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Improved results on planar Klein-Gordon-Maxwell system with critical exponential growth

  • Lixi Wen EMAIL logo and Peng Jin EMAIL logo
Published/Copyright: August 23, 2024

Abstract

This work is concerned with the following Klein-Gordon-Maxwell system:

Δ u + V ( x ) u ( 2 ω + ϕ ) ϕ u = f ( u ) , x R 2 , Δ ϕ = ( ω + ϕ ) u 2 , x R 2 ,

where ω > 0 is a constant, u , ϕ : R 2 R , V C ( R 2 , R ) , and f C ( R , R ) obeys exponential critical growth in the sense of the Trudinger-Moser inequality. We give some new sufficient conditions on f , specifically related to exponential growth, to obtain the existence of nontrivial solutions. Our results improve and extend the previous results. In particular, we give a more precise estimation than the ones in the existing literature about the minimax level.

MSC 2010: 35J20; 35J62; 35Q55

1 Introduction

This study is concerned with the following Klein-Gordon-Maxwell system:

(1.1) Δ u + V ( x ) u ( 2 ω + ϕ ) ϕ u = f ( u ) , x R 2 , Δ ϕ = ( ω + ϕ ) u 2 , x R 2 ,

where ω > 0 is a constant, u , ϕ : R 2 R , V : R 2 R , and f : R R is of critical exponential growth in the sense of Trudinger-Moser inequality which is stated in the later part of this study. It is well known that this system arises from describing the interaction between the nonlinear Klein-Gordon field and the electrostatic field. The unknowns of system (1.1) u and ϕ are the fields associated with the particle and electric potential, respectively. For more details and physical backgrounds, we refer the readers to [5,6].

In recent years, the majority of the literature focuses on the study of the special problem

(1.2) u + [ m 0 2 ( ω + ϕ ) 2 ] u = u q 2 u , x R 3 , ϕ = ( ω + ϕ ) u 2 , x R 3 ,

and the first result is of the study of Benci and Fortunat [6] by using variational methods. When 0 < ω < m o and 4 < q < 6 , they showed the existence of infinitely many radially symmetric solutions for Problem (1.2). D’Aprile and Mugnai [21] complement the results in the case 0 < ω < ( q 2 ) 2 m 0 and 2 < q 4 . In another paper [22], they also proved that (1.2) only has a trivial solution when 0 < q 2 or q 6 by a Pohozaev-type argument. Based on the work of [6,21], Azzollini and Pomponio in [2] proved that Problem (1.2) has nontrivial solution if one of the following conditions holds:

  1. 4 q < 6 and 0 < ω < m 0 ;

  2. 2 < q < 4 and 0 < ω < ( q 2 ) ( 6 q ) m 0 .

Subsequently, the existence range of ( m o , ω ) for 2 < q < 4 was further extended in [3,32].

Furthermore, in (1.2), if m 0 2 ω 2 is replaced by a potential V ( x ) and u q 2 u by a continuous nonlinearity f ( x , u ) , it reduces to the following Klein-Gordon-Maxwell system:

(1.3) Δ u + V ( x ) u ( 2 ω + ϕ ) ϕ u = f ( x , u ) , x R 3 , Δ ϕ = ( ω + ϕ ) u 2 , x R 3 .

Recently, Carrião et al. [11] studied the existence of ground state solutions of the corresponding critical problem and supposing that V is positive and periodic. After that, multiplicity results for (1.3) were obtained by He [24] and Li and Tang [28], extending the results obtained by Carrião et al. [11]. Cunha [19] also studied System (1.3), proving that the existence of nontrivial solution under potential function V satisfies the periodic condition. When V is radially symmetric, Chen and Song [13] obtained two solutions of the nonhomogeneous version of system (1.3). For other related results, we can refer [9,15,16,20,26,31] and references therein.

Unlike Klein-Gordon-Maxwell systems on R 3 , there are very few papers dealing with (1.1). In this study, as in [1,12], we are also interested in a borderline case of the Sobolev embedding theorems, commonly known as the Trudinger-Moser case. Different from the case when N 3 , the case N = 2 is very special, as clearly the Sobolev exponent 2 * becomes and the corresponding Sobolev embedding yields H 1 ( R 2 ) L s ( R 2 ) for all s [ 1 , + ) , but H 1 ( R 2 ) L ( R 2 ) . In this case, the Trudinger-Moser inequality in R 2 can be seen as a substitute of the Sobolev inequality, which was first established by Cao in [10], see also [4,14], and reads as follows:

Lemma 1.1

  1. If α > 0 and u H 1 ( R 2 ) , then

    R 2 ( e α u 2 1 ) d x < ,

  2. if u H 1 ( R 2 ) , u 2 2 1 , u 2 M < , and α < 4 π , then there exists a constant C ( M , α ) , which depends only on M and α , such that

    R 2 ( e α u 2 1 ) d x C ( M , α ) .

Based on Lemma 1.1, we say f ( t ) has critical exponential growth at t = ± if it verifies

  1. f C ( R , R ) and there exists α 0 > 0 such that

    (1.4) lim t f ( t ) e α t 2 = 0 for all α > α 0

    and

    (1.5) lim t f ( t ) e α t 2 = + for all α < α 0 .

In the present work, we will restrict our discussion to (1.1) for the case that f ( u ) has critical exponential growth which is more involved since the embedding of the Sobolev space H 1 ( R 2 ) into the Orlicz space associated with the function φ ( s ) = exp ( 4 π s 2 ) 1 is not compact.

It seems that the first result dealing with Klein-Gordon-Maxwell systems with critical exponential growth is due to Albuquerque and Li [1]. Replacing f ( u ) by K ( x ) f ( u ) with K C ( R 2 , R ) , by using the Mountain Pass theorem, they obtained the existence of nontrivial solutions for (1.1) when V and K satisfy

  1. V C ( R 2 , R ) is positive and radial, and there exist constants a 0 , a > 2 such that

    lim ̲ x 0 V ( x ) x a 0 > 0 , lim ̲ x V ( x ) x a > 0 ,

  2. K C ( R 2 , R ) is positive and radial, and there exist constants b 0 > 2 and b < a such that

    lim ¯ x 0 K ( x ) x b 0 < , lim ¯ x K ( x ) x b < ,

and f satisfies (F1) and the following assumptions:
  1. f ( t ) = o ( t ) as t 0 ;

  2. there exists a constant μ 0 4 such that f ( t ) t μ 0 F ( t ) 0 for all t R ; and

  3. there exist ϱ 0 > 0 large enough and κ 0 > 2 such that F ( t ) ϱ 0 t κ 0 for all t R .

To overcome the difficulty arising from the critical growth of Trudinger-Moser type, (F4 ) is very crucial in [1]. In fact, using (F4 ), it is easy to show that the Mountain pass level c * can be sufficiently small, provided that ϱ 0 > 0 is sufficiently large. Note that (F4 ) is a global condition which requires f ( t ) to be super-linear for all t 0 and the coefficient ϱ 0 > 0 should be large. It seems to be reasonable for large t since f ( t ) is of critical exponential growth at infinity under (F1), but it puts a strict restriction on f ( t ) when t is small. From this point of view, the result in [1] does not reveal the essential features of the exponential growth given in (F1).

Recently, Chen et al. [12] obtained the existence of nontrivial solutions for (1.1), where V satisfies the following assumptions:

  1. V C ( R 2 , R ) , V ( x ) = V ( x ) for all x R 2 and inf u H 1 ( R 2 ) \ { 0 } R 2 [ u 2 + V ( x ) u 2 ] d x R 2 u 2 d x > 0 ;

  2. V C 1 ( R 2 , R ) and there exists a constant β 0 0 such that V ( x ) + 1 2 V ( x ) x β 0 for every x R 2 ;

and f satisfies (F1)–(F2) and the following assumptions:
  1. there exists a constant μ > 2 such that f ( t ) t μ F ( t ) 0 for all t R ;

  2. lim ̲ t t 2 F ( t ) e α 0 t 2 κ > 4 α 0 2 ρ 2 , where ρ > 0 such that ρ 2 [ max x ρ V ( x ) + ω 2 ] < 2 ; and

  3. there exist M 0 > 0 and t 0 > 0 such that

    F ( t ) M 0 f ( t ) , t t 0 .

Simultaneously, assumption ( F 4 ) plays a crucial role in [12] to obtain the following estimation:

c * < 2 π α 0 ,

which is essential to prove the nontriviality of the solution of (1.1). By using the extremal function sequences related to the Moser-Trudinger inequality, they give the optimal threshold value under which the Cerami sequence weakly converges to some nontrivial solution of (1.1). Their approach is totally different from that of [1] which depends on the sufficient large parametric perturbation.

Motivated by [12,18,27,29,34], in the present study, we shall further study the existence of nontrivial solutions for (1.1) by using some new analytical approaches. In detail, we introduce a new assumption:

  1. lim ̲ t t 2 F ( t ) e α 0 t 2 κ > 2 e α 0 2 min r > 0 r 2 e 1 2 r 2 ( V r + ω 2 ) , where V r max x r V ( x ) .

To state our results, in addition to (F1)–(F4), we continue to introduce the following assumptions:

  1. V C ( R 2 , R ) , inf u H 1 ( R 2 ) \ { 0 } R 2 [ u 2 + V ( x ) u 2 ] d x R 2 u 2 d x > 0 and there exists a constant r > 0 such that

    lim y meas { x R 2 : x y r , V ( x ) M } = 0 , M > 0 ;

  2. lim ̲ t t 2 F ( t ) e α 0 t 2 κ > λ + ω 2 α 0 2 , where λ > 0 is a constant; and

  3. F ( t ) 0 for all t R , and there exist μ > 2 and θ > 0 such that

    f ( t ) t μ F ( t ) + θ t 2 0 , t R .

Now, our main results can be stated as follows.

Theorem 1.2

Assume that V and f satisfy (V1), (V2) with β 0 > 0 , and (F1)–(F5). Let 2 < μ 3 in (F3). Then, (1.1) has a nontrivial solution if one of the following cases holds:

  1. μ = 3 and ω > 0 ; and

  2. 2 < μ < 3 and 0 < ω < β 0 ( μ 2 ) ( 4 μ 0 ) ( 3 μ 0 ) .

Theorem 1.3

Assume that V and f satisfy (V1), (V2) with β 0 = 0 , and (F1)–(F5). Let 3 < μ < 4 in (F3). Then, (1.1) has a nontrivial solution for all ω > 0 .

Theorem 1.4

Assume that V and f satisfy (V1) and (F1)–(F5). Let μ = 4 in (F3). Then, (1.1) has a nontrivial solution for all ω > 0 .

Theorem 1.5

Assume that V and f satisfy (V3), (F1), (F2), (F3 ), (F4), and (F5). Then, (1.1) has a nontrivial solution for all ω > 0 .

Remark 1.6

Clearly, (F4) is weaker than ( F 4 ) used in [12]. In fact, if ρ 0 > 0 such that ρ 0 2 [ max x ρ 0 V ( x ) + ω 2 ] < 2 , then

2 e α 0 2 min r > 0 r 2 e 1 2 r 2 ( V r + ω 2 ) 2 e 1 2 ρ 0 2 ( V ρ 0 + ω 2 ) e α 0 2 ρ 0 2 < 2 e e α 0 2 ρ 0 2 = 2 α 0 2 ρ 0 2 .

In our arguments, we adopt the approach used by Chen et al. [18] to estimate precisely the mountain pass level, see Lemma 2.3.

When V ( x ) = λ > 0 for all x R 2 in (1.1), it reduces to the following Klein-Gordon-Maxwell system:

(1.6) Δ u + λ u ( 2 ω + ϕ ) ϕ u = f ( u ) , x R 2 , Δ ϕ = ( ω + ϕ ) u 2 , x R 2 .

Corollary 1.7

  1. Assume that f satisfies (F1)–(F3), ( F4 ), and (F5). Let 2 < μ 3 in (F3). Then, (1.6) has a nontrivial solution if one of the following cases holds:

    1. μ = 3 and ω > 0 ; and

    2. 2 < μ < 3 and 0 < ω < β 0 ( μ 2 ) ( 4 μ 0 ) ( 3 μ 0 ) .

  2. Assume that f satisfies (F1)–(F3), ( F4 ), and (F5). Let 3 < μ < 4 in (F3). Then, (1.6) has a nontrivial solution for all ω > 0 .

  3. Assume that f satisfies (F1)–(F3), (F4 ), and (F5). Let μ = 4 in (F3). Then, (1.6) has a nontrivial solution for all ω > 0 .

The study is organized as follows. In Section 2, we give the variational setting and preliminaries. We give the proofs of Theorems 1.21.5 and Corollary 1.7 in Sections 3.

Throughout the study, we make use of the following notations:

  • H 1 ( R 2 ) denotes the usual Sobolev space equipped with the standard norm;

  • L s ( R 2 ) ( 1 s < ) denotes the Lebesgue space with the norm u s = R 2 u s d x 1 s ;

  • For any x R 2 and r > 0 , B r ( x ) { y R 2 : y x < r } and B r = B r ( 0 ) ;

  • C 1 , C 2 , denote positive constants possibly different in different places.

2 Variational framework and preliminaries

Under (V1) or (V3), we define the Hilbert space

E u H 1 ( R 2 ) : R 2 [ u 2 + V ( x ) u 2 ] d x < +

endowed with the norm u ( u , u ) induced by the scalar product

( u , v ) R 2 ( u v + V ( x ) u v ) d x 1 2 .

Then, the embedding E H 1 ( R 2 ) is continuous, i.e., there exists γ 0 > 0 such that

(2.1) γ 0 R 2 ( u 2 + u 2 ) d x u 2 , u E .

Under assumptions (V1) (or (V3)), (F1), and (F2), due to the variational characteristic of (1.1), its weak solutions ( u , ϕ ) E × H 1 ( R 2 ) are critical points of the functional given by

(2.2) J ( u , ϕ ) = 1 2 R 2 [ u 2 + V ( x ) u 2 ϕ 2 ( 2 ω + ϕ ) ϕ u 2 ] d x R 2 F ( u ) d x .

In order to avoid the difficulty originated by the strongly indefiniteness of the functional J , we apply a reduction method developed by [7], reducing the study of J to the study of a one variable functional that does not present such a strongly indefinite nature. To this end, we need the following technical results.

According to [1], for any u H 1 ( R 2 ) , there exists a unique ϕ = ϕ u H 1 ( R 2 ) which solves the equation

(2.3) Δ ϕ + u 2 ϕ = ω u 2 .

Moreover, the map I : u H 1 ( R 2 ) ϕ u H 1 ( R 2 ) is continuously differentiable, and

(2.4) ω ϕ u 0 on the set { x R 2 : u ( x ) 0 } .

It is easy to see that

(2.5) R 2 ϕ u 2 d x = R 2 ω ϕ u u 2 d x R 2 ϕ u 2 u 2 d x .

By the definition of J and using (2.5), we let

(2.6) Φ ( u ) J ( u , ϕ u ) = 1 2 R 2 [ u 2 + V ( x ) u 2 ] d x R 2 1 2 ω ϕ u u 2 + F ( u ) d x , u E .

Under assumptions (F1) and (F2), fix α > α 0 , we know that for any ε > 0 and any q 0 , there exists C α , ε , q > 0 such that

(2.7) f ( t ) ε t + C α , ε , q ( e α t 2 1 ) t q , t R ,

and

(2.8) F ( t ) ε t 2 + C α , ε , q ( e α t 2 1 ) t q + 1 , t R .

In view of Lemma 1.1, we can demonstrate that Φ C 1 ( E , R ) , and

(2.9) Φ ( u ) , v = R 2 ( u v + V ( x ) u v ) d x R 2 [ ( 2 ω + ϕ u ) ϕ u u + f ( u ) ] v d x , u , v E .

Following [6], ( u , ϕ u ) E × H 1 ( R 2 ) is a critical point for J if and only if u E is a critical point of Φ ; moreover, every critical point u E is a weak solution to (1.1), i.e., satisfy (1.1) with ϕ = ϕ u .

When V satisfies the radial condition (V1), to handle the lack of compactness, we restrict the working space to radial functions, in other words, we will consider the relevant problems in the following space:

E r { u E : u ( x ) = u ( x ) , x R 2 } ,

which is compactly embedded in L s ( R 2 ) for s > 2 .

To prove Theorems 1.2 and 1.3, we apply Jeanjean’s monotonicity trick [25], i.e., an approximation procedure to obtain a bounded (PS)-sequence for Φ , instead of starting directly from an arbitrary (PS)-sequence. For this purpose, for every λ [ 1 2 , 1 ] , we introduce a family of functional on E r defined by

(2.10) Φ λ ( u ) 1 2 u 2 1 2 R 2 ω ϕ u u 2 d x λ R 2 F ( u ) d x .

When V satisfies the coercive condition (V3), in view of [8, Lemma 3.1], [35], the embedding E L s ( R 2 ) is compact for any s 2 .

Lemma 2.1

[33] Assume that (V1), (F1), and (F2) hold. If u is a critical point of Φ restricted to E r , then u is a critical point of Φ on E .

Lemma 2.2

[12] Assume that (V1) and (V2) hold. Let ( u , ϕ u ) be a solution of (1.1) such that u , ϕ u H 1 ( R 2 ) and F ( u ) L 1 ( R 2 ) . Then, we have the following Pohozaev-type identity:

(2.11) P λ ( u ) R 2 V ( x ) + 1 2 V ( x ) x u 2 d x R 2 ( 2 ω + ϕ u ) ϕ u u 2 d x 2 λ R 2 F ( u ) d x = 0 .

Inspired by [17], to overcome the difficulty arising from the critical growth of Trudinger-Moser-type, with (F4), we control the minimax level in a suitable range where we can recover the compactness.

To this end, as in [23], we define Moser-type functions θ n ( x ) supported in B ρ ( 0 ) as follows:

(2.12) θ n ( x ) = 1 2 π log n , 0 x ρ n ; log ( ρ x ) log n , ρ n x ρ ; 0 , x ρ ,

where ρ > 0 satisfying ρ 2 e 1 2 ρ 2 ( V ρ + ω 2 ) = min r > 0 r 2 e 1 2 r 2 ( V r + ω 2 ) and V r max x r V ( x ) . Through a direct computation, we have

(2.13) R 2 θ n 2 d x = B ρ θ n 2 d x = ρ 2 δ n

and

(2.14) θ n 2 = R 2 ( θ n 2 + V ( x ) θ n 2 ) d x B ρ θ n 2 d x + V ρ B ρ θ n 2 d x = 1 + ρ 2 V ρ δ n ,

where

(2.15) δ n 1 4 log n 1 4 n 2 log n 1 2 n 2 > 0 .

for n 2 .

Lemma 2.3

Assume that (V1), (F1), (F2), and (F4) hold. Then, there exists n ¯ N such that

(2.16) max t 0 Φ λ ( t θ n ¯ ) < 2 π α 0 , λ [ 1 2 , 1 ] .

Proof

Note that κ > 2 e α 0 2 min r > 0 r 2 e 1 2 r 2 ( V r + ω 2 ) in (F4) and ρ 2 e 1 2 ρ 2 ( V ρ + ω 2 ) = min r > 0 r 2 e 1 2 r 2 ( V r + ω 2 ) , we can choose ε > 0 such that

(2.17) κ ε > 2 e α 0 2 ρ 2 e 1 2 ρ 2 ( V ρ + ω 2 )

and

(2.18) κ ε > 4 ( 1 + ε ) 3 2 e ( 2 ε ) ρ 2 ( V ρ + ω 2 ) 4 ( 2 ε ) e α 0 2 ρ 2 .

Using (F4), we know that there exists t ε > 0 such that

(2.19) t 2 F ( t ) ( κ ε ) e α 0 t 2 , t t ε .

By (2.13) and (2.14), we have

(2.20) Φ λ ( t θ n ) = t 2 2 θ n 2 t 2 2 R 2 ω ϕ t θ n θ n 2 d x λ R 2 F ( t θ n ) d x t 2 2 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] λ R 2 F ( t θ n ) d x .

There are four possible cases as follows. From now, in the sequel, all inequalities hold for large n N .

Case (i) t 0 , 2 π α 0 . Then, it follows from (F3), (2.20), and (2.13) that

Φ λ ( t θ n ) t 2 2 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] λ R 2 F ( t θ n ) d x , 1 + δ n ρ 2 ( V ρ + ω 2 ) 2 t 2 π α 0 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] = π α 0 + π ρ 2 ( V ρ + ω 2 ) 4 α 0 log n + O 1 log 2 n < 2 π α 0 .

Clearly, there exists n ¯ N such that (2.16) holds.

Case (ii) t 2 π α 0 , 4 π α 0 . Then, t θ n ( x ) t ε for x B ρ n and for large n N , it follows from (F3), (2.12), and (2.19) that

(2.21) R 2 F ( t θ n ) d x B ρ n F ( t θ n ) d x B ρ n ( κ ε ) e α 0 t 2 θ n 2 t 2 θ n 2 d x ( κ ε ) α 0 2 log n B ρ n e α 0 t 2 θ n 2 d x = π ( κ ε ) α 0 ρ 2 2 n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 2 n 2 log n 1 2 1 n ( 2 π ) 1 α 0 t 2 s 2 2 s d s π ( κ ε ) α 0 ρ 2 2 n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 2 n 2 log n 1 2 1 n ( 2 π ) 1 α 0 t 2 s 2 d s π ( κ ε ) α 0 ρ 2 2 n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 4 π α 0 t 2 ( n ( 2 π ) 1 α 0 t 2 n ( 4 π ) 1 α 0 t 2 ) π ( κ ε ) α 0 ρ 2 n 2 log n e ( 2 π ) 1 α 0 t 2 log n O 1 n log n .

It follows from (2.13), (2.20), and (2.21) that

(2.22) Φ ( t θ n ) t 2 2 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] λ R 2 F ( t θ n ) d x t 2 2 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] λ π ( κ ε ) α 0 ρ 2 n 2 log n e ( 2 π ) 1 α 0 t 2 log n + O 1 n log n φ n ( t ) + O 1 n log n .

Let t n > 0 such that φ n ( t n ) = 0 . Then,

(2.23) 1 + ρ 2 ( V ρ + ω 2 ) δ n = ( k ε ) α 0 2 ρ 2 2 n 2 e ( 2 π ) 1 α 0 t n 2 log n ,

which implies

(2.24) t n 2 = 4 π α 0 1 + log [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log λ ( κ ε ) α 0 2 ρ 0 2 2 log n .

Combining (2.23) and (2.24), we have

(2.25) φ n ( t ) φ n ( t n ) = [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] 2 t n 2 π α 0 log n + O 1 log 2 n .

From (2.15), (2.17), and (2.24), we have

(2.26) [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] t n 2 = [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] 4 π α 0 1 + log [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log 1 2 ( κ ε ) α 0 2 ρ 0 2 2 log n = 4 π α 0 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log 1 2 ( κ ε ) α 0 2 ρ 2 2 log n + O 1 log 2 n = 4 π α 0 1 + ρ 2 ( V ρ + ω 2 ) 2 log 1 2 ( κ ε ) α 0 2 ρ 2 4 log n + O 1 log 2 n .

Hence, by (2.17), (2.22), and (2.26), we have

(2.27) Φ λ ( t θ n ) = 2 π α 0 1 + ρ 2 ( V ρ + ω 2 ) 2 log 1 2 ( κ ε ) α 0 2 ρ 2 4 log n π α 0 log n + O 1 log 2 n = 2 π α 0 π α 0 log n log 1 2 ( κ ε ) α 0 2 ρ 2 1 2 ρ 2 ( V ρ + ω 2 ) + 1 + O 1 log 2 n = 2 π α 0 π α 0 log n log 1 2 ( κ ε ) α 0 2 ρ 2 e e 1 2 ρ 2 ( V ρ + ω 2 ) + O 1 log 2 n < 2 π α 0 .

Clearly, in this case, (2.27) implies that there exists n ¯ N such that (2.16) holds.

Case (iii) t 4 π α 0 , 4 π α 0 ( 1 + ε ) . Then, t θ n ( x ) t ε for x B ρ n and for large n N , it follows from (F3) and (2.19) that

(2.28) R 2 F ( t θ n ) d x B ρ n ( κ ε ) e α 0 t 2 θ n 2 t 2 θ n 2 d x ( κ ε ) α 0 2 ( 1 + ε ) log n B ρ n e α 0 t 2 θ n 2 d x = π ( κ ε ) α 0 ρ 2 2 ( 1 + ε ) n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 2 n 2 log n 1 2 1 n ( 2 π ) 1 α 0 t 2 s 2 2 s d s π ( κ ε ) α 0 ρ 2 2 ( 1 + ε ) n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 2 log n 1 ε 1 n [ ( 1 ε ) ( 2 π ) 1 α 0 t 2 + 2 ε ] s d s π ( κ ε ) α 0 ρ 2 2 ( 1 + ε ) n 2 log n e ( 2 π ) 1 α 0 t 2 log n + 1 1 + ε e [ ( 1 ε ) ( 2 π ) 1 α 0 t 2 + 2 ε ] log n O 1 n 2 ε 2 log n π ( κ ε ) α 0 ρ 2 ( 1 + ε ) 3 2 n 2 ε log n e ( 2 ε ) ( 4 π ) 1 α 0 t 2 log n O 1 n 2 ε 2 log n .

It follows from (2.13), (2.20), and (2.28) that

(2.29) Φ λ ( t θ n ) t 2 2 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] λ R 2 F ( t θ n ) d x t 2 2 [ 1 + ρ 2 ( V ρ + ω 2 ) δ n ] λ π ( κ ε ) α 0 ρ 2 ( 1 + ε ) 3 2 n 2 ε log n e ( 2 ε ) ( 4 π ) 1 α 0 t 2 log n + O 1 n 2 ε 2 log n ψ n ( t ) + O 1 n 2 ε 2 log n .

Let t ˆ n > 0 such that ψ n ( t ˆ n ) = 0 . Then,

(2.30) 1 + ρ 2 ( V ρ + ω 2 ) δ n = λ ( 2 ε ) ( k ε ) α 0 2 ρ 2 2 ( 1 + ε ) 3 2 n 2 ε e ( 2 ε ) ( 4 π ) 1 α 0 t ˆ n log n ,

which implies

(2.31) t ˆ n 2 = 4 π α 0 1 + log [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log [ λ ( κ ε ) ( 2 ε ) α 0 2 ρ 2 ] + log [ 2 ( 1 + ε ) 3 2 ] ( 2 ε ) log n 4 π α 0 1 + log [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log [ 1 2 ( κ ε ) ( 2 ε ) α 0 2 ρ 2 ] + log [ 2 ( 1 + ε ) 3 2 ] ( 2 ε ) log n .

Combining (2.29) with (2.31), we have

(2.32) ψ n ( t ) ψ n ( t ˆ n ) = t ˆ n 2 2 [ 1 + ρ 2 ( V ρ + ω 2 ) δ n ] 2 π α 0 ( 2 ε ) log n + O 1 log 2 n .

It follows from (2.18), (2.31), and (2.32) that

(2.33) Φ λ ( t θ n ) t ˆ n 2 2 [ 1 + ρ 2 ( V ρ + ω 2 ) δ n ] 2 π α 0 ( 2 ε ) log n + O 1 log 2 n 2 π α 0 [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] + [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] ( 2 ε ) log n [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] log 2 ( 1 + ε ) 3 2 1 2 ( κ ε ) ( 2 ε ) α 0 2 ρ 2 ( 2 ε ) log n 2 π α 0 ( 2 ε ) log n + O 1 log 2 n = 2 π α 0 1 + ρ 2 ( V ρ + ω 2 ) 4 log n + log 2 ( 1 + ε ) 3 2 1 2 ( κ ε ) ( 2 ε ) α 0 2 ρ 2 ( 2 ε ) log n 1 ( 2 ε ) log n + O 1 log 2 n = 2 π α 0 + 2 π α 0 log 2 ( 1 + ε ) 3 2 e 2 ε 4 ρ 2 ( V ρ + ω 2 ) 1 2 ( κ ε ) ( 2 ε ) e α 0 2 ρ 2 ( 2 ε ) log n + O 1 log 2 n < 2 π α 0 .

Clearly, in this case, (2.33) implies that there exists n ¯ N such that (2.16) holds.

Case (iv) t 4 π α 0 ( 1 + ε ) , + . Then, t θ n ( x ) t ε for x B ρ n and for large n N , it follows from (2.4), (2.10), (2.13), (2.14), and (2.19) that, for large n

Φ λ ( t θ n ) = t 2 2 θ n 2 t 2 2 R 2 ω ϕ t θ n θ n 2 d x λ R 2 F ( t θ n ) d x 1 + δ n ρ 2 ( V ρ + ω 2 ) 2 t 2 π 2 ( κ ε ) ρ 2 n 2 t 2 log n e α 0 2 π t 2 log n 1 + δ n ρ 2 ( V ρ + ω 2 ) 2 t 2 π 2 ( κ ε ) ρ 2 t 2 log n exp 2 α 0 4 π t 2 1 log n 4 π [ 1 + δ n ρ 2 ( V ρ + ω 2 ) ] 2 α 0 ( 1 + ε ) α 0 π ( κ ε ) ρ 2 4 ( 1 + ε ) log n e 2 ε log n π α 0 , t 4 π α 0 ( 1 + ε ) , + ,

where to obtain the third inequality, we use the fact that the function

1 + δ n ρ 2 ( V ρ + ω 2 ) 2 t 2 π 2 ( κ ε ) ρ 2 t 2 log n exp 2 α 0 4 π t 2 1 log n

is increasing on t 4 π α 0 ( 1 + ε ) , + for large n N , since its stagnation points tend to 4 π α 0 as n . The above three cases show that there exists n ¯ N such that (2.16) holds.□

Lemma 2.4

[12] Assume that (V1), (F1), and (F2) hold. Then, there exists positive constants σ 0 and T 0 independent of λ such that for all λ [ 1 2 , 1 ] ,

(2.34) max { Φ λ ( 0 ) , Φ λ ( T 0 θ n ¯ ) } < σ 0 c λ inf γ Γ ˜ max t [ 0 , 1 ] Φ λ ( γ ( t ) ) ,

where θ n ¯ is given by Lemma 2.3 and

(2.35) Γ ˜ λ = { γ C ( [ 0 , 1 ] , E r ) : γ ( 0 ) = 0 , γ ( 1 ) = T 0 θ n ¯ } .

Lemma 2.5

[12] Assume that (V1), (F1), (F2), and (F4) hold. Then, c λ < 2 π α 0 for all λ [ 1 2 , 1 ] , where c λ is given by (2.34).

Lemma 2.6

[23,30] Assume that (F1) and (F2) hold. Let v n v ¯ in E and

(2.36) R 2 f ( v n ) v n d x K 0

for some constant K 0 > 0 . Then, for any ϕ C 0 ( R 2 )

(2.37) lim n R 2 f ( v n ) ϕ d x = R 2 f ( v ¯ ) ϕ d x .

Lemma 2.7

Assume that V satisfies (V1), and f satisfies (F1), (F2), (F4), and (F5). Then, c λ is nonincreasing on λ [ 1 2 , 1 ] . Moreover, for almost every λ [ 1 2 , 1 ] , there exists u λ E r \ { 0 } such that

(2.38) Φ λ ( u λ ) = 0 a n d Φ λ ( u λ ) c λ .

Proof

Using Lemma 2.4 and applying Jeanjean’s monotonicity trick [25], c λ is nonincreasing on λ [ 1 2 , 1 ] , and for almost every λ [ 1 2 , 1 ] , there exists a sequence { u n ( λ ) } E r (for simplicity, we denote it by { u n } instead of { u n ( λ ) } ) such that

(2.39) u n 2 C 0 , Φ λ ( u n ) c λ > 0 , Φ λ ( u n ) E r * 0 .

Then, there exists u λ E r such that, up to a subsequence, u n u λ in E r , u n u λ in L s ( R 2 ) for s > 2 , and u n u λ a.e. in R 2 . Now, we divide the proof into the following three steps.

Step 1. Proved that Φ λ ( u λ ) = 0 . Since Φ λ ( u n ) , u n 0 , by (2.4) and (2.39), we have

(2.40) 1 2 R 2 f ( u n ) u n d x λ R 2 f ( u n ) u n d x = R 2 [ u n 2 + V ( x ) u 2 ] d x R 2 [ ( 2 ω + ϕ u n ) ϕ u n u n ] d x + o ( 1 ) C 1 .

Hence, for any ϕ C 0 ( R 2 ) , we can deduce from Lemma 2.6 that

(2.41) Φ λ ( u λ ) , ϕ = lim n Φ λ ( u n ) , ϕ = 0 .

This implies that Φ λ ( u λ ) = 0 .

Step 2. Proved that u λ 0 . Arguing by the contradiction, suppose that u λ = 0 . For any given ε ( 0 , 2 M 0 C 1 t 0 ) , we choose M ε > 2 M 0 C 1 ε , then it follows from (F5) and (2.40) that

(2.42) u n M ε F ( u n ) d x M 0 u n M ε f ( u n ) d x M 0 M ε u n M ε f ( u n ) u n d x < ε .

Note that u n 2 2 C 0 γ 0 , where γ 0 > 0 is the embedding constant given by (2.1). Using (F2) and (F3), we can choose N ε ( 0 , 1 ) such that

(2.43) u n N ε f ( u n ) u n d x ε γ 0 C 0 u n N ε u n 2 d x ε γ 0 C 0 u n 2 2 < ε

and

(2.44) u n N ε F ( u n ) d x ε γ 0 C 0 u n N ε u n 2 d x ε γ 0 C 0 u n 2 2 < ε .

By (F1), we have

(2.45) N ε u n M ε F ( u n ) d x C 2 u n 3 3 = o ( 1 ) , N ε u n 1 f ( u n ) u n d x C 3 u n 3 3 = o ( 1 ) .

Due to the arbitrariness of ε ( 0 , 2 M 0 C 1 t 0 ) , (2.42), (2.44), and (2.45) imply that

(2.46) R 2 F ( u n ) d x = o ( 1 )

and

(2.47) u n 1 f ( u n ) u n d x = o ( 1 ) .

From (2.4), (2.10), (2.39), and Lemma 2.5, we deduce that there exists ε ¯ > 0 such that

(2.48) u n 2 2 c λ + o ( 1 ) 4 π α 0 ( 1 3 ε ¯ ) + o ( 1 ) .

Now, we choose q ( 1 , 2 ) such that

(2.49) ( 1 + ε ¯ ) ( 1 3 ε ¯ ) q 1 ε ¯ < 1 .

By (F1), there exists C 4 > 0 such that

(2.50) f ( t ) q C 4 [ e α 0 ( 1 + ε ¯ ) q t 2 1 ] , t 1 .

Then, (2.48), (2.49), (2.50), and Lemma 1.1(ii) imply that

(2.51) u n 1 f ( u n ) q d x C 4 R 2 [ e α 0 ( 1 + ε ¯ ) q u n 2 1 ] d x = C 4 R 2 [ e α 0 ( 1 + ε ¯ ) q u n 2 ( u n u n ) 2 1 ] d x C 5 .

Let q = q ( q 1 ) . By (2.51) and the Hölder inequality, we have

(2.52) u n 1 f ( u n ) u n d x u n 1 f ( u n ) q d x 1 q u n q = o ( 1 ) .

Then, from (2.47) and (2.52), we have

(2.53) R 2 f ( u n ) u n d x = o ( 1 ) .

By (2.10), (2.39), (2.46), and (2.53), we have

c λ + o ( 1 ) = Φ λ ( u n ) 1 2 Φ λ ( u n ) , u n = 1 2 ϕ u n 2 2 + λ R 2 1 2 f ( u n ) u n F ( u n ) d x o ( 1 ) ,

which contradicts the fact c λ > 0 . Therefore, u λ 0 .

Step 3. Proved that Φ λ ( u λ ) c λ . For any given ε ( 0 , M 0 t 0 ) , it follows from (F5) and (2.40) that

(2.54) u n M 0 ε 1 F ( u n ) d x M 0 u n M 0 ε 1 f ( u n ) d x ε u n M 0 ε 1 f ( u n ) u n d x 2 C 1 ε .

By (F1) and (F2), there exists δ ε > 0 such that

(2.55) u n δ ε F ( u n ) d x ε u n δ ε u n 2 d x ε u n 2 2 C 0 γ 0 ε .

Similarly to (2.54) and (2.55), we have

(2.56) u λ M 0 ε 1 F ( u λ ) d x 2 C 1 ε

and

(2.57) u λ δ ε F ( u λ ) d x C 0 γ 0 ε .

Then, (2.54) and (2.56) give

(2.58) u n M 0 ε 1 F ( u n ) F ( u λ ) d x u n M 0 ε 1 F ( u n ) d x + u n M 0 ε 1 F ( u λ ) d x u n M 0 ε 1 F ( u n ) d x + u λ M 0 ε 1 F ( u λ ) d x + o ( 1 ) 4 C 1 ε + o ( 1 ) .

Combining (2.55) with (2.57), one obtains

(2.59) u n δ ε F ( u n ) F ( u λ ) d x u n δ ε F ( u n ) d x + u n δ ε F ( u λ ) d x u n δ ε F ( u n ) d x + u λ δ ε F ( u λ ) d x + o ( 1 ) 2 C 0 γ 0 ε + o ( 1 ) .

For ε > 0 small enough, using (F1) and (F2), we can choose a constant C ε > 0 such that

(2.60) F ( t ) C ε t 3 , δ ε t M 0 ε 1 .

Noting that u n u λ in L 3 ( R 2 ) , by Lemma [33, Lemma A.1], there exists w 0 L 3 ( R 2 ) such that

(2.61) u n ( x ) w 0 ( x ) , u λ ( x ) w 0 ( x ) , a.e. x R 2 .

It is easy to check that there exists R ε > 0 such that

(2.62) R 2 \ B R ε F ( u λ ) d x < ε , C ε R 2 \ B R ε w 0 3 d x < ε .

By (2.60), (2.61), and (2.62), we have

(2.63) ( δ ε u n M 0 ε 1 ) ( R 2 \ B R ε ) F ( u n ) F ( u λ ) d x ( δ ε u n M 0 ε 1 ) ( R 2 \ B R ε ) F ( u n ) d x + R 2 \ B R ε F ( u λ ) d x C ε R 2 \ B R ε w 0 3 d x + R 2 \ B R ε F ( u λ ) d x < 2 ε .

Note that both (F1) and (F2) imply

(2.64) F ( u n ) F ( u λ ) χ { δ ε u n M 0 ε 1 } ( x ) C ε w 0 3 + F ( u λ ) , x B R ε .

Since u n u λ a.e. in R 2 , it follows from (2.64) and Lebesgue’s control convergent theorem that

(2.65) lim n ( u n M 0 ε 1 ) B R ε F ( u n ) F ( u λ ) d x = 0 .

Due to the arbitrariness of ε > 0 , from (2.58), (2.59), (2.63), and (2.65), we have

(2.66) lim n R 2 F ( u n ) F ( u λ ) d x = 0 ,

which implies that

(2.67) lim n R 2 F ( u n ) d x = R 2 F ( u λ ) d x .

By (2.4), (2.10), (2.39), (2.67), Fatou’s lemma, and the weak lower semicontinuity of the norm, we have

(2.68) Φ λ ( u λ ) = 1 2 u λ 2 1 2 R 2 ω ϕ u λ u λ 2 d x λ R 2 F ( u λ ) d x 1 2 lim ̲ n u n 2 + 1 2 lim ̲ n R 2 ω ( ϕ u n ) u n 2 d x λ lim n R 2 F ( u n ) d x = lim ̲ n 1 2 u n 2 1 2 R 2 ω ϕ u n u n 2 d x λ R 2 F ( u n ) d x = lim ̲ n Φ λ ( u n ) = c λ .

This shows that (2.38) holds.□

Lemma 2.8

[12] Assume that (V1), (F1), and (F2) hold. If u ¯ λ E r \ { 0 } satisfies Φ λ ( u ¯ λ ) = 0 for λ [ 1 2 , 1 ] , then there exists σ 0 > 0 , independent of λ , such that u ¯ λ σ 0 .

3 Proofs of results

Proof of Theorem 1.2

In view of Lemmas 2.2 and 2.7, there exist two sequences of { u λ n } E r (denoted it by { u n } ) and { λ n } [ 1 2 , 1 ] such that

(3.1) λ n 1 , Φ λ n ( u n ) = 0 , Φ λ n ( u n ) c λ n c 1 2 , P λ n ( u n ) = 0 .

Let 2 < μ 3 in (F3). By (V2), (F3), (2.4), (2.10), (2.11), and (3.1), we have

(3.2) c 1 2 c λ n Φ λ n ( u n ) = Φ λ n ( u n ) 1 2 Φ λ n ( u n ) , u n + μ 2 4 P λ n ( u n ) = μ 2 4 R 2 V ( x ) + 1 2 V ( x ) x u n 2 d x + R 2 3 μ 2 ω ϕ u n + 4 μ 4 ϕ u n 2 u n 2 d x + λ n ( μ + 2 ) 4 μ R 2 ( f ( u n ) u n μ F ( u n ) ) d x R 2 μ 2 4 β 0 + 3 μ 2 ω ϕ u n + 4 μ 4 ϕ u n 2 u n 2 d x β 0 ( μ 2 ) ( 4 μ ) ( 3 μ ) 2 ω 2 4 ( 4 μ ) u n 2 2 ,

where the last inequality follows from the fact that

μ 2 4 β 0 + 3 μ 2 ω t + 4 μ 4 t 2 β 0 ( μ 2 ) ( 4 μ ) ( 3 μ ) 2 ω 2 4 ( 4 μ ) , ω t 0 .

When μ = 3 , ω > 0 , or 2 < μ < 3 , 0 < ω < β 0 ( μ 2 ) ( 4 μ 0 ) 3 μ 0 , one has β 0 ( μ 2 ) ( 4 μ ) ( 3 μ ) 2 ω 2 > 0 . Therefore, (3.2) yields the boundedness of { u n 2 } . Using (F3), (2.4), (2.10), (2.11), and (3.1) again, we obtain

(3.3) c 1 2 c λ n Φ λ n ( u n ) = Φ λ n ( u n ) 1 μ Φ λ n ( u n ) , u n = 1 2 1 μ u n 2 1 2 R 2 ω ϕ u n u n 2 d x + 1 μ R 2 ( 2 ω + ϕ u n ) ϕ u n u n 2 d x + R 2 1 μ f ( u n ) u n F ( u n ) d x 1 2 1 μ u n 2 4 μ 2 μ ω 2 u n 2 2 ,

which, together with the boundedness of { u n 2 } , implies that { u n } is bounded in E r .

Hence, there exists u ¯ E r such that, up to a subsequence, u n u ¯ in E r , u n u ¯ in L s ( R 2 ) for s ( 2 , ) , and u n u ¯ a.e. in R 2 . By (F3) and Φ λ n ( u n ) , u n = 0 , we can easily check that

(3.4) R 2 f ( u n ) u n d x C 1 .

For any ϕ C 0 ( R 2 ) , we can deduce from (3.1) and Lemma 2.6 that

lim n Φ λ n ( u n ) , ϕ = Φ ( u ¯ ) , ϕ .

This implies that Φ ( u ¯ ) E r = 0 . Applying Lemma 2.1, we have Φ ( u ¯ ) = 0 .

Next, we prove that u ¯ 0 . Arguing by contradiction, suppose that u ¯ = 0 . Since Φ λ n ( u n ) , u n = 0 , it follows from (2.5) and (2.53) that

(3.5) u n 2 u n 2 + R 2 ( 2 ϕ u n 2 + ϕ u n 2 u n 2 ) d x = λ n R 2 f ( u n ) u n d x = o ( 1 ) .

From (3.1) and Lemma 2.8, we obtain u n σ 0 > 0 , which contradicts (3.5). Hence, u ¯ 0 . The proof is completed.□

Proof of Theorem 1.3

Similarly, in view of Lemmas 2.2 and 2.7, there exist two sequences of { u λ n } E r (denoted by { u n } ) and { λ n } [ 1 2 , 1 ] such that

(3.6) λ n 1 , Φ λ n ( u n ) = 0 , Φ λ n ( u n ) c λ n c 1 2 , P λ n ( u n ) = 0 .

By (V2), (F3), (2.10), (2.11), and (3.1), we have

(3.7) c 1 2 c λ n Φ λ n ( u n ) = Φ λ n ( u n ) 1 2 ( μ 2 ) Φ λ n ( u n ) , u n + 4 μ 4 ( μ 2 ) P λ n ( u n ) = μ 3 2 ( μ 2 ) R 2 [ u n 2 + V ( x ) u n 2 ] d x + 4 μ 4 ( μ 2 ) R 2 V ( x ) + 1 2 V ( x ) x u n 2 d x + 1 4 R 2 ϕ u n 2 u n 2 d x + λ n μ 4 μ ( μ 2 ) R 2 [ f ( u n ) u n μ F ( u n ) ] d x μ 3 2 ( μ 2 ) u n 2 .

Therefore, { u n } is bounded in E r . Then, there exists u ¯ E r such that, up to a subsequence, u n u ¯ in E r , u n u ¯ in L s ( R 2 ) for s > 2 , and u n u ¯ a.e. in R 2 . Proceeding as in the proof of Theorem 1.2, we can deduce that u ¯ 0 and Φ ( u ¯ ) E r = 0 . Applying Lemma 2.1, we have Φ ( u ¯ ) = 0 . The proof is completed.□

Proof of Theorem 1.4

By (2.8), we have for some constants α > α 0 and C 2 > 0

(3.8) F ( t ) γ 0 4 t 2 + C 2 ( e α t 2 1 ) t 3 , t R ,

where γ 0 > 0 is given by (2.1). In view of Lemma 1.1(ii), we have

(3.9) R 2 ( e 2 α u 2 1 ) d x = R 2 ( e 2 α u 2 ( u u ) 2 1 ) d x C ( γ 0 , 2 π ) , u π α .

From (3.8), (3.9), and the Sobolev embedding theorem, we obtain

(3.10) R 2 F ( u ) d x γ 0 4 u 2 2 + C 2 R 2 ( e α u 2 1 ) u 3 d x γ 0 4 u 2 2 + C 2 R 2 ( e 2 α u 2 1 ) d x 1 2 u 6 3 1 4 u 2 + C 3 u 3 , u π α .

Then, (2.4), (2.10), and (3.10) give

(3.11) Φ ( u ) 1 4 u 2 C 3 u 3 , u π α .

Therefore, there exist κ 0 > 0 and 0 < ρ < π α such that

(3.12) Φ ( u ) κ 0 , u S { u E r : u = ρ } .

Using (F1)–(F3), we can check easily that there exists k 1 , k 2 > 0 such that

(3.13) F ( t ) k 1 t μ k 2 t 2 , t R .

For θ n ¯ , which is given by Lemma 2.3, from (2.13), (2.14), and (3.13), we have

Φ ( t θ n ¯ ) = t 2 2 θ n ¯ 2 1 2 R 2 ω ϕ t θ n ¯ ( t θ n ¯ ) 2 d x R 2 F ( t θ n ¯ ) d x t 2 2 ( 1 + ω 2 ) θ n ¯ 2 2 k 1 t μ θ n ¯ μ μ + k 2 t 2 θ n ¯ 2 2 .

Hence, lim t Φ ( t θ n ¯ ) = , and so there exists T 0 > 0 such that Φ ( T 0 θ n ¯ ) < 0 .

Applying the Mountain pass theorem, we can deduce that there exists a sequence { u n } E r such that

(3.14) Φ ( u n ) c > 0 , Φ ( u n ) E r * ( 1 + u n ) 0 ,

where

c = inf γ Γ max t [ 0 , 1 ] Φ ( γ ( t ) ) , Γ = { γ C ( [ 0 , 1 ] , E r ) : γ ( 0 ) = 0 , γ ( 1 ) = T 0 θ n ¯ } .

In view of Lemma 2.3, we have c < 2 π α 0 . By (F3), (2.6), (2.9), and (3.14), we have

c = Φ ( u n ) + o ( 1 ) = Φ ( u n ) 1 4 Φ ( u n ) , u n = 1 4 R 2 [ u n 2 + V ( x ) u n 2 ] d x + 1 4 R 2 ϕ u n 2 u n 2 d x + R 2 1 4 f ( u n ) u n F ( u n ) d x 1 4 u n 2 .

This shows that { u n } is bounded in E r . Then, there exists u ¯ E r such that, up to a subsequence, u n u ¯ in E r , u n u ¯ in L s ( R 2 ) for s ( 2 , ) , and u n u ¯ a.e. in R 2 . Proceeding as in the proof of Lemma 2.7, we can deduce that u ¯ 0 , and Φ ( u ¯ ) E r = 0 . Applying Lemma 2.1, we have Φ ( u ¯ ) = 0 . The proof is completed.□

Proof of Theorem 1.5

Note that (F1), (F2), and (F3 ) yield that there exist k 3 , k 4 > 0 such that

(3.15) F ( t ) k 3 t μ + θ μ 2 t 2 k 4 t 2 , t R .

From (2.13), (2.14), and (3.15), we have

Φ ( t θ n ¯ ) = t 2 2 θ n ¯ 2 1 2 R 2 ω ϕ t θ n ¯ ( t θ n ¯ ) 2 d x R 2 F ( t θ n ¯ ) d x t 2 2 ( 1 + ω 2 ) θ n ¯ 2 2 k 3 t μ θ n ¯ μ μ θ μ 2 t 2 θ n ¯ 2 2 + k 4 t 2 θ n ¯ 2 2 ,

hence lim t Φ ( t θ n ¯ ) = . Similarly to the proof of Theorem 1.4, we can deduce that there exists a sequence { u n } E such that

(3.16) Φ ( u n ) c > 0 , Φ ( u n ) E * ( 1 + u n ) 0 ,

where

c = inf γ Γ max t [ 0 , 1 ] Φ ( γ ( t ) ) , Γ = { γ C ( [ 0 , 1 ] , E ) : γ ( 0 ) = 0 , γ ( 1 ) = T 0 θ n ¯ } .

In view of Lemma 2.3, we have c < 2 π α 0 . Now, we show that { u n } is bounded in E . To this end, arguing by contradiction, suppose that u n . Let v n = u n u n , then v n = 1 . Passing to a subsequence, we may assume that v n v in E , v n v in L s ( R 2 ) for s 2 , and v n v a.e. in R 2 . Since ω ϕ u n 0 , it follows from (F3), (2.6), (2.9), and (3.16) that

(3.17) c + o ( 1 ) = Φ ( u n ) 1 μ Φ ( u n ) , u n = 1 2 1 μ u n 2 + R 2 2 μ 1 2 ω ϕ u n + 1 μ ϕ u n 2 u n 2 d x + R 2 1 μ f ( u n ) u n F ( u n ) d x 1 2 1 μ u n 2 R 2 2 μ ω 2 + θ μ u n 2 d x .

Multiplying (3.17) by 1 u n 2 , we deduce from μ > 2 and v n v in L 2 ( R 2 ) that

(3.18) 2 μ ω 2 + θ μ v 2 2 = 2 μ ω 2 + θ μ lim n v n 2 2 1 2 1 μ > 0 .

This shows that v 0 . For a.e. x { y R 2 : v ( y ) 0 } , we have lim n u n ( x ) = . Using (F1)-(F3), it is easy to check that lim t F ( t ) t 2 = . Thus, it follows from (2.6) and (3.16) and Fatou’s lemma that

(3.19) 0 = lim n c + o ( 1 ) u n 2 = lim n Φ ( u n ) u n 2 = 1 2 lim ̲ n ω 2 u n 2 R 3 ϕ u n u n 2 d x lim ̲ n R 2 F ( u n ) u n 2 v n 2 d x 1 2 + lim ̲ n ω 2 u n 2 2 2 u n 2 R 2 lim ̲ n F ( u n ) u n 2 v n 2 d x = ,

which is a contradiction. Hence, { u n } is bounded in E . Then, there exists u ¯ E such that, up to a subsequence, u n u ¯ in E , u n u ¯ in L s ( R 2 ) for s 2 , and u n u ¯ a.e. in R 2 . Proceeding as in the proof of Lemma 2.7, we can deduce that u ¯ 0 and Φ ( u ¯ ) = 0 . The proof is completed.□

Proof of Corollary 1.7

Let g ( r ) r 2 e 1 2 r 2 ( λ + ω 2 ) . Then, it is easy to see that

g ( r ) = 2 r 3 e 1 2 r 2 ( λ + ω 2 ) + r 2 e 1 2 r 2 ( λ + ω 2 ) r ( λ + ω 2 ) .

Let r 0 > 0 such that g ( r 0 ) = 0 . Then, r 0 2 = λ + ω 2 2 . Hence, g ( r ) achieves its minimum at r 0 > 0 , i.e.,

min r > 0 g ( r ) = g ( r 0 ) = r 0 2 e 1 2 r 0 2 ( λ + ω 2 ) = λ + ω 2 2 e .

Note that

κ > λ + ω 2 α 0 2 = 2 e α 0 2 λ + ω 2 2 e = 2 e α 0 2 r 0 2 e 1 2 r 0 2 ( λ + ω 2 ) = 2 e α 0 2 min r > 0 r 2 e 1 2 r 2 ( λ + ω 2 ) ,

then similar to the proof of Theorems 1.21.4, we can prove Corollary 1.7.□

Acknowledgements

The authors thank Professor Xianhua Tang for his guidance and advice in the process of writing this article.

  1. Funding information: Authors state no funding involved.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. LXW wrote the first draft of this article and PJ revised it and wrote the final version.

  3. Conflict of interest: The authors state no conflict of interest.

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Received: 2023-12-27
Revised: 2024-05-08
Accepted: 2024-06-10
Published Online: 2024-08-23

© 2024 the author(s), published by De Gruyter

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

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