]}. In general, for any mN, and 0<α1, Dm+αx(t)=Dm+1I1αx(t),withDom(Dm+α)={x(t):Dm(I1αx)AC[0,1]}.

    For 0<α1, the generalized Caputo derivative is defined by Dαx(t):=Dα(x(t)x0),withDom(Dα)={x(t):I1α(xx0)AC[0,1]}, where x0=x(0).

    We studied initial value problems on the interval [0,1], but all results in this paper are also valid for any interval [0,T].

    Let us give some properties of Dα,DαC,Dα, and Iα that we will use frequently in this paper.

    Proposition 1.1. a)Iα:Lp[0,1]Lp[0,1],1p is a bounded operator for all α>0 ([36], Proposition 3.2).

    b)Iα:C[0,1]C[0,1] is a bounded operator for all α>0 ([36], Proposition 3.2), where C[0,1] denotes the normed space of all continuous functions defined on the interval [0,1], with the norm ||x||=maxt[0,1]|x(t)|.

    c) If α,β>0 and xL1[0,1], then IαIβx(t)=Iα+βx(t) for a.e. t[0,1]. Moreover, if α+β1 and xL1[0,1], then IαIβx(t)=Iα+βx(t) for all t[0,1] ([36], Lemma 3.4).

    d) For 0<α1 and mN, the operator Dm+α is the left inverse of the operator Im+α in L1[0,1], i.e., Dm+αIm+αx(t)=x(t), for all xL1[0,1] ([36], Lemma 4.2).

    e) Let 0<α<1. The operator Dα is the left inverse of the operator Iα in the space C[0,1], i.e., DαIαx(t)=x(t), for all xC[0,1] ([36], Lemma 4.5).

    f) If 0<α<1 and x(t)AC[0,1], then Dαx(t)=Dα(x(t)x(0))=DαCx(t) ([36], Proposition 4.4). Hence, the operator DαC is the left inverse of the operator Iα in the space AC[0,1], i.e., DαCIαx(t)=x(t), for all xAC[0,1].

    But, in general, DαC is not a right inverse for Iα. However, the following formula holds.

    IαDαCx(t)=x(t)x(0),forallxAC[0,1].

    In general, fractional differential equations include the Caputo fractional derivative, generalized Caputo fractional derivative, and Riemann-Liouville fractional derivative. Initial and boundary value problems containing Riemann-Liouville fractional derivatives were studied by many authors (see [3,6,7,12,20,22,38] and references therein). The application areas of Riemann-Liouville fractional differential equations have been gradually expanding in recent years (see [26,27]).

    In the past decades, different aspects of Caputo fractional differential equations were studied by a number of researchers. For example, initial value problems were studied in [13,15,23,28,36,37]. The articles [2,4,5,8,16,25,30,34,35] deal with boundary value problems for Caputo fractional differential equations. However, none of these articles contain the fractional p-Laplacian.

    Chen et al. [9] studied the existence of antiperiodic solutions for the Lienard-type p-Laplacian equation. Chen et al. [10] studied the solvability of periodic boundary value problem for the fractional p-Laplacian equation in the following form:

    DβCϕp(DαCx(t))=f(t,x(t),DαCx(t)),x(0)=x(T),DαCx(0)=DαCx(T),fC([0,T]×R),0<α,β1.

    The existence of solutions for the anti-periodic boundary value problem of a fractional p-Laplacian equation was studied by Chen et al.[11]. They studied the following problem:

    DβCϕp(DαCx(t))=f(t,x(t)),x(0)=x(1),DαCx(0)=DαCx(1),fC([0,T]×R),0<α,β1.

    Some anti-periodic problems were also considered in [19,30].

    Note that, none of these articles contain singularity. Since we are dealing with singular initial value problems, we would like to give a brief overview of some related results.

    In [3], Agarwal et al. studied the existence of positive solutions for the following singular fractional boundary value problem with the Riemann-Liouville fractional derivative:

    Dαu(t)+f(t,u(t),Dμu(t))=0,1<α<2,0<μα1,u(0)=u(1)=0,f is singular at 0.

    In [4], Agarwal et al. proved the existence of positive solutions to the singular fractional boundary value problem with the Caputo fractional derivative:

    DαCu(t)+f(t,u(t),u,DμCu(t))=0,1<α<2,0<μ<1,u(0)=0,u(1)=0,f is singular at 0.

    Webb [36] studied initial value problems for Caputo fractional differential equations with singular nonlinearities in the forms:

    Dm+αCu(t)=tγf(t,u(t)),0<α<1,0γ<α,u(0)=u0,...,um(0)=um,DmuAC[0,T],fC([0,T]×R).

    and

    D1+αu(t)=tγf(t,u(t),DβCu(t)),0γ<α<1,0<β1,u(0)=u0,u(0)=u1,fC([0,T]×R×R).

    The proof of the existence of a solution to these problems is based on the Leray-Schauder fixed point theorem and Gronwall type inequalities. Gronwall inequalities are widely used to obtain a priori bounds (see [18,29,37,38,39]).

    We note that, for the main concepts used in this article, we generally followed [11,36,37].

    This paper consists of three sections. The first section includes the introduction and preliminary information. The second section includes a precise mathematical analysis of the equivalence of the fractional differential equations and Volterra integral equations studied in this paper. The existence of solutions to initial value problems is discussed in Section 3.

    In this section, we establish a relationship between Caputo fractional differential equations and Volterra integral equations.

    Theorem 2.1. Let fC([0,1]×R) and 0γ<α,β1. If a function x(t) satisfies the initial value problem

    {DβCϕp(DαCx(t))=tγf(t,x(t)),x(0)=x0,DαCx(0)=xα,x0,xαR,xAC[0,1],DαCxAC[0,1], (2.1)

    then x(t) satisfies the Volterra integral equation

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))+x1]+x0, (2.2)

    where x1=ϕp(xα), p,q>1,and1/p+1/q=1.

    Proof. The condition xAC[0,1] implies xDom(DαC) and DαCxAC[0,1] implies that ϕp(DαCx)AC[0,1]. Consequently, ϕp(DαCx)Dom(DβC). This means that under these conditions the problem (2.1) is well-defined. Here, for the inclusion ϕp(DαCx)AC[0,1], we have used the composition rule: If the functions F:[c,d]R and G:[a,b][c,d] are absolutely continuous, then F(G(t)) is also absolutely continuous.

    Let x(t) satisfy (2.1). Since DβCϕp(DαCx(t))L1, we can apply Iβ to (2.1). Then

    IβDβCϕp(DαCx(t))=Iβ(tγf(t,x(t)))

    or the same

    IβI1βD(ϕp(DαCx(t)))=Iβ(tγf(t,x(t))).

    D(ϕp(DαCx(t))L1, then by Proposition 1.1 c) we have

    ID(ϕp(DαCx(t)))=Iβ(tγf(t,x(t))).

    Hence, by using ϕp(DαCx)AC[0,1] we get

    ϕp(DαCx(t))ϕp(DαCx(0))=Iβ(tγf(t,x(t)))

    and

    DαCx(t)=ϕq[Iβ(tγf(t,x(t)))+x1], (2.3)

    where x1=ϕp(DαCx(0))=ϕp(xα).

    Applying the operator Iα to both sides of (2.3) and using Proposition 1.1 f) we get the Volterra integral equation (2.2):

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))+x1]+x0.

    Thus, (2.1) implies (2.2). But in general (2.2) does not imply (2.1). More precisely, if xC[0,1] satisfies the Volterra integral equation (2.2) then, in general, it does not follow that xAC[0,1] and DαCxAC[0,1]. However, if we use the generalized Caputo derivative Dα instead of the Caputo derivative DαC then we can show that they are equivalent. The equivalence result is given in Theorem 2.2.

    Additionally, we note that Iα:AC[0,1]AC[0,1] and consequently

    AC[0,1]=Dom(DαC)Dom(Dα)={x(t):I1α(xx0)AC[0,1]}.

    On the other hand, if 0<α<1 and xDom(DαC), then Dαx(t)=DαCx(t) (Proposition 1.1 f)). This means that the operator Dα is an extension of the operator DαC.

    Now we give the equivalence result.

    Theorem 2.2. Let fC([0,1]×R) and 0γ<α,β1. A function xC[0,1] satisfies the Volterra integral equation

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))+x1]+x0,x0,x1R (2.4)

    if and only if xC[0,1] is a solution of the following initial value problem

    {Dβϕp(Dαx(t))=tγf(t,x(t)),x(0)=x0,Dαx(0)=xα,xDom(Dα),ϕp(Dαx)Dom(Dβ),DαxC[0,1], (2.5)

    where xα=ϕq(x1).

    Proof. Let xC[0,1] satisfy the Volterra integral equation (2.4). By using the substitution s=σt we can write

    Iβ(tγf(t,x(t)))=1Γ(β)t0(ts)β1sγf(s,x(s))ds=
    tβγΓ(β)10(1σ)β1σγf(tσ,x(tσ))dσ.

    This equality together with the conditions fC([0,1]×R) and 0γ<β1 yields

    ϕq(Iβ(tγf(t,x(t))))C[0,1]. (2.6)

    It follows from (2.6) that Iαϕq[Iβ(tγf(t,x(t)))+x1]|t=0=0 and consequently x(0)=x0. Applying I1α to (2.4) and using Proposition 1.1. c) we get

    I1α(x(t)x0)=Iϕq[Iβ(tγf(t,x(t)))+x1]C1[0,1]AC[0,1].

    This means that xDom(Dα). By Proposition 1.1 d)

    Dαx(t)=Dα(x(t)x0)=DαIαϕq[Iβ(tγf(t,x(t)))+x1]=ϕq[Iβ(tγf(t,x(t)))+x1]. (2.7)

    So, DαxC[0,1] and Dαx(0)=ϕq(x1)=xα. Applying ϕp to both sides of (2.7) and using the fact that tγf(t,x(t))L1 we obtain

    ϕp(Dαx(t))=Iβ(tγf(t,x(t)))+x1ϕp(Dαx(t))ϕp(Dαx(0))=Iβ(tγf(t,x(t)))I1β[ϕp(Dαx(t))ϕp(Dαx(0))]=I(tγf(t,x(t)))AC[0,1]ϕp(Dαx)Dom(Dβ). (2.8)

    It follows from the second equation in (2.8) that

    Dβϕp(Dαx(t))=tγf(t,x(t)).

    Conversely, assume that the conditions xDom(Dα), ϕp(Dαx)Dom(Dβ), and DαxC[0,1] are satisfied and Eq (2.5) holds. By using these conditions, the definition of Dβ and Proposition 1.1, we obtain from Eq (2.5) that

    DI1β[ϕp(Dαx(t))x1]=tγf(t,x(t))(by the conditions)I1β[ϕp(Dαx(t))x1]=I(tγf(t,x(t)))(by applyingIβ)I[ϕp(Dαx(t))x1]=Iβ+1(tγf(t,x(t)))(taking derivative)ϕp(Dαx(t))=Iβ(tγf(t,x(t)))+x1(by applyingϕq)Dαx(t)=ϕq[Iβ(tγf(t,x(t)))+x1].

    Now, using the definition of Dα and the condition xDom(Dα) we obtain that

    DI1α(x(t)x0)=ϕq[Iβ(tγf(t,x(t)))+x1]I1α(x(t)x0)=Iϕq[Iβ(tγf(t,x(t)))+x1]I(x(t)x0)=Iα+1ϕq[Iβ(tγf(t,x(t)))+x1].

    Taking the derivative from the last equation we get

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))+x1]+x0.

    Finally, we compare the problems (2.1) and (2.5).

    Theorem 2.3. Let 0γ<α,β<1. If a function x(t) is a solution of the problem with Caputo derivative:

    {DβCϕp(DαCx(t))=tγf(t,x(t)),x(0)=x0,DαCx(0)=xα,x0,xαR,xAC[0,1],DαCxAC[0,1],

    then it is a solution of the problem with generalized Caputo derivative:

    {Dβϕp(Dαx(t))=tγf(t,x(t)),x(0)=x0,Dαx(0)=xα,xDom(Dα),ϕp(Dαx)Dom(Dβ),DαxC[0,1].

    Conversely, if a function x(t) is a solution of the problem with the generalized Caputo derivative and additionally xAC[0,1],DαCxAC[0,1], then it is a solution of the problem with the Caputo derivative.

    Proof. This fact immediately follows from the definition of the domain of the operators DαC,Dα, and Proposition 1.1. f), i.e., if 0<α<1 and x(t)AC[0,1] then Dαx(t)=Dα(x(t)x(0))=DαCx(t).

    Our starting initial value problem is:

    DβCϕp(DαCx(t))=tγf(t,x(t)),0γ<α,β<1,x(0)=0,DαCx(0)=0,xAC[0,1],DαCxAC[0,1]. (3.1)

    We consider the homogeneous initial value problem because by using the substitution x(t)=y(t)+xααΓ(α)tα+x0 one can transform the non-homogeneous initial conditions x(0)=x0,DαCx(0)=xα into the homogeneous conditions.

    Theorem 2.3 gives us the basis for defining a generalized solution concept as follows.

    Definition 3.1. A function x(t) is called a generalized solution to the problem (3.1) if it is a solution to the following problem.

    {Dβϕp(Dαx(t))=tγf(t,x(t)),0γ<α,β<1,x(0)=0,Dαx(0)=0,xDom(Dα),ϕp(Dαx)Dom(Dβ),DαxC[0,1]. (3.2)

    By Theorem 2.2, the problem (3.2) is equivalent to the Volterra integral equation:

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))],xC[0,1]. (3.3)

    For this reason, we will study the problem (3.3) instead of (3.2).

    The main theorem regarding the existence of a solution to the Volterra integral equation is as follows.

    Theorem 3.1. Let f be continuous on [0,1]×R and there exist nonnegative functions a,bC[0,1] such that |f(t,u)|a(t)+b(t)|u|p1, for all t[0,1] and uR. If

    Γ(1γ)bΓ(α+1)p1Γ(1γ+β)<1, (3.4)

    then the Volterra integral equation (3.3) has a solution in C[0,1]. So, the problem (3.1) has a generalized solution in C[0,1].

    Proof. We use Schaefer's fixed point theorem to show the existence of a solution to problem (3.3). A version of Schaefer's fixed point theorem is as follows:

    Let X be a Banach space. If

    ⅰ) T:XX is a continuous compact operator,

    ⅱ) the set 0λ1{xX:x(t)=λTx(t)} is bounded,

    then T has a fixed point in X.

    We have

    x(t)=Iαϕq[Iβ(tγf(t,x(t)))].

    Denoting Tx(t)=Iαϕq[Iβ(tγf(t,x(t)))] we can write this equation in the following form

    Tx(t)=x(t).

    Now, we need to show that T has a fixed point in C[0,1].

    Tx(t)=1Γ(α)t0(ts)α1ϕq[1Γ(β)s0(sτ)β1τγf(τ,x(τ))dτ]ds.

    ⅰ) We have to prove that T:C[0,1]C[0,1] is a continuous compact operator. First, let us show that T:C[0,1]C[0,1] is continuous, i.e.,

    xn(t)x(t)inC[0,1]Txn(t)Tx(t)inC[0,1].

    A convergent sequence in a normed space is bounded. Hence,

    xn(t)x(t)inC[0,1]xn(t)M,for alltC[0,1].

    Since the function f is continuous on [0,1]×[M,M], it is uniformly continuous on this compact set. It means that for ε>0 there exists nεN such that

    |f(t,xn(t))f(t,x(t))|<ε,for allnnε,and allt[0,1].

    By the definition of T,

    Txn(t)Tx(t)=Iα[ϕq(Iβ(tγf(t,xn(t))))ϕq(Iβ(tγf(t,x(t))))].

    Let

    yn(t)=Iβ(tγf(t,xn(t)))=1Γ(β)t0(ts)β1sγf(s,xn(s))ds,
    y(t)=Iβ(tγf(t,x(t)))=1Γ(β)t0(ts)β1sγf(s,x(s))ds.

    The function f is bounded on [0,1]×[M,M]. Hence |f(t,s)|L,for all(t,s)[0,1]×[M,M].

    Then

    yn,yLΓ(β)B(β,1γ)

    and

    |yn(t)y(t)|εΓ(β)B(β,1γ)for allnnε,and allt[0,1],

    where B(ν,μ)=10(1s)ν1sμ1ds=Γ(ν)Γ(μ)Γ(ν+μ) for ν>0,μ>0.

    Finally, it follows from the continuity of the function ϕp and continuity of the operator Iα:C[0,1]C[0,1] that the operator T is continuous.

    Now we prove that the operator Tx(t)=Iαϕq[Iβ(tγf(t,x(t)))] is compact. For this, we need to show that for any bounded set ΩC[0,1] the set ¯T(Ω) is compact. By the Arzela-Ascoli theorem, ¯T(Ω) is compact if and only if

    a) T(Ω) is bounded in C[0,1] (it is the same that T(Ω) is uniformly bounded).

    b) T(Ω) is equicontinuous.

    We first show that T(Ω) is bounded in C[0,1]. Boundedness of T(Ω) means that

    Tx(t)C,xΩ,

    where C does not depend on x. This is a trivial fact. However, we will give a short proof. Since Ω is bounded, we have xC for all xΩ. From the continuity of f it follows that it is uniformly continuous on [0,1]×[C,C]. Then |f(t,x(t)|M for all xΩ and all t[0,1]. Hence,

    |Iβ(tγf(t,x(t)))|=|1Γ(β)t0(ts)β1sγf(s,x(s))ds|MΓ(β)B(β,1γ). (3.5)

    On the other hand, Iα:C[0,1]C[0,1] is a bounded operator and

    Iαx(t)1Γ(α+1)x(t. (3.6)

    Then, by using (3.5), (3.6), and the monotonicity of sq1, we obtain that

    Tx(t)=Iαϕq[Iβ(tγf(t,x(t)))]1Γ(α+1)[MΓ(β)B(β,1γ)]q1.

    This means that T(Ω) is uniformly bounded.

    b) Now we show that T(Ω) is equicontinuous. To prove this, we use the same technique as in [11] (see theorem 3.1 in [11]).

    Let 0t1<t21 and xΩ. By (3.5) we have

    |ϕq[Iβ(tγf(t,x(t)))]|(MΓ(β)B(β,1γ))q1. (3.7)

    Then, using (3.7) we get

    |Tx(t2)Tx(t1)|=1Γ(α)|t20(t2s)α1ϕq[Iβ(sγf(s,x(s)))]dst10(t1s)α1ϕq[Iβ(sγf(s,x(s)))]ds|=1Γ(α)|t10((t2s)α1(t1s)α1)ϕq[Iβ(sγf(s,x(s)))]ds+t2t1(t2s)α1ϕq[Iβ(sγf(s,x(s)))]|1Γ(α+1)(MΓ(β)B(β,1γ))q1[tα1+tα2+2(t2t1)α].

    This inequality together with uniform continuity of tα on [0,1] yields that T(Ω) is equicontinuous.

    ⅱ) Let us prove that the set 0λ1{xC[0,1]:x(t)=λTx(t)} is bounded. Let x(t)=λTx(t),λ(0,1]. By the condition of the theorem |f(t,u)|a(t)+b(t)|u|p1, for all t[0,1] and all uR. Then

    |x(t)|Tx(t)|=1Γ(α)|t0(ts)α1ϕq[1Γ(β)s0(sτ)β1τγf(τ,x(τ))dτ]ds|.

    Setting τ=σs, we obtain that

    |x(t)|1Γ(α)|t0(ts)α1ϕq[sβγΓ(β)10(1σ)β1σγf(σs,x(σs))dσ]ds|1Γ(α)|t0(ts)α1[B(β,1γ)Γ(β)(a+bxp1)]q1ds|=1Γ(α+1)[Γ(1γ)Γ(1γ+β)(a+bxp1)]q1,

    where we have used B(β,1γ)=Γ(β)Γ(1γ)Γ(1γ+β). Since (p1)(q1)=1, we obtain from the last inequality that

    xp11Γ(α+1)p1Γ(1γ)Γ(1γ+β)(a+bxp1)

    and consequently,

    bxp1a+bxp1Γ(1γ)bΓ(α+1)p1Γ(1γ+β). (3.8)

    If the set 0λ1{xC[0,1]:x(t)=λTx(t)} were unbounded then there would be a sequence in this set that converges to infinity. Then taking limit as x in (3.8) and using the condition of the theorem Γ(1γ)bΓ(α+1)p1Γ(1γ+β)<1, we get the following contradiction

    1Γ(1γ)bΓ(α+1)p1Γ(1γ+β)<1.

    Hence, the set 0λ1{xC[0,1]:x(t)=λTx(t)} is bounded and, by Schaefer's theorem, the operator T has a fixed point.

    Using the methods applied in Theorem 3.1, we obtain the following result for the smallest eigenvalue of the fractional p-laplacian operator.

    Corollary 3.1. Let μ1 be the smallest eigenvalue of the following eigenvalue problem:

    Dβϕp(Dαx(t))=μ|x|p2x,μ>0,p>1,0<α,β1,x(0)=0,Dαx(0)=0.

    Then

    μ1Γ(α+1)p1Γ(β+1).

    Proof. Let 0<μ1<Γ(α+1)p1Γ(β+1) be an eigenvalue of the given problem. Then,

    Dβϕp(Dαx(t))=μ1|x|p2x,x(0)=0,Dαx(0)=0.

    By Theorem 2.2, this problem is equivalent to the Volterra integral equation:

    x(t)=Iαϕq[Iβ(μ1|x|p2x)],xC[0,1].

    According to Theorem 3.1, we have a=0,b=μ1 and by the assumption 0<μ1<Γ(α+1)p1Γ(β+1) the condition

    bΓ(α+1)p1Γ(1+β)=μ1Γ(α+1)p1Γ(1+β)<1

    is satisfied. Then under these conditions we have proved in Theorem 3.1 that the set 0λ1{xC[0,1]:x(t)=λTx(t)} is bounded, where Tx(t)=Iαϕq[Iβ(μ1|x|p2x)]. But, since μ1 is an eigenvalue, the set x(t)=λTx(t)} is unbounded for λ=1. This is a contradiction. Consequently, μ1Γ(α+1)p1Γ(β+1).

    Finally, we give an example as an application of Theorem 3.1.

    Example 3.1. If 0<b<12 then the following problem has a solution in C[0,1].

    D12ϕ3(D12x(t))=t1/3(1+bx2(t)),x(0)=0,Dαx(0)=0.

    Let us check the conditions of Theorem 3.1. In this case

    f(t,x(t))=1+bx2(t),p=3,γ=13,α=β=12 and 0<γ<α,β<1. Then

    Γ(1γ)bΓ(α+1)p1Γ(1γ+β)=Γ(23)bΓ(32)2Γ(76).

    Γ(23)=1,35411..,Γ(32)=0,88622..,Γ(76)=0,92771.. and

    Γ(23)bΓ(32)2Γ(76)<1,4b(0,88)20,92<2b<1.

    Consequently, the condition

    Γ(1γ)bΓ(α+1)p1Γ(1γ+β)<1

    is satisfied and the problem has a solution.

    The main subject of this paper was the initial value problems for Caputo fractional differential equations with singular nonlinearities involving the p-Laplacian operator. The most important difference of this paper from other studies on this subject is that the equation

    DβCϕp(DαCx(t))=tγf(t,x(t)),0γ<αβ1

    contains the singular term tγ.

    In general, Volterra integral equations are used to solve problems given for fractional differential equations. However, their equivalence is not always shown in the literature. In this article, the equivalence of such problems was discussed in full detail. We would like to emphasize that we benefited from the techniques of Webb [36] in this subject.

    For the case f=μ|x|p2x,γ=0, a result related to the lower bound of the eigenvalues is given as Corollary 3.1.

    By the methods, applied in the article, similar results can be obtained for higher order equations and different initial and boundary value problems.

    The author declares that he has not used Artificial Intelligence (AI) tools in the creation of this article.

    The author declares that he has no competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.



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