Startseite Infinitely many solutions for quasilinear Schrödinger equations with sign-changing nonlinearity without the aid of 4-superlinear at infinity
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Infinitely many solutions for quasilinear Schrödinger equations with sign-changing nonlinearity without the aid of 4-superlinear at infinity

  • Mustapha Khiddi EMAIL logo und Lakbir Essafi
Veröffentlicht/Copyright: 22. November 2022
Veröffentlichen auch Sie bei De Gruyter Brill

Abstract

In this article, we will prove the existence of infinitely many solutions for a class of quasilinear Schrödinger equations without assuming the 4-superlinear at infinity on the nonlinearity. We achieve our goal by using the Fountain theorem.

MSC 2010: 35J20; 35J60; 35Q55

1 Introduction

In this article, we are concerned with quasilinear Schrödinger equation of the form

(P) Δ u + V ( x ) u Δ ( u 2 ) u = λ g ( x , u ) in R N , u H 1 ( R N ) ,

where g ( x , u ) C ( R N × R , R ) is q -superlinear with 2 < q < min ( 4 , p ) , 2 < p < 2 2 2 = 2 N N 2 , N 3 , λ is positive parameter, and V ( x ) C ( R N , R + ) L ( R N ) .

We would like to mention that quasilinear Schrödinger equation (P) arises in various branches of mathematical physics and has been derived as models of several physics phenomenon see [1,2,3]. The quasilinear problem has been studied extensively in recent years with a huge variety of conditions on the potential V and the nonlinearity g , see, for example, [4,5, 6,7,8] and references therein.

Alves et al. [9] assumed nonlinearity to be 4-superlinear and that the potential V is continuous and satisfies:

( V 1 ) V C ( R N , R ) and V ( x ) V ( 0 ) > 0 , for all x R N .

( V 2 ) lim x + V ( x ) = V and V ( x ) V , for all x R N .

Under these two conditions, they combined the variational method with perturbation arguments to obtain a solution. In [10,11], the authors obtained the multiplicity of solutions and made an assumption on V and also the nonlinearity to be 4-superlinear, where they have used variational methods. Most of the works requires the condition on the nonlinearity to be 4-superlinear in that sense p > 4 . This condition that makes the application of the Fountain theorem is fluid. Indeed, if 4 < p < 2 2 , the boundedness of the Cerami sequences of the energy functional associated with the problem (P) follows by compactness embedding of H 1 ( R N ) into L p 2 ( R N ) . So, the energy functional satisfies the Cerami condition for any c . Consequently, we obtain a set of unbounded of critical values. In contrast, if the nonlinear term g ( x , u ) changes its sign and without the aid of condition 4 < p < 2 2 , we will have difficulties to obtain the delamination of the Cerami sequences, as well as the convergence, to complete the hypotheses of the Fountain theorem [12]. To the best of our knowledge, there are only a few recent papers that deal with the nonlinearity to be not 4-superlinear and the sign-changing potential case for problem (P). Motivated by papers [10,11], in the present article, we shall consider problem (P) with more generally on the nonlinearity of g ( x , u ) . Namely, when powers of the nonlinearity are 2 < q < min ( 4 , p ) , 2 < p < 2 2 and can be allowed to sign-changing. By a dual approach and Fountain theorem, we establish the existence of infinitely many solutions. We state below our hypotheses on the potential V and on the nonlinearity g ( x , u ) in the subcritical case:

  1. V C ( R N , R ) L ( R N ) and 0 < V 0 = inf x R N V ( x ) .

  2. ( V ( x ) ) 1 L 1 ( R N ) .

  3. g C ( R N × R , R ) such that g ( x , u ) = g ( x , u ) for all x , u R N × R , there exit positive functions ξ L 4 q ( R N ) L 2 ( R N ) , 4 q is the conjugate of 4 q and τ L ( R N ) L 2 ( R N ) such that

    g ( x , t ) ξ ( x ) t q 1 + τ ( x ) t p 1 , ( x , t ) R N × R .

  4. Let G ( x , t ) = 0 t g ( x , u ) d u , lim t + G ( x , t ) t q = + uniformly in x , 2 < q < 4 and there exists r 0 > 0 such that G ( x , t ) 0 ( x , t ) R N × R , t r 0 .

  5. There exists c 0 > 0 and δ 0 > 0 such that

    G ( x , t ) t 2 p p 2 c 0 G ˜ ( x , t ) for all ( x , t ) R N × R \ [ δ 0 , δ 0 ] ,

    where G ˜ ( x , t ) = 1 μ g ( x , t ) t G ( x , t ) is the positive function and q < μ < p .

The main result is the following:

Theorem 1.1

Let 2 < p < 2 2 and 2 < q < min ( 4 , p ) . Suppose that ( V 1 ) , ( V 2 ) , ( G 1 ) , ( G 2 ) , and ( G 3 ) are satisfied. Then for all λ > 0 , the problem (P) has infinitely many solutions.

Remark 1

Example of nonlinearity meets all requirements of Theorem 1.1.

Let 2 < q = 2.5 < 4 , 4 < p = 4.5 < 2 2 , and q < μ = 4 < p .

V ( x ) = e x 2 + 1 and g ( x , u ) = 1 + sin 2 x 1 + x N u 5 2 u 1 + cos 2 x 1 + x N u 1 2 u .

We take

ξ ( x ) = 2 1 + x 2 L 8 3 ( R N ) L 2 ( R N ) , τ ( x ) = 1 + sin 2 x 1 + x 2 L ( R N ) L 2 ( R N ) .

So,

g ( x , u ) 2 1 + x N u 3 2 + 1 + sin 2 x 1 + x N u 7 2 ,

by the definition given for G ( x , u ) , we have

G ( x , u ) = 2 9 1 + sin 2 x 1 + x N u 9 2 2 5 1 + cos 2 x 1 + x N u 5 2 1 5 ( x , u ) R N × R , u 1 ,

as well as

G ˜ ( x , u ) = 3 20 1 + cos 2 x 1 + x N u 5 2 + 1 36 1 + sin 2 x 1 + x N u 9 2 0 ,

G ( x , u ) u 2 9 5 G ˜ ( x , u ) 36 2 9 9 5 > 0 .

Notation. In this article, we make use of the following notation:

  • C , C 0 , C 1 , C 2 , denote positive (possibly different) constants.

  • For 2 s < , the usual Lebesgue space is endowed with the norm

    u s = u L s ( R N ) R N u ( x ) s d x 1 / s , u L s ( R N ) .

  • H 1 ( R N ) = { u L 2 ( R N ) : u L 2 ( R N ) } endowed with the norm

    u H 1 ( R N ) = R N u ( x ) 2 + u 2 d x 1 / 2 .

  • S denotes the best constant that verifies

    (1) S = inf u H 1 ( R N ) \ { 0 } R N u 2 d x 1 / 2 R N u 2 d x 1 / 2 .

  • For any s ( 2 , 2 ) , S s denote by

    (2) S s inf u H 1 ( R N ) \ { 0 } u H 1 ( R N ) u s .

2 Reformulation of the problem

From the embedding results, it is well known that H 1 ( R N ) L p ( R N ) is continuous for 2 p 2 and is compact when 2 p < 2 . For more general facts about the embedded, we refer the reader to the previous articles [13,14].

To find solutions of (P), we will use a variational approach. Hence, we will associate a suitable functional to our problem. More precisely, the Euler-Lagrange functional related to problem (P) is given by J λ : H 1 ( R N ) R defined as follows:

J λ ( u ) = 1 2 R N u 2 + V ( x ) u 2 d x + 1 4 R N ( u 2 ) 2 d x λ R N G ( x , u ) d x .

Since,

1 4 R N ( u 2 ) 2 d x = R N u 2 u 2 d x ,

we obtain

J λ ( u ) = 1 2 R N ( 1 + 2 u 2 ) u 2 d x + 1 2 R N V ( x ) u 2 d x λ R N G ( x , u ) d x .

Because of the term R N u 2 u 2 d x , the functional J λ is not defined on space H 1 ( R N ) . For this, we follow the argument developed by Liu et al. [15] (see also [4]), and we change the variables v = f 1 ( u ) , where f is defined by

f ( t ) = 1 ( 1 + 2 f 2 ( t ) ) 1 / 2 on [ 0 , + ) , f ( t ) = f ( t ) on [ 0 , + ) .

Let us assemble some properties of the change of variables f : R R , which will be used in the continued from the article.

Lemma 2.1

The function f ( t ) and its derivative satisfy the following properties:

  1. f is uniquely defined, C and invertible;

  2. f ( t ) 1 for all t R ;

  3. f ( t ) t for all t R ;

  4. f ( t ) t 1 as t 0 ;

  5. f ( t ) t 1 / 2 2 1 / 4 as t + ;

  6. f ( t ) / 2 t f ( t ) f ( t ) for all t > 0 ;

  7. f 2 ( t ) / 2 t f ( t ) f ( t ) f 2 ( t ) for all t R ;

  8. f ( t ) 2 1 / 4 t 1 / 2 for all t R ;

  9. There exists a positive constant C such that

    f ( t ) C t , if t 1 , C t 1 / 2 , if t 1 ;

  10. For each α > 0 , there exists a positive constant C ( α ) such that

    f 2 ( α t ) C ( α ) f 2 ( t ) ;

  11. f ( t ) f ( t ) 1 / 2 .

Proof

Proofs may be found in [4,8].□

Under this change the functional J λ will be

I λ ( v ) = 1 2 R N v 2 + V ( x ) f 2 ( v ) d x λ R N G ( x , f ( v ) ) d x .

Obviously, I λ C 1 ( H 1 ( R N ) , R ) and

I λ ( v ) , w = R N v w d x + R N V ( x ) f ( v ) f ( v ) w d x λ R N g ( x , f ( v ) ) f ( v ) w d x ,

for any w H 1 ( R N ) .

Moreover, the critical points of I λ are the weak solutions of the following equation:

(3) Δ v = 1 1 + 2 f ( v ) 2 ( g ( x , f ( v ) ) V ( x ) f ( v ) ) in R N .

Considering that if v is a critical point of the functional I λ , so u = f ( v ) is a critical point of the functional J λ , that is, u = f ( v ) is a solution of problem (P).

3 The boundedness of the Cerami sequences

Lemma 3.1

Let r 0 > 0 . There exists k 0 > 0 such that

R N v 2 + V ( x ) f 2 ( v ) d x k 0 v H 1 ( R N ) 2 , v H 1 ( R N ) : v H 1 ( R N ) r 0 .

Proof

The proof of analogous results can be found in [4,8].□

Definition 1

Let X be a Banach space with the norm and F : X R a function.

A Gâteaux differentiable function F satisfies the Cerami condition locally at c ( ( C ) c -condition for short) if any sequence { v n } X such that F ( v n ) c and ( 1 + v n ) F ( v n ) 0 , as n + , has a convergent subsequence.

The first step for the ( C ) c -condition to hold is bounded.

Lemma 3.2

Assume that V and G ( x , v ) satisfy ( V 1 ) , ( V 2 ) , ( G 1 ) , ( G 2 ) , ( G 3 ) , and { v n } is ( C ) c -condition for I λ , then { v n } is bounded in X for any λ > 0 .

Proof

We have

(4) I λ ( v n ) c and ( 1 + v n H 1 ( R N ) ) I λ ( v n ) 0 in H 1 ( R N ) .

We claim that there exists C > 0 such that

(5) R N v n 2 + V ( x ) f 2 ( v n ) d x v n f 2 C .

Next we use Lemma 3.1 to deduce that { v n } is bounded in H 1 ( R N ) .

Assume by contradiction that v n f + .

We have from (4)

c + o n ( 1 ) = 1 2 R N ( v n 2 + V ( x ) f 2 ( v n ) ) d x λ R N G ( x , f ( v n ) ) d x ,

so,

R N G ( x , f ( v n ) ) d x = c + o n ( 1 ) λ + 1 2 λ v n f 2

and

(6) lim n + R N G ( x , f ( v n ) ) v n f 2 d x = 1 2 λ > 0 .

Then

(7) 0 < 1 2 λ = lim n + R N G ( x , f ( v n ) ) v n f 2 d x lim n + R N G ( x , f ( v n ) ) v n f 2 d x .

Let ω n = f ( v n ) v n f . Clearly, ω n H 1 ( R N ) 1 .

Up to a subsequence, we may assume that

ω n ω in H 1 ( R N ) .

Because H 1 ( R N ) L r ( R N ) is compact for all 2 r < 2 , this implies

ω n ω in L r ( R N ) , 2 r < 2 .

ω n ( x ) ω ( x ) a.e on R N .

Since μ 4 and (4), this implies that

c + 1 I λ ( v n ) 1 μ I λ ( v n ) , f ( v n ) f ( v n ) = R N 1 2 1 μ 1 + 4 f 2 ( v n ) 1 + 2 f 2 ( v n ) v n 2 d x + 1 2 1 μ R N V ( x ) f 2 ( v n ) d x + λ R N 1 μ g ( x , f ( v n ) ) f ( v n ) G ( x , f ( v n ) ) d x λ R N G ˜ ( x , f ( v n ) ) d x ,

then

(8) R N G ˜ ( x , f ( v n ) ) d x c + 1 λ .

Let us consider

Ω n ( a , b ) = { x R N / 0 < a f ( v n ) < b } .

By ( G 2 ) and ( G 3 ) , there exists δ 2 > 0 such that

G ˜ ( x , f ( v n ( x ) ) ) > C p p 2 c 0 r ( q 2 ) p p 2 , x Ω n ( r , ) , r > δ 2 .

Moreover, by (8), we can write

0 C p p 2 c 0 r ( q 2 ) p p 2 meas ( Ω n ( r , ) ) Ω n ( r , ) G ˜ ( x , f ( v n ) ) d x R N G ˜ ( x , f ( v n ) ) d x c + 1 λ .

Thus, this implies that

0 meas ( Ω n ( r , ) ) c 0 ( c + 1 ) λ C p p 2 r ( q 2 ) p p 2 .

Then

(9) meas ( Ω n ( r , ) ) 0 as r + .

We have 1 < 2 < 2 , there exists 0 < ξ < 1 such that 2 = ξ + 2 ( 1 ξ ) . By ( f 4 ) , the Hölder inequality and Sobolev inequality, thus, we can write

Ω n ( r , ) ( v n 1 ) f 2 ( v n ) d x Ω n ( r , ) ( v n 1 ) f ( v n ) d x ξ Ω n ( r , ) ( v n 1 ) f ( v n ) 2 d x 1 ξ Ω n ( r , ) ( v n 1 ) V 1 ( x ) d x ξ 2 Ω n ( r , ) ( v n 1 ) V ( x ) f 2 ( v n ( x ) ) d x ξ 2 Ω n ( r , ) ( v n 1 ) v n 2 d x 1 ξ

(10) V 0 ξ 2 ( meas ( Ω n ( r , ) ) ) ξ 2 v n f ξ S 2 ( 1 ξ ) R N v n 2 d x 2 ( 1 ξ ) 2 V 0 ξ 2 ( meas ( Ω n ( r , ) ) ) ξ 2 v n f ξ S 2 ( 1 ξ ) v n f 2 ( 1 ξ ) V 0 ξ 2 ( meas ( Ω n ( r , ) ) ) ξ 2 v n f 2 .

By using the fact that ( a + b ) κ a κ + b κ , if a , b 0 and κ [ 0 , 1 ] , ( f 4 ) and (10), we obtain

(11) Ω n ( r , ) v n d x κ = Ω n ( r , ) ( v n < 1 ) v n d x + Ω n ( r , ) ( v n 1 ) v n d x κ ( meas ( Ω n ( r , ) ) ) κ + 1 C κ Ω n ( r , ) ( v n 1 ) f 2 ( v n ) d x κ ( meas ( Ω n ( r , ) ) ) κ + V 0 ξ κ 2 C κ ( meas ( Ω n ( r , ) ) ) ξ κ 2 v n f 2 κ .

Since 2 < p < 2 2 , there exists 0 < θ < 1 such that p 2 = θ + 2 ( 1 θ ) .

By using (11) and considering κ = 2 p , we can write

(12) 1 v n f 2 , Ω n ( r , ) v n p 2 d x 2 p 1 v n f 2 Ω n ( r , ) v n d x 2 θ p Ω n ( r , ) v n 2 d x 2 ( 1 θ ) p S 2 2 ( 1 θ ) p v n f 2 Ω n ( r , ) v n d x 2 θ p Ω n ( r , ) v n 2 d x 2 ( 1 θ ) p S 2 2 ( 1 θ ) p v n f 1 + 2 θ p ( meas ( Ω n ( r , ) ) ) 2 θ p + S 2 2 ( 1 θ ) p V 0 ξ θ p C 2 θ p v n f ( meas ( Ω n ( r , ) ) ) ξ θ p .

Thus, it follows from v n f > 1 , ( G 3 ) , (8), (9), and (12) that

(13) Ω n ( r , ) G ( x , f ( v n ) ) v n f 2 d x = R N G ( x , f ( v n ) ) f 2 ( v n ) p p 2 d x p 2 p Ω n ( r , ) f ( v n ) p d x 2 p 1 v n f 2 2 1 / 4 c 0 p 2 p R N G ˜ ( x , f ( v n ) ) d x p 2 p Ω n ( r , ) v n p 2 d x 2 p 1 v n f 2

(13) 2 1 / 4 c 0 p 2 p S 2 2 ( 1 θ ) p v n f 1 + 2 θ p c + 1 λ p 2 p ( meas ( Ω n ( r , ) ) ) 2 θ p + 2 1 / 4 c 0 p 2 p V 0 ξ θ p C 2 θ p S 2 2 ( 1 θ ) p v n f c + 1 λ p 2 p ( meas ( Ω n ( r , ) ) ) ξ θ p 2 1 / 4 c 0 p 2 p S 2 2 ( 1 θ ) p c + 1 λ p 2 p ( meas ( Ω n ( r , ) ) ) 2 θ p + 2 1 / 4 c 0 p 2 p V 0 ξ θ p C 2 θ p S 2 2 ( 1 θ ) p c + 1 λ p 2 p ( meas ( Ω n ( r , ) ) ) ξ θ p 0 ,

as r uniformly in n .

By ( G 2 ) and ω n L 2 ( R N ) ω L 2 ( R N ) , as n + , we find that for fix r > 0

(14) Ω n ( 0 , r ) G ( x , f ( v n ) ) v n f 2 d x Ω n ( 0 , r ) ξ ( x ) q f ( v n ) q + τ ( x ) p f ( v n ) p v n f 2 d x r q 1 q v n f ξ L 2 ( R N ) ω n L 2 ( R N ) + r p 1 p v n f τ L 2 ( R N ) ω n L 2 ( R N ) 0 , as n .

Combining (7), (13), and (14), we have

0 < 1 2 λ R N G ( x , f ( v n ) ) v n f 2 d x = Ω n ( 0 , r ) G ( x , f ( v n ) ) v n f 2 d x + Ω n ( r , ) G ( x , f ( v n ) ) v n f 2 d x 0 , as n , r .

This is inconsistent with (6). Hence, there exists M > 0 such that v n f M for all n N . Then by Lemma 3.1, the sequence { v n } is bounded in H 1 ( R N ) .□

Lemma 3.3

Assume that V and G ( x , v ) satisfy ( V 1 ) , ( V 2 ) , ( G 1 ) , ( G 2 ) , and ( G 3 ) . Then, the functional I λ satisfies the ( C ) c -condition for all c > 0 and for all λ > 0 .

Proof

Consider a ( C ) c -sequence { v n } for I λ , that is,

I λ ( v n ) c and ( 1 + v n H 1 ( R N ) ) I λ ( v n ) 0 in H 1 ( R N ) .

From Lemma 3.2 { v n } is bounded in H 1 ( R N ) . Going if necessary to a subsequence, we can assume that

(15) v n v , in H 1 ( R N ) , v n v , in L r ( R N ) , 2 r < 2 .

From ( V 2 ) , we conclude that x > R V 1 d x 0 as R + .

(16) x > R v n v d x x > R V 1 d x 1 / 2 x > R V ( x ) v n v 2 d x 1 / 2 x > R V 1 d x 1 / 2 v n v X M x > R V 1 d x 1 / 2 0 , as R + uniformly in n .

Moreover, since v n v in L loc 1 ( R N ) , x R v n v d x 0 , as n + , which implies that

(17) R N v n v d x 0 , as n + .

By using (17), p 2 = θ + 2 ( 1 θ ) for some 0 < θ < 1 and v n v H 1 ( R N ) 2 ( 1 θ ) M for some M > 0 , we can write

(18) v n v L p 2 ( R N ) p 2 = R N v n v p 2 d x R N v n v d x θ R N v n v 2 d x 1 θ S 2 ( 1 θ ) R N v n v d x θ R N ( v n v ) 2 d x 2 ( 1 θ ) 2 S 2 ( 1 θ ) R N v n v d x θ v n v H 1 ( R N ) 2 ( 1 θ ) M S 2 ( 1 θ ) R N v n v d x θ 0 as n + .

With similar arguments in Lemma 3.1, we have

There exists C > 0 such that

(19) R N ( v n v ) 2 + V ( x ) ( f ( v n ) f ( v n ) f ( v ) f ( v ) ) ( v n v ) d x C v n v H 1 ( R N ) 2 .

Because v n v in H 1 ( R N ) and I λ ( v n ) 0 in H 1 ( R N ) as n + , we have

I λ ( v n ) I λ ( v ) , v n v 0 as n + .

By the Hölder inequality, (15), (18), (19), ( f 2 ) , ( G 1 ) , 2 < p < 2 2 , and 2 < q < min ( 4 , p ) , we can write

o n ( 1 ) = I λ ( v n ) I λ ( v ) , v n v = R N ( v n v ) 2 + V ( x ) ( f ( v n ) f ( v n ) f ( v ) f ( v ) ) ( v n v ) d x λ R N ( g ( x , f ( v n ) ) f ( v n ) g ( x , f ( v ) ) f ( v ) ) ( v n v ) d x C v n v H 1 ( R N ) 2 λ R N ( ξ ( x ) f ( v n ) q 1 + τ ( x ) f ( v n ) p 1 + ξ ( x ) f ( v ) q 1 + τ ( x ) f ( v ) p 1 ) v n v d x C v n v H 1 ( R N ) 2 λ R N ξ ( x ) v n q 2 1 + τ ( x ) v n p 2 1 + ξ ( x ) v q 2 1 + τ ( x ) v p 2 1 v n v d x C v n v H 1 ( R N ) 2 λ ξ L 4 q ( R N ) v n L 2 ( R N ) q 2 2 + v L 2 ( R N ) q 2 2 v n v L 2 ( R N ) λ τ v n L p 2 ( R N ) p 2 2 + v L p 2 ( R N ) p 2 2 v n v L p 2 ( R N ) k 0 v n v H 1 ( R N ) 2 + o n ( 1 ) .

Therefore, v n v H 1 ( R N ) 0 as n + . Hence, I λ satisfies the ( C ) c -condition for any c > 0 . This completes the proof.□

4 Proof of the main result

In this section, we show the existence of infinitely many solutions via the Fountain theorem [12].

Let X be a Banach and separable. Since X is separable (see [16]), there exist { e n } n N X and { f n } n N X with

X = span { e n } n = 1 ¯ , X = span { f n } n = 1 ¯

f i , e j = 1 if i = j , 0 if i j ,

where , is the duality pairing between X and X .

Let X j = span { e j } , Y n = j = 0 n X j , Z n = j = n X j ¯ .

Lemma 4.1

([12] Fountain theorem). Consider an even functional I λ C ( X , R ) . If, for every k N , there exist ρ k > r k > 0 such that

  1. a k = max { I λ ( v ) : v Y k , v X = ρ k } 0 ,

  2. b k = inf { I λ ( v ) : v Z k , v X = ρ k } as k + ,

  3. I λ satisfies the ( C ) c -condition for every c > 0 .

Then I λ has an unbounded sequence of critical values.

Proof of Theorem 1.1

The functional I λ is even, I λ C ( H 1 ( R N ) , R ) . By Lemma 3.3 I λ satisfies the ( C ) c -condition for every c > 0 . We only need to verify I λ satisfying (1) and (2) of Lemma 4.1.

For k 1 . Denote

α k = sup v Z k , v H 1 ( R N ) = 1 v L p 2 ( R N ) .

Then we have α k 0 as k + . In fact, suppose to the contrary that there exist ε 0 > 0 , k 0 N and the sequence { v k } in Z k such that v k H 1 ( R N ) = 1 and v k L p 2 ( R N ) ε 0 for all k k 0 . Since the sequence { v k } is bounded in H 1 ( R N ) , there exist v H 1 ( R N ) such that v k v in X as n + and

f j , v = lim k + f j , v k = 0 for j = 1 , 2 , .

Hence, we obtain v = 0 . However, we have

ε 0 lim k + v k L p 2 ( R N ) = v L p 2 ( R N ) = 0 ,

which provides a contradiction.

Fix r > 0 .

Since α k 0 as k + ,

(20) n 0 > 0 : k n 0 0 < α k < k 0 S p 2 p 2 2 1 / 4 λ τ 2 p r 4 p p .

For all v Z k such that v H 1 ( R N ) = r . It follows from the assumptions ( G 1 ) , ( f 8 ) , 2 < p < 2 2 , 2 < q < min ( 4 , p ) , and Lemma 3.1, we have

I λ ( v ) 1 2 R N v 2 + V ( x ) f 2 ( v ) d x 2 1 / 4 λ q R N ξ ( x ) v q 2 d x 2 1 / 4 λ p R N τ ( x ) v p 2 d x 1 2 R N v 2 + V ( x ) f 2 ( v ) d x 2 1 / 4 λ q ξ L 4 q ( R N ) v L 2 ( R N ) q 2 2 1 / 4 λ p τ v L p 2 ( R N ) p 2 1 2 k 0 v H 1 ( R N ) 2 2 1 / 4 λ q S 2 q 2 ξ L 4 q ( R N ) v H 1 ( R N ) q 2 2 1 / 4 λ S p 2 p 2 p α k p 2 τ v H 1 ( R N ) p 2 1 2 k 0 r 2 2 1 / 4 λ q S 2 q 2 ξ L 4 q ( R N ) r q 2 2 1 / 4 λ S p 2 p 2 p α k p 2 τ r p 2 k 0 1 2 1 p r 2 2 1 / 4 λ q S 2 q 2 ξ L 4 q ( R N ) r q 2 + , as r + ,

which implies (2).

For the first item (1), let us suppose that this condition is not meant for a certain k . So, there exists a sequence { v n } in Y k such that

(21) v n H 1 ( R N ) + , and I λ ( v n ) 0 .

We consider ω n = f ( v n ) v n H 1 ( R N ) . Then it is obvious that ω n H 1 ( R N ) 1 . As Y k is of finite dimension, there exists ω Y k with ω 0 such that

ω n ω strongly in Y k and w n ( x ) w ( x ) a.e on R N as n + .

Let

Ω 2 = { x R N : ω ( x ) 0 } .

Then meas ( Ω 2 ) > 0 .

For all x Ω 2 , we have

f ( v n ( x ) ) = ω n ( x ) v n H 1 ( R N ) + , as n + .

For r 0 > 0 , we have Ω 2 Ω n ( r 0 , ) for large n .

Hence, for all x Ω 2 and by ( G 2 ) , we have

lim n + G ( x , f ( v n ) ) v n H 1 ( R N ) 2 = lim n + G ( x , f ( v n ) ) f 2 ( v n ( x ) ) w n ( x ) 2 d x = lim n + G ( x , f ( v n ) ) f ( v n ( x ) ) q w n ( x ) q 2 d x = + .

Thus, it follows from (14), (21), and Fatou’s lemma that

0 liminf n + I λ ( v n ) v n H 1 ( R N ) 2 1 2 liminf n + λ 1 v n X 2 Ω n ( 0 , r 0 ) G ( x , f ( v n ) ) d x Ω n ( r 0 , ) G ( x , f ( v n ) ) f 2 ( v n ) ω n 2 d x 1 2 + r 0 q 1 q ξ L 2 ( R N ) ω L 2 ( R N ) + r 0 p 1 p τ L 2 ( R N ) ω L 2 ( R N ) λ R N liminf n + G ( x , f ( v n ) ) f 2 ( v n ) χ Ω n ( r 0 , ) ω n 2 d x = ,

which provides a contradiction. So, I λ satisfies (2). All the assumptions of Lemma 4.1 are satisfied. Therefore, this concludes the proof of Theorem 1.1.□

Acknowledgments

The authors would like to thank the anonymous referee for his comments that helped us improve this article.

  1. Funding information: The authors state that no funding is involved.

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

  3. Ethical approval: The conducted research is not related to either human or animal use.

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Received: 2021-01-04
Revised: 2022-09-20
Accepted: 2022-09-29
Published Online: 2022-11-22

© 2022 the author(s), published by De Gruyter

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

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