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Infinitely many weak solutions for a fractional Schrödinger equation

Wei Dong1*, Jiafa Xu2 and Zhongli Wei23

Author Affiliations

1 Department of Mathematics, Hebei University of Engineering, Handan 056038, Hebei, China

2 School of Mathematics, Shandong University, Jinan 250100, Shandong, China

3 Department of Mathematics, Shandong Jianzhu University, Jinan 250101, Shandong, China

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Boundary Value Problems 2014, 2014:159  doi:10.1186/s13661-014-0159-6

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/159


Received:22 February 2014
Accepted:13 June 2014
Published:12 July 2014

© 2014 Dong et al.; licensee Springer

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Abstract

In this paper we are concerned with the fractional Schrödinger equation ( Δ ) α u + V ( x ) u = f ( x , u ) , x R N , where 0 < α < 1 , N > 2 α , ( Δ ) α stands for the fractional Laplacian of order α, V is a positive continuous potential, and f is a continuous subcritical nonlinearity. We obtain the existence of infinitely many weak solutions for the above problem by the fountain theorem in critical point theory.

Keywords:
fractional Laplacian; subcritical nonlinearity; fountain theorem; weak solution

Introduction

In this paper we consider the following fractional Schrödinger equation:

( Δ ) α u + V ( x ) u = f ( x , u ) , x R N , (1)
where 0 < α < 1 , N > 2 α , ( Δ ) α stands for the fractional Laplacian of order α, and the potential V : R N R is a continuous function satisfying

(V) 0 < inf x R N V ( x ) = V 0 < lim inf | x | V ( x ) = V < .

The nonlinearity f : R N × R R is a continuous function, satisfying the subcritical condition.

(H1) There exist d 1 > 0 , d 2 > 0 and p ( 2 , 2 α ) such that

| f ( x , s ) | d 1 | s | + d 2 | s | p 1 , ( x , s ) R N × R , (2)
where 2 α = 2 N N 2 α is the fractional critical exponent.

Recently, there have appeared plenty of works on the fractional Schrödinger equations; for example, see [[1]–[11]] and the references therein. In [[1]], Shang and Zhang considered the critical fractional Schrödinger equation

ε 2 α ( Δ ) α u + V ( x ) u = | u | 2 α 2 u + λ f ( u ) , x R N , (3)
where ε and λ are positive parameters, V and f satisfy (V) and (H1), respectively. They obtained the result that (1.2) has a nonnegative ground state solution and investigated the relation between the number of solutions and the topology of the set where V attains its minimum for all sufficiently large λ and small ε. In [[2]], Shang et al. considered the existence of nontrivial solutions for (1.2) with f ( u ) = | u | q 2 u , where 2 < q < 2 α .

In [[8]], Hua and Yu studied the critical fractional Laplacian equation

{ ( Δ ) α 2 u = | u | 2 α 2 u + μ u in  Ω , u = 0 on  Ω , (4)
where 0 < α < 2 , Ω R N , N > ( 1 + 2 ) α is a bounded domain. They obtained the result that the problem (1.3) possesses a nontrivial ground state solution for any μ > 0 .

In [[7]], Secchi investigated the existence of radially symmetric solutions for (1.1) replacing f ( t , u ) by g ( u ) , where g satisfies the following conditions.

(g1) g : R R is of class C 1 , γ for some γ > max { 0 , 1 2 α } , and odd,

(g2) < lim inf t 0 + g ( t ) t lim sup t 0 + g ( t ) t = m < 0 ,

(g3) < lim sup t + g ( t ) t 2 α 1 0 ,

(g4) for some ξ > 0 such that G ( ξ ) = 0 ξ g ( t ) d t > 0 .

Inspired by the mentioned papers, we first establish a compact embedding lemma via a fractional Gagliardo-Nirenberg inequality. Then by virtue of the fountain theorem in critical point theory, we get two existence results of infinitely many weak solutions for (1.1).

Preliminary results

In this section we offer some preliminary results which enable us to obtain the main existence theorems. First, we collect some useful facts of the fractional order Sobolev spaces.

For any 0 < α < 1 , the fractional Sobolev space H α ( R N ) is defined by

H α ( R N ) = { u L 2 ( R N ) : | u ( x ) u ( y ) | | x y | N + 2 α 2 L 2 ( R N × R N ) } , (5)
endowed with the norm
u H α ( R N ) = ( R N | u | 2 d x + R 2 N | u ( x ) u ( y ) | 2 | x y | N + 2 α d x d y ) 1 2 , (6)
where [ u ] H α ( R N ) = ( R 2 N | u ( x ) u ( y ) | 2 | x y | N + 2 α d x d y ) 1 / 2 is the so-called Gagliardo semi-norm of u. Let S be the Schwartz space of rapidly decaying C functions in R N , for any u S and α ( 0 , 1 ) , and let ( Δ ) α be defined as
( Δ ) α u ( x ) = k N , α P.V. R N u ( x ) u ( y ) | x y | N + 2 α d y = k N , α lim ε 0 C B ε ( x ) u ( x ) u ( y ) | x y | N + 2 α d y . (7)
The symbol P.V. stands for the Cauchy principal value, and k N , α is a dimensional constant that depends on N and α, precisely given by k N , α = ( R N 1 cos ζ 1 | ζ | N + 2 α d ζ ) 1 .

Indeed, the fractional Laplacian ( Δ ) α can be viewed as a pseudo-differential operator of symbol | ξ | 2 α , as stated in the following.

Lemma 2.1

(see [[12]])

Let α ( 0 , 1 ) and ( Δ ) α : S L 2 ( R N ) be the fractional Laplacian operator defined by (2.1). Then for any u S ,

( Δ ) α u ( x ) = F 1 ( | ξ | 2 α ( F u ) ) ( x ) , ξ R N , (8)
whereis the Fourier transform, i.e.,
F ( ϕ ) ( ξ ) = 1 ( 2 π ) N / 2 R N exp { 2 π i ξ x } ϕ ( x ) d x . (9)

Now we can see that an alternative definition of the fractional Sobolev space H α ( R N ) via the Fourier transform is as follows:

H α ( R N ) = { u L 2 ( R N ) : R N ( 1 + | ξ | 2 α ) | F u | 2 d ξ < + } . (10)
It can be proved that
2 k N , α 1 R N | ξ | 2 α | F u | 2 d ξ = 2 k N , α 1 ( Δ ) α 2 u L 2 ( R N ) 2 = [ u ] H α ( R N ) 2 . (11)
As a result, the norms on H α ( R N ) ,
u u H α ( R N ) , u ( u L 2 ( R N ) 2 + ( Δ ) α 2 u L 2 ( R N ) 2 ) 1 2 , u ( u L 2 ( R N ) 2 + R N | ξ | 2 α | F u | 2 d ξ ) 1 2 , (12)
are all equivalent.

In this paper, in view of the presence of potential V ( x ) , we consider its subspace

E = { u H α ( R N ) : R N V ( x ) u 2 d x < } . (13)
We define the norm in E by
u E = ( R N ( | ξ | 2 α u ˆ 2 + u ˆ 2 ) d ξ + R N V ( x ) u 2 d x ) 1 2 , (14)
where u ˆ = F ( u ) . Moreover, by [[6]], E is a Hilbert space with the inner product
u , v E = R N ( | ξ | 2 α u ˆ ( ξ ) v ˆ ( ξ ) + u ˆ ( ξ ) v ˆ ( ξ ) ) d ξ + R N V ( x ) u ( x ) v ( x ) d x , u , v E . (15)
Note that by (2.2) and (2.3), together with the condition (V), we know that E is equivalent to the norm
u = ( R N ( | ( Δ ) α 2 u | 2 + V ( x ) u 2 ) d x ) 1 2 . (16)
The corresponding inner product is
u , v = R N ( ( Δ ) α 2 u ( x ) ( Δ ) α 2 v ( x ) + V ( x ) u ( x ) v ( x ) ) d x . (17)
Throughout out this paper, we will use the norm in E.

Definition 2.2

We say that u E is a weak solution of (1.1), if

R N ( ( Δ ) α 2 u ( Δ ) α 2 ϕ + V ( x ) u ϕ ) d x = R N f ( x , u ) ϕ d x , ϕ E . (18)

Lemma 2.3

(see [[7]] and [[12]])

Eis continuously embedded into L p ( R N ) for p [ 2 , 2 α ] and compactly embedded into L loc p ( R N ) for p [ 2 , 2 α ) .

Lemma 2.4

Eis compactly embedded into L p ( R N ) for p [ 2 , 2 α ) with 2 α = 2 N N 2 α .

Proof

By [[4]], we know E is compactly embedded into L 2 ( R N ) , i.e., if there exists a sequence { u n } E and u 0 E such that u n u 0 weakly in E, passing to a subsequence if necessary, we have u n u 0 strongly in L 2 ( R N ) . Therefore, we only consider p ( 2 , 2 α ) . In order to do this, we need the following fractional Gagliardo-Nirenberg inequality, see [[13], Corollary 2.3]. Let 1 p , p 2 < , 0 < s < p < , 0 < α < N and 1 < p 1 < N / α . Then

u ( x ) L p ( R N ) η s p ( Δ ) α / 2 u ( x ) L p 1 ( R N ) s p u ( x ) L p 2 ( R N ) 1 s p (19)
with
s ( 1 p 1 α N ) + p s p 2 = 1 and η = 2 α π α / 2 Γ ( ( N α ) / 2 ) Γ ( ( N + α ) / 2 ) ( Γ ( N ) Γ ( N / 2 ) ) α / N . (20)
Note that the dimension N > 2 α , we can take p 1 = p 2 = 2 , and then s ( 1 2 α N ) + p s 2 = 1 whence s = ( p 2 ) N 2 α ( 0 , p ) as p ( 2 , 2 s ) . Consequently, from (2.5) we have
u L p ( R N ) η s p ( Δ ) α / 2 u L 2 ( R N ) s p u L 2 ( R N ) 1 s p . (21)
Furthermore, note that (2.4); we see that
u L p ( R N ) η s p u s p u L 2 ( R N ) 1 s p . (22)
Then by (2.6) and E L 2 ( R N ) , we find
u n u 0 L p ( R N ) η s p u n u 0 s p u n u 0 L 2 ( R N ) 1 s p η s p ( u n s p + u 0 s p ) u n u 0 L 2 ( R N ) 1 s p 0 . (23)
Therefore, E is compactly embedded into L p ( R N ) for p [ 2 , 2 α ) with 2 α = 2 N N 2 α , as required. This completes the proof. □

The functional associated with (1.1) is defined by

J ( u ) = 1 2 R N ( | ( Δ ) α 2 u | 2 + V ( x ) u 2 ) d x R N F ( x , u ) d x , u E , (24)
where F ( x , u ) = 0 u f ( x , s ) d s .

Now, we list our assumptions on f and F.

(H2) lim | s | F ( x , s ) | s | 2 = + uniformly for x R N .

(H3) There exist d 3 > 0 and φ 1 > 0 such that R N φ 1 ( x ) d x < + and

t f ( x , t ) 2 F ( x , t ) s f ( x , s ) 2 F ( x , s ) + d 3 φ 1 , 0 < t < s  or  s < t < 0 , x R N . (25)

(H4) F ( x , s ) 0 for ( x , s ) R N × R .

(H5) There exist β > 2 , r 0 > 0 such that

β F ( x , s ) + s f ( x , s ) 0 , | s | r 0 , uniformly for  x R N . (26)

(H6) F ( x , s ) = F ( x , s ) for all ( x , s ) R N × R .

Remark 2.5

(1) Let F ( x , s ) = s 2 ln ( | s | + 1 ) , for all x R N and s R . Then (H1), (H2), (H4), and (H6) hold. Moreover, we easily have f ( x , s ) = 2 s ln ( | s | + 1 ) + s 3 [ | s | ( | s | + 1 ) ] 1 and s f ( x , s ) 2 F ( x , s ) = | s | 3 ( | s | + 1 ) 1 , so (H3) is satisfied.

However, we can see that F ( x , s ) does not satisfy the Ambrosetti-Rabinowitz condition (see [[6], ( f 4 )]):

(AR) there is a constant μ > 2 such that

0 < μ F ( x , s ) s f ( x , s ) for all  x R N  and  s R { 0 } . (27)

Indeed, s f ( x , s ) μ F ( x , s ) = s 2 [ ( 2 μ ) ln ( | s | + 1 ) + | s | ( | s | + 1 ) 1 ] 0 is impossible for all x R N and s R { 0 } .

(2) Let β > 2 and F ( x , s ) = | s | β ln ( | s | + 1 ) , for all x R N and s R . Then (H1), (H2), (H4), and (H6) hold. Moreover, from β F ( x , s ) + s f ( x , s ) = | s | β + 1 ( | s | + 1 ) 1 , and (H5) holds.

Note that from Theorem 4 in [[14]] we have (H4) and (H5) imply (H2).

Lemma 2.6

(see [[15], Lemma 1])

Let (V) and (H1) hold. Then J C 1 ( E , R ) and its derivative

( J ( u ) , ϕ ) = R N ( ( Δ ) α 2 u ( Δ ) α 2 ϕ + V ( x ) u ϕ ) d x R N f ( x , u ) ϕ d x , u , ϕ E . (28)
Moreover, the critical points ofJare weak solutions of (1.1).

To complete the proofs of our theorems, we need the following critical point theorems in [[16]–[19]].

Definition 2.7

Let ( X , ) be a real Banach space, J C 1 ( X , R ) . We say that J satisfies the ( C c ) condition if any sequence { u n } X such that J ( u n ) c and J ( u n ) ( 1 + u n ) 0 as n has a convergent subsequence.

Let X be a Banach space equipped with the norm and X = j N X j ¯ , where dim X j < for any j N . Set Y k = j = 1 k X j and Z k = j = k X j ¯ .

Lemma 2.8

Let ( X , ) be a real reflexive Banach space, J C 1 ( X , R ) satisfies the ( C c ) condition for any c > 0 andJis even. If for each sufficiently large k N , there exist ρ k > r k > 0 such that the following conditions hold:

(i) a k : = inf { u Z k , u = r k } J ( u ) + as k ,

(ii) b k : = max { u Y k , u = ρ k } J ( u ) 0 ,

then the functionalJhas an unbounded sequence of critical values, i.e., there exists a sequence { u k } X such that J ( u k ) = 0 and J ( u k ) + as k .

In the following, we will introduce a variant fountain theorem by Zou [[16]]. Let X and the subspace Y k and Z k be defined above. Consider the following C 1 -functional J λ : X R defined by

J λ ( u ) : = A ( u ) λ B ( u ) , λ [ 1 , 2 ] . (29)

Lemma 2.9

If the functional J λ satisfies

(T1) J λ maps bounded sets to bounded sets uniformly for λ [ 1 , 2 ] , and, moreover, J λ ( u ) = J λ ( u ) for all ( λ , u ) [ 1 , 2 ] × X ,

(T2) B ( u ) 0 for all u X ; moreover, A ( u ) or B ( u ) as u ,

(T3) there exist r k > ρ k > 0 such that

a k ( λ ) : = inf u Z k , u = ρ k J λ ( u ) > b k ( λ ) : = max u Y k , u = r k J λ ( u ) , λ [ 1 , 2 ] , (30)
then
a k ( λ ) ζ k ( λ ) = inf γ Γ k max u B k J λ ( γ ( u ) ) , λ [ 1 , 2 ] , (31)
where B k = { u Y k : u r k } and Γ k = { γ C ( B k , X ) : γ  is odd , γ | B k = id } . Moreover, for a.e. λ [ 1 , 2 ] , there exists a sequence { u n k ( λ ) } n = 1 such that
sup n u n k ( λ ) < , J λ ( u n k ( λ ) ) 0 and J λ ( u n k ( λ ) ) ζ k ( λ ) as  n . (32)

Remark 2.10

As mentioned in [[6]], E is a Hilbert space. Let { e j } be an orthonormal basis of E and define X j : = span { e j } , Y k : = j = 1 k X j , and Z k : = j = k + 1 X j ¯ , k N . Clearly, E = j N X j ¯ with dim X j < for all j N .

Existence of weak solutions for (1.1)

Theorem 3.1

Assume that (V), (H1)-(H4), and (H6) hold. Then (1.1) has infinitely many weak solutions { u k } satisfying

1 2 R N ( | ( Δ ) α 2 u k | 2 + V ( x ) u k 2 ) d x R N F ( x , u k ) d x + as  k . (33)

Proof

We first prove that J satisfies the ( C c ) condition for any c > 0 . Let { u n } E be a ( C c ) sequence, i.e.,

J ( u n ) c > 0 , J ( u n ) ( 1 + u n ) 0 when  n , (34)
which implies that
c = J ( u n ) + o ( 1 ) , J ( u n ) u n = o ( 1 ) as  n . (35)

In what follows, we shall show that { u n } is bounded. Otherwise, up to a subsequence, { u n } is unbounded in E, and we may assume that u n as n . We define the sequence { w n } by w n = u n u n , n = 1 , 2 ,  . Clearly, { w n } E and w n = 1 for any n. Going over, if necessary, to a subsequence, we may assume that

w n w  weakly in  E , w n w  strongly in  L p ( R N )  for  p [ 2 , 2 s ) , w n ( x ) w ( x )  a.e.  x R N . (36)
Suppose that w 0 in E. Dividing by u n 2 in both sides of (2.7), noting that J ( u n ) c , we obtain
R N F ( x , u n ) u n 2 d x = 1 2 + o ( u n 2 ) < + . (37)
On the other hand, denote Ω : = { x R N : w ( x ) 0 } , by (H2), for all x Ω , and we find
F ( x , u n ) u n 2 = F ( x , u n ) | u n | 2 | u n | 2 u n 2 = F ( x , u n ) | u n | 2 | w n | 2 + when  n . (38)
If | Ω | > 0 , using Fatou’s lemma, we obtain
R N F ( x , u n ) u n 2 d x + as  n . (39)
This contradicts (3.4). Hence, Ω has zero measure, i.e., w = 0 a.e. in R N . Let t n [ 0 , 1 ] such that
J ( t n u n ) = max t [ 0 , 1 ] J ( t u n ) . (40)
Then we claim J ( t n u n ) is bounded. If t n = 0 , J ( 0 ) = 0 ; if t n = 1 , J ( t n u n ) = J ( u n ) c . Therefore, J ( t n u n ) is bounded when t n = 0 , 1 . If 0 < t n < 1 for n large enough
R N ( ( Δ ) α 2 t n u n ( Δ ) α 2 t n u n + V ( x ) t n u n t n u n ) d x R N f ( x , t n u n ) t n u n d x = ( J ( t n u n ) , t n u n ) = t n d d t | t = t n J ( t u n ) = 0 . (41)
Consequently, by (H3), noting that (3.1) and (3.2) hold, we have
J ( t u n ) J ( t n u n ) 1 2 ( J ( t n u n ) , t n u n ) = 1 2 R N f ( x , t n u n ) t n u n d x R N F ( x , t n u n ) d x R N [ ( 1 2 u n f ( x , u n ) F ( x , u n ) ) + d 3 2 φ 1 ( x ) ] d x = J ( u n ) 1 2 ( J ( u n ) , u n ) + R N d 3 2 φ 1 ( x ) d x d 4 , t [ 0 , 1 ] , (42)
where d 4 is a positive constant. But fixing any m > d 4 , we let w ¯ n = 2 m u n u n = 2 m w n . Note that from (H1) we see that there exist d 5 > 0 , d 6 > 0 such that
F ( x , u ) d 5 | u | 2 + d 6 | u | p , ( x , u ) R N × R . (43)
Then by (3.3) we have
lim n R N F ( x , w ¯ n ) d x lim n R N ( d 5 | w ¯ n | 2 + d 6 | w ¯ n | p ) d x = 0 . (44)
Then for n large enough,
J ( t n u n ) J ( 2 m u n u n ) = J ( w ¯ n ) = m R N F ( x , w ¯ n ) d x m . (45)
This also contradicts (3.5).

Now the sequence { u n } is bounded, as required. Next, we verify that { u n } has a convergent subsequence. Without loss of generality, we assume that

u n u  weakly in  E , u n u  strongly in  L p ( R N )  for  p [ 2 , 2 s ) . (46)
Combining this with (H1) and the Hölder inequality, we see
| R N [ f ( x , u n ) f ( x , u ) ] ( u n u ) d x | R N [ d 1 ( | u n | + | u | ) + d 2 ( | u n | p 1 + | u | p 1 ) ] | u n u | d x d 1 ( u n 2 + u 2 ) u n u 2 + d 2 ( u n p p 1 + u p p 1 ) u n u p 0 as  n . (47)
Consequently,
u n u 2 = ( J ( u n ) J ( u ) , u n u ) + R N [ f ( x , u n ) f ( x , u ) ] ( u n u ) d x 0 (48)
with the fact that ( J ( u n ) J ( u ) , u n u ) 0 when n . Therefore, we prove that J satisfies the ( C c ) condition for any c > 0 .

Clearly, J ( u ) = J ( u ) by (H6). It remains to prove that the conditions (i) and (ii) of Lemma 2.8 hold. Let β r ( k ) : = sup { u Z k , u = 1 } u r with r [ 2 , 2 α ) , where Z k is defined in Remark 2.10. Then by Lemma 3.8 of [[19]], β r ( k ) 0 as k for the fact that E L r ( R N ) .

Now for u Z k with u = r k = ( β 2 ( k ) + β p ( k ) ) 1 , we obtain

J ( u ) 1 2 u 2 R N ( d 5 | u | 2 + d 6 | u | p ) d x = 1 2 u 2 d 5 u 2 2 d 6 u p p 1 2 u 2 d 5 β 2 2 ( k ) u 2 d 6 β p p ( k ) u p 1 2 r k 2 d 5 d 6 as  k . (49)
Hence,
a k : = inf u Z k , u = r k J ( u ) + as  k . (50)

Next we shall prove that, for any finite dimensional subspace X E , we have

J ( u ) as  u , u X . (51)
Suppose the contrary. For some sequence { u n } X with u n , there is a M > 0 such that J ( u n ) M for all n N . Put v n = u n u n and then v n = 1 . Up to a subsequence, assume that v n v weakly in E. Since dim X < , v n v X in E, v n v a.e. on R N , and v = 1 . Denote Ω : = { x R N : v ( x ) 0 } , then meas ( Ω ) > 0 and for a.e. x Ω , lim n | u n ( x ) | . It follows from (2.7) that
lim n R N F ( x , u n ) d x u n 2 = lim n 1 2 u n 2 J ( u n ) u n 2 d 7 with a constant  d 7 > 0 . (52)
But, for large n, on account of F being nonnegative, (H2) and Fatou’s Lemma enable us to obtain
lim n R N F ( x , u n ) d x u n 2 lim n Ω F ( x , u n ) v n 2 u n 2 d x lim inf n Ω F ( x , u n ) v n 2 u n 2 d x Ω lim inf n F ( x , u n ) v n 2 u n 2 d x = Ω lim inf n F ( x , u n ) u n 2 [ χ Ω ( x ) ] v n 2 d x (53)
as n . This contradicts (3.10). Consequently, (3.9) holds, as required. Note that dim Y k < in Remark 2.10, and there exist positive constants d k such that
J ( u ) 0 , for each  u Y k  and  u d k . (54)
Combining this and (3.8), we can take ρ k : = max { d k , r k + 1 } , and thus b k : = max { u Y k , u = ρ k } J ( u ) 0 . Until now, we have proved the functional J satisfies all the conditions of Lemma 2.8. Hence, J has an unbounded sequence of critical values, i.e., there exists a sequence { u k } E such that J ( u k ) = 0 and J ( u k ) + as k . This completes the proof. □

We prove that there exists a 0 = a 0 ( r 0 ) > 0 ( r 0 is determined in (H5)) such that

| β F ( x , s ) + s f ( x , s ) | a 0 | s | 2 , | s | r 0 ,  for all  x R N . (55)
Indeed, by (3.6) we see
| F ( x , s ) | d 5 | s | 2 + d 6 | s | p ( d 5 + d 6 r 0 p 2 ) | s | 2 , | s | r 0 , x R N . (56)
This, together with (H1), implies that
| β F ( x , s ) + s f ( x , s ) | β | F ( x , s ) | + | s f ( x , s ) | β ( d 5 + d 6 r 0 p 2 ) | s | 2 + | s ( d 1 | s | + d 2 | s | p 1 ) | ( d 1 + β d 5 + ( d 2 + β d 6 ) r 0 p 2 ) | s | 2 , if  | s | r 0 . (57)
Clearly, (3.12) holds true with a 0 = d 1 + β d 5 + ( d 2 + β d 6 ) r 0 p 2 . In the following theorem, we make the following assumption instead of (V):

(V′): V C ( R N , R ) , inf x R N V ( x ) = V 0 [ 1 2 a 0 + 1 ( β 2 1 ) ] 1 > 0 , where a 0 in (3.12), β in (H5).

Especially, by (V′), we obtain

u 2 2 1 2 a 0 + 1 ( β 2 1 ) u 2 , u E . (58)

Now, we define a class of functionals on E by

J λ ( u ) = 1 2 u 2 λ R N F ( x , u ) d x = A ( u ) λ B ( u ) , λ [ 1 , 2 ] . (59)
It is easy to know that J λ C 1 ( E , R ) for all λ [ 1 , 2 ] and the critical points of J 1 correspond to the weak solutions of problem (1.1). Note that J 1 = J , where J is the functional defined in (2.7).

Theorem 3.2

Assume that (V′), (H1), and (H4)-(H6) hold. Then (1.1) possesses infinitely many weak solutions.

Proof

We first prove that there exist a positive integer k 1 and two sequences r k > ρ k as k such that

a k ( λ ) = inf u Z k , u = ρ k J λ ( u ) > 0 , k k 1 , (60)
b k ( λ ) = max u Y k , u = r k J λ ( u ) < 0 , k N , (61)
where Y k and Z k are defined in Remark 2.10.

Step 1. We claim that (3.14) is true.

By (3.6) and (H4) we have

J λ ( u ) = 1 2 u 2 λ R N F ( x , u ) d x 1 2 u 2 2 R N F ( x , u ) d x 1 2 u 2 2 R N ( d 5 | u | 2 + d 6 | u | p ) d x 1 2 u 2 2 d 5 u 2 2 2 d 6 u p p . (62)
Since E L r ( R N ) with r [ 2 , 2 α ) , and from Theorem 3.1 we have
J λ ( u ) 1 2 u 2 2 d 5 β 2 2 ( k ) u 2 2 d 6 β p p ( k ) u p . (63)
Let ρ k = 1 β 2 ( k ) + β p ( k ) as k . Then there exists k 1 such that 1 2 ρ k 2 2 d 5 2 d 6 > 0 , k k 1 . Therefore,
a k ( λ ) = inf u Z k , u = ρ k J λ ( u ) 1 2 ρ k 2 2 d 5 2 d 6 > 0 , k k 1 . (64)

Step 2. We show that (3.15) is true.

We apply the method in Lemma 2.6 of [[20]] to verify the claim. First, we prove that there exists ε > 0 such that

meas ( x R N : | u ( x ) | ε u ) ε , u X { 0 } , X E  and  dim X < . (65)
There would otherwise exist a sequence { u n } n N X { 0 } such that
meas ( x R N : | u n ( x ) | u n n ) < 1 n , n N . (66)
For each n N , let v n : = u n u n X . Then v n = 1 , n N and
meas ( x R N : | v n ( x ) | 1 n ) < 1 n , n N . (67)
Passing to a subsequence if necessary, we may assume v n v 0 in E for some v 0 X since X is of finite dimension. We easily find v 0 = 1 . Consequently, there exists a constant σ 0 > 0 such that
meas ( x R N : | v 0 ( x ) | σ 0 ) σ 0 . (68)
Indeed, if not, then we have
meas ( x R N : | v 0 ( x ) | 1 n ) = 0 , n N , (69)
which implies
0 R N | v 0 ( x ) | 3 d x v 0 2 2 n 0 as  n . (70)
This leads to v 0 = 0 , contradicting v 0 = 1 . In view of E L p ( R N ) and the equivalence of any two norms on X , we have
R N | v n v 0 | 2 d x 0 as  n . (71)
For every n N , denote
N : = { x R N : | v n ( x ) | < 1 n } and N c : = { x R N : | v n ( x ) | 1 n } , (72)
and N 0 : = { x R N : | v 0 ( x ) | σ 0 } , where σ 0 is defined by (3.20). Then for n large enough, by (3.20), we see
meas ( N N 0 ) meas ( N 0 ) meas ( N c ) σ 0 1 n 3 σ 0 4 . (73)
Consequently, for n large enough, we find
R N | v n v 0 | 2 d x N N 0 | v n v 0 | 2 d x 1 4 N N 0 | v 0 | 2 d x N N 0 | v n | 2 d x ( σ 0 2 4 1 n 2 ) meas ( N N 0 ) 9 σ 0 3 64 > 0 . (74)
This contradicts (3.22). Therefore, (3.17) holds. For the ε given in (3.17), we let
N u : = { x R N : | u ( x ) | ε u } , u X { 0 } . (75)
Then by (3.17), we find
meas ( N u ) ε , u X { 0 } . (76)
As is well known, (H5) implies (H2), and hence for any k N , there is a constant S k > 0 such that
F ( x , u ) | u | 2 ε 3 , | u | S k , (77)
where ε is determined in (3.17). Therefore,
J λ ( u ) = 1 2 u 2 λ R N F ( x , u ) d x 1 2 u 2 N u | u | 2 ε 3 d x ( 1 2 1 ) u 2 . (78)
Now for any k N , if we take r k > max { ρ k , S k ε } , so u = r k is large enough, we have
b k ( λ ) = max u Y k , u = r k J λ ( u ) < 0 , k N . (79)

Step 3. Clearly, J λ C 1 ( E , R ) implies that J λ maps bounded sets to bounded sets uniformly for λ [ 1 , 2 ] . In view of (H6), J λ ( u ) = J λ ( u ) for all ( λ , u ) [ 1 , 2 ] × E . Thus the condition (T1) of Lemma 2.9 holds. Besides, the condition (T2) of Lemma 2.9 holds for the fact that A ( u ) = 1 2 u 2 as u and B ( u ) 0 since F ( x , u ) 0 . Evidently, Step 1 and Step 2 imply that the condition (T3) of Lemma 2.9 also holds for all k k 1 . Consequently, Lemma 2.9 implies that for any k k 1 and a.e. λ [ 1 , 2 ] , there exists a sequence { u n k ( λ ) } n = 1 such that

sup n u n k ( λ ) < , J λ ( u n k ( λ ) ) 0 and J λ ( u n k ( λ ) ) ζ k ( λ ) as  n , (80)
where
B k = { u Y k : u r k } , Γ k = { γ C ( B k , W ) : γ  is odd , γ | B k = id } , ζ k ( λ ) = inf γ Γ k max u B k J λ ( γ ( u ) ) , λ [ 1 , 2 ] . (81)
Furthermore, we easily have ζ k ( λ ) [ a ¯ k , ζ ¯ k ] , k k 1 , where ζ ¯ k : = max u B k J λ ( γ ( u ) ) and a ¯ k : = 1 2 ρ k 2 2 d 5 2 d 6 as k .

Claim 1. { u n k ( λ ) } n = 1 E possesses a strong convergent subsequence in E, a.e. λ [ 1 , 2 ] and k k 1 . In fact, by the boundedness of { u n k ( λ ) } n = 1 , passing to a subsequence, as n , we may assume u n k ( λ ) u k ( λ ) in E. By the method of Theorem 3.1, we easily prove that u n k ( λ ) u k ( λ ) strongly in E.

Thus, for each k k 1 , we can choose λ l 1 such that for the sequence { u n k ( λ l ) } n = 1 we have obtained a convergent subsequence, and passing again to a subsequence, we may assume

lim n u n k ( λ l ) = u l k  in  E , l N  and  k k 1 . (82)
Thus we obtain
J λ l ( u l k ) = 0 and J λ l ( u l k ) [ a ¯ k , ζ ¯ k ] , l N  and  k k 1 . (83)

Claim 2. { u l k } is bounded in E and has a convergent subsequence with the limit u k E for all k k 1 . For convenience, we set u l k = u l for all l N . Consequently, (3.12) and (H5) imply that

β J λ l ( u l ) ( J λ l ( u l ) , u l ) = ( β 2 1 ) u l 2 + λ l R N [ β F ( x , u l ( x ) ) + f ( x , u l ( x ) ) u l ( x ) ] d x = ( β 2 1 ) u l 2 + λ l | u l | r 0 [ β F ( x , u l ( x ) ) + f ( x , u l ( x ) ) u l ( x ) ] d x + λ l | u l | > r 0 [ β F ( x , u l ( x ) ) + f ( x , u l ( x ) ) u l ( x ) ] d x ( β 2 1 ) u l 2 λ l | u l | r 0 | β F ( x , u l ( x ) ) + f ( x , u l ( x ) ) u l ( x ) | d x ( β 2 1 ) u l 2 λ l a 0 u l 2 2 ( β 2 1 ) u l 2 2 a 0 u l 2 2 ( β 2 1 ) u l 2 2 a 0 2 a 0 + 1 ( β 2 1 ) u l 2 = 1 2 a 0 + 1 ( β 2 1 ) u l 2 . (84)
Therefore, { u l } l = 1 is bounded in E. By Claim 1, we see that { u l } l = 1 has a convergent subsequence, which converges to an element u k W for all k k 1 .

Hence, passing to the limit in (3.24), we see J 1 ( u k ) = 0 and J 1 ( u k ) [ a ¯ k , ζ ¯ k ] , l N and k k 1 . Since a ¯ k as k , we get infinitely many nontrivial critical points of J 1 = J . Therefore (1.1) possesses infinitely many nontrivial solutions by Lemma 2.9. This completes the proof. □

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

WD and JX obtained the results in a joint research. All the authors read and approved the final manuscript.

Acknowledgements

The authors are highly grateful for the referees’ careful reading of this paper and their comments. Supported by the NNSF-China (11371117), Shandong Provincial Natural Science Foundation (ZR2013AM009) and Hebei Provincial Natural Science Foundation (A2012402036).

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