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Article

Infinitely Many Solutions for the Fractional p&q-Laplacian Problems in RN

1
College of Intelligent Systems Science and Engineering, Harbin Engineering University, Harbin 150001, China
2
College of Mathematical Sciences, Harbin Engineering University, Harbin 150001, China
*
Authors to whom correspondence should be addressed.
Symmetry 2022, 14(12), 2486; https://doi.org/10.3390/sym14122486
Submission received: 26 October 2022 / Revised: 15 November 2022 / Accepted: 17 November 2022 / Published: 23 November 2022
(This article belongs to the Section Mathematics)

Abstract

:
In this paper, we consider the following class of the fractional p & q -Laplacian problem: ( Δ ) p s u + ( Δ ) q s u + V ( x ) ( | u | p 2 u + | u | q 2 u ) + g ( x ) | u | r 2 u = K ( x ) f ( x , u ) + h ( u ) , x R N , V : R N R + is a potential function, and h : R R is a perturbation term. We studied two cases: if f ( x , u ) is sublinear, by means of Clark’s theorem, which considers the symmetric condition about the functional, we get infinitely many solutions; if f ( x , u ) is superlinear, using the symmetric mountain-pass theorem, infinitely many solutions can be obtained.

1. Introduction

In recent years, the fractional Laplacian problem has gained wide applications in numerous fields of science(see [1,2,3]). Xiang et al. [4] got two solutions for the Kirchhoff-type problem involving the non-local fractional p-Laplacian ( Δ ) p s u . Torres et al. [5] established an existence theorem for a nontrivial solution for the following equation:
( ) p s u + V ( x ) | u | p 2 = f ( x , u ) in R N ,
when the potential V C ( R N ) satisfies
( V ) inf R N V ( x ) V 0 > 0 and μ { x R N : V ( x ) M } < + , M > 0 ,
where μ denotes the Lebesgue measure in R N . Many authors studied the fractional Laplacian equation; see, for instance, [6,7,8,9,10,11,12,13,14,15,16,17]. The nonlocal operator ( Δ ) m s is the fractional m-Laplacian operator, defined as
( Δ ) m s u ( x ) = 2 lim ϵ 0 R N \ B ε ( x ) | u ( x ) u ( y ) | m 2 ( u ( x ) u ( y ) ) | x y | N + s m d y , x R N ,
where m ( p , q ) , s ( 0 , 1 ) , 2 p q < N s .
Recently, the fractional p & q -Laplacian equation has been gaining more and more attention. For instance, Chen and Bao [18] applied a version of the symmetric mountain pass lemma in [19] and adapted some ideas developed by Pucci et al. [20] and Xiang et al. [4] to obtain multiplicity results for the following equation:
( Δ ) p s u + ( Δ ) q s u + a ( x ) | u | p 2 u + b ( x ) | u | q 2 u + μ ( x ) | u | r 2 u = λ h ( x ) | u | m 2 u .
Vincenzo [21] obtained a nontrivial solution by using the mountain-pass theorem [22] in the following equation and a suitable version of the Lions’compactness result [23]:
( Δ ) p s u + ( Δ ) q s u + | u | p 2 u + | u | q 2 u + | u | q * s 2 u = λ h ( x ) f ( x ) , x R N .
In terms of the existence of solutions of the fractional p & q -Laplacian equation, there are only a few results [5,24,25,26].
Motivated by the above papers, the purpose of this paper was to study the following problem:
( Δ ) p s u + ( Δ ) q s u + V ( x ) ( | u | p 2 u + | u | q 2 u ) + g ( x ) | u | r 2 u = K ( x ) f ( x , u ) + h ( u ) , x R N ,
where m { p , q } , s ( 0 , 1 ) , 2 p r q < N s . We assume for f ( x , t ) , K ( x ) , V ( x ) , g ( x ) and h ( u ) that:
( h 1 ) h ( u ) C ( R , R ) , h is odd; there exist δ > 0 , C > 0 , 1 γ < p q , such that f ( x , t ) C ( R N × [ δ , δ ] , R ) ; f is odd in u; | f ( x , u ) | C | u | γ 1 ; and lim u 0 F ( x , u ) | u | p = + uniformly in some ball B r ( x 0 ) R N , where F ( x , u ) = 0 u f ( x , s ) d s .
( h 2 ) There exist β 0 > 0 , such that h ( u ) β 0 , f C ( R N + 1 , R ) ; there exist constants c 1 , c 2 > 0 and q 1 ( q , q s * ) , q s * = N q N s q such that
| f ( x , u ) | c 1 | u | q 1 + c 2 | u | q 1 1 , ( x , u ) R N × R .
( h 3 ) h ( u ) , f ( x , u ) is an odd function according to u, that is, h ( u ) = h ( u ) , u R N , f ( x , u ) = f ( x , u ) , ( x , u ) R N × R .
( h 4 ) H ( u ) 1 c q ρ q 4 p , where 0 < ρ < 1 , c q = 2 q 1 , H ( u ) = 0 u h ( t ) d t , lim | u | | F ( x , u ) | | u | q = , a . e . x R N , and there exists r 0 0 such that
F ( x , u ) 0 , ( x , u ) R N × R , | u | r 0 .
( h 5 ) There exist μ > q , 1 μ u h ( u ) H ( u ) 0 , for u R , 1 μ u f ( x , u ) F ( x , u ) 0 .
( V 1 ) K , g C ( R N , R + ) , V C ( R N , R ) , V ( x ) α 1 and K ( x ) β 1 for some α 1 > 0 , β 1 > 0 and, M : = K p p γ V γ p γ L 1 ( R N ) .
( V 2 ) g , v C ( R N , R + ) : there exists constant V 0 > 0 , such that for all x R N , V ( x ) V 0 .
( V 3 ) There exists r > 0 , α > N ( m 1 ) , such that for any b > 0 ,
lim | y | μ ( { x R N | V ( x ) | x | α b } B r ( y ) ) = 0 ,
where μ ( · ) is the Lebesgue measure in R N .
The Ambrosetti–Rabinowitz (AR) condition ( h 3 ) plays an important role in showing that any Palais–Smale sequence is bounded in the work. As shown in [27], condition ( h 3 ) is somewhat weaker than the local growth condition. However, there are some functions which do not satisfy condition ( h 3 ) ; for example, f ( x , u ) = | u | q 2 u ln ( 1 + | u | ) .
The main purpose of this paper is to generalize the main results of [28,29]. In the next section, we will give some preliminary results and some notation, due to the presence of the general function, and the hypotheses on potentials V in a more refined form are necessary, such as in Lemma 5. We present the variational framework of the problem and compactness results, which will be useful for the next sections. In Section 3, we prove Theorems 1 and 2, in the proof of Theorem 1 we adapt some arguments that can be found in [28].

2. Preliminaries

In this section, we collect the notation and introduce a new function space needed in our approach, providing some of its properties. For any Banach space ( E ( Ω ) , · ) , we will denote
B ρ = { u E : u u 0 ρ } ( u 0 E , ρ > 0 ) .
The Gagliardo seminorm is defined by
[ u ] s , m = ( R 2 N | u ( x ) u ( y ) | m | x y | N + s m d x d y ) 1 m , m { p , q } .
The W s , m ( R N ) space is defined by
W s , m ( R N ) = { u L m ( R N ) , [ u ] s , m < } ,
and it is equipped with the norm
u s , m = ( u m m + [ u ] s , m m ) 1 m , u s = ( R N | u | s d s ) 1 s .
Let us introduce the W V s , m ( R N ) space
W V s , m ( R N ) = { u W s , m ( R N ) , R N V ( x ) | u | m d x < + } ,
with the corresponding norm
u W V s , m = [ u ] s , m m + u m , V m 1 m , u m , V m = R N V ( x ) | u | m d x .
In order to obtain solution of problem (2), we will work in the following linear subspace:
E = W V s , p ( R N ) W V s , q ( R N )
which is a uniformly convex Banach space (similar to [15]), endowed with the norm
u E = m = p , q u W V s , m = u W V s , p + u W V s , q .
Let φ E be fixed, and denote by B φ the linear functional on E defined by
D φ , m ( v ) = R 2 N | φ ( x ) φ ( y ) | m 2 ( φ ( x ) φ ( y ) ) ( v ( x ) v ( y ) ) | x y | N + s m d x d y ,
D φ ( v ) = m = p , q D φ , m ( v ) = D φ , p ( v ) + D φ , q ( v ) for all v E .
We define Φ : E E * as
Φ ( u ) , v = D φ ( v ) + m = p , q R N V ( x ) | u ( x ) | m 2 u ( x ) d x ,
where E * denotes the dual space of E and · , · denotes the pairing between E and E * . Suppose u is a weak solution of problem (2); then,
Φ ( u ) , v = R N K ( x ) f ( x , u ) v d x + R N h ( u ) v d x ,
for all v E .
Lemma 1
([15]). Let ( V 2 ) hold. If ν [ m , m s * ) , m { p , q } , m s * = N m N s m , then the embeddings W V s , m ( R N ) W s , m ( R N ) L ϑ ( R N ) are continuous with
min { 1 , V 0 } u W s , m ( R N ) m u W V s , m ( R N ) m
for all u W V s , m ( R N ) . Moreover, for any R > 0 , ϑ [ 1 , m s * ) , the embedding W V s , m ( R N ) L ϑ ( B R ( 0 ) ) is compact.
Proposition 1.
Suppose I C 1 ( E , R 1 ) is a functional on Banach space E, dual space ( E * , · E * )
(1) 
I satisfies the C c condition for c R 1 , if for any sequence { x n } E with
I ( x n ) c , I ( x n ) E * ( 1 + x n E ) 0 .
(2) 
I satisfies the ( P S ) c condition for c R 1 , if for any sequence { x n } E with
I ( x n ) c , I ( x n ) 0 i n E * .
Wang and Liu [28] give an abstract setting of Clark’s theorem which relies on the symmetric condition of the functional.
Lemma 2
([28]). Suppose I C 1 ( E , R 1 ) is a functional on real Banach space E, I ( 0 ) = 0 and I ( u ) = I ( u ) for all u E , if I satisfies the ( P S ) c condition and is bounded from below. There exists a subspace E k of E and ρ k > 0 such that sup E k S ρ I < 0 , where dim E = k , S ρ = { u E u = ρ } ; then, at least one of the following conclusions holds.
(i) 
There exists a sequence of critical points u k satisfying I ( u k ) < 0 for all k and u k 0 as k .
(ii) 
There exists r > 0 such that for any 0 < a < r there exists a critical point u such that u = a and I ( u ) = 0 .
The following the classical symmetric version of mountain-pass theorem was established in [29,30,31].
Lemma 3
([29,30,31]). Suppose I C 1 ( E , R 1 ) is a functional on Banach space E, E = X Y , dim E < , dim Y < , I ( 0 ) = 0 and I ( u ) is even for all u E , if I satisfies the C c condition for all c > 0 , and
I
there exist constants ρ , α > 0 such that I | B ρ X α ;
I
for any finite dimensional subspace E ˜ E , there is R = R ( E ˜ ) > 0 such that I ( u ) 0 on E ˜ \ B R ;
then I possesses an unbounded sequence of critical values.
Lemma 4
([8]). For any s > 0 , the space C 0 ( R N ) of smooth functions with compact support is dense in W s , p ( R N ) .
Lemma 5.
Suppose that ( V 2 ) and ( V 3 ) are fulfilled. If { u j } is a bounded sequence in E, then there exists u E L ϑ ( R N ) such that up to a subsequence,
u j u s t r o n g l y i n L ϑ ( R N )
as j , for any ϑ [ 1 , m s * ) .
Proof. 
We divide our proof in two steps.
Since { u j } j is a bounded sequence in W V s , m , up to a subsequence, there exist a positive constant C and u W V s , m . Let v n = u n u , such that v n 0 ( n ) in W V s , m and v n W V s , m C .
Step 1. We first consider the case ϑ = 1 . The embedding theorem in bounded domains implies that u j u strongly in L ( B R ) . To estimate B R c | u n | d x , let us first choose ( y i ) i R N , such that R N i = 1 B r ( y i ) , and each x R N is covered by at most 2 N such balls. Set A b ( y ) : = { x R N : V ( x ) | x | α b } B r ( y ) and C b ( y ) : = { x R N : V ( x ) | x | α > b } B r ( y ) . For R > 2 r , we get
B R c | v n | d x | y i | R r B r ( y i ) | v n | d x = | y i | R r A b ( y i ) | v n | d x + | y i | R r C b ( y i ) | v n | d x .
Then,
| y i | R r A b ( y i ) | v n | d x | y i | R r A b ( y i ) | v n | m s * d x 1 m s * A b ( y i ) | 1 | m s * m s * 1 d x m s * 1 m s * | y i | R r v n L m * ( B r ( y i ) ) sup | y i | R r [ μ ( A b ( y i ) ) ] m s * 1 m s * 2 N v n L m * ( B R 2 r c ) sup | y i | R r [ μ ( A b ( y i ) ) ] m s * 1 m s * 2 N C m * sup | y i | R r [ μ ( A b ( y i ) ) ] m s * 1 m s * ,
where C m * is a positive constant depending on m * .
Set D n : = { x R N : | x | α | v n | m 1 1 } ; with α > N ( m 1 ) , we have
| y i | R r C b ( y i ) | v n | d x = | y i | R r C b ( y i ) D n c | v n | d x + | y i | R r C b ( y i ) D n | v n | d x | y i | R r C b ( y i ) D n c | x | α | v n | | x | α d x + | y i | R r C b ( y i ) D n | x | α m 1 d x
| y i | R r C b ( y i ) | x | α | v n | m d x + | y i | R r C b ( y i ) | x | α m 1 d x 2 N b B R 2 r c V ( x ) | v n | m d x + B R 2 r c | x | α m 1 d x 2 N b u W V s , m m + 2 N ( R 2 r ) α m 1 N 2 N C m b + 2 N ( R 2 r ) α m 1 N .
Then,
B R c | v n | d x 2 N C m * sup | y i | R r [ μ ( A b ( y i ) ) ] m s * 1 m s * + 2 N C m b + 2 N ( R 2 r ) α m 1 N .
Choose b > 0 such that 2 N b C m < ε 3 , so for such a fixed b > 0 , there exists R > 0 such that
2 N C m * sup | y i | R r [ μ ( A b ( y i ) ) ] m s * 1 m s * < ε 3
and
2 N ( R 2 r ) α m 1 N < ε 3 .
Thus, u j u strongly in L ( R N ) .
Step 2. For 1 < ϑ < m s * , δ = 1 ϑ m s * ϑ m s * 1 ( 0 , 1 ) such that 1 ϑ = δ + 1 δ m s * and
v n L ϑ ( R N ) v n L ( R N ) δ v n L m s * ( R N ) 1 δ 0 , as n ,
since { v n } n bounded in L m s * ( R N ) . □

3. Main Results

In the following, we construct proper functional and get some properties.
Choose f ^ ( x , t ) C ( R N × R , R ) and h ( u ) C ( R , R ) to be odd, and define
f ^ ( x , t ) = f ( x , t ) , x R N and | t | a / 2 , f ^ ( x , t ) = f ( x , a / 2 ) δ / 2 ( t a ) , x R N and a / 2 < t < a , f ^ ( x , t ) = f ( x , a / 2 ) δ / 2 ( t + a ) , x R N and a < t < a / 2 , f ^ ( x , t ) = 0 , x R N and | t | a .
h ^ ( u ) = h ( u ) , | u | a , h ^ ( u ) = 0 , | u | > a .
Now, in order to prove the existence of solutions of problem (1), we consider
( Δ ) p s u + ( Δ ) q s u + V ( x ) ( | u | p 2 u + | u | q 2 u ) + g ( x ) | u | r 2 u = K ( x ) f ^ ( x , u ) + h ^ ( u ) , x R N ,
and its associated functional
I ( u ) = u W V s , p p p + u W V s , q q q + u r , g r r R N K ( x ) F ^ ( x , u ) d x R N H ^ ( u ) d x ,
where F ^ ( x , u ) = 0 u f ^ ( x , t ) d t , H ^ ( u ) = 0 u h ^ ( t ) d t , u r , g = ( R N g ( x ) | u | r 2 u d x ) 1 r .
Lemma 6.
Assume that f and V satisfy ( h 1 ) and ( V 1 ) . Then, the functional I ( u ) satisfies the ( P S ) c condition in E.
Proof. 
It is easy see that I ( u ) C 1 ( E , R ) , I ( u ) is even and I ( 0 ) = 0 . Then, we have
R N K ( x ) | F ^ ( x , u ) | d x C 1 R N K ( x ) | u | γ = C 1 M L 1 ( R N ) p γ p V | u | p L 1 ( R N ) γ p C 2 u W V s , p γ .
In addition,
| R N H ^ ( u ) d x | = | | u | a H ^ ( u ) d x + | u | > a H ^ ( u ) d x | | u | a | H ^ ( u ) | d x C .
Therefore,
I ( u ) u W V s , p p p + u W V s , q q q + u r , g r r C 2 u W V s , p γ C , u E ,
when u E , for u W V s , p and u W V s , q —at least one of them approaches infinity—then I ( u ) is coercive and bounded below.
Let { u n } E be a ( P S ) c sequence for I ( u ) , that is,
I ( u n ) c , I ( u n ) 0 in E * .
As { u n } is bounded in E, we may assume that { u n } u in E. From Lemma 1, suppose that
u n ( x ) u ( x ) a . e . in B R ( 0 ) , u n u in L ϑ ( B R ( 0 ) ) ,
{ u n } and u are bounded in L ϑ ( R N ) , where ϑ [ 1 , m s * ) .
To prove that { u n } converges strongly to u in E, for any R > 0 , we have
R N K ( x ) | f ^ ( x , u n ) f ^ ( x , u ) | | u n u | d x c R N \ B R ( 0 ) K ( x ) ( | u n | γ + | u | γ ) d x + c B R ( 0 ) ( | u n | γ 1 + | u | γ 1 ) ( u n u ) d x c ( V | u n | p L 1 ( R N \ B R ( 0 ) ) γ p + V | u | p L 1 ( R N \ B R ( 0 ) ) γ p ) M L 1 ( R N \ B R ( 0 ) ) p γ p + c ( u n L γ ( B R ( 0 ) γ 1 + u n L γ ( B R ( 0 ) ) γ 1 ) u n u L γ ( B R ( 0 ) c 1 M L 1 ( R N \ B R ( 0 ) ) p γ p + c 1 u n u L γ ( B R ( 0 ) ,
where c 1 and c are positive constants, which implies
lim n R N K ( x ) | f ^ ( x , u n ) f ^ ( x , u ) | | u n u | d x = 0 .
From this and the Lebesgue dominated convergence theorem, for | u | > a , h ^ ( u ) = 0 , we have
lim n R N | h ^ ( u n ) h ^ ( u ) | | u n u | d x = 0 .
By u n u in E and I ( u ) = 0 in E * , as n , we can get that
I ( u n ) I ( u ) , u n u = [ D u n ( u n u ) D u ( u n u ) ] + m = p , q R N V ( x ) ( | u n | m 2 u n | u | m 2 u ) ( u n u ) d x + R N g ( x ) ( | u n | r 2 u n | u | r 2 u ) ( u n u ) d x R N K ( x ) | f ^ ( x , u n ) f ^ ( x , u ) | | u n u | d x R N | h ^ ( u n ) h ^ ( u ) | | u n u | d x = o ( 1 ) .
After that, according to the inequalities
| ξ η | m c m ( | ξ | m 2 ξ | η | m 2 η ) ( ξ η ) , for m 2 , ξ , η R N ,
where c m is a positive constant depending on m.
By ( V 1 ), we have
R N g ( x ) ( | u n | r 2 u n | u | r 2 u ) ( u n u ) d x R N g ( x ) | u n u | r d x 0 .
As n
D u n ( u n u ) D u ( u n u ) c m = p , q R N | u n ( x ) u n ( y ) u ( x ) + u ( y ) | m | x y | N + s m .
Similarly, by ( V 1 ) , as n
m = p , q R N V ( x ) ( | u n | m 2 u n ) | u | m 2 u ) ( u n u ) d x m = p , q c m u n u m , V m .
In conclusion, by (3)–(8), we have u n u E 0 as n , as needed. Then, I satisfies the ( P S ) c condition. □
Theorem 1.
Assume that f, h and V satisfy ( h 1 ) and ( V 1 ) . Then, the problem (2) has infinitely many solutions { u k } such that u k L 0 as k .
Proof. 
By ( h 1 ) , we have that for any M 0 > 0 , there exists δ = δ ( M 0 ) > 0 ; if u C 0 ( B r ( x 0 ) ) and | u | < δ , then
F ^ ( x , u ( x ) ) M 0 | u ( x ) | p .
For any k N , if E k is a k dimensional subspace of E, such that any norms in E k are equivalent, there exist d 1 , d 2 > 0 such that
d 1 u E u r , g d 2 u E , d 1 u E u p d 2 u E , u E k .
Let ρ k be small enough. Then, for any u E k S ρ k , there is a constant C k > 0 , such that C k ρ k < | u | < δ , where S ρ k = { u E | u E = ρ k } ; thus,
I ( u ) = u W V s , p p p + u W V s , q q q + u r , g r r R N K ( x ) F ^ ( x , u ) d x R N H ^ ( u ) d x 2 u E p p + d 2 u E r r M 0 u p p + C ( 2 p + d 2 r M 0 C p ) ρ k p + C < 0 .
Using Lemma 5, we can obtain infinitely many solutions { u k } for problem (3).
Finally, let u be a solution of (2). Multiply both sides by | u T | ν u T ( x ) , where T > 0 , u T ( x ) = max { T , min { u ( x ) , T } } , and then
m = p , q R 2 N | u ( x ) u ( y ) | m 2 ( u ( x ) u ( y ) ) ( | u T | ν u T ( x ) | u T | ν u T ( y ) ) | x y | N + s m d x d y + m = p , q R N V ( x ) ( | u | m 2 u | u T | ν u T ( x ) ) d x + R N g ( x ) ( | u | r 2 u | u T | ν u T ( x ) ) d x = R N f ^ ( x , u ) | u T | ν u T ( x ) d x + R N h ^ ( u ) | u T | ν u T ( x ) d x .
Taking into account the definition of u T , we have
m = p , q R 2 N | u ( x ) u ( y ) | m 2 ( u ( x ) u ( y ) ) ( | u T | ν u T ( x ) | u T | ν u T ( y ) ) | x y | N + s p d x d y m = p , q R 2 N | u T ( x ) u T ( y ) | m 2 ( u T ( x ) u T ( y ) ) ( | u T | ν u T ( x ) | u T | ν u T ( y ) ) | x y | N + s m d x d y .
Using the following inequalities ([32])
| d b | m 2 ( d b ) ( w ( d ) w ( b ) ) | W ( d ) W ( b ) | m ,
where w : R R is an increasing function, W ( t ) = 0 t ( w ( τ ) ) 1 / m d τ , we can see that
R 2 N | u T ( x ) u T ( y ) | m 2 ( u T ( x ) u T ( y ) ) ( | u T | ν u T ( x ) | u T | ν u T ( y ) ) | x y | N + s m d x d y ( ν + 1 ) m m ( ν + m ) m R 2 N | | u T ( x ) | ν m u M ( x ) | u T ( y ) | ν m u T ( y ) | m | x y | N + s m d x d y .
By ( h 1 ) , we have f ^ ( x , t ) < c | t | γ 1 ; then,
( ν + 1 ) q q ( ν + q ) q R 2 N | u T ( x ) | ν q u T ( x ) | u T ( y ) | ν q u T ( y ) | q | x y | N + s q d x d y C R N | u T ( x ) | ν + γ d x + C R N | u T ( x ) | ν + 1 d x C 1 R N | u T ( x ) | ν + γ 1 d x
( γ 1 = γ , when R N | u T ( x ) | ν + γ d x R N | u T ( x ) | ν + 1 d x ; γ 1 = 1 , otherwise).
Together with Lemma 1, it is implied that
u T L ( ν + q ) N / ( N s q ) ( R N ) ( D 1 ( ν + q ) ) q / ( ν + q ) u T L ν + γ 1 ( R N ) ( ν + γ 1 ) / ( ν + q ) ,
for some D 1 1 independent of u and ν .
Set ν 0 = q s * γ 1 and ν k = ( ν k 1 + q ) N N s q γ 1 ; that is, ν k = ( q s * / q ) k + 1 1 ( q s * / q ) 1 ν 0 , for k = 1 , 2 , · · · . It is easy to see that ν k + , as k + . From the last inequality, an iterating process leads to
u T L ν k + 1 + γ 1 ( R N ) exp ( i = 0 k q ln ( D 1 ( ν i + q ) ) ν i + q ) u T L q s * ( R N ) ν k ,
where ν k = i = 0 k ( ν i + γ 1 ) / ( ν i + q ) .
Letting T + , we deduce
u L ( R N ) exp ( i = 0 q ln ( D 1 ( ν i + q ) ) ν i + q ) u L q s * ( R N ) ν ,
where ν = i = 0 ( ν i + γ 1 ) / ( ν i + q ) with 0 < ν < 1 and exp ( i = 0 q ln ( D 1 ( ν i + q ) ) ν i + q ) is a positive number. The proof is completed. □
Remark 1.
There are functions f satisfying condition of Theorem 1, which are applicable to indefinite problems, such as problems on periodic solutions of first-order Hamiltonian systems [28].
We will work space E with a sequence of finite dimensional subspaces. We know that C 0 ( R N ) is separable, and by Lemma 4, C 0 ( R N ) is dense in W s , m ( R N ) , so E W s , m ( R N ) is separable. Note that E is a reflexive Banach space. Then, from ([33]), there are { e i } i N E and { e i * } i N E * such that E = s p a n { e i : i N } ¯ , E * = s p a n { e i * : i N } ¯ , and
e n , e m = 1 , i = j , 0 , i j .
For k = 1 , 2 , , let X k = s p a n { e 1 , , e k } , Y k = s p a n { e k , e k + 1 , } ¯ . We first give a preliminary Lemma 7.
Lemma 7.
If ( V 2 ) and ( V 3 ) hold, then we have that
β k ( ϑ ) : = lim u Z k , u E = 1 u ϑ 0 , k ,
where m ϑ < m s * .
Proof. 
It is obvious that 0 < β k + 1 ( ϑ ) β k ( ϑ ) , so that β k ( ϑ ) β ( ϑ ) 0 , k . For every k 0 , there exists u k Z k , such that u k E = 1 and u k ϑ > β k ( ϑ ) / 2 . As E is reflexive, such that, up to a subsequence, u k u E . Since e i * , u k = 0 , if k > i , then lim k e i * , u k = e i * , u = 0 for all i N . By definition of Z k , u k 0 in E, Lemma 1 implies that u k 0 in L ϑ ( R N ) . Thus, we have that β ( ϑ ) = 0 . □
Theorem 2.
If K ( x ) = 1 , x R N —and assume that f, g and V satisfy ( h 2 ) , ( h 3 ) , ( h 4 ) , ( h 5 ) , ( V 2 ) and ( V 3 ) —then problem (1) has infinitely many solutions.
Proof. 
Let J ( u ) be the energy functional associated with (1) defined by
J ( u ) = u W V s , p p p + u W V s , q q q + u r , g r r R N K ( x ) F ( x , u ) d x R N H ( u ) d x ,
where F ( x , u ) = 0 u f ( x , t ) d t .
The proof of Theorem 2 is divided into three steps as follows.
Step 1. J satisfies C c condition.
Let u n E be a C c sequence for J ( u ) ; that is, J ( u n ) c , J ( u n ) E * ( 1 + u E ) 0 , which shows that
c = J ( u n ) + o ( 1 ) , J ( u n ) , u n = o ( 1 ) ,
where o ( 1 ) 0 as n .
Just suppose that u n E + ; then we have the following two cases:
(1)
u n W V s , p .
(2)
u n W V s , p is bounded, u n W V s , q .
In either case, observe that for n large enough
c + 1 J ( u n ) 1 μ J ( u n ) , u n = ( 1 p 1 μ ) u n W V s , p p + ( 1 q 1 μ ) u n W V s , q q + ( 1 r 1 μ ) u n r , g r + 1 μ R N f ( x , u n ) u n d x 1 μ R N μ F ( x , u n ) d x + 1 μ R N h ( u n ) u n d x 1 μ R N μ H ( u ) d x ( 1 p 1 μ ) u n W V s , p p + ( 1 q 1 μ ) u n W V s , q q ,
which is a contradiction; thus, { u n } is bounded in E, up to a subsequence, still denoted by { u n } n . From Lemma 5, we have
u n u in E , u n u in L ϑ ( R N ) , u n u a . e . in R N ,
where ϑ [ m , m s * ) .
By ( h 2 ) , c 1 , c 2 > 0 and q 1 ( q , q s * ) such that
| f ( x , u ) | c 1 | u | q 1 + c 2 | u | q 1 1 , ( x , u ) R N × R .
Using the Hölder inequality, we get
R N | f ( x , u n ) f ( x , u ) | | u n u | d x c 1 u n q q 1 u n u q q + c 2 u n q 1 q 1 1 u n u q 1 q 1 + c 1 u q q 1 u n u q q + c 2 u q 1 q 1 1 u n u q 1 q 1 ,
which implies that
lim n R N | f ( x , u n ) f ( x , u ) | | u n u | d x = 0
and
lim n R N | h ( u n ) h ( u ) | | u n u | d x = lim n R N ( | h ( u n ) | + | h ( u ) | ) | u n u | = 0 .
Obviously
J ( u n ) J ( u ) , u n u 0 .
By a similar fashion as (4)–(8), we can get that
u n u E 0 , n .
Then, J ( u ) satisfies the C c condition.
Step 2. For any finite dimensional subspace E ˜ E , then J ( u ) , u n E , u E ˜ .
Arguing indirectly, assume that for some sequence { u n } E ˜ with u n E , we have the two cases:
(1)
u n W V s , p .
(2)
u n W V s , p is bounded, u n W V s , q .
In the first case, for n sufficiently large, we have that u n W V s , p q p 1 ; hence
u n W V s , p p + u n W V s , q q u n W V s , p q + u n W V s , q q u n E q ;
then
lim n + u n W V s , p p p + u n W V s , q q q u n E q lim n + 1 p ( u n W V s , p q + u n W V s , q q ) u n E q lim n + 1 p u n E q u n E q = 1 p .
In the second case, u n W V s , q q u n E q , we have
lim n + u n W V s , p p p + u n W V s , q q q u n E q lim n + u n W V s , p p p + u n W V s , q q q u W V s , q q = 1 q < 1 p .
It is well known that any norms in finite-dimensional subspace E ˜ are equivalent. Thus, there exist d 1 , d 2 > 0 such that
d 1 u E u r , g d 2 u E , u E ˜ .
There is M > 0 , such that J ( u n ) M . Let v n = u n / u n E ; then, v n E = 1 , q r s < q s * , we have v n s r s v n = r s . Passing to a subsequence, we suppose that v n v in E, and then v n v E ˜ , v n v a . e . on R N . Let A : = { x R N , v ( x ) 0 } and Ω n ( a , b ) = { x R N : a | u n ( x ) | < b } . If v 0 , then m e a s ( A ) > 0 . For a . e . x A , we have lim n + | u n ( x ) | = . Hence, A Ω n ( r 0 , ) for large n N , we have
lim n + M u n E q lim n + J ( u n ) u n E q = lim n + u n W V s , p p p + u n W V s , q q q + u n r , g r r R N F ( x , u n ) d x R N H ( u n ) d x u n E q 1 p lim n + [ R N F ( x , u n ) u n E q d x ] + lim n + d 2 u n E r r + c q ρ q 4 p u n E q 1 p lim n + [ Ω n ( 0 , r 0 ) F ( x , u n ) u n E q d x + Ω n ( r 0 , ) F ( x , u n ) u n E q d x ] .
It follows from ( h 2 ) , ( h 4 ) and Fadou’s Lemma that
1 p lim n + [ Ω n ( 0 , r 0 ) F ( x , u n ) u n E q d x + Ω n ( r 0 , ) F ( x , u n ) u n E q d x ] 1 p + lim sup n + ( C 1 q + C 2 q 1 r 0 q 1 q ) R N | v n | q d x lim inf n + Ω n ( r 0 , ) F ( x , u n ) d x | u n | q | v n | q d x 1 p + ( C 1 q + C 2 q 1 r 0 q 1 q ) r q q lim inf n + Ω n ( r 0 , ) F ( x , u n ) | u n | q | v n | q d x 1 p + ( C 1 q + C 2 q 1 r 0 q 1 q ) r q q R N lim inf n + | F ( x , u n ) | | u n | q [ χ Ω n ( r 0 , ) ( x ) ] | v n | q d x = ,
but lim n + M u n E q = 0 , which is a contradiction.
Step 3. ρ , α > 0 , such that J | B ρ Z m α .
By Lemma 7, we have
β k ( s ) : = lim u Z k , u E = 1 u s 0 , k .
We can find an integer m 1 , such that
u q q q c q 4 p c 1 u E q , u q 1 q 1 q 1 c q 4 p c 2 u E q 1 , u Z m .
For u Z m , by ( h 2 ), we have
J ( u ) = u W V s , p p p + u W V s , q q q + u r , g r r R N F ( x , u ) d x R N H ( u ) d x c q u E q p c 1 q R N | u | q d x c 2 q 1 R N | u | q 1 d x c q ρ q 4 p c q u E q p c q u E q 4 p c q u E q 1 4 p c q ρ q 4 p c q u E q 2 p c q ρ q 4 p = c q ρ q 4 p = α > 0 .
Let Y = Y m and X = Z m , based on the above steps, by Lemma 4; then, the problem (1) possesses infinitely many nontrivial solutions. □
Remark 2.
Our hypotheses are similar to those employed by Tang [29]. There are functions V satisfying ( V 1 ) and not satisfying (V). Condition ( V 1 ), which is weaker than (V), was introduced by Bartsch et al. [34].

4. Conclusions

We studied the existence of infinitely many solutions for the fractional p & q -Laplacian problem involving the vanishing behavior at infinity.
In Theorem 1, function f does not satisfy the classical A R condition. As in the standard case of the fractional p & q -Laplacian, the main difficulty is to prove the Palais–Smale ( P S ) condition for the energy functional associated with the problem.
In Theorem 2, F ( x , t ) is allowed to be sign-changing, and the related energy functional dissatisfies the ( P S ) condition. To overcome this problem, the energy functional satisfying the Cerami condition was proved through giving some assumptions on the nonlinearity term f. Then, we exploited the domain decomposition technique and the classical symmetric version of the mountain-pass theorem in the proof of the existence of infinitely many weak solutions.
All these theorems extend some classical results of fractional p & q -Laplacian equations to the nonlocal condition. Clark’s theorem is an important result in critical point theory which relies on the symmetric condition of Euler–Lagrange functional. The present paper was devoted to establishing a new variant of the Clark’s theorem for nonsmooth functionals which do not satisfy the Palais–Smale condition. In the future, we will try to use a similar non-symmetric condition of the Clark’s theorem for solving the problem.

Author Contributions

Writing—original draft, writing—review editing, methodology, L.W.; conceptualization, L.W. and K.C.; investigation, J.S.; supervision, B.G. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Wang, L.; Chi, K.; Shen, J.; Ge, B. Infinitely Many Solutions for the Fractional p&q-Laplacian Problems in RN. Symmetry 2022, 14, 2486. https://doi.org/10.3390/sym14122486

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Wang L, Chi K, Shen J, Ge B. Infinitely Many Solutions for the Fractional p&q-Laplacian Problems in RN. Symmetry. 2022; 14(12):2486. https://doi.org/10.3390/sym14122486

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Wang, Liyan, Kun Chi, Jihong Shen, and Bin Ge. 2022. "Infinitely Many Solutions for the Fractional p&q-Laplacian Problems in RN" Symmetry 14, no. 12: 2486. https://doi.org/10.3390/sym14122486

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