Existence of ground state solutions for critical quasilinear Schrödinger equations with steep potential well

Yan-Fang Xue; Xiao-Jing Zhong; Chun-Lei Tang

doi:10.1515/ans-2022-0030

Open Access Published by De Gruyter November 5, 2022

Existence of ground state solutions for critical quasilinear Schrödinger equations with steep potential well

Yan-Fang Xue , Xiao-Jing Zhong and Chun-Lei Tang

From the journal Advanced Nonlinear Studies

https://doi.org/10.1515/ans-2022-0030

Abstract

We study the existence of solutions for the quasilinear Schrödinger equation with the critical exponent and steep potential well. By using a change of variables, the quasilinear equations are reduced to a semilinear one, whose associated functionals satisfy the geometric conditions of the Mountain Pass Theorem for suitable assumptions. The existence of a ground state solution is obtained, and its concentration behavior is also considered.

Keywords: quasilinear Schrödinger equation; ground state solutions; critical exponents; steep potential well

MSC 2010: 35A15; 35D30; 35J62

1 Introduction and main result

We are concerned with the existence of solitary wave solutions for the following quasilinear Schrödinger equation:

(1.1) − Δ u + a ( x ) u − γ u 2 1 + u 2 Δ ( 1 + u 2 ) = k ( x , u ) , x ∈ R N .

This equation is related to the Schrödinger equation of the form

(1.2) i ∂ ψ ∂ t = − Δ ψ + W ( x ) ψ − k ( x , ψ ) − Δ ρ ( ∣ ψ ∣ 2 ) ρ ′ ( ∣ ψ ∣ 2 ) ψ ,

where W : R N → R is a given potential, and ρ is a real function. The form of (1.2) has been derived as models of several physical phenomena corresponding to various types of ρ ( s ) . Seeking solutions of the type stationary waves, namely, the solutions of the form ψ ( t , x ) = exp ( − i E t ) u ( x ) , E ∈ R , and u is a real function, equation (1.2) can be reduced to the corresponding equation of elliptic type

(1.3) − Δ u + a ( x ) u − Δ ρ ( u 2 ) ρ ′ ( ∣ u ∣ 2 ) u = k ( x , u ) , x ∈ R N ,

where a ( x ) = W ( x ) − E is the new potential function. If we take ρ ( s ) = s , we obtain the superfluid film equation in plasma physics as follows:

(1.4) − Δ u + a ( x ) u − Δ ( u 2 ) u = k ( x , u ) , x ∈ R N .

If we set ρ ( s ) = 1 + s , we obtain equation (1.1), which models the self-channeling of a high-power ultrashort laser in the matter (see [9]). For more physical motivations and more references dealing with applications, we can refer to [14,17] and references therein.

Recent studies have been focused on problem (1.4) (see [8,14,17,20] and references therein), but there are few results on problem (1.1). Here, we establish the existence of ground state solutions for problem (1.1). The main mathematical difficulty with problem (1.1) is caused by the second-order derivatives γ u 2 1 + u 2 Δ ( 1 + u 2 ) , and the natural functional corresponding to problem (1.1) may be not well defined in the space H 1 ( R N ) . To overcome this difficulty, various arguments have been developed, such as a change of variables (see [6,7,12,19] and references therein) and the perturbation method (see [11]). More precisely, in [7], under some appropriate assumptions on the nonlinear term, they established the existence of a positive solution. The method is based on a change of variables, a monotonicity trick developed by Jeanjean, and an a priori estimate. By using the same change of variable and variational argument, Shen and Wang studied problem (1.1) with critical growth and obtained positive solutions in [19]. In [11], by using the perturbation method that was initially proposed in [15], a positive ground state solution has been obtained.

Enlightened by [16,19,21], we study the existence of solutions for problem (1.1) with steep potential well a ( x ) = λ V ( x ) and k ( x , u ) = h ( u ) + ∣ u ∣ 2 ∗ − 2 u , namely, the following quasilinear Schrödinger equation:

(1.5) − Δ u + λ V ( x ) u − γ u 2 1 + u 2 Δ ( 1 + u 2 ) = h ( u ) + ∣ u ∣ 2 ∗ − 2 u , x ∈ R N ,

where N ≥ 3 , γ , λ > 0 , the potential V ( x ) satisfies the following conditions:

( V 1 ) V ( x ) ∈ C ( R N , R ) , V ( x ) ≥ 0 for every x ∈ R N .

( V 2 ) There exists V 0 > 0 such that M 0 ≔ { x ∈ R N : V ( x ) ≤ V 0 } is nonempty and has a finite measure.

( V 3 ) Ω ≔ int V − 1 ( 0 ) is nonempty.

This kind of assumption was first introduced by Bartsch and Wang [3] in dealing with the semilinear Schrödinger equation. Then, many results on this kind of potential were obtained, and we refer the readers to [2,5,13,21,23,25] and references therein. The potential well λ V ( x ) represents a potential well whose depth is controlled by λ > 0 .

We assume that the nonlinearity h : R → R is continuous and satisfies the following conditions:

( h 1 ) lim s → 0 h ( s ) s = 0 .

( h 2 ) lim s → + ∞ h ( s ) s 2 ∗ − 1 = 0 .

( h 3 ) There exists θ < μ < 2 ∗ such that 0 < μ H ( s ) ≤ s h ( s ) , where H ( s ) = ∫ 0 s h ( t ) d t ,

θ = 4 , if N = 3 , 2 , if N ≥ 4 .

Now, we state our main existence results.

Theorem 1.1

Suppose that conditions ( h 1 )–( h 3 ) and ( V 1 )–( V 3 ) hold. Then, for λ large, problem (1.5) possesses a ground state solution if 0 < γ < γ ∗ , where

γ ∗ = 16 ( μ − 2 ) ( μ − 4 ) 2 , if μ < 4 , + ∞ , if μ ≥ 4 .

Note that under our assumptions, for λ large enough, the following Dirichlet problem is a kind of limit problem:

(1.6) − Δ u − γ u 2 1 + u 2 Δ ( 1 + u 2 ) = h ( u ) + ∣ u ∣ 2 ∗ − 2 u , x ∈ Ω ,

where Ω = int V − 1 ( 0 ) . Next, we give a result related to problem (1.6).

Theorem 1.2

Assume that u λ n are the solutions obtained in Theorem 1.1 and Ω is defined by ( V 3 ) , then u λ n → u ˆ as λ n → ∞ , where u ˆ ∈ H 0 1 ( Ω ) is a nontrivial solution of problem (1.6).

Remark 1.3

For steep potential well, there are many results for several different equations. For example, the classical Schrödinger equation is discussed in [2, 5,21,23,25], the nonlinear Kirchhoff-type equation with steep potential well is considered in [16], the Schrödinger-Poisson system is researched in [24], and equation (1.4) with steep potential well is discussed in [10]. However, there is no result on equation (1.5) with this kind of potential. To our knowledge, it is the first time to study the existence and concentration behavior of ground state solutions for critical quasilinear Schrödinger equation (1.5) with steep potential well.

Remark 1.4

We denote ∫ R N h ( x ) d x as ∫ R N h ( x ) for simplicity.

Notation: In this article, we use the following notations.

E ≔ u ∈ H 1 ( R N ) : ∫ R N V ( x ) u 2 < + ∞ is the Hilbert space endowed with the norm
‖ u ‖ 2 = ∫ R N ( ∣ ∇ u ∣ 2 + V ( x ) u 2 ) .
For λ > 0 , we also define the norm
‖ u ‖ λ 2 = ∫ R N ( ∣ ∇ u ∣ 2 + λ V ( x ) u 2 ) .
It is clear that the two norms are equal.
L s ( R N ) is the usual Banach space endowed with the norm
∣ u ∣ s s = ∫ R N ∣ u ∣ s , ∀ s ∈ [ 1 , + ∞ ) .
B r ( y ) ≔ { x ∈ R N : ∣ x − y ∣ < r } .
C , C 0 , C 1 , ⋯ denote various positive (possibly different) constants.

2 Some preliminary results

We note that the solutions of problem (1.5) are the critical points of the functional

J λ ( u ) = 1 2 ∫ R N 1 + γ u 2 2 ( 1 + u 2 ) ∣ ∇ u ∣ 2 + 1 2 ∫ R N λ V ( x ) u 2 − ∫ R N H ( u ) − 1 2 ∗ ∫ R N u 2 ∗ .

Variational methods cannot be applied directly to find weak solutions of problem (1.5), since the natural associated functional J λ ( u ) is not well defined, in general, in the space E . To overcome this difficulty, we borrow an idea from Shen and Wang [18]. We use the change of variables v ≔ F ( u ) = ∫ 0 u f ( t ) d t , where f is defined by

(2.1) f ( t ) = 1 + γ t 2 2 ( 1 + t 2 ) .

After the change of variables from J λ , we obtain a new variational functional

I λ ( v ) = 1 2 ∫ R N ( ∣ ∇ v ∣ 2 + λ V ( x ) ∣ F − 1 ( v ) ∣ 2 ) − ∫ R N H ( F − 1 ( v ) ) − 1 2 ∗ ∫ R N ∣ F − 1 ( v ) ∣ 2 ∗ .

Since f is a nondecreasing positive function, we obtain ∣ F − 1 ( v ) ∣ ≤ ∣ v ∣ f ( 0 ) = ∣ v ∣ . From this and the conditions of h , it is clear that I λ is well defined in E and I λ ∈ C 1 ( E , R ) (see [9,18,19] for details). Now, we give another equation

(2.2) − div 1 + γ u 2 2 ( 1 + u 2 ) ∇ u + λ V ( x ) u + γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 = h ( u ) + ∣ u ∣ 2 ∗ − 2 u ,

which is equivalent to (1.5). In fact, we only need to show that

− div 1 + γ u 2 2 ( 1 + u 2 ) ∇ u + γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 = − Δ u − γ u 2 1 + u 2 Δ ( 1 + u 2 ) .

By a direct calculation, we obtain

− div 1 + γ u 2 2 ( 1 + u 2 ) ∇ u + γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 = − div ∇ u − γ u 2 2 ( 1 + u 2 ) div ∇ u − ∇ u ⋅ ∇ γ u 2 2 ( 1 + u 2 ) + γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 = − Δ u − γ u 2 2 ( 1 + u 2 ) Δ u − γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 = − Δ u − γ u 2 1 + u 2 u 1 + u 2 div ∇ u + ∇ u ⋅ ∇ u 1 + u 2 = − Δ u − γ u 2 1 + u 2 Δ ( 1 + u 2 ) .

If u is a weak solution of problem (1.5), then it is also a weak solution of (2.2) and should satisfy

(2.3) ∫ R N 1 + γ u 2 2 ( 1 + u 2 ) ∇ u ⋅ ∇ φ + γ u 2 ( 1 + u 2 ) 2 ∣ ∇ u ∣ 2 φ + λ V ( x ) u φ − h ( u ) φ − ∣ u ∣ 2 ∗ − 1 φ = 0 ,

for all φ ∈ C 0 ∞ ( R N ) . Let φ = ψ f ( u ) , then it can be checked that (2.3) is equivalent to the following equality:

(2.4) ⟨ I λ ′ ( v ) , ψ ⟩ = ∫ R N ∇ v ⋅ ∇ ψ + λ V ( x ) F − 1 ( v ) f ( F − 1 ( v ) ) ψ − h ( F − 1 ( v ) ) f ( F − 1 ( v ) ) ψ − ∣ F − 1 ( v ) ∣ 2 ∗ − 1 f ( F − 1 ( v ) ) ψ = 0 .

Therefore, in order to find the solutions of problem (1.5), it suffices to study the existence of solutions of the following equation:

− Δ v + λ V ( x ) F − 1 ( v ) f ( F − 1 ( v ) ) = h ( F − 1 ( v ) ) f ( F − 1 ( v ) ) + ∣ F − 1 ( v ) ∣ 2 ∗ − 1 f ( F − 1 ( v ) ) , x ∈ R N .

Now, we summarize the properties of F − 1 , f , which have been proved in [19].

Lemma 2.1

The function F − 1 and f satisfy the following properties:

1 ≤ f ( t ) ≤ 2 + γ 2 for all t ∈ R ;
1 ≤ F − 1 ( t ) f ( F − 1 ( t ) ) t ≤ 4 + 2 γ − 2 4 + 2 γ γ for all t ∈ R , t ≠ 0 ;
2 2 + γ ∣ t ∣ ≤ ∣ F − 1 ( t ) ∣ ≤ ∣ t ∣ for all t ∈ R ;
F − 1 ( t ) t → 1 as t → 0 ;
F − 1 ( t ) t → 2 2 + γ as t → ∞ ;
0 ≤ f ′ ( t ) t f ( t ) ≤ 1 + 4 − 2 4 + 2 γ γ for all t ∈ R .

Definition 2.2

Let E be a real Banach space, I ∈ C 1 ( E , R ) , and c ∈ R . The function I satisfies the ( P S ) c condition if any sequence { u n } ⊂ E such that

I ( u n ) → c , I ′ ( u n ) → 0 ,

has a convergent subsequence.

Lemma 2.3

[26] Assume that k : R N × R → R is continuous and there exists a constant C > 0 such that

lim s → 0 k ( x , s ) s ≤ C lim s → + ∞ k ( x , s ) s 2 ∗ − 1 = 0 .

Let { v n } ⊂ E be a bounded sequence and v ⊂ E with v n ⇀ v in E , then

lim n → + ∞ ∫ R N K ( x , v n ) − ∫ R N K ( x , v ) − ∫ R N K ( x , v n − v ) = 0 ,

where K ( x , v ) = ∫ 0 v k ( x , s ) d s .

Lemma 2.4

[22] Let E be a real Banach space and suppose that I ∈ C 1 ( E , R ) satisfies

max { I ( 0 ) , I ( e ) } ≤ ξ < η ≤ inf ‖ u ‖ = ρ I ( u ) ,

for some ξ < η , ρ > 0 , and e ∈ E with ‖ e ‖ > ρ . Let c ≥ η be characterized by c = inf γ ∈ Γ max t ∈ [ 0 , 1 ] I ( γ ( t ) ) , where Γ = { γ ∈ C ( [ 0 , 1 ] , E ) : γ ( 0 ) = 0 , γ ( 1 ) = e } is the set of continuous paths joining 0 and e . If I satisfies the ( P S ) c condition, then c is a critical value of I.

3 Proof of Theorems 1.1 and 1.2

Lemma 3.1

Conditions ( V 1 )–( V 3 ) and ( h 1 )–( h 3 ) hold. Then, there exist ρ > 0 and η > 0 such that inf ‖ v ‖ = ρ I λ ( v ) ≥ η .

Proof

By ( h 1 ) and ( h 2 ) , for ε > 0 sufficiently small, there exists a constant C ε > 0 such that

(3.1) H ( s ) ≤ ε ∣ s ∣ 2 + C ε ∣ s ∣ 2 ∗ .

By using ( V 1 ) , ( V 2 ) , and the Hölder and Sobolev inequalities, we have

(3.2) ∫ R N ∣ v ∣ 2 = ∫ M 0 ∣ v ∣ 2 + ∫ R N \ M 0 ∣ v ∣ 2 ≤ ∣ M 0 ∣ 2 ∗ − 2 2 ∗ ∣ v ∣ 2 ∗ 2 + 1 V 0 ∫ R N \ M 0 V ( x ) v 2 ≤ ∣ M 0 ∣ 2 ∗ − 2 2 ∗ S − 1 ∣ ∇ v ∣ 2 2 + 1 V 0 λ ∫ R N λ V ( x ) v 2 ≤ C 1 ‖ v ‖ λ 2 ,

where C 1 = max ∣ M 0 ∣ 2 ∗ − 2 2 ∗ S − 1 , 1 V 0 λ , S is the best constant for the Sobolev embedding and ∣ M 0 ∣ denotes the Lebesgue measure of the set M 0 . By (3.1), (3.2), Lemma 2.1-(3), and the Sobolev embedding inequality, we have

I λ ( v ) = 1 2 ∫ R N [ ∣ ∇ v ∣ 2 + λ V ( x ) ∣ F − 1 ( v ) ∣ 2 ] − ∫ R N H ( F − 1 ( v ) ) − 1 2 ∗ ∫ R N ∣ F − 1 ( v ) ∣ 2 ∗ ≥ 1 2 ∫ R N ∣ ∇ v ∣ 2 + 1 2 + γ ∫ R N λ V ( x ) ∣ v ∣ 2 − ε ∫ R N ∣ v ∣ 2 − C ε ∫ R N ∣ v ∣ 2 ∗ − 1 2 ∗ ∫ R N ∣ v ∣ 2 ∗ ≥ 1 2 + γ ‖ v ‖ λ 2 − ε C 1 ‖ v ‖ λ 2 − C ε + 1 2 ∗ S − 2 ∗ ∕ 2 ∣ ∇ v ∣ 2 2 ∗ ≥ 1 2 + γ − ε C 1 ‖ v ‖ λ 2 − C ε + 1 2 ∗ S − 2 ∗ ∕ 2 ‖ v ‖ λ 2 ∗ .

We choose ε small enough such that 1 2 + γ − ε C 1 > 0 . Therefore, we conclude that there is ρ > 0 small enough, such that I λ ( v ) > 0 whenever ‖ v ‖ λ ≤ ρ , v ≠ 0 . Then, there exists η > 0 such that for any ‖ v ‖ λ = ρ , one has I λ ( v ) ≥ η > 0 . □

Lemma 3.2

Suppose that conditions ( V 1 )–( V 3 ) and ( h 1 )–( h 3 ) are satisfied. Then, there exists e ∈ E with ‖ e ‖ λ > ρ , such that I λ ( e ) < 0 , where ρ is given by Lemma 3.1.

Proof

We choose some φ ∈ C 0 ∞ ( R N , [ 0 , 1 ] ) , with supp φ = B ¯ 1 , where B ¯ 1 is the closed unit ball in R N . We will prove that I λ ( t φ ) → − ∞ as t → ∞ , which will prove the result if we take e = t φ with t large enough. In fact, by Lemma 2.1-(3) and ( h 3 ) , one has

I λ ( t φ ) = 1 2 ∫ R N ∣ ∇ ( t φ ) ∣ 2 + 1 2 ∫ R N λ V ( x ) ∣ F − 1 ( t φ ) ∣ 2 − ∫ R N H ( F − 1 ( t φ ) ) − 1 2 ∗ ∫ R N ∣ F − 1 ( t φ ) ∣ 2 ∗ ≤ t 2 2 ∫ R N ∣ ∇ φ ∣ 2 + t 2 2 ∫ R N λ V ( x ) ∣ φ ∣ 2 − t 2 ∗ 2 ∗ 2 2 + γ 2 ∗ ∫ R N ∣ φ ∣ 2 ∗ → − ∞ as t → + ∞ . □

Lemma 3.3

Suppose ( V 1 )–( V 3 ) and ( h 1 )–( h 3 ) hold, 0 < γ < γ ∗ . Then, there exists { v n } ⊂ E such that I λ ( v n ) → c , I λ ′ ( v n ) → 0 , and { v n } is bounded in E .

Proof

It follows from Lemmas 3.1 and 3.2 that, there exists a ( P S ) sequence { v n } for I λ . We only need to prove that { v n } is bounded. Let { v n } ⊂ E be an arbitrary ( P S ) sequence for I λ at level c > 0 , namely

(3.3) I λ ( v n ) = 1 2 ∫ R N ∣ ∇ v n ∣ 2 + 1 2 ∫ R N λ V ( x ) ∣ F − 1 ( v n ) ∣ 2 − ∫ R N H ( F − 1 ( v n ) ) − 1 2 ∗ ∫ R N ∣ F − 1 ( v n ) ∣ 2 ∗ = c + o n ( 1 ) ,

and for any φ ∈ E ,

⟨ I λ ′ ( v n ) , φ ⟩ = ∫ R N ∇ v n ⋅ ∇ φ + λ V ( x ) F − 1 ( v n ) f ( F − 1 ( v n ) ) φ − ∫ R N h ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) φ − ∫ R N ∣ F − 1 ( v n ) ∣ 2 ∗ − 2 F − 1 ( v n ) f ( F − 1 ( v n ) ) φ = o n ( 1 ) .

Choosing φ = φ n = F − 1 ( v n ) f ( F − 1 ( v n ) ) , from Lemma 2.1-(1)(3), we obtain ∣ φ n ∣ ≤ C ∣ v n ∣ and

∣ ∇ φ n ∣ = 1 + F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ∇ v n ≤ C ∣ ∇ v n ∣ .

Thus, φ n ∈ E and ⟨ I λ ′ ( v n ) , φ n ⟩ = o n ( 1 ) . Recalling that { v n } ⊂ E is a ( P S ) sequence, by ( h 3 ) and Lemma 2.1-(3), we obtain

(3.4) μ c + o n ( 1 ) = μ I λ ( v n ) − ⟨ I λ ′ ( v n ) , φ n ⟩ = ∫ R N μ − 2 2 − F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ∣ ∇ v n ∣ 2 + μ − 2 2 ∫ R N λ V ( x ) ∣ F − 1 ( v n ) ∣ 2 + 2 ∗ − μ 2 ∗ ∫ R N ∣ F − 1 ( v n ) ∣ 2 ∗ + ∫ R N [ h ( F − 1 ( v n ) ) F − 1 ( v n ) − μ H ( F − 1 ( v n ) ) ] ≥ ∫ R N μ − 2 2 − F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ∣ ∇ v n ∣ 2 + μ − 2 2 + γ ∫ R N λ V ( x ) v n 2 .

By Lemma 2.1-(6), one has

μ − 2 2 − F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ≥ μ − 2 2 − 1 − 4 − 2 4 + 2 γ γ ≔ μ − 4 2 + l ( γ ) .

If μ ≥ 4 , γ > 0 , we obtain μ − 4 2 ≥ 0 , l ( γ ) > 0 . Then,

(3.5) μ − 2 2 − F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ≥ μ − 4 2 + l ( γ ) > 0 .

If 2 < μ < 4 , 0 < γ < γ ∗ = 16 ( μ − 2 ) ( μ − 4 ) 2 , we obtain inf γ > 0 l ( γ ) = 4 − μ 2 . Then,

(3.6) μ − 2 2 − F − 1 ( v n ) f ′ ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ≥ μ − 4 2 + l ( γ ) > 0 .

Combining (3.4), (3.5), and (3.6), one obtains that ‖ v n ‖ λ is bounded.□

Up to now, we establish that the ( P S ) sequence { v n } is bounded in E . We may obtain, up to a subsequence, v n ⇀ v in E , v n → v in L loc q ( R N ) ( 2 ≤ q < 2 ∗ ) , and v n ( x ) → v ( x ) a.e. in R N . A routine computation shows that, for any ϕ ∈ C 0 ∞ ( R N ) ,

⟨ I λ ′ ( v n ) − I λ ′ ( v ) , ϕ ⟩ = ∫ R N ∇ ( v n − v ) ⋅ ∇ ϕ + λ ∫ R N V ( x ) F − 1 ( v n ) f ( F − 1 ( v n ) ) − F − 1 ( v ) f ( F − 1 ( v ) ) ϕ − ∫ R N h ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) − h ( F − 1 ( v ) ) f ( F − 1 ( v ) ) ϕ − ∫ R N ∣ F − 1 ( v n ) ∣ 2 ∗ − 2 F − 1 ( v n ) f ( F − 1 ( v n ) ) − ∣ F − 1 ( v ) ∣ 2 ∗ − 2 F − 1 ( v ) f ( F − 1 ( v ) ) ϕ .

Let Θ = supp ϕ , then v n → v in L s ( Θ ) ( 2 ≤ s < 2 ∗ ) . By Lemma A.1 in [22], there is w s ( x ) ∈ L s ( Θ ) , such that for every n ∈ N and a.e. x ∈ Θ , ∣ v n ( x ) ∣ ≤ w s ( x ) . Thus, for a.e. x ∈ Θ , one has

F − 1 ( v n ) f ( F − 1 ( v n ) ) − F − 1 ( v ) f ( F − 1 ( v ) ) → 0 , as n → + ∞ , h ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) − h ( F − 1 ( v ) ) f ( F − 1 ( v ) ) → 0 , as n → + ∞ ,

and

∣ F − 1 ( v n ) ∣ 2 ∗ − 2 F − 1 ( v n ) f ( F − 1 ( v n ) ) − ∣ F − 1 ( v ) ∣ 2 ∗ − 2 F − 1 ( v ) f ( F − 1 ( v ) ) → 0 , as n → + ∞ .

Moreover, by Lemma 2.1-(1)(3), we obtain

F − 1 ( v n ) f ( F − 1 ( v n ) ) ϕ ≤ ∣ F − 1 ( v n ) ϕ ∣ ≤ ∣ v n ϕ ∣ ≤ ∣ w 2 ∣ ⋅ ∣ ϕ ∣ ∈ L 1 ( R N )

and

∣ F − 1 ( v n ) ∣ 2 ∗ − 2 F − 1 ( v n ) f ( F − 1 ( v n ) ) ϕ ≤ ∣ v n ∣ 2 ∗ − 1 ⋅ ∣ ϕ ∣ ≤ ∣ w 2 ∗ − 1 ∣ 2 ∗ − 1 ⋅ ∣ ϕ ∣ ∈ L 1 ( R N ) .

From ( h 1 ) and ( h 2 ) , one can deduce

h ( F − 1 ( v n ) ) f ( F − 1 ( v n ) ) ϕ ≤ ( ε ∣ F − 1 ( v n ) ∣ + C ε ∣ F − 1 ( v n ) ∣ 2 ∗ − 1 ) ϕ f ( F − 1 ( v n ) ) = ε F − 1 ( v n ) f ( F − 1 ( v n ) ) ϕ + C ε ∣ F − 1 ( v n ) ∣ 2 ∗ − 2 F − 1 ( v n ) f ( F − 1 ( v n ) ) ϕ ≤ ε ∣ w 2 ∣ ⋅ ∣ ϕ ∣ + C ε ∣ w 2 ∗ − 1 ∣ 2 ∗ − 1 ⋅ ∣ ϕ ∣ ∈ L 1 ( R N ) .

It is clear that

⟨ I λ ′ ( v n ) − I λ ′ ( v ) , ϕ ⟩ = o n ( 1 ) , ⟨ I λ ′ ( v n ) , ϕ ⟩ = o n ( 1 ) .

Then, it follows from the Lebesgue-dominated theorem that, ⟨ I λ ′ ( v ) , ϕ ⟩ = 0 , i.e., v is a critical point of I λ .

Given ε > 0 , we consider the function w ε : R N → R defined by

w ε ( x ) = C ( N ) ε ( N − 2 ) ∕ 2 ( ε 2 + ∣ x ∣ 2 ) ( N − 2 ) ∕ 2 ,

where

C ( N ) = [ N ( N − 2 ) ] ( N − 2 ) ∕ 4 .

We observe that { w ε } is a family of functions in which the infimum that defines the best constant S for the Sobolev embedding D 1 , 2 ( R N ) ↪ L 2 ∗ ( R N ) is attained.

By a similar computation to that in [1,4], we have

(3.7) ∫ R N \ B 1 ( 0 ) ∣ ∇ w ε ∣ 2 = O ( ε N − 2 ) , as ε → 0 + .

(3.8) ∫ R N ∣ ∇ w ε ∣ 2 = ∫ R N w ε 2 ∗ = S N ∕ 2 .

(3.9) ∫ B 1 ( 0 ) ∣ ∇ w ε ∣ 2 ≤ ∫ B 1 ( 0 ) w ε 2 ∗ .

Let ϕ ∈ C 0 ∞ ( Ω , [ 0 , 1 ] ) be a cut-off function satisfying ϕ ≡ 1 in B 1 ( 0 ) and ϕ ≡ 0 in R N \ B 2 ( 0 ) , where Ω ≔ int V − 1 ( 0 ) . Define

u ε = ϕ w ε , v ε = u ε ∣ u ε ∣ 2 ∗ .

Then, by (3.7)–(3.9), as ε → 0 , we have

(3.10) ∫ R N ∣ ∇ v ε ∣ 2 = S + O ( ε N − 2 ) , if N ≥ 3 ,

and

(3.11) ∣ v ε ∣ 2 2 = O ( ε ) , if N = 3 , O ( ε 2 ∣ ln ε ∣ ) , if N = 4 , O ( ε 2 ) , if N ≥ 5 .

Lemma 3.4

The minimax level c satisfies

c < 1 N 2 + γ 2 N 2 S N 2 .

Proof

It suffices to show that there exists v 0 ∈ E ⧹ { 0 } such that

max t ≥ 0 I λ ( t v 0 ) < 1 N 2 + γ 2 N 2 S N 2 .

Since lim t → ∞ I λ ( t v ε ) = − ∞ and I λ ( t v ε ) > 0 for t > 0 small enough, there exists t ε > 0 such that I λ ( t ε v ε ) = max t > 0 I λ ( t v ε ) . We claim that there are constants T 1 and T 2 such that 0 < T 1 < t ε < T 2 . First, we prove that t ε is bounded from below by a positive constant. Otherwise, if t ε → 0 as ε → 0 , we have t ε v ε → 0 . Therefore, 0 < c ≤ max t ≥ 0 I λ ( t v ε ) → 0 , which is a contradiction. Second, if t ε → + ∞ as ε → 0 , then, similar to the proof of Lemma 3.2, we can obtain 0 < c ≤ I λ ( t ε v ε ) → − ∞ , which is a contradiction. Hence, there is T 2 > 0 such that t ε < T 2 for ε small enough.

Now, by ( V 3 ) , Lemma 2.1-(3), and (3.7)–(3.11), we observe that

I λ ( t ε v ε ) = 1 2 ∫ Ω ∣ ∇ ( t ε v ε ) ∣ 2 + λ 2 ∫ Ω V ( x ) ∣ F − 1 ( t ε v ε ) ∣ 2 − ∫ Ω H ( F − 1 ( t ε v ε ) ) − 1 2 ∗ ∫ Ω ∣ F − 1 ( t ε v ε ) ∣ 2 ∗ ≤ t ε 2 2 ∫ Ω ∣ ∇ v ε ∣ 2 − ∫ Ω H ( F − 1 ( t ε v ε ) ) − t ε 2 ∗ 2 ∗ 2 2 + γ 2 ∗ .

Denote

g ( t ) ≔ t 2 2 ∣ ∇ v ε ∣ 2 2 − t 2 ∗ 2 ∗ 2 2 + γ 2 ∗ .

It is very standard to obtain that g ( t ) achieves its maximum at

t 0 = 2 + γ 2 N ∕ 2 ∣ ∇ v ε ∣ 2 2 2 ∗ − 2

and

(3.12) g ( t 0 ) = 1 N 2 + γ 2 N 2 ∣ ∇ v ε ∣ 2 N ∕ 2 .

Then, it follows from (3.10) and (3.12) that

I λ ( t ε v ε ) ≤ 1 N 2 + γ 2 N 2 [ S + O ( ε N − 2 ) ] N ∕ 2 − ∫ Ω H ( F − 1 ( t ε v ε ) ) .

Using the following inequality

( a + b ) r ≤ a r + r ( a + b ) r − 1 b , a > 0 , b > 0 r ≥ 1 ,

we have

(3.13) I λ ( t ε v ε ) ≤ 1 N 2 + γ 2 N 2 S N ∕ 2 + O ( ε N − 2 ) − ∫ Ω H ( F − 1 ( t ε v ε ) ) .

By ( h 3 ) , the definition of v ε and since F − 1 ( t ) is increasing, there exists a constant δ > 0 such that

(3.14) H ( F − 1 ( t ε v ε ) ) ≥ C ∣ F − 1 ( t ε v ε ) ∣ μ ≥ C ∣ F − 1 ( T 1 v ε ) ∣ μ ≥ C ∣ F − 1 ( δ ε 2 − N 2 ) ∣ μ

for ∣ x ∣ ≤ ε . Then, by (3.14) and Lemma 2.1-(3), one has

(3.15) ∫ Ω H ( F − 1 ( t ε v ε ) ) ≥ C ∫ B ε ( 0 ) ∣ F − 1 ( δ ε 2 − N 2 ) ∣ μ ≥ C 0 ε ( 2 − N ) μ 2 + N .

Therefore, by (3.13) and (3.15), we have

I λ ( t ε v ε ) ≤ 1 N 2 + γ 2 N 2 S N ∕ 2 + O ( ε N − 2 ) − C 0 ε ( 2 − N ) μ 2 + N .

Since ( 2 − N ) μ 2 + N < N − 2 when N ≥ 3 , μ > θ , we can obtain our result.□

Lemma 3.5

Assume that the conditions of Theorem 1.1 hold, { v n } is a ( P S ) c sequence for I λ with

c < 1 N 2 + γ 2 N 2 S N 2 .

Then, for λ > 0 large enough, { v n } possesses a strongly convergent subsequence.

Proof

Denote

k ( x , s ) = λ V ( x ) s − F − 1 ( s ) f ( F − 1 ( s ) ) + h ( F − 1 ( s ) ) f ( F − 1 ( s ) ) + ∣ F − 1 ( s ) ∣ 2 ∗ − 1 f ( F − 1 ( s ) ) − 2 2 + γ 2 ∗ s 2 ∗ − 1

and

K ( x , s ) = ∫ 0 s k ( x , τ ) d τ = 1 2 λ V ( x ) [ s 2 − ∣ F − 1 ( s ) ∣ 2 ] + H ( F − 1 ( s ) ) + ∣ F − 1 ( s ) ∣ 2 ∗ 2 ∗ − 2 2 + γ 2 ∗ s 2 ∗ 2 ∗ .

Then, the functional I λ can be rewritten as

I λ ( v ) = 1 2 ∫ R N ( ∣ ∇ v ∣ 2 + λ V ( x ) v 2 ) − ∫ R N K ( x , v ) − 2 2 + γ 2 ∗ 1 2 ∗ ∫ R N v 2 ∗ .

The functions k ( x , s ) and K ( x , s ) enjoy the following properties under the assumptions ( V 1 )–( V 3 ) and ( h 1 )–( h 3 ).

(3.16) lim s → 0 k ( x , s ) s = 0 , lim s → + ∞ k ( x , s ) s 2 ∗ − 1 = 0 .

We are going to prove (3.16). From ( h 2 ) , Lemma 2.1-(4)(5), and the fact that f ( t ) = 1 + γ t 2 2 ( 1 + t 2 ) , F − 1 ( s ) = t , we have

k ( x , s ) s = λ V ( x ) 1 − F − 1 ( s ) s ⋅ 1 f ( F − 1 ( s ) ) + h ( F − 1 ( s ) ) s ⋅ 1 f ( F − 1 ( s ) ) + ∣ F − 1 ( s ) ∣ 2 ∗ − 2 f ( F − 1 ( s ) ) ⋅ F − 1 ( s ) s − 2 2 + γ 2 ∗ s 2 ∗ − 2 → 0 , as s → 0 ,

and

k ( x , s ) s 2 ∗ − 1 = λ V ( x ) 1 s 2 ∗ − 2 − F − 1 ( s ) s ⋅ 1 s 2 ∗ − 2 ⋅ 1 f ( F − 1 ( s ) ) + h ( F − 1 ( s ) ) s 2 ∗ − 1 ⋅ 1 f ( F − 1 ( s ) ) + ∣ F − 1 ( s ) ∣ 2 ∗ − 1 s 2 ∗ − 1 ⋅ 1 f ( F − 1 ( s ) ) − 2 2 + γ 2 ∗ → 0 , as s → + ∞ .

Applying Lemma 3.3, { v n } is bounded. Then, by Lemma 2.3, one has

(3.17) lim n → + ∞ ∫ R N K ( x , v n ) − ∫ R N K ( x , v ) − ∫ R N K ( x , v n − v ) = 0 .

Setting w n = v n − v , the Brezis-Lieb lemma leads to

(3.18) ‖ v n ‖ λ 2 = ‖ w n ‖ λ 2 + ‖ v ‖ λ 2 + o n ( 1 ) , ∣ v n ∣ 2 ∗ 2 ∗ = ∣ w n ∣ 2 ∗ 2 ∗ + ∣ v ∣ 2 ∗ 2 ∗ + o n ( 1 ) .

Next, we prove that w n → 0 in E . Now, there are two cases that may occur:

( i ) lim n → ∞ sup y ∈ R N ∫ B r ( y ) w n 2 d x > 0 ;

( i i ) lim n → ∞ sup y ∈ R N ∫ B r ( y ) w n 2 d x = 0 .

If the case (i) holds, then there exists a positive constant δ > 0 such that

lim n → ∞ sup y ∈ R N ∫ B r ( y ) w n 2 d x = δ > 0 .

From the weakly lower semicontinuous of the norm and the definition of w n , we infer that

(3.19) ‖ w n ‖ λ = ‖ v n − v ‖ λ ≤ ‖ v n ‖ λ + ‖ v ‖ λ ≤ ‖ v n ‖ λ + liminf n → ∞ ‖ v n ‖ λ .

From Lemma 3.3, there exists a constant C such that

(3.20) limsup n → ∞ ‖ v n ‖ λ ≤ C .

Combining (3.19) with (3.20), we deduce

(3.21) limsup n → ∞ ‖ w n ‖ λ ≤ 2 C .

Set D R ≔ { x ∈ R N \ B R : V ( x ) ≥ V 0 } . Let λ ≥ 16 C 2 δ V 0 , we obtain

(3.22) limsup n → ∞ ∫ D R w n 2 ≤ limsup n → ∞ 1 λ V 0 ∫ D R λ V ( x ) w n 2 ≤ limsup n → ∞ 1 λ V 0 ‖ w n ‖ λ 2 ≤ 4 C 2 λ V 0 ≤ δ 4 .

Let A R ≔ { x ∈ R N \ B R : V ( x ) < V 0 } , then from ( V 2 ) and the Hölder and Sobolev inequalities, one obtains

(3.23) limsup n → ∞ ∫ A R w n 2 ≤ limsup n → ∞ ∫ A R w n p 2 ∕ p ∫ A R 1 ( p − 2 ) ∕ p ≤ limsup n → ∞ C 1 ‖ w n ‖ λ 2 ∣ A R ∣ ( p − 2 ) ∕ p ≤ 4 C 2 C 1 ∣ A R ∣ ( p − 2 ) ∕ p → 0 , as R → + ∞ ,

where p ∈ ( 2 , 2 ∗ ] . From w n → 0 in L loc p ( R N ) with p ∈ [ 2 , 2 ∗ ) , we infer, as R → + ∞ ,

δ = lim n → ∞ sup y ∈ R N ∫ B r ( y ) w n 2 ≤ limsup n → ∞ ∫ R N w n 2 = limsup n → ∞ ∫ B R w n 2 + ∫ B R c w n 2 = limsup n → ∞ ∫ D R w n 2 + ∫ A R w n 2 ≤ δ 4 ,

which is a contradiction.

If case (ii) holds, it follows from the Lions lemma that w n → 0 in L p ( R N ) with p ∈ ( 2 , 2 ∗ ) . By (3.16), we infer that for any ε > 0 , there exists C ε > 0 such that

∣ K ( x , w n ) ∣ ≤ ε ( ∣ w n ∣ 2 + ∣ w n ∣ 2 ∗ ) + C ε ∣ w n ∣ p , ∣ k ( x , w n ) w n ∣ ≤ ε ( ∣ w n ∣ 2 + ∣ w n ∣ 2 ∗ ) + C ε ∣ w n ∣ p .

Thus, one obtains

(3.24) ∫ R N K ( x , w n ) = o n ( 1 ) , ∫ R N k ( x , w n ) w n = o n ( 1 ) .

It follows from (3.17), (3.18), and (3.24) that

(3.25) I λ ( v n ) − I λ ( v ) = 1 2 ( ‖ v n ‖ λ 2 − ‖ v ‖ λ 2 ) − ∫ R N [ K ( x , v n ) − K ( x , v ) ] − 2 2 + γ 2 ∗ 1 2 ∗ ∫ R N ( v n 2 ∗ − v 2 ∗ ) = 1 2 ‖ w n ‖ λ 2 − 2 2 + γ 2 ∗ 1 2 ∗ ∫ R N w n 2 ∗ + o n ( 1 )

and

(3.26) ⟨ I λ ′ ( v n ) , v n ⟩ − ⟨ I λ ′ ( v ) , v ⟩ = ‖ v n ‖ λ 2 − ‖ v ‖ λ 2 − ∫ R N [ k ( x , v n ) v n − k ( x , v ) v ] − 2 2 + γ 2 ∗ ∫ R N [ v n 2 ∗ − v 2 ∗ ] = ‖ w n ‖ λ 2 − 2 2 + γ 2 ∗ ∫ R N w n 2 ∗ + o n ( 1 ) .

We may assume that ‖ w n ‖ λ 2 → l , then by (3.26) and the fact that ⟨ I λ ′ ( v n ) , v n ⟩ − ⟨ I λ ′ ( v ) , v ⟩ → 0 , we obtain

2 2 + γ 2 ∗ ∫ R N w n 2 ∗ → l .

Next, we are going to claim that l = 0 . Since v is a critical point of I λ , it satisfies

⟨ I λ ′ ( v ) , F − 1 ( v ) f ( F − 1 ( v ) ) ⟩ = 0 .

Then, we can infer

μ I λ ( v ) = μ I λ ( v ) − ⟨ I λ ′ ( v ) , F − 1 ( v ) f ( F − 1 ( v ) ) ⟩ = ∫ R N μ − 2 2 − F − 1 ( v ) f ′ ( F − 1 ( v ) ) f ( F − 1 ( v ) ) ∣ ∇ v ∣ 2 + μ − 2 2 ∫ R N λ V ( x ) ∣ F − 1 ( v ) ∣ 2 + 2 ∗ − μ 2 ∗ ∫ R N ∣ F − 1 ( v ) ∣ 2 ∗ + ∫ R N [ h ( F − 1 ( v ) ) F − 1 ( v ) − μ H ( F − 1 ( v ) ) ] .

Since 2 < μ < 2 ∗ , it follows from the above equality, (3.5), (3.6), and ( h 3 ) that

(3.27) I λ ( v ) > 0 .

Combining (3.25) and (3.27) with the fact that I λ ( v n ) → c , we deduce

(3.28) 1 2 − 1 2 ∗ l ≤ c .

By the definition of S , we can obtain that

S ∫ R N w n 2 ∗ 2 ∕ 2 ∗ ≤ ∫ R N ∣ ∇ w n ∣ 2 ≤ ‖ w n ‖ λ 2 ,

namely,

(3.29) S 2 + γ 2 2 ∗ l 2 ∕ 2 ∗ ≤ l .

Either l = 0 or l ≥ 2 + γ 2 N S N 2 .

Assume l ≥ 2 + γ 2 N S N 2 , we obtain from (3.28) that

c ≥ 1 2 − 1 2 ∗ l ≥ 1 N 2 + γ 2 N 2 S N 2 ,

which is contrary to Lemma 3.4. Therefore, l = 0 , i.e., ‖ w n ‖ λ → 0 . The proof is completed.□

Proof of Theorem 1.1

By Lemmas 3.1 and 3.2, I λ satisfies the mountain path geometry, and we can obtain a ( P S ) c sequence { v n } . Then, by Lemma 3.5, the sequence { v n } has a strong convergent subsequence. So, we can obtain a nontrivial solution by the Mountain Pass theorem.

Finally, we try to find the least energy solution. Define

m λ = inf { I λ ( u ) : u ∈ E , u ≠ 0 , I λ ′ ( u ) = 0 } .

We claim that m λ > 0 . By the definition of m λ , there exists a minimizing sequence { u n } for m λ , i.e., I λ ( u n ) → m λ , I λ ′ ( u n ) = 0 . It follows from (3.16) that, for any ε > 0 , there is C ε > 0 such that

(3.30) ∣ k ( x , s ) ∣ ≤ ε ∣ s ∣ + C ε ∣ s ∣ 2 ∗ − 1 .

Similar to the proof of (3.2), one can obtain

(3.31) ∫ R N ∣ u n ∣ 2 ≤ C 1 ‖ u n ‖ λ 2 .

Since ⟨ I λ ′ ( u n ) , u n ⟩ = 0 , combining the Sobolev inequality with (3.30) and (3.31), we obtain

‖ u n ‖ λ 2 = ∫ R N k ( x , u n ) u n + 2 2 + γ 2 ∗ u n 2 ∗ ≤ ∫ R N ( ε ∣ u n ∣ 2 + C ε ∣ u n ∣ 2 ∗ ) + 2 2 + γ 2 ∗ ∫ R N ∣ u n ∣ 2 ∗ ≤ ε C 1 ‖ u n ‖ λ 2 + C 2 ∫ R N ∣ ∇ u n ∣ 2 2 ∗ 2 ≤ ε C 1 ‖ u n ‖ λ 2 + C 3 ‖ u n ‖ λ 2 ∗ ,

where C 1 , C 2 , and C 3 are some positive constants. Take ε = 1 2 C 1 , we know that

( 1 − ε C 1 ) ‖ u n ‖ λ 2 = 1 2 ‖ u n ‖ λ 2 ≤ C 3 ‖ u n ‖ λ 2 ∗ .

Then, we obtain

‖ u n ‖ λ ≥ 1 2 C 3 1 2 ∗ − 2 > 0 .

Similar to the proof of (3.4), we have

μ m λ + o n ( 1 ) = μ I λ ( u n ) − ⟨ I λ ′ ( u n ) , F − 1 ( u n ) f ( F − 1 ( u n ) ) ⟩ ≥ ∫ R N μ − 2 2 − F − 1 ( u n ) f ′ ( F − 1 ( u n ) ) f ( F − 1 ( u n ) ) ∣ ∇ u n ∣ 2 + μ − 2 2 + γ ∫ R N λ V ( x ) u n 2 .

Thanks to (3.5) and (3.6), there is δ > 0 such that

μ − 2 2 − F − 1 ( u n ) f ′ ( F − 1 ( u n ) ) f ( F − 1 ( u n ) ) ≥ δ > 0 .

Thus, one obtains

μ m λ + o n ( 1 ) ≥ min δ , μ − 2 2 + γ ‖ u n ‖ λ 2 ≥ min { δ , μ − 2 2 + γ } 1 2 C 3 2 2 ∗ − 2 > 0 ,

which implies m λ > 0 .

Similar to the proof of Lemmas 3.3 and 3.5, we can obtain that { u n } is bounded and there exists u ≠ 0 such that u n ⇀ u , I λ ′ ( u ) = 0 . Hence, by ( h 3 ) and the Fatou lemma, we obtain

μ m λ ≤ μ I λ ( u ) − ⟨ I λ ′ ( u ) , F − 1 ( u ) f ( F − 1 ( u ) ) ⟩ = ∫ R N μ − 2 2 − F − 1 ( u ) f ′ ( F − 1 ( u ) ) f ( F − 1 ( u ) ) ∣ ∇ u ∣ 2 + μ − 2 2 ∫ R N λ V ( x ) ∣ F − 1 ( u ) ∣ 2 + ∫ R N [ h ( F − 1 ( u ) ) F − 1 ( u ) − μ H ( F − 1 ( u ) ) ] + 2 ∗ − μ 2 ∗ ∫ R N ∣ F − 1 ( u ) ∣ 2 ∗ ≤ liminf n → ∞ ( μ I λ ( u n ) − ⟨ I λ ′ ( u n ) , F − 1 ( u n ) f ( F − 1 ( u n ) ) ⟩ ) = μ m λ .

Therefore, u ≠ 0 satisfies I λ ( u ) = m λ and I λ ′ ( u ) = 0 . The proof is completed.□

Proof of Theorem 1.2

Let v λ be the ground state solution obtained in Theorem 1.1, then we can obtain that I λ ( v λ ) = m λ < 1 N 2 + γ 2 N ∕ 2 S N ∕ 2 ≔ c ˆ , I λ ′ ( v λ ) = 0 . Define v n = v λ n , then there exists a sequence { v n } such that I λ n ( v n ) = m λ n < c ˆ , I λ n ′ ( v n ) = 0 . As in the proof of Lemma 3.3, we obtain

(3.32) ‖ v n ‖ λ n ≤ C m λ n ≤ C c ˆ .

Hence, { v n } is bounded in E . Then, we may obtain, up to a subsequence, v n ⇀ v ˆ in E . Since

(3.33) ∫ D R v n 2 ≤ 1 λ n V 0 ∫ D R λ n V ( x ) v n 2 ≤ C 2 c ˆ 2 λ n V 0 → 0 , as λ n → + ∞ .

From the Hölder and Sobolev inequalities, (3.23), (3.32), and (3.33), one obtains

∫ B R c v n p = ∫ B R c ∣ v n ∣ 2 2 ∗ − p 2 ∗ − 2 ∫ B R c ∣ v n ∣ 2 ∗ p − 2 2 ∗ − 2 ≤ ∫ D R ∣ v n ∣ 2 + ∫ A R ∣ v n ∣ 2 2 ∗ − p 2 ∗ − 2 S − N 4 ( p − 2 ) ∫ B R c ∣ ∇ v n ∣ 2 N 4 ( p − 2 ) ≤ C 2 c ˆ 2 λ n V 0 + C 2 ∣ A R ∣ ( p − 2 ) ∕ p 2 ∗ − p 2 ∗ − 2 C 3 ‖ v n ‖ λ N 2 ( p − 2 ) → 0 ,

as R → + ∞ , λ n → + ∞ , and uniformly in n , where p ∈ ( 2 , 2 ∗ ) . Thus, as λ n → + ∞ ,

∫ B R c ∣ ∣ v n ∣ p − ∣ v ˆ ∣ p ∣ ≤ ∫ B R c ∣ v n ∣ p + ∫ B R c ∣ v ˆ ∣ p → 0 , as R → + ∞ .

Since v n → v ˆ in L loc p ( R N ) ( 2 ≤ p < 2 ∗ ) , we deduce

∫ ∣ x ∣ < R ∣ v n ∣ p → ∫ ∣ x ∣ < R ∣ v ˆ ∣ p .

Therefore, v n → v ˆ in L p ( R N ) as λ n → + ∞ . Let w n = v n − v ˆ , as the proof of Lemma 3.5, we may derive ‖ w n ‖ λ → 0 . Then, together with Lemma 3.3 and (3.32), we obtain

(3.34) λ n ∫ R N V ( x ) ∣ v n ∣ 2 ≤ ‖ v n ‖ λ n 2 ≤ C 2 c ˆ 2 .

Combining (3.34) with the Fatou lemma, we obtain ∫ R N V ( x ) ∣ v ˆ ∣ 2 = 0 . Thus, by ( V 3 ) , we deduce that v ˆ = 0 a.e. in x ∈ R N \ Ω and v ˆ ∈ H 0 1 ( Ω ) . The proof is complete.□

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Acknowledgment

The authors would like to express sincere thanks to the referees and the handling editor whose careful reading of the manuscript and valuable comments greatly improve the original manuscript.

Funding information: This research was funded by National Natural Science Foundation of China (Nos. 11971393, 11901499, and 11801465) and Nanhu Scholar Program for Young Scholars of XYNU (No. 201912).
Conflict of interest: Authors state no conflict of interest.
Data availability statement: No data, models, or code were generated or used during the study.

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Received: 2021-08-20

Revised: 2022-09-05

Accepted: 2022-09-20

Published Online: 2022-11-05

This work is licensed under the Creative Commons Attribution 4.0 International License.

Existence of ground state solutions for critical quasilinear Schrödinger equations with steep potential well

Abstract

1 Introduction and main result

Theorem 1.1

Theorem 1.2

Remark 1.3

Remark 1.4

2 Some preliminary results

Lemma 2.1

Definition 2.2

Lemma 2.3

Lemma 2.4

3 Proof of Theorems 1.1 and 1.2

Lemma 3.1

Proof

Lemma 3.2

Proof

Lemma 3.3

Proof

Lemma 3.4

Proof

Lemma 3.5

Proof

Proof of Theorem 1.1

Proof of Theorem 1.2

Acknowledgment

References

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