Home Mountain-pass-type solutions for Schrödinger equations in R2 with unbounded or vanishing potentials and critical exponential growth nonlinearities
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Mountain-pass-type solutions for Schrödinger equations in R2 with unbounded or vanishing potentials and critical exponential growth nonlinearities

  • Xiaoyan Lin EMAIL logo , Xianhua Tang and Ning Zhang
Published/Copyright: May 7, 2024

Abstract

In this article, we consider the existence of solutions for nonlinear elliptic equations of the form

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

where the nonlinear term f ( s ) has critical exponential growth which behaves like e α s 2 , the radial potentials V , Q : R + R are unbounded, singular at the origin or decaying to zero at infinity. By combining the variational methods, Trudinger-Moser inequality, and some new approaches to estimate precisely the minimax level of the energy functional, we prove the existence of a Mountain-pass-type solution for the above problem under some weak assumptions.

MSC 2010: 35J20; 58E50

1 Introduction

In the present study, we consider the existence of solutions for nonlinear elliptic equations of the form

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

where V , Q , and f satisfy the following assumptions:

  1. V C ( ( 0 , + ) , ( 0 , + ) ) , and there exist a 0 , a > 2 such that

    limsup r 0 + V ( r ) r a 0 < + , liminf r + V ( r ) r a > 0 ;

  2. Q C ( ( 0 , + ) , ( 0 , + ) ) , and there exists b 0 > 2 such that

    limsup r 0 + Q ( r ) r b 0 < + , limsup r + Q ( r ) V ( r ) = 0 ;

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

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

    and

    (1.3) lim t f ( t ) e α t 2 = + , for all  α < α 0 ;

  4. f ( t ) = o ( t ) as t 0 ;

  5. t f ( t ) > 0 for all t 0 , and there exist M 0 > 0 and t 0 > 0 such that

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

In the above assumptions, (V1) and (Q1) imply that the potentials V , Q : R + R are unbounded, singular at the origin or decaying to zero at infinity.

When V ( x ) V ˆ > 0 and Q ( x ) 1 , we derive the following autonomous equation from (1.1):

(1.4) Δ u + V ˆ u = f ( u ) , x R 2 .

There are a lot of work on the existence of solutions for (1.4) in the literature, for example, see [26,8,12,1517], in which besides (F1)–(F3) and the following well-known Ambrosetti-Rabinowitz (AR)-type condition:

  1. there exists θ > 2 such that

    f ( t ) t θ F ( t ) > 0 , t 0 .

Some additional growth conditions on the nonlinearity f are also required. In particular, two different types of growth conditions are usually used in the above-mentioned work. The first one [5] prescribes the growth of f near the origin:
  1. there exists p > 2 such that f ( t ) t λ t p for all t R , where

    λ > α 0 ( p 2 ) 4 π p p 2 2 C p p ,

    where

    (1.5) C p inf u H 1 ( R 2 ) u 2 2 + V ˆ u 2 2 u p .

The second growth condition [15] prescribes an asymptotic behavior at infinity:
  1. there holds

    (1.6) lim t f ( t ) t e α 0 t 2 > 2 e V ˆ α 0 .

In recent articles [911], (1.6) has been weaken. However, there seems to be no better condition than (1.5) in the literature.

To study the existence of solutions of (1.1) via the classical Mountain-pass theorem, we will frame the variational study of (1.1) in the weighted space:

E u L loc 2 ( R 2 ) : u is radial , u L 2 ( R 2 ) and R 2 V ( x ) u 2 d x < + .

And we define

( u , v ) R 2 [ u v + V ( x ) u v ] d x , u , v E ; u 2 ( u , u ) u E .

Then E is a Hilbert space on the above inner product ( , ) . Furthermore, for s [ 1 , + ) , we define

L s ( R 2 , Q ) u : R 2 R : u is mensurable , R 2 Q ( x ) u s d x < .

Ambrosetti et al. [7] and do Ó et al. [13] studied equation (1.1) by assuming that V and Q satisfy the following assumption:

  1. there are A 1 , A 2 , A 3 > 0 , 0 < β 1 < 2 , and β 2 > 2 such that

    A 1 1 + x β 1 V ( x ) A 2 , 0 < Q ( x ) A 3 1 + x β 2 .

Obviously, when V and Q are radial, ( V Q ) implies (V1) and (Q1). Instead of (Q1) by using the following assumption:
  1. Q C ( ( 0 , + ) , ( 0 , + ) ) , and there exist b 0 > 2 and b < ( a 2 ) 2 such that

    limsup r 0 + Q ( r ) r b 0 < + , limsup r + Q ( r ) r b < + .

Albuquerque et al. [1] proved the following theorem.

Theorem 1.1

[1] Assume that V and Q satisfy (V1) and (Q1 ), and that f satisfies (F1)–(F3), AR-type condition, and the following assumption:

  1. there exists p > 2 such that F ( t ) μ p t p for all t R , where

    μ > α 0 ( p 2 ) α p p 2 2 S p p ,

    where α min { 4 π , 2 π ( 2 + b 0 ) } , and

    (1.7) S τ inf u E \ { 0 } u R 2 Q ( x ) u τ d x 1 τ , τ [ 2 , + ) .

Then (1.1) has a positive weak solution.

Obviously, ( F 4 ) in Theorem 1.1 extends (H4); moreover, it is very crucial to overcome the obstacles caused by the critical exponential term (see [1, Lemma 3.3]). AR-type condition plays a crucial role to prove the boundedness of (Palais-Smale) sequences for the energy functional associated (1.1).

Since b < ( a 2 ) 2 and a + 2 > 0 in (V1) and ( Q 1 ), it is easy to verify that (V1) and ( Q 1 ) imply (Q1). However, there are many functions V ( r ) and Q ( r ) satisfying (V1) and (Q1) but not ( Q 1 ). For example,

  1. Let V ( r ) r θ 1 and Q ( r ) r θ 2 . Then V and Q satisfy (V1) and (Q1) when θ 1 > θ 2 > 2 but not ( Q 1 ) when θ 2 + 1 > θ 1 2 ;

  2. Let V ( r ) log ( 1 + r ) and Q ( r ) r σ . Then V and Q satisfy (V1) and (Q1) when σ ( 2 , 0 ] but not ( Q 1 ) when σ [ 1 , 0 ] .

In the present study, we shall develop some delicate analyses to control the Mountain-pass minimax level c * by a fine threshold ( 2 + b 0 ) π α 0 under weaker assumptions on f . Furthermore, we will use the stronger version of Trudinger-Moser inequality established in [14] to restore the compactness for Cerami sequence under the fine threshold.

To state our main results, we first introduce the following assumptions:

  1. there exists p > 2 such that F ( t ) μ p t p for all t R , where

    μ > α 0 ( p 2 ) 2 π ( 2 + b 0 ) p p 2 2 S p p ;

  2. there exist q > p > 2 , μ > 0 , λ > 0 , and ν [ 0 , 1 ] such that

    (1.8) F ( t ) μ p t p + λ q t q , t R

    and

    (1.9) ( p 2 ) ν p ( p 2 ) 2 p μ 2 ( p 2 ) + ( q 2 ) ( 1 ν ) q ( q 2 ) 2 q λ 2 ( q 2 ) S 2 4 ( q p ) ( p 2 ) ( q 2 ) < ( 2 + b 0 ) π α 0 S p 2 p p 2 ;

  3. there exist q > p > 2 , μ > 0 , λ > 0 , and ν ( 0 , 1 ] such that (1.8) and the following inequality hold:

    (1.10) μ > ν p 2 1 1 ν λ 2 q q q 2 ( q 2 ) α 0 S p 2 p p 2 2 q ( 2 + b 0 ) π S 2 4 ( q p ) ( p 2 ) ( q 2 ) p 2 2 α 0 ( p 2 ) 2 π ( 2 + b 0 ) p p 2 2 S p p ;

  4. t f ( t ) 2 F ( t ) for all t R .

It is easy to see that (1.10) is equivalent to (1.9). For equation (1.4), we give the following assumption:

  1. there exist q > p > 2 , ν ( 0 , 1 ] , μ > 0 , and λ > 0 such that (1.8) and the following inequality hold:

    (1.11) μ > ν p 2 1 1 ν λ 2 q q q 2 ( q 2 ) α 0 C p 2 p p 2 4 π q V ˆ 2 ( q p ) ( p 2 ) ( q 2 ) p 2 2 α 0 ( p 2 ) 4 π p p 2 2 C p p .

Our main results are the following two theorems.

Theorem 1.2

Assume that V , Q , and f satisfy (V1), (Q1), and (F1)–(F4). Then (1.1) has a nontrivial solution u ¯ . Furthermore, if (F7) also holds, then u ¯ is a Mountain-pass-type solution.

Theorem 1.3

Assume that V , Q , and f satisfy (V1), (Q1), (F1)–(F3), and (F5) (or (F6)). Then (1.1) has a nontrivial solution u ¯ . Furthermore, if (F7) also holds, then u ¯ is a Mountain-pass-type solution.

Since

V ˆ u 2 u 2 2 + V ˆ u 2 2 , u H 1 ( R 2 ) ,

S 2 V ˆ , so we can derive (1.10) with V ( x ) V ˆ from (1.9). Therefore, we have the following theorem by the proof [5, Corollary 1.5].

Theorem 1.4

Assume that V ˆ > 0 and f satisfy (F1)–(F3) and (H6). Then (1.4) has a positive weak solution u ¯ .

Example 1.5

Set

F ( t ) = κ ( e t 2 1 t 2 ) 1 + t 4 , κ > 0 .

Then

(1.12) f ( t ) = F ( t ) = 2 κ t ( t 2 1 ) 2 ( e t 2 1 ) + 4 κ t 5 ( 1 + t 4 ) 2 .

It is easy to see that f ( t ) satisfies (F1)–(F3) and (F7) with α 0 = 1 . Furthermore, we have

F ( t ) κ t 4 8 + κ t 12 8 ! , t R .

This shows that (1.8) holds with p = 4 , q = 12 , μ = κ 2 , and λ = 12 κ 8 ! . Hence, if

κ > S 4 4 2 π ( 2 + b 0 ) ,

then (F4) holds. In addition, (1.11) equivalent to

κ 2 > ν 1 ( 1 ν ) 6 5 8 ! 12 κ 1 5 5 C 4 4 2 π V ˆ 4 5 C 4 4 8 π

for some ν ( 0 , 1 ) , which can be derived from the following inequality:

(1.13) κ > min 0 < s < 32 ( π V ˆ ) 4 3125 ( 4 7 ! ) C 4 16 1 s 1 ( 4 7 ! ) 1 5 5 C 4 16 5 2 ( π V ˆ ) 4 5 s 6 5 C 4 4 4 π .

Therefore, if (1.13) holds, then f ( t ) defined by (1.12) also satisfies (H4), but does not satisfy (H5). It is worth to point out that a continuous function f ( t ) has the above-mentioned behavior, which was constructed in [17, Proposition 2.9], expressed in complicated series form.

Example 1.6

Set

(1.14) ζ ( t ) t 2 log 1 + 1 t 2 , t 0 ; 0 , t = 0

and

F ( t ) = ζ ( t ) + κ ( e t 2 1 t 2 ) 1 + t 2 , κ > 0 .

Then

(1.15) f ( t ) = F ( t ) = 2 κ t 3 e t 2 ( 1 + t 2 ) 2 + 2 t ( 1 + t 2 ) log 1 + 1 t 2 + 1 ( 1 + t 2 ) log 2 1 + 1 t 2 , t 0 ; 0 , t = 0 .

It is easy to see that f ( t ) satisfies (F1)–(F3) and (F7) with α 0 = 1 . Furthermore, we have

F ( t ) κ t 4 3 , t R .

This shows that (F4) holds with p = 4 , μ = 4 κ 3 , and

κ > 3 S 4 4 16 π ( 2 + b 0 ) .

However, f ( t ) does not satisfy AR-type condition for any θ > 2 .

The article is organized as follows. In Section 2, we give the variational setting and some preliminary lemmas. Section 3 is devoted to the estimation of the energy functional associated with (1.1) where a fine threshold is established by a careful analysis. In Section 4, we complete the proofs of Theorems 1.2 and 1.3.

Throughout the sequel, we denote the usual Lebesgue space with norm u s = R 2 u s d x 1 s by L s ( R 2 ) , where 1 s < , B r { x R 2 : x < r } for all r > 0 , and C denotes different positive constants in different places.

2 Variational framework and preliminaries

Lemma 2.1

[14] Assume that (V1) and (Q1) hold. Then the embeddings E L s ( R 2 , Q ) are continuous and compact for all 2 s < . Therefore, there exists γ s > 0 such that

(2.1) R 2 Q ( x ) u s d x 1 s γ s u , u E , 2 s < .

Lemma 2.2

[14] Assume that V and Q satisfy (V1) and (Q1). Then the following conclusions hold.

  1. If α > 0 and u E , then

    R 2 Q ( x ) ( e α u 2 1 ) d x < ;

  2. If α < 2 π ( 2 + b 0 ) , then there exists a constant C > 0 such that

    sup u E , u 1 R 2 Q ( x ) ( e α u 2 1 ) d x C .

In view of Lemmas 2.1 and 2.2, it is easy to see that the energy functional associated with problem (1.1) Φ : E R by

(2.2) Φ ( u ) = 1 2 R 2 [ u 2 + V ( x ) u 2 ] d x R 2 Q ( x ) F ( u ) d x

is well defined under (V1), (Q1), (F1), and (F2), and by using standard arguments, we can show that Φ C 1 ( E , R ) with derivative given by

(2.3) Φ ( u ) , φ = R 2 [ u φ + V ( x ) u φ ] d x R 2 Q ( x ) f ( u ) φ d x

for all φ E . Now we say that u E is a weak solution to problem (1.1), if for all φ C 0 ( R 2 ) it holds that Φ ( u ) , φ = 0 .

Finally, let

X u L loc 2 ( R 2 ) : u L 2 ( R 2 ) and R 2 V ( x ) u 2 d x < + .

Then X is a Hilbert space when endowed with inner product

( u , v ) X = R 2 [ u v + V ( x ) u v ] d x .

Lemma 2.3

[14] Assume that (V1), (Q1), (F1), and (F2) hold. If u ¯ is a critical point of Φ restricted to E, then u ¯ is a critical point of Φ on X .

Lemma 2.4

Assume that (V1), (Q1), (F1), and (F2) hold. Then there exists a sequence { u n } E satisfying

(2.4) Φ ( u n ) c * , Φ ( u n ) ( 1 + u n ) 0 ,

where c * is given by

(2.5) c * = inf γ Γ max t [ 0 , 1 ] Φ ( γ ( t ) ) ,

Γ = { γ C ( [ 0 , 1 ] , E ) : γ ( 0 ) = 0 , Φ ( γ ( 1 ) ) < 0 } .

Proof

Let γ s be defined by (2.1). By (F1) and (F2), there exists C 1 > 0 such that

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

In view of Lemma 2.2(ii), we have

(2.7) R 2 Q ( x ) ( e 4 α 0 u 2 1 ) d x C 2 , u π ( b 0 + 2 ) 3 α 0 .

From (2.1), (2.6), and (2.7), we obtain

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

Hence, it follows from (2.2) and (2.8) that

(2.9) Φ ( u ) = 1 2 u 2 R 2 Q ( x ) F ( u ) d x 1 4 u 2 C 3 u 3 , u π ( b 0 + 2 ) 3 α 0 .

Therefore, there exist κ 0 > 0 and 0 < ρ 0 < π ( b 0 + 2 ) 3 α 0 such that

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

Now we choose w 0 E \ { 0 } , it is easy to show that lim t Φ ( t w 0 ) = due to (F1). Hence, we can choose T > 0 such that e T w 0 { u E : u > ρ 0 } and Φ ( e ) < 0 , then in view of the Mountain-pass lemma, we deduce that there exists a sequence { u n } E satisfying (2.4).□

Lemma 2.5

Assume that (V1), (Q1), and (F1)–(F3) hold. Then any sequence { u n } satisfying (2.4) is bounded in E .

Proof

To prove the boundedness of { u n } , arguing by contradiction, suppose that u n as n . Let v n = u n u n . Then 1 = v n 2 . In view of Lemma 2.1, without loss of generality, we may assume

(2.11) R 2 Q ( x ) v n 2 d x R 2 Q ( x ) v ¯ 2 d x

for some v ¯ E . By (F1)–(F3), there exists C 1 > 0 such that

(2.12) f ( t ) t 4 F ( t ) + C 1 t 2 0 , t R .

From (2.2), (2.3), (2.4), and (2.12), one has

c * + o ( 1 ) = Φ ( u n ) 1 4 Φ ( u n ) , u n = 1 4 u n 2 + 1 4 R 2 Q ( x ) [ f ( u n ) u n 4 F ( u n ) ] d x 1 4 u n 2 C 1 4 R 2 Q ( x ) u n 2 d x ,

which together with (2.11) implies

(2.13) R 2 Q ( x ) v ¯ 2 d x = lim n R 2 Q ( x ) v n 2 d x 1 C 1 .

Hence from (F1), (2.2), (2.4), and (2.13), we have

1 2 = lim n R 2 Q ( x ) F ( u n ) u n 2 v n 2 d x = + .

This contradiction shows that { u n } is bounded in E .□

Lemma 2.6

[14] Assume that (Q1), (F1)–(F3) hold. Let u n u ¯ in E and

(2.14) R 2 Q ( x ) f ( u n ) u n d x K 0

for some constant K 0 > 0 . Then there hold:

  1. for every φ E C 0 ( R 2 )

    (2.15) lim n R 2 Q ( x ) f ( u n ) φ d x = R 2 Q ( x ) f ( u ¯ ) φ d x ;

  2. (2.16) lim n R 2 Q ( x ) F ( u n ) d x = R 2 Q ( x ) F ( u ¯ ) d x .

3 Minimax estimates

In this section, we give a precise estimation about the minimax level c * defined by Lemma 2.4.

Lemma 3.1

Assume that (V1), (Q1), and (F1)–(F4) hold. Then

(3.1) c * < ( 2 + b 0 ) π α 0 .

Proof

By the definition of S p , there exists { w n } E such that

(3.2) S p p w n p R 2 Q ( x ) w n p d x < S p p + 1 n , n N .

From (F4) and (2.2), we have

(3.3) Φ ( t w n ) = t 2 2 w n 2 R 2 Q ( x ) F ( t w n ) d x t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x .

Set

(3.4) t n w n 2 μ R 2 Q ( x ) w n p d x 1 ( p 2 ) .

Then it follows that

(3.5) t n 2 w n 2 = μ t n p R 2 Q ( x ) w n p d x .

From (3.2), (3.3), (3.4), and (3.5), we have

(3.6) Φ ( t w n ) t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x t n 2 2 w n 2 μ t n p p R 2 Q ( x ) w n p d x = p 2 2 p μ 2 ( p 2 ) w n p R 2 Q ( x ) w n p d x 2 p 2 = p 2 2 p μ 2 ( p 2 ) S p 2 p p 2 + o ( 1 ) .

By (F4), (2.5), and (3.6), there exists n ¯ N such that

(3.7) c * max t 0 Φ ( t w n ¯ ) < ( 2 + b 0 ) π α 0 .

This shows (3.1) holds.□

Lemma 3.2

Assume that (V1), (Q1), (F1)–(F3), and (F5) hold. Then

(3.8) c * < ( 2 + b 0 ) π α 0 .

Proof

By the definition of S p , there exists { w n } E such that (3.2) holds. From (1.7) and the Hölder inequality, one has

(3.9) R 2 Q ( x ) w n p d x R 2 Q ( x ) w n 2 d x q p q 2 R 2 Q ( x ) w n q d x p 2 q 2 ( S 2 2 w n 2 ) q p q 2 R 2 Q ( x ) w n q d x p 2 q 2 ,

which implies

(3.10) R 2 Q ( x ) w n q d x ( S 2 2 w n 2 ) p q p 2 R 2 Q ( x ) w n p d x q 2 p 2 .

From (1.8), (2.2), and (3.10), we have

(3.11) Φ ( t w n ) = t 2 2 w n 2 R 2 Q ( x ) F ( t w n ) d x t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x λ t q q R 2 Q ( x ) w n q d x t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x λ t q q ( S 2 2 w n 2 ) p q p 2 R 2 Q ( x ) w n p d x q 2 p 2 = ν t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x + ( 1 ν ) t 2 2 w n 2 λ t q q ( S 2 2 w n 2 ) p q p 2 R 2 Q ( x ) w n p d x q 2 p 2 = ϕ n ( t ) + ψ n ( t ) ,

where

(3.12) ϕ n ( t ) ν t 2 2 w n 2 μ t p p R 2 Q ( x ) w n p d x

and

(3.13) ψ n ( t ) ( 1 ν ) t 2 2 w n 2 λ t q q ( S 2 2 w n 2 ) p q p 2 R 2 Q ( x ) w n p d x q 2 p 2 .

Set

(3.14) t n ν w n 2 μ R 2 Q ( x ) w n p d x 1 p 2

and

(3.15) t ˜ n 1 ν λ S 2 2 ( q p ) ( p 2 ) 1 q 2 w n 2 R 2 Q ( x ) w n p d x 1 p 2 .

Then it follows from (3.2), (3.11)–(3.14), and (3.15) that

(3.16) Φ ( t w n ) ϕ n ( t ) + ψ n ( t ) ϕ n ( t n ) + ψ n ( t ˜ n ) = ν ( p 2 ) 2 p t n 2 w n 2 + ( 1 ν ) ( q 2 ) 2 q t ˜ n 2 w n 2 = ( p 2 ) ν p ( p 2 ) 2 p μ 2 ( p 2 ) + ( q 2 ) ( 1 ν ) q ( q 2 ) 2 q λ 2 ( q 2 ) S 2 4 ( q p ) ( p 2 ) ( q 2 ) w n p R 2 Q ( x ) w n p d x 2 p 2 = ( p 2 ) ν p ( p 2 ) 2 p μ 2 ( p 2 ) + ( q 2 ) ( 1 ν ) q ( q 2 ) 2 q λ 2 ( q 2 ) S 2 4 ( q p ) ( p 2 ) ( q 2 ) S p 2 p p 2 + o ( 1 ) .

By (1.9), (2.5), and (3.16), there exists n ¯ N such that

(3.17) c * max t 0 Φ ( t w n ¯ ) < ( 2 + b 0 ) π α 0 .

This shows (3.8) holds.□

4 Proofs of main results

In this section, we give the proofs of Theorems 1.2 and 1.3.

Proof of Theorem 1.2

Applying Lemmas 2.4 and 2.5, we deduce that there exists a sequence { u n } E satisfying (2.4) and u n C 1 for some constant C 1 > 0 . It follows from (2.3) and (2.4) that

(4.1) R 2 Q ( x ) f ( u n ) u n d x C 2 .

Since u n C 1 , by Lemma 2.1, we may thus assume, passing to a subsequence if necessary, that u n u ¯ in E , u n u ¯ in L s ( R 2 , Q ) for s [ 2 , ) and u n u ¯ a.e. on R 2 . If u ¯ = 0 , then u n 0 in L s ( R 2 , Q ) for s [ 2 , ) . Hence by (Q1), Lemma 2.6, we have

(4.2) lim n R 2 Q ( x ) F ( u n ) d x = 0 .

Hence, it follows from (2.2), (2.4), (3.1), and (4.2) that

(4.3) u n 2 2 c * + o ( 1 ) 2 ( b 0 + 2 ) π α 0 ( 1 3 ε ¯ ) + o ( 1 ) .

Lemma 3.1 implies that ε ¯ > 0 . Now we choose p 1 ( 1 , 2 ) such that

(4.4) ( 1 + ε ¯ ) ( 1 3 ε ¯ ) p 1 1 ε ¯ < 1 .

By (F1), there exists C 3 > 0 such that

(4.5) f ( t ) p 1 C 3 [ e α 0 ( 1 + ε ¯ ) p 1 t 2 1 ] , t 1 .

It follows from (4.3), (4.4), (4.5), and Lemma 2.2(ii) that

(4.6) u n 1 Q ( x ) f ( u n ) p 1 d x C 3 R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) p 1 u n 2 1 ] d x = C 3 R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) p 1 u n 2 ( u n u n ) 2 1 ] d x C 4 .

Let p 2 = p 1 ( p 1 1 ) . Then p 2 > 2 . Hence it follows from (F1), (F2), u n 0 in L s ( R 2 , Q ) for s 2 , (4.6) and the Hölder inequality that

(4.7) R 2 Q ( x ) f ( u n ) u n d x = u n < 1 Q ( x ) f ( u n ) u n d x + u n 1 Q ( x ) f ( u n ) u n d x C 5 u n < 1 Q ( x ) u n 2 d x + u n 1 Q ( x ) f ( u n ) p 1 d x 1 p 1 u n 1 Q ( x ) u n p 2 d x 1 p 2 = o ( 1 ) .

From (2.2), (2.3), (2.4), (4.2), and (4.7), we have

(4.8) c * + o ( 1 ) = Φ ( u n ) 1 2 Φ ( u n ) , u n = R 2 Q ( x ) 1 2 f ( u n ) u n F ( u n ) d x = o ( 1 ) .

This contradiction shows that u ¯ 0 . In view of Lemma 2.6, we have

(4.9) lim n R 2 Q ( x ) f ( u n ) φ d x = R 2 Q ( x ) f ( u ¯ ) φ d x , φ E C 0 ( R 2 ) .

By (2.4) and (4.9), it is easy to deduce that Φ ( u ¯ ) = 0 . Furthermore, Lemma 2.3 shows that u ¯ is a nontrivial weak solution to problem (1.1).

Next, we prove that Φ ( u ¯ ) = c * . In fact, it is sufficient to prove that

(4.10) A 2 lim n u n 2 = u ¯ 2 .

By (2.3), (2.4), and the fact that Φ ( u ¯ ) = 0 , we obtain

(4.11) 0 = lim n Φ ( u n ) , u n = lim n u n 2 R 2 Q ( x ) f ( u n ) u n d x = A 2 lim n R 2 Q ( x ) f ( u n ) u n d x

and

(4.12) R 2 [ u ¯ φ + V ( x ) u ¯ φ ] d x = R 2 Q ( x ) f ( u ¯ ) φ d x , φ E C 0 ( R 2 ) .

Note that u ¯ E , there exists { φ n } E C 0 ( R 2 ) such that

(4.13) R 2 Q ( x ) φ n u ¯ 2 d x = o ( 1 ) .

Using (4.13) together with (F1), (F2), and Lemma 2.2(i), we obtain

(4.14) R 2 Q ( x ) f ( u ¯ ) ( φ n u ¯ ) d x R 2 Q ( x ) f ( u ¯ ) φ n u ¯ d x C 6 R 2 Q ( x ) [ u ¯ + ( e 2 α 0 u ¯ 2 1 ) ] φ n u ¯ d x C 6 R 2 Q ( x ) u ¯ 2 d x 1 2 + R 2 Q ( x ) ( e 4 α 0 u ¯ 2 1 ) d x 1 2 R 2 Q ( x ) φ n u ¯ 2 d x 1 2 = o ( 1 ) .

It follows from (4.12) and (4.14) that

(4.15) u ¯ 2 = R 2 [ u ¯ 2 + V ( x ) u ¯ 2 ] d x = lim n R 2 [ u ¯ φ n + V ( x ) u ¯ φ n ] d x = lim n R 2 Q ( x ) f ( u ¯ ) φ n d x = R 2 Q ( x ) f ( u ¯ ) u ¯ d x ,

which, together with (4.11), implies

(4.16) lim n R 2 Q ( x ) [ f ( u n ) u n f ( u ¯ ) u ¯ ] d x = A 2 u ¯ 2 .

In view of Lemma 2.6, we have

(4.17) lim n R 2 Q ( x ) F ( u n ) d x = R 2 Q ( x ) F ( u ¯ ) d x .

From (F7), (2.2), (2.3), (2.4), (4.16), and (4.17), we have

(4.18) c * = lim n Φ ( u n ) 1 2 Φ ( u n ) , u n = 1 2 lim n R 2 Q ( x ) [ f ( u n ) u n 2 F ( u n ) ] d x = 1 2 lim n R 2 Q ( x ) f ( u n ) u n d x R 2 Q ( x ) F ( u ¯ ) d x 1 2 lim n R 2 Q ( x ) [ f ( u n ) u n f ( u ¯ ) u ¯ ] d x = 1 2 ( A 2 u ¯ 2 ) = 1 2 lim n u n u ¯ 2 .

Hence, there exist ε ¯ > 0 and n 1 N such that

(4.19) u n u ¯ 2 4 π α 0 ( 1 3 ε ¯ ) , n n 1 .

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

(4.20) ( 1 + ε ¯ ) 2 ( 1 3 ε ¯ ) q 1 2 1 ε ¯ < 1 .

It follows from (4.19), (4.20), the Young’s inequality, and Lemma 2.2(ii) that

(4.21) R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) q 1 u n 2 1 ] d x R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) 2 ε ¯ 1 q 1 u ¯ 2 e α 0 ( 1 + ε ¯ ) 2 q 1 ( u n u ¯ ) 2 1 ] d x q 1 1 q 1 R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) 2 ε ¯ 1 q 1 2 ( q 1 1 ) 1 u ¯ 2 1 ] d x + 1 q 1 R 2 Q ( x ) [ e α 0 ( 1 + ε ¯ ) 2 q 1 2 ( u n u ¯ ) 2 1 ] d x C 7 .

Let q 2 = q 1 ( q 1 1 ) . Then for any ε > 0 , by (F1), (F2), (4.21), and the Hölder inequality, we obtain

(4.22) R 2 Q ( x ) f ( u n ) ( u n u ¯ ) d x R 2 Q ( x ) [ u n + C 8 ( e α 0 ( 1 + ε ¯ ) u n 2 1 ) ] u n u ¯ d x R 2 Q ( x ) u n 2 d x 1 2 R 2 Q ( x ) u n u ¯ 2 d x 1 2 + C 8 R 2 Q ( x ) ( e α 0 ( 1 + ε ¯ ) q 1 u n 2 1 ) d x 1 q 1 R 2 Q ( x ) u n u ¯ q 2 d x 1 q 2 o ( 1 ) ,

which implies

(4.23) R 2 Q ( x ) f ( u n ) ( u n u ¯ ) d x = o ( 1 ) .

Therefore, it follows from (2.3), (2.4), and (4.23) that

o ( 1 ) = Φ ( u n ) , u n u ¯ = R 2 [ u n ( u n u ¯ ) + V ( x ) u n ( u n u ¯ ) ] d x R 2 Q ( x ) f ( u n ) ( u n u ¯ ) d x = u n 2 R 2 ( u n u ¯ + V ( x ) u n u ¯ ) d x + o ( 1 ) = A 2 u ¯ 2 + o ( 1 ) ,

which implies that A 2 = u ¯ 2 .□

By using Lemma 3.2 instead of Lemma 3.1, we can prove Theorem 1.3 by the same arguments as the proof of Theorem 1.2.

  1. Funding information: This work was partially supported by the National Natural Science Foundation of China (No: 11971485) and Scientific Research Fund of Hunan Provincial Education Department (18A452) of China [18A452].

  2. Conflict of interest: Authors state no conflict of interest.

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Received: 2023-09-12
Revised: 2023-11-24
Accepted: 2024-03-20
Published Online: 2024-05-07

© 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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