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An eigenvalue problem related to the variable exponent double-phase operator

  • In this paper, we studied a double-phase eigenvalue problem with large variable exponents. Let λ1(pn(),qn()) be the first eigenvalues and un be the first eigenfunctions, normalized by unHn=1. Under some assumptions on the variable exponents pn() and qn(), we showed that λ1(pn(),qn()) converges to Λ, un converges to u uniformly in the space Cα(Ω)(0<α<1) and u is a nontrivial viscosity solution to a Dirichlet -Laplacian problem. Even in the case where the variable exponents reduce to the constant exponents, our work is the first one dealing with a double-phase eigenvalue problem with large exponents.

    Citation: Lujuan Yu, Beibei Wang, Jianwei Yang. An eigenvalue problem related to the variable exponent double-phase operator[J]. AIMS Mathematics, 2024, 9(1): 1664-1682. doi: 10.3934/math.2024082

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  • In this paper, we studied a double-phase eigenvalue problem with large variable exponents. Let λ1(pn(),qn()) be the first eigenvalues and un be the first eigenfunctions, normalized by unHn=1. Under some assumptions on the variable exponents pn() and qn(), we showed that λ1(pn(),qn()) converges to Λ, un converges to u uniformly in the space Cα(Ω)(0<α<1) and u is a nontrivial viscosity solution to a Dirichlet -Laplacian problem. Even in the case where the variable exponents reduce to the constant exponents, our work is the first one dealing with a double-phase eigenvalue problem with large exponents.



    An expedient feature of the p-Laplacian eigenvalue problem is that the eigenfunctions may be multiplied by constant factors (in other words, the fact that if u is an eigenfunction, so is ku). Unfortunately, the p(x)-Laplacian eigenvalue problem does not possess this expedient property. It is important to stress that the loss of the property under consideration is not only a consequence of the dependence on x, but it can also occur in presence of unbalanced growth. For example, the double phase operator (that does not depend on x)

    div(|u|p2u+μ(x)|u|q2u), (1.1)

    loses this property. In this paper we are interested in considering that the operator has both peculiarities: It depends on x and it is unbalanced.

    Let ΩRN(N2) be a bounded domain with Lipschitz boundary Ω. This article studies an eigenvalue problem coming from the minimization of the Rayleigh quotient:

    ||u||Hn||u||Hn, (1.2)

    among all uW1,Hn0(Ω),u0. These functions belong to an appropriate Musielak-Orlicz Sobolev space with variable exponents; see its definition in section two. The function a:ˉΩ[0,+) is a C1 differentiable function.

    Put

    Kn(u):=||u||Hn,kn(u):=||u||Hn,Sn(u):=Ω[pn(x)|uKn(u)|pn(x)2+qn(x)a(x)|uKn(u)|qn(x)2]dxΩ[pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2]dx,Hn:=tpn(x)+a(x)tqn(x) (1.3)

    and define the first eigenvalue as

    λ1(pn(),qn())=infuW1,Hn0(Ω){0}||u||Hn||u||Hn. (1.4)

    By a similar proof of Proposition 3.1 in [1], we can show that the following equation

    div[(pn(x)|uKn(u)|pn(x)2+qn(x)a(x)|uKn(u)|qn(x)2)uKn(u)]=λ(pn(),qn())Sn(u)ukn(u)(pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2),uW1,Hn0(Ω) (1.5)

    is the Euler-Lagrange equation corresponding to the minimization of the Rayleigh quotient (1.2), where λ(pn(),qn())=λ1(pn(),qn()).

    Here, we impose the following hypotheses on the variable exponents pn(x) and qn(x).

    (H1): Assume that pn(x) and qn(x) are two sequences of C1 functions in ¯Ω, qn()>pn() for every n1 and

    pn(x),qn(x)+,uniformly for allxΩ, (1.6)
    pn(x)pn(x)ξ1(x),uniformly for allxΩ, (1.7)
    qn(x)qn(x)ξ2(x),uniformly for allxΩ. (1.8)

    (H2): The following two quotients are bounded, namely,

    lim supn+p+npnk1,lim supn+q+nqnk2, (1.9)

    where for a function g we denote

    g=minx¯Ωg(x),g+=maxx¯Ωg(x).

    (H3): We also assume that

    pn>1,qn>1,q+npn<1+1N, (1.10)

    then we can find a positive and continuous function θ(0<θ<+), such that

    limnqn(x)pn(x)=θ(x) (1.11)

    uniformly for all xΩ.

    The differential operator in (1.5) is the double-phase operator with variable exponents, which can be given by

    div(|u|pn(x)2u+μ(x)|u|qn(x)2u). (1.12)

    This operator is the classical double phase operator (1.1) when pn(x) and qn(x) are constants. Moreover, special cases of (1.12), studied extensively in the literature, occur when inf¯Ωμ>0 (the weighted (q(x),p(x))-Laplacian) or when μ0 (the p(x)-Laplacian).

    The energy functional related to the double-phase operator (1.12) is given by

    Ω|u|pn(x)+μ(x)|u|qn(x)dx, (1.13)

    whose integrand switches two different elliptic behaviors. The integral functional (1.13) was first introduced by Zhikov [2,3,4,5], who obtained that the energy density changed its ellipticity and growth properties according to the point in order to provide models for strongly anisotropic materials. Moreover, double phase differential operators (1.12) and corresponding energy functionals (1.13) have several physical applications. We refer to the works of [6] on transonic flows, [7] on quantum physics and [8] on reaction diffusion systems. Finally, we mention a recent paper that is very close to our topic. For related works dealing with the double phase eigenvalue problems, we refer to the works of Colasuonno-Squassina [9], who proved the existence and properties of related variational eigenvalues. By using the Rayleigh quotient of two norms of Musielak-Orlicz space, the author of this paper has defined the eigenvalue, which has the same properties as the p-Laplace operator. Recently, Liu-Papageorgiou has considered an eigenvalue problem for the Dirichlet (p,q())Laplacian by using the Nehari method (see [10]), a nonlinear eigenvalue problem for the Dirichlet (p,q)Laplacian with a sign-changing Carathéodory reaction (see [11]) and a nonlinear eigenvalue problem driven by the anisotropic (p(),q())Laplacian (see [12]). Motivated by [9], Yu[13] discuss the asymptotic behavior of an eigenvalue for the double phase operator. However, to the author's knowledge, the eigenvalue problem for variable exponents double phase operator has remained open. Our article fits into this general field of investigation.

    Assume that δ:Ω[0,) is the distance function, which is given by

    δ(x):=dist(x,Ω)=infyΩ|xy|.

    This function is a Lipschitz continuous function. For all xΩ, we get |δ|=1. Define

    Λ:=infφW1,0(Ω){0}φL(Ω)φL(Ω). (1.14)

    It is known from the paper [1] that

    Λ=δL(Ω)δL(Ω)=1maxxΩ{dist(x,Ω)}. (1.15)

    Define

    u:=Ni,j=1(u)xi(u)xj(u)xixj,k(u):=||u||L(Ω)=esssupxΩ|u|, (1.16)
    K(u):=||u||L(Ω)=esssupxΩ|u|, (1.17)

    and

    k(u):=||u||L(Ω)=esssupxΩ|u|.

    The following are the main results of this paper.

    Theorem 1.1. Let uC(Ω) be a weak solution of problem (1.5), then it is also a viscosity solution of the problem (3.2).

    Theorem 1.2. Let hypotheses (H1)–(H3) be satisfied, λ1(pn(),qn()) and Λ be defined by (1.4) and (1.14), respectively. In addition, assume that un normalized by unHn=1 is the positive first eigenfunction, then,

    (1)

    limnλ1(pn(),qn())=Λ; (1.18)

    (2) there exists a nonnegative function u such that uCα(Ω){0} and ||u||L(Ω)=1;

    (3) we can extract a subsequence, which is still denoted by un, such that

    unu, (1.19)

    in the space Cα(Ω), where α(0<α<1) is a constant;

    (4) we can obtain that the function u(x) is a nontrivial viscosity solution of the problem

    {min{Λu+|u|,Λ(u)θ(x0)+|u|,u[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ2(x0)}=0,xΩ,u=0,xΩ. (1.20)

    To the best of our knowledge, this is the first work dealing with the double phase eigenvalue problem (1.5). The rest of this paper is organized as follows. In section two, we collect some notations and facts about the Musielak-Orlicz space LH(Ω) and W1,H0(Ω), which will be used in this paper. Section three and section four are devoted to prove Theorems 1.1 and 1.2, respectively.

    In this section, we recall some known results about the Musielak-Orlicz spaces LH(Ω) and W1,H0(Ω). For more detail, please see references [9,14,15,16,17].

    We follow the notation of [9]. Let N(Ω) denote the set of all generalized N-functions. Let us introduce the nonlinear function H:Ω×[0,+)[0,+) defined as

    H(x,t):=tp(x)+a(x)tq(x),forall(x,t)Ω×[0,+),

    with 1<p(x)<q(x) and 0a()L1(Ω). It is clear that HN(Ω) is a locally integrable and generalized N-function. In addition, it fulfills the Δ2 condition, namely,

    H(x,2t)2q+H(x,t).

    Therefore, in correspondence to H, we define the Musielak-Orlicz space LH(Ω) as

    LH(Ω):={u:ΩRmeasurable:ρH(u)<+},

    which can be equipped with the norm

    uH:=inf{γ>0:ρH(u/γ)1},

    where

    ρH(u):=ΩH(x,|u|)dx,

    which is called H-modular.

    Similarly, we can define the Musielak-Orlicz Sobolev spaces. The space W1,H(Ω) is given by

    W1,H(Ω)={uLH(Ω)suchthat|u|LH(Ω)},

    with the norm

    u1,H:=uH+uH.

    We denote by W1,H0(Ω) the completion of C0(Ω) in W1,H(Ω). With these norms, the spaces LH(Ω), W1,H(Ω) and W1,H0(Ω) are separable, reflexive and uniformly convex Banach spaces.

    From Proposition 2.16 (ⅱ) in [18], if

    q+p<1+1N,

    then the following Poincarˊe-type inequality

    uHCuH

    holds for all uW1,H0(Ω), where C is a positive constant independent of u. Therefore, in this paper, we equip W1,H0(Ω) with the equivalent norm uH for all uW1,H0(Ω).

    Proposition 2.1. [18] If uLH(Ω) and ρH(u) is the H-modular, then the following properties hold.

    (1) Ifu0, then uH=λ if, and only if, ϱH(uλ)=1;

    (2) uH<1 (=1;>1) if, and only if, ϱH(u)<1 (=1;>1);

    (3) If uH1, then uq+HρH(u)upH;

    (4) If uH1, then upHρH(u)uq+H;

    (5) uH0 if, and only if, ρH(u)0;

    (6) uH0 if, and only if, ρH(u)0.

    Given uC(Ω)W1,Hn0(Ω) and ϕC2(Ω). Define

    pn(x)ϕ:=div(|ϕ|pn(x)2ϕ)=|ϕ|pn(x)4{|ϕ|2ϕ+(pn(x)2)ϕ+|ϕ|2ln(|ϕ|)ϕpn},
    qn(x)ϕ:=div(|ϕ|qn(x)2ϕ)=|ϕ|qn(x)4{|ϕ|2ϕ+(qn(x)2)ϕ+|ϕ|2ln(|ϕ|)ϕqn},

    and

    ϕ:=Ni,j=1ϕxiϕxj2ϕxixj,

    where ϕ is the -Laplacian.

    Here, we are now in a position to give the following definition of weak solutions to problem (1.5).

    Definition 3.1. We call uW1,Hn0(Ω){0} a weak solution of problem (1.5) if

    Ω(pn(x)|uK(u)|pn(x)2+qn(x)a(x)|uK(u)|qn(x)2)uvK(u)dx=λ(pn(),qn())S(u)Ω(pn(x)|uk(u)|pn(x)2+qn(x)a(x)|uk(u)|qn(x)2)uvk(u)dx (3.1)

    is satisfied for all test functions vW1,Hn0(Ω). If u0, we say that λ(pn(),qn()) is an eigenvalue of (1.5) and that u is an eigenfunction corresponding to λ(pn(),qn()).

    In (1.5), we replace u by ϕ and keep Sn, Kn and kn unchanged, then

    {pn(x)(K(u))1pn(x)pn(x)ϕqn(x)a(x)(K(u))1qn(x)qn(x)ϕqn(x)(K(u))1qn(x)|ϕ(x)|qn(x)2ϕ(x)a(x)(K(u))1pn(x)|ϕ|pn(x)2ϕ(x)pn(x)a(x)(K(u))1qn(x)|ϕ(x)|qn(x)2ϕ(x)qn(x)+pn(x)(K(u))1pn(x)ln(K(u))|ϕ(x)|pn(x)2ϕ(x)pn(x)+qn(x)a(x)(K(u))1qn(x)ln(K(u))|ϕ(x)|qn(x)2ϕ(x)qn(x)λ(pn(),qn())S(u)(pn(x)(k(u))1pn(x)|ϕ|pn(x)2ϕ+qn(x)a(x)(k(u))1qn(x)|ϕ(x)|qn(x)2ϕ(x))=0,xΩ,ϕ=0,xΩ.

    We first recall the definition of viscosity solutions. Assume we are given a continuous function

    F:RN×R×RN×S(N)R,

    where S(N) denotes the set of N×N symmetric matrices.

    Consider the problem

    F(x,u,u,D2u)=0, (3.2)

    where

    F(x,u,u,D2u)=pn(x)(K(u))1pn(x){|u|pn(x)4[|u|2u+(pn(x)2)u+|u|2ln(|u|)upn(x)]}qn(x)a(x)(K(u))1qn(x){|u|qn(x)4[|u|2u+(qn(x)2)u+|u|2ln(|u|)uqn(x)]}qn(x)(K(u))1qn(x)|u|qn(x)2ua(x)(K(u))1pn(x)|u|pn(x)2upn(x)a(x)(K(u))1qn(x)|u|qn(x)2uqn(x)+pn(x)(K(u))1pn(x)ln(K(u))|u|pn(x)2upn(x)+qn(x)a(x)(K(u))1qn(x)ln(K(u))|u|qn(x)2uqn(x)λ(pn(),qn())S(u)(pn(x)(k(u))1pn(x)|u|pn(x)2u+qn(x)a(x)(k(u))1qn(x)|u|qn(x)2u). (3.3)

    Definition 3.2. Assume that x0Ω, uC(Ω), ψC2(Ω) and φC2(Ω).

    (1) Let u(x0)=ψ(x0) and suppose that uψ attains its strict maximum value at x0. If

    F(x0,ψ(x0),ψ(x0),D2ψ(x0))0

    for all of such x0, then the function u is said to be a viscosity subsolution of Eq (3.2).

    (2) Let u(x0)=φ(x0) and suppose that uφ attains its strict minimum value at x0. If

    F(x0,φ(x0),φ(x0),D2φ(x0))0

    for all of such x0, then the function u is said to be a viscosity supersolution of Eq (3.2).

    (3) If u is both a subsolution and a supersolution of the problem (3.2), then u is a viscosity solution of the problem (3.2).

    Proof of Theorem 1.1. Claim: u is a viscosity supersolution of (3.2).

    Let x0Ω and φC2(Ω). Assume that u(x0)=φ(x0) and the function uφ obtains its strict minimum value at the point x0. Our goal is to show that

    F(x0,u(x0),φ(x0),D2φ(x0))0. (3.4)

    If

    F(x0,u(x0),φ(x0),D2φ(x0))<0,

    then by continuity there exists a positive constant r such that B(x0,2r)Ω, u>φ in this ball, except for the point x0 and

    F(x,u(x),φ(x),D2φ(x))<0,

    for all xB(x0,2r). Thus, if xB(x0,r), we have

    div[(pn(x)|φ(x)Kn(u)|pn(x)2+qn(x)a(x)|φ(x)Kn(u)|qn(x)2)φ(x)Kn(u)]λ(pn(),qn())Sn(u)(pn(x)|u(x)kn(u)|pn(x)2+qn(x)a(x)|u(x)kn(u)|qn(x)2)u(x)kn(u)<0.

    If xB(x0,r), the minimum value of the function uφ is defined as m. Let Φ(x):=φ(x)+m2. Note that m>0 and the above inequality still holds if the function φ(x) is replaced by Φ(x), namely,

    div[(pn(x)|Φ(x)Kn(u)|pn(x)2+qn(x)a(x)|Φ(x)Kn(u)|qn(x)2)Φ(x)Kn(u)]λ(pn(),qn())Sn(u)(pn(x)|u(x)kn(u)|pn(x)2+qn(x)a(x)|u(x)kn(u)|qn(x)2)u(x)kn(u)<0. (3.5)

    Define η(x):=(Φu)+0, then if xB(x0,r), we have η(x)0.

    Let

    Ω1={x|xB(x0,r)andΦ(x)>u(x)}.

    We multiply (3.5) by the function η(x) and integrate over B(x0,r), then the inequality

    Ω1(pn(x)|ΦKn(u)|pn(x)2+qn(x)a(x)|ΦKn(u)|qn(x)2)ΦKn(u)(Φu)dxΩ1λ(pn(),qn())Sn(u)(pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2)ukn(u)(Φu)dx<0 (3.6)

    is true.

    If we define

    η1(x)={(Φu)+,xB(x0,r),0,xΩB(x0,r),

    and use η1(x) as a test function in (3.1), then we get

    Ω1(pn(x)|uKn(u)|pn(x)2+qn(x)a(x)|uKn(u)|qn(x)2)uKn(u)(Φu)dxΩ1λ(pn(),qn())Sn(u)(pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2)ukn(u)(Φu)dx=0. (3.7)

    Subtracting (3.7) from (3.6), we arrive at

    Ω1pn(x)(|ΦKn(u)|pn(x)2ΦKn(u)|uKn(u)|pn(x)2uKn(u))(Φu)dx+Ω1qn(x)a(x)(|ΦKn(u)|qn(x)2ΦKn(u)|uKn(u)|qn(x)2uKn(u))(Φu)dx<0. (3.8)

    The first integral is nonnegative due to the elementary inequality

    |a|p2a|b|p2b,ab0, (3.9)

    which holds for all p>1. Here, we take p=pn(x). We get a contradiction. Hence, (3.4) holds. Similarly, we conclude that u is a viscosity subsolution of (3.2) and we omit the details.

    Let nN be large enough such that pnr>N, which results in W1,Hn0(Ω)W1,r0(Ω) (see Proposition 2.16 (1) of Blanco, Gasiński, Harjulehto and Winkert [18]). It follows that un are continuous functions. The reason is that the space W1,r0(Ω)↪↪Cα(Ω), 0<α<1. Moreover, it is known (see [9]) that for each nN fixed, we have un>0.

    In order to prove Theorem 1.2, we only need to prove the following conclusions.

    Lemma 4.1. Let h:¯Ω(1,) be a given continuous function, then

    ||v|s|1sp(x)s||v||H, (4.1)

    for all vW1,H0(Ω) and s(1,p).

    Proof. Since v||v||HH=1, it follows from Proposition 2.1 that

    Ω[(|v|||v||H)p(x)+a(x)(|v|||v||H)q(x)]dx=1. (4.2)

    Thus,

    Ω[(|v|||v||H)s]p(x)sdxp(x)s1. (4.3)

    Invoking Proposition 2.1 again, we conclude that

    |(|v|||v||H)s|p(x)s1,

    which implies (4.1).

    Lemma 4.2. If uL(Ω), then we have

    limnkn(u)=k(u). (4.4)

    Proof. Step1: To show that the following inequality holds,

    lim supnkn(u)k(u). (4.5)

    If kn(u)k(u), the above inequality is true. Thus, we can assume that kn(u)>k(u), and since qn(x)>pn(x)>1, we have

    1=(Ω|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1pn[Ω(k(u)kn(u))pn(x)+a(x)(k(u)kn(u))qn(x)dx]1pn[Ω(k(u)kn(u))pn+a(x)(k(u)kn(u))pndx]1pn=k(u)kn(u)(|Ω|+Ωa(x)dx)1pn,

    which implies (4.5) holds.

    Step2: To show that the following inequality holds,

    lim infnkn(u)k(u). (4.6)

    Case1: k(u)=0. It is easy to find that (4.6) holds.

    Case2: k(u)>0. Given ε>0, there exists a nonempty set ΩεΩ such that, for all xΩε, |u|>k(u)ε. Ignoring those indices n that kn(u)k(u)ε, we have

    1=(Ω|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1pn(Ωε|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1pn(Ωε|k(u)εkn(u)|pn(x)+a(x)|k(u)εkn(u)|qn(x)dx)1pn(Ωε|k(u)εkn(u)|pn+a(x)|k(u)εkn(u)|pndx)1pn=k(u)εkn(u)(|Ωε|+Ωεa(x)dx)1pn,

    which gives

    lim infnkn(u)k(u)ε.

    The arbitrariness of ε implies that (4.6) is true. Consequently, (4.4) holds.

    Remark 4.1. If |u|L(Ω), we can argue as Lemma 4.2 to obtain that

    limnKn(u)=K(u). (4.7)

    Lemma 4.3. If the assumptions of Theorem 1.2 hold, then

    (1) (1.18) holds;

    (2) there exists a nonnegative function u such that uCα(Ω){0} and ||u||L(Ω)=1;

    (3) we can extract a subsequence, which is still denoted by un, such that

    unu

    in the space Cα(Ω), where α(0<α<1) is a constant.

    Proof. Assume for simplicity that the following inequality holds

    Ωdx=1.

    Step 1: To show that,

    lim supnλ1(pn(),qn())Λ. (4.8)

    Inserting u(x)=δ(x) into (1.4) gives

    λ1(pn(),qn())||δ||Hn||δ||Hn.

    Note that by Lemma 4.2 and Remark 4.1, we have

    lim supnλ1(pn(),qn())δL(Ω)δL(Ω)=Λ.

    Step 2: We now claim that uW1,0(Ω).

    Since (4.8) holds, for all nN sufficiently large, we can assume that λ1(pn(),qn())Λ+1. Thus, we have

    Λ+1λ1(pn(),qn())=||un||Hn||un||Hn=||un||Hn.

    Note that the sequence {||un||Hn} is bounded.

    Let r[1,) be arbitrary. We can find an integer nr, for all nnr, such that pn()r and

    W1,Hn0(Ω)W1,r0(Ω)↪↪Lr(Ω).

    Hence, the sequence {un} is bounded in the reflexive Banach space W1,r0(Ω). We can find a subsequence, still defined by {un}, and a function uW1,r0(Ω), such that unu in W1,r0(Ω) and unu in Lr(Ω).

    Define

    sn(x):=pn(x)pn(x)r,xΩ,

    and it follows that

    s+n=pnpnr,sn=p+np+nr

    and

    |1|sn(x)max{|Ω|1s+n,|Ω|1sn}. (4.9)

    Using Hölder's inequality and the above inequality, we have

    (Ω|un|rdx)1r(1sn+rpn)|1|1rsn(x)||un|r|1rpn(x)r(1sn+rpn)max{|Ω|1s+n,|Ω|1sn}1r||un|r|1rpn(x)r. (4.10)

    Thus, (4.1) and (4.10) ensure that

    unLr(Ω)2(1+|Ω|)||un||Hn2(1+|Ω|)Λ+1. (4.11)

    We choose an arbitrary positive real number r1 such that B(x,r1)Ω, where the point xΩ is a Lebesgue point such that |u|L1(Ω), then we find that

    1|B(x,r1)|B(x,r1)|u(y)|dylim infn1|B(x,r1)|B(x,r1)|un(y)|dylim infn|B(x,r1)|1r||un||Lr(Ω)|B(x,r)|1r2(1+|Ω|)(Λ+1). (4.12)

    Passing to the limit as r+ in the above inequality, gives

    1|B(x,r1)|B(x,r1)|u(y)|dy2(1+|Ω|)(Λ+1).

    Letting r10+ in the above inequality, gives

    |u(x)|2(1+|Ω|)(Λ+1),

    for a.e. xΩ, which implies that uL(Ω), as claimed.

    Step 3: We want to prove that unu in Cα(Ω) (0<α<1) and uL(Ω)=1.

    Keeping in mind that r[1,) is an arbitrary constant, we can assume that r>N. Therefore, this combined with the fact that W1,r0(Ω)↪↪Cα(Ω)(0<α<1) implies that there exists a nonnegative function uCα(Ω){0}, such that unu in Cα(Ω) and un converges uniformly to u in Ω. Given ε(0,1), we can find a constant NεN such that

    |un(x)u(x)|<ε, (4.13)

    for all xΩ,nNε. It follows that

    [ρHn(unu)]1pn=[Ω|unu|pn(x)+a(x)|unu|qn(x)dx]1pn[Ωεpn(x)+a(x)εqn(x)dx]1pnε[Ω(1+a(x))dx]1pn[Ω(1+a(x))dx]1pn (4.14)

    and

    [ρHn(unu)]1q+nεpnq+n[Ω(1+a(x))dx]1q+n[Ω(1+a(x))dx]1q+n, (4.15)

    for all nNε. Letting n in (4.14) and (4.15) yields

    limn[ρHn(unu)]1pn=limn[ρHn(unu)]1q+n=0. (4.16)

    Thus, the inequality

    |||un||HnuL(Ω)||||un||Hn||u||Hn|+|||u||HnuL(Ω)|||unu||Hn+|||u||HnuL(Ω)|{[ρHn(unu)]1pn+[ρHn(unu)]1q+n}+|||u||HnuL(Ω)|.{[ρHn(unu)]1pn+[ρHn(unu)]1q+n}+|kn(u)uL(Ω)|

    holds. In view of Lemma 4.2 and (4.16), we can get

    uL(Ω)=limn||un||Hn=1. (4.17)

    Step 4: To show that lim infnλ1(pn(),qn())Λ.

    Since unu in W1,r0(Ω), ||un||Hn=1 and the inequality (4.11) holds, we have

    uLr(Ω)lim infnunLr(Ω)lim infnunHn=lim infnλ1(pn(),qn()).

    Letting r and using Proposition 7 in [19] and equality (4.17), we get

    ΛuL(Ω)uL(Ω)lim infnλ1(pn(),qn()). (4.18)

    Thus, (4.8) and (4.18) imply that (1.18) holds. The proof is complete.

    Remark 4.2. We can again argue with Step 3 to obtain

    uL(Ω)=limn||un||Hn. (4.19)

    The function u(x) also has the following property.

    Lemma 4.4. If the assumptions of Theorem 1.2 hold, we can deduce that u(x) is a nontrivial viscosity solution of the problem (1.20).

    Proof. For the first part we only need to show that u is a viscosity subsolution of (1.20). Let x0Ω and ψC2(Ω). Assume that uψ attains its strict maximum value of zero at x0, namely, u(x0)ψ(x0)=0.

    Claim: We want to show that

    max{Λψ(x0)|ψ(x0)|,(ψ(x0))θ(x0)K(u)|ψ(x0)|,ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ2(x0)}0. (4.20)

    By Lemma 4.3, we know that the convergence of un to u in Ω is uniform. Therefore, there exists a sequence {xn}Ω such that xnx0 (as n), un(xn)=ψ(xn) and unψ attains its strict maximum value at xn.

    Employing Theorem 1.1, it turns out that for any nN large enough, un are continuous viscosity solutions of (1.5) with λ(pn(),qn())=λ1(pn(),qn()). Thus, we have

    pn(xn)(Kn(un))1pn(xn)|ψ(xn)|pn(xn)4{|ψ(xn)|2ψ(xn)+(pn(xn)2)ψ(xn)+[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)pn(xn)}qn(xn)a(xn)(Kn(un))1qn(xn)|ψ(xn)|qn(xn)4{|ψ(xn)|2ψ(xn)+(qn(xn)2)ψ(xn)+[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)qn(xn)}qn(xn)(Kn(un))1qn(xn)|ψ(xn)|qn(xn)2ψ(xn)a(xn)(Kn(un))1pn(xn)|ψ(xn)|pn(xn)2ψ(xn)pn(xn)a(xn)(Kn(un))1qn(xn)|ψ(xn)|qn(xn)2ψ(xn)qn(xn)λ1(pn(),qn())Sn(un)pn(xn)(kn(un))1pn(xn)|ψ(xn)|pn(xn)2ψ(xn)λ1(pn(),qn())Sn(un)qn(xn)a(xn)(kn(un))1qn(xn)|ψ(xn)|qn(xn)2ψ(xn)0. (4.21)

    Case 1: ψ(x0)=u(x0)>0.

    Continuing (4.21), for nN sufficiently large, we have |ψ(xn)|>0. Let us assume the assertion is not true, then by (4.21) and continuity, we have ψ(x0)0. This leads to a contradiction.

    Dividing both sides of (4.21) by

    pn(xn)(pn(xn)2)(Kn(un))1pn(xn)|ψ(xn)|pn(xn)4,

    we see that the following inequality holds

    |ψ(xn)|2ψ(xn)pn(xn)2ψ(xn)[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)pn(xn)pn(xn)2qn(xn)pn(xn)|ψ(xn)Kn(un)|qn(xn)pn(xn){a(xn)|ψ(xn)|2ψ(xn)pn(xn)2+a(xn)(qn(xn)2pn(xn)2)ψ(xn)+|ψ(xn)|2ψ(xn)a(xn)pn(xn)2+a(xn)|ψ(xn)|2pn(xn)2ψ(xn)qn(xn)qn(xn)+a(xn)[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)qn(xn)pn(xn)2}|ψ(xn)|2pn(xn)ψ(xn)pn(xn)pn(xn)2(λ1(pn(),qn()))3Sn(un)|λ1(pn(),qn())ψ(xn)ψ(xn)|pn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2+(λ1(pn(),qn()))3Sn(un)qn(xn)pn(xn)a(xn)|ψ(xn)|2ψ(xn)pn(xn)2[(|ψ(xn)|kn(un))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|]pn(xn)40. (4.22)

    Now, letting n, we deduce that

    |ψ(xn)|2ψ(xn)pn(xn)2ψ(xn)ψ(x0),[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)pn(xn)pn(xn)2[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ1(x0),qn(xn)pn(xn){a(xn)|ψ(xn)|2ψ(xn)pn(xn)2+a(xn)(qn(xn)2pn(xn)2)ψ(xn)+|ψ(xn)|2ψ(xn)a(xn)pn(xn)2+a(xn)[ln(|ψ(xn)|)ln(Kn(un))]|ψ(xn)|2ψ(xn)qn(xn)pn(xn)2+a(xn)|ψ(xn)|2pn(xn)2ψ(xn)qn(xn)qn(xn)}θ2(x0)a(x0){ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ2(x0)},|ψ(xn)|2pn(xn)ψ(xn)pn(xn)pn(xn)20.

    Taking the lower limit in inequality (4.22) and employing the limits above, we have

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))θ2(x0)a(x0){ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ2(x0)}{ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ1(x0)}=(|ψ(x0)K(u)|lim infn(qn(xn)pn(xn))θ2(x0)a(x0)+1)ψ(x0)[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)(ξ1(x0)+θ2(x0)a(x0)|ψ(x0)K(u)|lim infn(qn(xn)pn(xn))ξ2(x0))(Λ)3lim infnSn(un)|λ1(pn(),qn())ψ(xn)ψ(xn)|pn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2+(Λ)3θ(x0)a(x0)lim infnSn(un)|ψ(xn)|2ψ(xn)pn(xn)2[(|ψ(xn)|kn(un))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|]pn(xn)40. (4.23)

    Note that by (4.17), (4.19) and u(x0)=ψ(x0)>0, we have

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))=|ψ(x0)Λk(u)|lim infn(qn(xn)pn(xn))|ψ(x0)Λu(x0)|lim infn(qn(xn)pn(xn))=|ψ(x0)Λψ(x0)|limn(qn(xn)pn(xn)) (4.24)

    and

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))=|ψ(x0)Λ(k(u))θ(x0)|lim infn(qn(xn)pn(xn))|ψ(x0)Λ(u(x0))θ(x0)|lim infn(qn(xn)pn(xn))=|ψ(x0)Λ(ψ(x0))θ(x0)|lim infn(qn(xn)pn(xn)). (4.25)

    Claim:

    Λψ(x0)|ψ(x0)|0. (4.26)

    Assume that Λψ(x0)>|ψ(x0)|, then (4.24) and (1.11) imply

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))=0 (4.27)

    and

    limn|λ1(pn(),qn())ψ(xn)ψ(xn)|(pn(xn)4)(qn(xn)4)=(Λψ(x0)|ψ(x0)|)1θ(x0)>1. (4.28)

    Thus, choosing ε>0 small enough, we have

    |λ1(pn(),qn())ψ(xn)ψ(xn)|(pn(xn)4)(qn(xn)4)1+ε, (4.29)

    for all nN sufficiently large. By (4.29), we get

    lim infn|λ1(pn,qn)ψ(xn)ψ(xn)|pn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2=lim infn(|λ1(pn,qn)ψ(xn)ψ(xn)|(pn(xn)4)(qn(xn)4))qn(xn)4qn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2qn(xn)4Rψ(x0)3limn(1+ε)qn(xn)4qn(xn)4=+. (4.30)

    From (4.23), (4.27) and (4.30), we see that

    {ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ1(x0)}+, (4.31)

    which is a contradiction. Hence, (4.26) holds.

    Claim:

    (ψ(x0))θ(x0)K(u)|ψ(x0)|0. (4.32)

    Suppose that the above inequality is not true, then we have

    limn[(ψ(xn)kn(un))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|](pn(xn)4)/(qn(xn)4)=limn[(ψ(xn))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|](pn(xn)4)/(qn(xn)4)=[(ψ(x0))θ(x0)K(u)|ψ(x0)|]1θ(x0)>1.

    Thus, choosing ε1>0 small enough, we have

    [(ψ(xn)kn(un))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|](pn(xn)4)/(qn(xn)4)1+ε1, (4.33)

    for all nN sufficiently large. We are led to

    lim infn[(|ψ(xn)|kn(un))(qn(xn)4)/(pn(xn)4)Kn(un)|ψ(xn)|]pn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2lim infn(1+ε1)qn(xn)4qn(xn)4|ψ(xn)|2ψ(xn)pn(xn)2qn(xn)4=θ(x0)ψ(x0)3limn(1+ε1)qn(xn)4qn(xn)4=+. (4.34)

    In view of (ψ(x0))θ(x0)K(u)|ψ(x0)|>0 and (4.25),

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))=0.

    Therefore, this fact along with (4.23) shows that (4.31) holds. This is a contradiction. Thus we deduce that (4.32) holds.

    Claim:

    ψ(x0)+[ln(|ψ(x0)|)ln(K(u))]|ψ(x0)|2ψ(x0)ξ2(x0)0. (4.35)

    Taking (4.24) and (4.26) into account, we have

    |ψ(x0)K(u)|lim infn(qn(xn)pn(xn))=+. (4.36)

    At the same time, by (4.25) and (4.32), we also deduce that (4.36) holds. If we assume that inequality (4.35) does not hold, then by (4.23) and (4.36), there is a contradiction. Thus, we deduce that (4.35) holds.

    Case 2: ψ(x0)=u(x0)=0.

    Note that if |ψ(x0)|=0 (in this case, we have ψ(x0) = 0), the inequality (4.20) trivially holds. Hence, let us assume that |ψ(x0)|>0, then |ψ(xn)|>0 for nN large enough. We can use very similar arguments as Case 1 to conclude that (4.20) holds. The same argument can be used in order to show that u is a viscosity supersolution of (1.20).

    By Lemmas 4.3 and 4.4, it follows that Theorem 1.2 holds.

    Remark 4.3. In the particular case where pn(x)=np(x) and qn(x)=nq(x), Theorems 1.1 and 1.2 are also true.

    In this paper, we studied a double-phase eigenvalue problem with large variable exponents. As we know, for p-Laplace operator eigenvalue problems, there is an important feature that if u is an eigenfunction, so is ku, where k is an arbitrary constant. However, the double-phase operator with variable exponents looses this property. To overcome the above mentioned shortcoming, we defined the eigenvalue by using the Rayleigh quotient of two norms of Musielak-Orlicz space. Moreover, in the particular case where pn()=pn and qn()=qn, Theorems 1.1 and 1.2 are also true (see [13]).

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

    This work was supported by the National Natural Science Foundation of China (No.12001196) and the Natural Science Foundation of Henan (No. 232300421143).

    The authors declare that they have no competing interests.



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