The purpose of this paper is to examine a class of elliptic problems that involve negative potentials a∈LN2(Ω) and critical nonlinearities. To discuss this, the well-known eigenvalue problem −Δ−a 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 a∈LN2(Ω) and critical nonlinearities. To discuss this, the well-known eigenvalue problem −Δ−a 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:
{−Δu−a(x)u=f(x,u) in Ω,u=0 in RN∖Ω, | (1.1) |
where a∈LN2(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 a∈LN2(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)=λk−V(x), where V∈LN2(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
{−Δu−λu=f(x,u)+|u|2∗−2u in Ω,u=0 in RN∖Ω, | (1.2) |
where f=0 in [13,16] and f is a lower order perturbation of the critical power |u|2∗−2u 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
{−Δu−a(x)u=f(x,u)+|u|2∗−2u in Ω,u=0 in RN∖Ω, | (1.3) |
where Ω is a bounded domain, N≥3, and 0<a∈LN2(Ω). We believe that it is an intriguing question to ask whether there exists a nontrivial solution for Eq (1.3) with a∈LN2(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 λ1≤0 or λ1>0, where λ1 denotes the eigenvalue for the elliptic linear operator −Δ−a with zero Dirichlet boundary condition. If λ1≤0, 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 −Δ−a.
In order to study the problem mentioned above, we shall consider its weak formulation, given by
∫Ω∇u(x)∇ϕ(x)dx−∫Ωa(x)u(x)ϕ(x)dx=∫Ωf(x,u(x))ϕ(x)dx+∫Ω|u(x)|2∗−2u(x)ϕ(x)dx, ∀ϕ, u∈H10(Ω), |
where the Hilbert space H10(Ω) is defined as the closure D(Ω) in H1(RN) with the scalar product
⟨u,v⟩=∫Ω∇u(x)∇v(x)dx, |
for any u,v∈H10(Ω) and the norm ‖u‖2 = ⟨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)=12‖u‖2−12∫Ωa(x)u2(x)dx−∫ΩF(x,u(x))dx−12∗|u|2∗2∗. |
Moreover, J∈C1(H10(Ω),R) and for any u, ϕ∈H10(Ω), we have
⟨J′(u),ϕ⟩=∫Ω∇u(x)∇ϕ(x)dx−∫Ωa(x)u(x)ϕ(x)dx−∫Ωf(x,u(x))ϕ(x)dx−∫Ω|u(x)|2∗−2u(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
{−Δu−a(x)u=λu in Ω,u=0 in RN∖Ω | (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,
−∞<λ1≤λ2≤...≤λk≤λk+1≤..., |
where
λk+1=minu∈Pk+1∖{0}‖u‖2−∫Ωa(x)u2(x)dx∫Ωu2(x)dx,k∈N, |
Pk+1={u∈H10(Ω)|⟨u,ej⟩=0, j=1,...,k}. If λ1≤0, for convenience, we set
−∞<λ1≤λ2≤...≤λk≤0<λk+1≤.... |
Corresponding, let
Y={e1,...,ek} and Z={u∈H10(Ω),⟨u,v⟩L2=0,v∈Y}, | (1.5) |
thereby, H10(Ω)=Y⊕Z. We need the following constants:
S=infu∈H10(Ω)∖{0}‖u‖2(∫Ω|u(x)|2∗dx)22∗, | (1.6) |
Sa=infu∈H10(Ω)∖{0}Sa(u), |
where
Sa(u)=‖u‖2−∫Ωa(x)|u(x)|2dx(∫Ω|u(x)|2∗dx)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×R⟶R satisfying conditions (f1)–(f4) (or (f′4)):
(f1) sup{|f(x,t)|: a.e. x∈Ω,|t|≤M}<+∞ for any M>0;
(f2) limt→0f(x,t)t=0 and lim|t|→+∞f(x,t)|t|2∗−1=0 uniformly in x∈Ω;
(f3) there exists α>2 such that
0<αF(x,t)≤f(x,t)t, for t≠0, |
where F(x,t)=∫t0f(x,τ)dτ;
(f4) limt→+∞F(x,t)t4=+∞, when N=3, uniformly in x∈Ω;
limt→+∞F(x,t)t2|lnt|=+∞, when N=4, uniformly in x∈Ω;
limt→+∞F(x,t)t2=+∞, when N≥5, uniformly in x∈Ω;
(f′4) there exists u0∈H10(Ω)∖{0} with u0≥0 a.e. in RN such that Sa(u)<S for any u∈U, where U=span{e1,e2,...,ek,u0}, k∈N.
Our main result is as follows:
Theorem 1. Suppose that conditions (f1)–(f3) and (f4) (or (f′4)) hold. Then, problem (1.3) admits a nontrivial solution u∈H10(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. x∈Ω and any t∈RN,
|f(x,t)|≤2∗ε|t|2∗−1+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 un→u a.e. in Ω, then un⇀u 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 un⇀u in H10(Ω), then
∫ΩF(x,un(x))dx→∫ΩF(x,u(x))dx. |
Proof. The proof is standard and omitted here.
Lemma 5. [7, Lemma 3.1] If a∈LN2(Ω), |Ω|<∞ and un⇀u∈H10(Ω), then
∫Ωa(x)|un(x)|2dx→∫Ωa(x)|u(x)|2dx. |
Lemma 6. [17, Lemma 2.14] If a∈LN2(Ω) and |Ω|<∞, then
λ1=infu∈H10(Ω),|u|2=1∫Ω(|∇u|2−a(x)u2(x))dx>−∞. | (2.1) |
Lemma 7. [11, Lemma 3.1] If a∈LN2(Ω) and |Ω|<∞, then
ˆδ=infu∈Z,‖u‖=1∫Ω(|∇u|2−a(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|2∗−2u 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−(N−2)2u∗(xε), x∈RN,uε(x)=η(x)Uε(x), x∈RN |
and
vε(x)=uε|uε|2∗, x∈RN, |
where η∈C∞(RN) is such that 0≤η≤1 in RN, η≡1 in Bδ and η≡0 in RN∖B2δ, with Bδ=B(0,δ) and δ>0 such that B4δ⊂Ω. Note that uε∈H10(Ω) for any ε>0. What's more, according to [9], one has
|vε|2∗=1, ∫RN|∇vε|2dx=S+O(εN−22), |
Υ(ε):=∫RN|vε|2dx={O(ε12), N=3,O(ε|lnε|),N=4,O(ε), N≥5, | (3.1) |
and
|uε|2∗≤2L 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 v∈H10(Ω)∖{0} such that
maxt≥0J(tv)<1NSN2. |
Proof. Obviously, J(u)≤J0(u), where
J0(u)=12‖u‖2−∫ΩF(x,u(x))dx−12∗|u|2∗2∗. |
Hence, it is sufficient to prove that maxt≥0J0(tv)<1NSN2. By (f1)–(f3), there exists tε>0 such that J0(tεvε)=maxt≥0J0(tvε). By dJ0(tvε)dt|t=tε=0 and noticing
|vε|2∗=1, ∫RN|∇vε|2dx=S+O(εN−22), |
we derive that
0=tε‖vε‖2−∫Ωf(x,tεvε)vεdx−tε2∗−1≤tε‖vε‖2−tε2∗−1, |
and it follows that
tε2∗−2≤‖vε‖2≤S+O(εN−22). |
This implies that tε≤C, where C is independent of ε(>0) with ε small.
We claim that tε≥C>0 for sufficiently small ε. If not, there exists a sequence εn→0 as n→∞ such that tεn→0 as n→∞. It follows that tεnvεn→0 in H10(Ω) as n→∞. Hence,
0<c0≤maxt≥0J0(tvεn)=J0(tεnvεn)≤12‖tεnvεn‖2→0, |
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, N≥5. | (3.2) |
Hence, for sufficiently small ε, we can conclude that
∫|x|<ε12F(x,vε)≥{CM∫|x|<ε12ε−1dx=CMε12, N=3,C2M∫|x|<ε12ε−1ln(Cε−12)dx=CMεln(Cε−12),N=4,CM∫|x|<ε12ε−N−22dx=CMε, N≥5. | (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
limε→0+∫|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)≤12‖tεvε‖2+12∫Ω|tεvε|2dx−∫ΩF(x,tεvε)dx−12∗∫Ω|tεvε|2∗dx≤maxt≥0(12‖tεvε‖2−12∗∫Ω|tεvε|2∗dx)+12∫Ω|tεvε|2dx−∫ΩF(x,tεvε)dx≤S2NN+O(εN−22)−∫|x|<ε12F(x,tεvε)−∫|x|≥ε12F(x,tεvε)dx+C∫Ω|vε|2dx≤1NS2N+O(εN−22)−∫|x|<ε12F(x,tεvε)dx+C∫Ω|vε|2dx<1NS2N. |
This completes the proof.
Lemma 9. For u∈H10(Ω)∖{0}, we have
supξ≥0(ξ22‖u‖2−ξ22∫Ωa(x)u2(x)dx−ξ2∗2∗|u|2∗2∗)=1NSN2a(u), | (3.5) |
where Sa is defined in (1.7).
Proof. Let M:[0,+∞)→R be the following function
M(ξ)=ξ22‖u‖2−ξ22∫Ωa(x)u2(x)dx−ξ2∗2∗|u|2∗2∗. |
Note that
M′(ξ)=ξ‖u‖2−ξ∫Ωa(x)u2(x)dx−ξ2∗−1|u|2∗2∗, |
and M′(ξ)≥0 if and only if
ξ≤ˉξ:=(‖u‖2−∫Ωa(x)u2(x)dx|u|2∗2∗)12∗−2. |
It follows that Mmax(ξ)=M(ˉξ). By accurate calculation,
supξ≥0M(ξ)=maxξ≥0M(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, a∈LN2(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 ˆu∈H10(Ω) such that, up to a subsequence,
‖un−ˆu‖→0, as n→+∞. | (4.1) |
Proof. By Eq (1.5), we can write un=yn+zn, where yn∈Y and zn∈Z. Let's set max{1α,12∗}<β<12, where α is defined in (f3). By αβ>1 and β≥12∗, as well as applying (f3), Lemmas 3, 6 and 7, we derive that
J(un)−β⟨J′(un),un⟩=(12−β)∫Ω(|∇un|2−a(x)u2n(x))dx+(β−12∗)|un|2∗2∗−∫ΩF(x,un(x))dx+∫Ωβf(x,un(x))un(x)dx≥(12−β)∫Ω(|∇un|2−a(x)u2n)dx+(β−12∗)|un|2∗2∗+(αβ−1)∫ΩF(x,un)dx=(12−β)∫Ω(|∇yn|2−a(x)y2n(x)+|∇zn|2−a(x)z2n(x))dx+(αβ−1)∫ΩF(x,un(x))dx+(β−12∗)|un|2∗2∗≥(12−β)(λ1|yn|22+ˆδ‖zn‖2)+(β−12∗)|un|2∗2∗+C1(αβ−1)|un|αα−C3≥(12−β)(λ1|yn|22+ˆδ‖zn‖2)+(β−12∗)|un|2∗2∗+C1(αβ−1)|yn|αα−C3≥(12−β)(λ1|yn|22+ˆδ‖zn‖2)+(β−12∗)|un|2∗2∗+C1C(αβ−1)|yn|α2−ˉC−C3≥(12−β)λ1|yn|22+(12−β)ˆδ‖zn‖2+C4|yn|22−C5=[C4+(12−β)λ1]|yn|22+[(12−β)ˆδ]‖zn‖2−C5. |
And then, it is easy to know
C+|yn|2+‖zn‖≥J(un)−β⟨J′(un),un⟩≥C6|yn|22+[(12−β)ˆδ]‖zn‖2−C5. |
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 un⇀ˆu in H10(Ω), because of the boundedness of {un} in H10(Ω). This convergence can be summarized as follows
{un⇀ˆu in H10(Ω),un→ˆu in L2(Ω),un(x)→ˆu(x) a.e. in Ω. | (4.2) |
In addition, we have the following convergence:
⟨un,v⟩→⟨ˆu,v⟩, ∀v∈H10(Ω) |
and by Lemma 5, we have
∫Ωa(x)u2ndx→∫Ωa(x)ˆu2dx, as n→+∞. |
We know {un} is bounded in L2∗(Ω) because of H10(Ω)↪L2∗(Ω), and |un|2∗−2un is bounded in L2∗2∗−1(Ω). We obtain that
|un|2∗−2un⇀|ˆu|2∗−2ˆu in L2∗2∗−1(Ω), as n→+∞. | (4.3) |
Thereby,
∫Ω|un(x)|2∗−2un(x)ϕ(x)dx→∫Ω|ˆu(x)|2∗−2ˆu(x)ϕ(x)dx, |
as n→+∞, ∀ϕ∈L2∗(Ω), and so,
∫Ω|un(x)|2∗−2un(x)ϕ(x)dx→∫Ω|ˆu(x)|2∗−2ˆu(x)ϕ(x)dx, |
as n→+∞, ∀ϕ∈H10(Ω). By Lemma 1, we obtain that f(x,un) is bounded in L2∗2∗−1(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 L2∗2∗−1(Ω), as n→∞. Then,
∫Ωf(x,un(x))ϕ(x)→∫Ωf(x,ˆu(x))ϕ(x), as n→+∞, ∀ϕ∈L2∗(Ω), |
so that
∫Ωf(x,un(x))ϕ(x)→∫Ωf(x,ˆu(x))ϕ(x), as n→+∞, ∀ϕ∈H10(Ω). |
By assumption, for any ϕ∈H10(Ω), ⟨J′(un),ϕ⟩→0, as n→∞. Therefore,
⟨ˆu,ϕ⟩−∫Ωa(x)ˆuϕdx−∫Ω(f(x,ˆu)ϕ−|ˆu|2∗−2ˆuϕ)dx=0,∀ϕ∈H10(Ω). |
Hence, ˆu is a solution of (1.3).
Step 2. We claim that the following equality holds true:
J(ˆu)=1N|ˆu|2∗2∗+12∫Ωf(x,ˆu(x))ˆu(x)dx−∫ΩF(x,ˆu(x))dx≥0. |
Noticing that
⟨J′(ˆu),ˆu⟩=‖ˆu‖2−∫Ωa(x)ˆu2(x)dx−∫Ωf(x,ˆu(x))ˆu(x)dx−|ˆu|2∗2∗=0 |
and
J(ˆu)=12‖ˆu‖2−12∫Ωa(x)ˆu2(x)dx−∫ΩF(x,ˆu(x))dx−12∗|ˆu|2∗2∗, |
by (f3), we obtain
J(ˆu)=1N|ˆu|2∗2∗+12∫Ω(f(x,ˆu(x))ˆu(x)−2F(x,ˆu(x)))dx≥0. |
Step 3. We claim that
J(un)=J(ˆu)+12‖un−ˆu‖2−12∗|un−ˆu|2∗2∗+o(1), as n→+∞. |
Recall J(un)=12‖un‖2−12∫Ωa(x)u2n(x)dx −∫ΩF(x,un(x))dx−12∗|un|2∗2∗. By Brezis-Lieb lemma, we also get
‖un−ˆu‖2=‖un‖2−‖ˆu‖2+o(1),|un|2∗2∗=|un−ˆu|2∗2∗+|ˆu|2∗2∗+o(1). | (4.4) |
As un⇀ˆu in H10(Ω) and by Lemma 5, we obtain
∫Ωa(x)ˆu2n(x)dx→∫Ωa(x)ˆu2(x)dx, as n→+∞. |
By Lemma 4, it follows that
∫ΩF(x,un(x))dx→∫ΩF(x,u(x))dx, as n→+∞. |
Hence,
J(un)=12‖un−ˆu‖2+12‖ˆu‖2−12∫Ωa(x)ˆu(x)2dx−∫ΩF(x,u(x))dx−12∗‖un−ˆu‖2∗2∗−12∗‖ˆu‖2∗2∗+o(1)=J(ˆu)+12‖un−ˆu‖2−12∗‖un−ˆu‖2∗2∗+o(1), as n→+∞. |
Step 4. We claim that ‖un−ˆu‖2=|un−ˆu|2∗2∗+o(1), as n→+∞ holds. By (4.2)–(4.4), we infer that
∫Ω(|un(x)|2∗−2un(x)−|ˆu(x)|2∗−2ˆu(x))(un(x)−ˆu(x))dx=∫Ω|un(x)|2∗dx−∫Ω|ˆu(x)|2∗dx+o(1)=∫Ω|un(x)−ˆu(x)|2∗dx+o(1), as n→∞. | (4.5) |
Moreover,
∫Ω(f(x,un(x))−f(x,ˆu(x)))(un(x)−ˆu(x))dx→0, as n→+∞. | (4.6) |
Easily to know that
⟨J′(un)−J′(ˆu),un−ˆu⟩=‖un−ˆu‖2−∫Ω(|un(x)|2∗−2un(x)−|ˆu(x)|2∗−2ˆu(x))(un(x)−ˆu(x))dx−∫Ω(f(x,un(x))−f(x,ˆu(x)))(un(x)−ˆu(x))dx−∫Ωa(x)(un(x)−ˆu(x))2dx=‖un−ˆu‖2−∫Ω|un(x)−ˆu(x)|2∗dx+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),un−ˆu⟩=⟨J′(un)−J′(ˆu),un−ˆu⟩, 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)+12‖un−ˆu(x)‖2−12∗|un−ˆu|2∗2∗+o(1)≥12‖un−ˆu‖2−12∗|un−ˆu|2∗2∗+o(1)≥1N‖un−ˆu‖2+o(1), as n→∞. | (4.9) |
Let ‖un−ˆu‖2→L, as n→∞. By Step 4, |un−ˆu|2∗2∗→L, as n→∞. In view of H10(Ω)↪L2∗(Ω) and the definition of S in (1.6), so L22∗S≤L, then L=0 or L≥SN2. If L≥SN2, by (4.9) there will be
c≥1NL≥1NSN2, |
which against the assumption. Thus, L=0, which implies un→ˆu in H10(Ω), as n→+∞. This completes the proof.
If λ1≤0, 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 u∈Z with ‖u‖=ρ, we have J(u)≥κ;
(ii) J(u)≤0 for any u∈Y;
(iii) there exists R0>ρ such that J(u)≤0 for any u∈F with ‖u‖≥R0, where F={u+tz|u∈Y,t>0,z∈Z∖{0}}.
Proof. Let u ∈Z, by (f2) and (f3), we have
J(u)=12∫Ω(|∇u|2−a(x)u2(x))dx−∫ΩF(x,u(x))dx−12∗∫Ω|u(x)|2∗dx≥ˆδ2‖u‖2−ε2|u|22−M(ε)|u|ςς−1+ε2∗|u|2∗2∗=12(ˆδ−εk2)‖u‖2−kςM(ε)‖u‖ς−(1+ε)S2∗22∗‖u‖2∗, |
where ς∈(2,2∗). Fixed ε such that ˆδ−εk2>0. For u∈Z, ‖u‖=ρ sufficiently small,
κ=infu∈H10(Ω),|u|2=ρJ(u)≥12(ˆδ−εk2)ρ2−kςM(ε)ρς−(1+ε)S2∗22∗ρ2∗>0. |
Now we will give the proof of (ii). Thanks to λk≤0, for any u∈Y=span{e1,...,ek},
J(u)=12‖u‖2−12∫Ωa(x)u2(x)dx−∫ΩF(x,u(x))dx−12∗∫Ω|u(x)|2∗dx≤12λk|u|22−12∗∫Ω|u(x)|2∗dx≤0. | (4.10) |
For any u∈F, we also have
J(u)=12‖u‖2−12∫Ωa(x)u(x)2dx−∫ΩF(x,u(x))dx−12∗∫Ω|u(x)|2∗dx≤12‖u‖2−C2∗‖u‖2∗2∗. |
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=infγm∈Γmmaxt∈[0,1]J(γm(t)), |
where
Γm={γm∈C([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 λ1≤0, we construct the functions z,ˉz∈Z as follows:
z=u0−Σki=1(∫Ωu0(x)ei(x)dx)ei, ˉz=z‖z‖, |
where u0 is defined in (f′4). Let c be the linking critical level of J, that is,
c=infγ∈Γsupu∈MJ(γ(u)) |
is the linking critical level of J, where
Γ={γ∈C(¯M,H10(Ω)):γ=id on ∂M} |
and
M={u=y+tˉz|‖u‖≤R0,t≥0,y∈Y}. |
Our goal is to show that c<1NSN2. Since F=Y⊕span{ˉz} is a linear space, we have
supu∈FJ(u)=supu∈F,ξ≠0J(|ξ|u|ξ|)=supu∈F,ξ>0J(ξu)≤supu∈F,ξ≥0J(ξu), | (4.11) |
so, in view of M⊂F, we obtain
c≤supu∈MJ(u)≤supu∈FJ(u)≤supu∈F,ξ≥0J(ξu). | (4.12) |
By (f3), J(u)≤12‖u‖2−12∫Ωa(x)u2(x)dx−12∗|u|2∗2∗. By Lemma 9, for any u∈H10∖{0},
supξ≥0J(ξu)≤1NSN2a(u). | (4.13) |
Therefore, combining (4.12), (4.13), and (f′4), we derive that
c≤supu∈F,ξ≥0J(ξu)≤1Nsupu∈FSN2a(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 a∈N2(Ω). Under some mild assumptions, we obtain an existence result. For the potential a∈N2(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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