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Research article

The existence of a nontrivial solution to an elliptic equation with critical Sobolev exponent and a general potential well

  • Received: 17 January 2025 Revised: 14 March 2025 Accepted: 21 March 2025 Published: 28 March 2025
  • MSC : 35A15, 35J60

  • The purpose of this paper is to examine a class of elliptic problems that involve negative potentials aLN2(Ω) and critical nonlinearities. To discuss this, the well-known eigenvalue problem Δa is considered. Under some mild assumptions, an existence result is obtained, which extends the existing results to the critical case.

    Citation: Ye Xue, Yongzhen Ge, Yunlan Wei. The existence of a nontrivial solution to an elliptic equation with critical Sobolev exponent and a general potential well[J]. AIMS Mathematics, 2025, 10(3): 7339-7354. doi: 10.3934/math.2025336

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  • The purpose of this paper is to examine a class of elliptic problems that involve negative potentials aLN2(Ω) and critical nonlinearities. To discuss this, the well-known eigenvalue problem Δa is considered. Under some mild assumptions, an existence result is obtained, which extends the existing results to the critical case.



    Over the years, with the aid of variational methods, for varying conditions of the potential and nonlinearity, the existence and multiplicity of solutions for elliptic equations have been extensively discussed in the literature, among which we highlight [1,4,5]. One interesting characteristic is that the potential function can be negative or indefinite, as shown in [8,11]. On the other hand, equations with critical growth raise interest; see [2,6,12].

    Li and Wang [11] established the existence result for the following equation:

    {Δua(x)u=f(x,u) in Ω,u=0 in RNΩ, (1.1)

    where aLN2(Ω) and f(x,u) is superlinear at u=0 and subcritical at u=. They established the existence result for the equation above without assuming that the Ambrosetti-Rabinowitz condition holds.

    Ke and Tang [8] studied (1.1) where aLN2(Ω) and g has super-linear but sub-critical growth. By introducing a new super-linear condition, they proved the existence and multiplicity of solutions. In [7], they gave the existence and multiplicity results for Eq (1.1) with a(x)=λkV(x), where VLN2(Ω), g is sublinear, and λk denotes the kth eigenvalue for the elliptic linear operator Δ+V(x) with zero Dirichlet boundary condition.

    The works [13,15,16] were devoted to studying the critical equation

    {Δuλu=f(x,u)+|u|22u in Ω,u=0 in RNΩ, (1.2)

    where f=0 in [13,16] and f is a lower order perturbation of the critical power |u|22u in [13,16]. In [10], they were concerned with the existence and bifurcation of nontrivial solutions for Eq (1.2) with λ=0 and f(x,u)=μg(x,u) with μ>0. The single solution results obtained by [10] extend the main results of [15].

    Inspired by the research mentioned above, this paper focuses on the existence of nontrivial solutions for the critical problem

    {Δua(x)u=f(x,u)+|u|22u in Ω,u=0 in RNΩ, (1.3)

    where Ω is a bounded domain, N3, and 0<aLN2(Ω). We believe that it is an intriguing question to ask whether there exists a nontrivial solution for Eq (1.3) with aLN2(Ω). Although there have been some works on this type of potential (see [3,8,11]), to the best of our knowledge, no attempt has been made to answer this question for the critical term. In this paper, we will provide an affirmative answer to this question. The result we obtained extends the results of [8] and [11] to the critical case.

    The main objective of this paper is to construct nontrivial solutions of (1.3) using variational techniques. Our strategy will depend on whether λ10 or λ1>0, where λ1 denotes the eigenvalue for the elliptic linear operator Δa with zero Dirichlet boundary condition. If λ10, we will use the Linking theorem to obtain a nontrivial solution. On the other hand, if λ1>0, the Mountain Pass theorem will be effective. In order to achieve this, we will consider the well-known eigenvalue problem Δa.

    In order to study the problem mentioned above, we shall consider its weak formulation, given by

    Ωu(x)ϕ(x)dxΩa(x)u(x)ϕ(x)dx=Ωf(x,u(x))ϕ(x)dx+Ω|u(x)|22u(x)ϕ(x)dx,  ϕ, uH10(Ω),

    where the Hilbert space H10(Ω) is defined as the closure D(Ω) in H1(RN) with the scalar product

    u,v=Ωu(x)v(x)dx,

    for any u,vH10(Ω) and the norm u2 = u,u.

    We can observe that Eq (1.3) is an Euler-Lagrange equation of the functional J:H10(Ω)R defined as follows

    J(u)=12u212Ωa(x)u2(x)dxΩF(x,u(x))dx12|u|22.

    Moreover, JC1(H10(Ω),R) and for any u, ϕH10(Ω), we have

    J(u),ϕ=Ωu(x)ϕ(x)dxΩa(x)u(x)ϕ(x)dxΩf(x,u(x))ϕ(x)dxΩ|u(x)|22u(x)ϕ(x)dx.

    Now we recall an eigenvalue problem (see [11,17]) related to the problem mentioned earlier. Let {λk} be the positive and increasing sequence of the eigenvalues of the following problem

    {Δua(x)u=λu in Ω,u=0 in RNΩ (1.4)

    and the sequence {ek} of the eigenfunctions corresponding to {λk} is an orthonormal basis of L2(Ω) and an orthogonal basis of H10(Ω). Moreover,

    <λ1λ2...λkλk+1...,

    where

    λk+1=minuPk+1{0}u2Ωa(x)u2(x)dxΩu2(x)dx,kN,

    Pk+1={uH10(Ω)|u,ej=0, j=1,...,k}. If λ10, for convenience, we set

    <λ1λ2...λk0<λk+1....

    Corresponding, let

    Y={e1,...,ek} and Z={uH10(Ω),u,vL2=0,vY}, (1.5)

    thereby, H10(Ω)=YZ. We need the following constants:

    S=infuH10(Ω){0}u2(Ω|u(x)|2dx)22, (1.6)
    Sa=infuH10(Ω){0}Sa(u),

    where

    Sa(u)=u2Ωa(x)|u(x)|2dx(Ω|u(x)|2dx)22. (1.7)

    In this subsection, we present the main result of the paper. We consider the nonlinear partial differential Eq (1.3) with a Caratheodory function f:Ω×RR satisfying conditions (f1)–(f4) (or (f4)):

    (f1) sup{|f(x,t)|: a.e. xΩ,|t|M}<+ for any M>0;

    (f2) limt0f(x,t)t=0 and lim|t|+f(x,t)|t|21=0 uniformly in xΩ;

    (f3) there exists α>2 such that

    0<αF(x,t)f(x,t)t, for t0,

    where F(x,t)=t0f(x,τ)dτ;

    (f4) limt+F(x,t)t4=+, when N=3, uniformly in xΩ;

    limt+F(x,t)t2|lnt|=+, when N=4, uniformly in xΩ;

    limt+F(x,t)t2=+, when N5, uniformly in xΩ;

    (f4) there exists u0H10(Ω){0} with u00 a.e. in RN such that Sa(u)<S for any uU, where U=span{e1,e2,...,ek,u0}, kN.

    Our main result is as follows:

    Theorem 1. Suppose that conditions (f1)–(f3) and (f4) (or (f4)) hold. Then, problem (1.3) admits a nontrivial solution uH10(Ω){0}.

    Remark 1. We note that the function a(x) and the critical exponent term pose natural difficulties in this problem. One difficulty is that the boundedness of the Palais-Smale sequence fails, and we need to apply certain inequalities to recover it. Another difficulty is to prove that the required level c is below the threshold. We use various techniques to overcome these difficulties. Our result extends the results of [8] and [11] to the critical case. Furthermore, if we set s=1 in the fractional problem, our result generalizes the single-solution results of [10] and [15].

    The structure of the remaining part of this paper is as follows. In Sections 2 and 3, we present several lemmas and estimates that are crucial for the proof of the main theorem. These results, based on the relevant lemmas from Sections 2 and 3, are then used in Section 4 to complete the proof of the main theorem.

    In this section, we will first present some relevant information that will be useful. Some of the lemmas provided are standard, and readers familiar with them may proceed directly to the estimation part.

    Lemma 1. Assume conditions (f1) and (f2) hold; then for any ε>0, there exists M=M(ε)>0 such that for a.e. xΩ and any tRN,

    |f(x,t)|2ε|t|21+M(ε)

    and

    |F(x,t)|ε|t|2+M(ε)|t|,

    where F(x,t) is defined as in (f3).

    Lemma 2. [17] Let 1<r<. If {un} is bounded in Lr(Ω) and unu a.e. in Ω, then unu in Lr(Ω).

    Lemma 3. [17] The following assertions are true:

    (a) The embedding H10(Ω)Lν(Ω) is compact for any ν[1,2).

    (b) The embedding H10(Ω)L2(Ω) is continuous.

    Lemma 4. Suppose that |Ω|<, and conditions (f1) and (f2) hold. If unu in H10(Ω), then

    ΩF(x,un(x))dxΩF(x,u(x))dx.

    Proof. The proof is standard and omitted here.

    Lemma 5. [7, Lemma 3.1] If aLN2(Ω), |Ω|< and unuH10(Ω), then

    Ωa(x)|un(x)|2dxΩa(x)|u(x)|2dx.

    Lemma 6. [17, Lemma 2.14] If aLN2(Ω) and |Ω|<, then

    λ1=infuH10(Ω),|u|2=1Ω(|u|2a(x)u2(x))dx>. (2.1)

    Lemma 7. [11, Lemma 3.1] If aLN2(Ω) and |Ω|<, then

    ˆδ=infuZ,u=1Ω(|u|2a(x)u2(x))dx>0, (2.2)

    where Z is defined in (1.5).

    Next, we will provide some estimates. Firstly, let us recall that the limiting problem

    u=|u|22u in RN

    admits a solution u in H1(RN) (see, for instance, [2,14]). Now, for any ε>0, let us consider the following functions: Uε, uε, and vε defined as

    Uε=ε(N2)2u(xε), xRN,uε(x)=η(x)Uε(x), xRN

    and

    vε(x)=uε|uε|2, xRN,

    where ηC(RN) is such that 0η1 in RN, η1 in Bδ and η0 in RNB2δ, with Bδ=B(0,δ) and δ>0 such that B4δΩ. Note that uεH10(Ω) for any ε>0. What's more, according to [9], one has

    |vε|2=1, RN|vε|2dx=S+O(εN22),
    Υ(ε):=RN|vε|2dx={O(ε12), N=3,O(ε|lnε|),N=4,O(ε), N5, (3.1)

    and

    |uε|22L for ε small enough, 

    where L is some positive constant.

    Known to all, for problem (1.3), the compactness condition holds true only within a suitable threshold related to the best critical Sobolev constant. Now we will deal with the problem.

    Lemma 8. Suppose that (f1)–(f4) hold. There is a vH10(Ω){0} such that

    maxt0J(tv)<1NSN2.

    Proof. Obviously, J(u)J0(u), where

    J0(u)=12u2ΩF(x,u(x))dx12|u|22.

    Hence, it is sufficient to prove that maxt0J0(tv)<1NSN2. By (f1)–(f3), there exists tε>0 such that J0(tεvε)=maxt0J0(tvε). By dJ0(tvε)dt|t=tε=0 and noticing

    |vε|2=1, RN|vε|2dx=S+O(εN22),

    we derive that

    0=tεvε2Ωf(x,tεvε)vεdxtε21tεvε2tε21,

    and it follows that

    tε22vε2S+O(εN22).

    This implies that tεC, where C is independent of ε(>0) with ε small.

    We claim that tεC>0 for sufficiently small ε. If not, there exists a sequence εn0 as n such that tεn0 as n. It follows that tεnvεn0 in H10(Ω) as n. Hence,

    0<c0maxt0J0(tvεn)=J0(tεnvεn)12tεnvεn20,

    where c0 denotes the Mountain Pass level of J0. This is a contradiction.

    According to condition (f3), for any M>0, there exists TM>0 such that for t>TM, we obtain that

    F(x,t){Mt4,       N=3,Mt2|lnt|,N=4,Mt2,       N5. (3.2)

    Hence, for sufficiently small ε, we can conclude that

    |x|<ε12F(x,vε){CM|x|<ε12ε1dx=CMε12,   N=3,C2M|x|<ε12ε1ln(Cε12)dx=CMεln(Cε12),N=4,CM|x|<ε12εN22dx=CMε,   N5. (3.3)

    More details about the estimate can be found in [9, Lemma 3.4]. By the arbitrariness of M and (3.1), we obtain

    limε0+|x|<ε12F(x,vε)dxΥ(ε)=+, (3.4)

    where Υ(ε) is defined in (3.1). Thus, by (3.4) and the fact that F(x,u)0, for sufficiently small ε, one has

    J0(tεvε)12tεvε2+12Ω|tεvε|2dxΩF(x,tεvε)dx12Ω|tεvε|2dxmaxt0(12tεvε212Ω|tεvε|2dx)+12Ω|tεvε|2dxΩF(x,tεvε)dxS2NN+O(εN22)|x|<ε12F(x,tεvε)|x|ε12F(x,tεvε)dx+CΩ|vε|2dx1NS2N+O(εN22)|x|<ε12F(x,tεvε)dx+CΩ|vε|2dx<1NS2N.

    This completes the proof.

    Lemma 9. For uH10(Ω){0}, we have

    supξ0(ξ22u2ξ22Ωa(x)u2(x)dxξ22|u|22)=1NSN2a(u), (3.5)

    where Sa is defined in (1.7).

    Proof. Let M:[0,+)R be the following function

    M(ξ)=ξ22u2ξ22Ωa(x)u2(x)dxξ22|u|22.

    Note that

    M(ξ)=ξu2ξΩa(x)u2(x)dxξ21|u|22,

    and M(ξ)0 if and only if

    ξˉξ:=(u2Ωa(x)u2(x)dx|u|22)122.

    It follows that Mmax(ξ)=M(ˉξ). By accurate calculation,

    supξ0M(ξ)=maxξ0M(ξ)=M(ˉξ)=1NSN2a(u).

    This concludes the proof.

    Firstly, we are going to prove that there exists a bounded Palais-Smale sequence for J in H10(Ω).

    Lemma 10. Suppose that (f1)–(f3) hold, aLN2(Ω) and |Ω|<. If c(,1NSN2) and {un} is sequence in H10(Ω) such that J(un)0and J(un)c, as n+, then there exists ˆuH10(Ω) such that, up to a subsequence,

    unˆu0, as n+. (4.1)

    Proof. By Eq (1.5), we can write un=yn+zn, where ynY and znZ. Let's set max{1α,12}<β<12, where α is defined in (f3). By αβ>1 and β12, as well as applying (f3), Lemmas 3, 6 and 7, we derive that

    J(un)βJ(un),un=(12β)Ω(|un|2a(x)u2n(x))dx+(β12)|un|22ΩF(x,un(x))dx+Ωβf(x,un(x))un(x)dx(12β)Ω(|un|2a(x)u2n)dx+(β12)|un|22+(αβ1)ΩF(x,un)dx=(12β)Ω(|yn|2a(x)y2n(x)+|zn|2a(x)z2n(x))dx+(αβ1)ΩF(x,un(x))dx+(β12)|un|22(12β)(λ1|yn|22+ˆδzn2)+(β12)|un|22+C1(αβ1)|un|ααC3(12β)(λ1|yn|22+ˆδzn2)+(β12)|un|22+C1(αβ1)|yn|ααC3(12β)(λ1|yn|22+ˆδzn2)+(β12)|un|22+C1C(αβ1)|yn|α2ˉCC3(12β)λ1|yn|22+(12β)ˆδzn2+C4|yn|22C5=[C4+(12β)λ1]|yn|22+[(12β)ˆδ]zn2C5.

    And then, it is easy to know

    C+|yn|2+znJ(un)βJ(un),unC6|yn|22+[(12β)ˆδ]zn2C5.

    Hence, {un} is bounded in H10(Ω) because of the boundedness of {yn} and {zn}, using the fact that dimY is finite.

    Now, in order to verify the PS-condition, we need to establish several results (Steps 1–5) whose proofs are standard. Now we give a brief statement. Let {un} be a sequence in H10(Ω) such that

    J(un)0 and J(un)c, as n+.

    Step 1. We aim to show that ˆu is a solution of (1.3). We assume that as n+, there exists ˆu in H10(Ω) such that unˆu  in H10(Ω), because of the boundedness of {un} in H10(Ω). This convergence can be summarized as follows

    {unˆu in H10(Ω),unˆu in L2(Ω),un(x)ˆu(x) a.e. in Ω. (4.2)

    In addition, we have the following convergence:

    un,vˆu,v, vH10(Ω)

    and by Lemma 5, we have

    Ωa(x)u2ndxΩa(x)ˆu2dx, as n+.

    We know {un} is bounded in L2(Ω) because of H10(Ω)L2(Ω), and |un|22un is bounded in L221(Ω). We obtain that

    |un|22un|ˆu|22ˆu in L221(Ω), as n+. (4.3)

    Thereby,

    Ω|un(x)|22un(x)ϕ(x)dxΩ|ˆu(x)|22ˆu(x)ϕ(x)dx,

     as n+, ϕL2(Ω), and so,

    Ω|un(x)|22un(x)ϕ(x)dxΩ|ˆu(x)|22ˆu(x)ϕ(x)dx,

     as n+, ϕH10(Ω). By Lemma 1, we obtain that f(x,un) is bounded in L221(Ω). It follows from (4.2) that f(x,un(x))f(x,ˆu(x)) a.e. in Ω, as n. Thus, by Lemma 2, we obtain that f(x,un(x))f(x,ˆu(x)) in L221(Ω), as n. Then,

    Ωf(x,un(x))ϕ(x)Ωf(x,ˆu(x))ϕ(x), as n+, ϕL2(Ω),

    so that

    Ωf(x,un(x))ϕ(x)Ωf(x,ˆu(x))ϕ(x), as n+, ϕH10(Ω).

    By assumption, for any ϕH10(Ω), J(un),ϕ0, as n. Therefore,

    ˆu,ϕΩa(x)ˆuϕdxΩ(f(x,ˆu)ϕ|ˆu|22ˆuϕ)dx=0,ϕH10(Ω).

    Hence, ˆu is a solution of (1.3).

    Step 2. We claim that the following equality holds true:

    J(ˆu)=1N|ˆu|22+12Ωf(x,ˆu(x))ˆu(x)dxΩF(x,ˆu(x))dx0.

    Noticing that

    J(ˆu),ˆu=ˆu2Ωa(x)ˆu2(x)dxΩf(x,ˆu(x))ˆu(x)dx|ˆu|22=0

    and

    J(ˆu)=12ˆu212Ωa(x)ˆu2(x)dxΩF(x,ˆu(x))dx12|ˆu|22,

    by (f3), we obtain

    J(ˆu)=1N|ˆu|22+12Ω(f(x,ˆu(x))ˆu(x)2F(x,ˆu(x)))dx0.

    Step 3. We claim that

    J(un)=J(ˆu)+12unˆu212|unˆu|22+o(1), as n+.

    Recall J(un)=12un212Ωa(x)u2n(x)dx ΩF(x,un(x))dx12|un|22. By Brezis-Lieb lemma, we also get

    unˆu2=un2ˆu2+o(1),|un|22=|unˆu|22+|ˆu|22+o(1). (4.4)

    As unˆu in H10(Ω) and by Lemma 5, we obtain

    Ωa(x)ˆu2n(x)dxΩa(x)ˆu2(x)dx, as n+.

    By Lemma 4, it follows that

    ΩF(x,un(x))dxΩF(x,u(x))dx, as n+.

    Hence,

    J(un)=12unˆu2+12ˆu212Ωa(x)ˆu(x)2dxΩF(x,u(x))dx12unˆu2212ˆu22+o(1)=J(ˆu)+12unˆu212unˆu22+o(1), as n+.

    Step 4. We claim that unˆu2=|unˆu|22+o(1), as n+ holds. By (4.2)–(4.4), we infer that

    Ω(|un(x)|22un(x)|ˆu(x)|22ˆu(x))(un(x)ˆu(x))dx=Ω|un(x)|2dxΩ|ˆu(x)|2dx+o(1)=Ω|un(x)ˆu(x)|2dx+o(1), as n. (4.5)

    Moreover,

    Ω(f(x,un(x))f(x,ˆu(x)))(un(x)ˆu(x))dx0, as n+. (4.6)

    Easily to know that

    J(un)J(ˆu),unˆu=unˆu2Ω(|un(x)|22un(x)|ˆu(x)|22ˆu(x))(un(x)ˆu(x))dxΩ(f(x,un(x))f(x,ˆu(x)))(un(x)ˆu(x))dxΩa(x)(un(x)ˆu(x))2dx=unˆu2Ω|un(x)ˆu(x)|2dx+o(1), (4.7)

    as n+. On the other hand, by the boundedness of {un} in H10(Ω) and Step 1, it follows that

    o(1)=J(un),unˆu=J(un)J(ˆu),unˆu, n. (4.8)

    Hence, from (4.7) and (4.8), we get the assertion of Step 4.

    Step 5. We conclude the proof of Lemma 10. As n+, by Steps 2–4, we derive that

    J(un)=J(ˆu)+12unˆu(x)212|unˆu|22+o(1)12unˆu212|unˆu|22+o(1)1Nunˆu2+o(1), as n. (4.9)

    Let unˆu2L, as n. By Step 4, |unˆu|22L, as n. In view of H10(Ω)L2(Ω) and the definition of S in (1.6), so L22SL, then L=0 or LSN2. If LSN2, by (4.9) there will be

    c1NL1NSN2,

    which against the assumption. Thus, L=0, which implies unˆu in H10(Ω), as n+. This completes the proof.

    If λ10, now we will show that J fits with the geometric assumptions of Rabinowitz's linking theorem, e.g., see Theorem 2.12 in [17].

    Lemma 11. Suppose that (f1)-(f3) hold. Then, for Y and Z defined in (1.5), we have

    (i) there are ρ,κ>0 such that for any uZ with u=ρ, we have J(u)κ;

    (ii) J(u)0 for any uY;

    (iii) there exists R0>ρ such that J(u)0 for any uF with uR0, where F={u+tz|uY,t>0,zZ{0}}.

    Proof. Let u Z, by (f2) and (f3), we have

    J(u)=12Ω(|u|2a(x)u2(x))dxΩF(x,u(x))dx12Ω|u(x)|2dxˆδ2u2ε2|u|22M(ε)|u|ςς1+ε2|u|22=12(ˆδεk2)u2kςM(ε)uς(1+ε)S222u2,

    where ς(2,2). Fixed ε such that ˆδεk2>0. For uZ, u=ρ sufficiently small,

    κ=infuH10(Ω),|u|2=ρJ(u)12(ˆδεk2)ρ2kςM(ε)ρς(1+ε)S222ρ2>0.

    Now we will give the proof of (ii). Thanks to λk0, for any uY=span{e1,...,ek},

    J(u)=12u212Ωa(x)u2(x)dxΩF(x,u(x))dx12Ω|u(x)|2dx12λk|u|2212Ω|u(x)|2dx0. (4.10)

    For any uF, we also have

    J(u)=12u212Ωa(x)u(x)2dxΩF(x,u(x))dx12Ω|u(x)|2dx12u2C2u22.

    Hence, by 2>2, J(u), as u+, thanks to the fact that in any finite-dimensional space all norms are equivalent.

    Proof of Theorem 1. We give the proof as follows for two cases. For the one case where λ1>0, we can apply the Mountain pass theorem to show the existence of a nontrivial solution for Eq (1.3). The proof follows the standard procedure, so we omit it here.

    Let us define the Mountain Pass level of J as

    cm=infγmΓmmaxt[0,1]J(γm(t)),

    where

    Γm={γmC([0,1],H10(Ω)):γm(0)=0,γm(1)=e}.

    By combining Lemmas 8 and 10, we can conclude that there exists a nontrivial solution for Eq (1.3).

    For the case where λ10, we construct the functions z,ˉzZ as follows:

    z=u0Σki=1(Ωu0(x)ei(x)dx)ei, ˉz=zz,

    where u0 is defined in (f4). Let c be the linking critical level of J, that is,

    c=infγΓsupuMJ(γ(u))

    is the linking critical level of J, where

    Γ={γC(¯M,H10(Ω)):γ=id on M}

    and

    M={u=y+tˉz|uR0,t0,yY}.

    Our goal is to show that c<1NSN2. Since F=Yspan{ˉz} is a linear space, we have

    supuFJ(u)=supuF,ξ0J(|ξ|u|ξ|)=supuF,ξ>0J(ξu)supuF,ξ0J(ξu), (4.11)

    so, in view of MF, we obtain

    csupuMJ(u)supuFJ(u)supuF,ξ0J(ξu). (4.12)

    By (f3), J(u)12u212Ωa(x)u2(x)dx12|u|22. By Lemma 9, for any uH10{0},

    supξ0J(ξu)1NSN2a(u). (4.13)

    Therefore, combining (4.12), (4.13), and (f4), we derive that

    csupuF,ξ0J(ξu)1NsupuFSN2a(u)1NSN2,

    thanks to F=Y span {ˉz}={e1,e2,...,ek,u0}. Then, combining Lemma 10 with Lemma 11, we deduce that there is a critical point u of J such that J(0)=0<κ<J(u)<1NSN2. This completes the proof.

    Despite much literature concerning the existence of nontrivial solutions for the critical problem, to our knowledge, there are no available results involving the potential aN2(Ω). Under some mild assumptions, we obtain an existence result. For the potential aN2(Ω), we would like to go further in this direction, not just in bounded Ω.

    Ye Xue: Conceptualization, Investigation, Methodology, Supervision, Validation, Writing-original draft, Writing-review and editing; Yongzhen Ge: Conceptualization, Investigation, Methodology, Supervision, Validation, Writing-original draft, Writing-review and editing; Yunlan Wei: Validation, Writing-original draft, Writing-review and editing. All authors have read and agreed to the published version of the manuscript.

    Ye Xue was supported by the National Natural Science Foundation of China (Grant No. 12301129), and the Applied Basic Research Support Project of Changzhou Science and Technology (Grant No. CJ20235014). Yunlan Wei was supported by the Natural Science Foundation of Shandong Province (Grant No. ZR2022QA057).

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

    The authors declare no conflicts of interest.



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