Least energy sign-changing solutions for Schrödinger-Poisson systems with potential well

Xiao-Ping Chen; Chun-Lei Tang

doi:10.1515/ans-2022-0021

Article Open Access

Least energy sign-changing solutions for Schrödinger-Poisson systems with potential well

Xiao-Ping Chen and Chun-Lei Tang

Published/Copyright: August 25, 2022

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From the journal Advanced Nonlinear Studies Volume 22 Issue 1

Abstract

In this article, we investigate the existence of least energy sign-changing solutions for the following Schrödinger-Poisson system

− Δ u + V ( x ) u + K ( x ) ϕ u = f ( u ) , x ∈ R 3 , − Δ ϕ = K ( x ) u 2 , x ∈ R 3 ,

where the functions V ( x ) , K ( x ) have finite limits as ∣ x ∣ → ∞ satisfying some mild assumptions. By combining variational methods with the global compactness lemma, we obtain a least energy sign-changing solution with exactly two nodal domains, and its energy is strictly larger than twice that of least energy solutions.

Keywords: Schrödinger-Poisson system; sign-changing solutions; variational methods

MSC 2010: 35A15; 35B38; 35J50

1 Introduction and main results

In this article, we are concerned with the following Schrödinger-Poisson system:

− Δ u + V ( x ) u + K ( x ) ϕ u = f ( u ) , x ∈ R 3 , − Δ ϕ = K ( x ) u 2 , x ∈ R 3 , ( SP )

where the functions V ( x ) , K ( x ) , and the nonlinearity f satisfy the following assumptions:

( V 0 ) V ∈ C ( R 3 , R ) , 0 < V ∞ ≔ lim ∣ x ∣ → ∞ V ( x ) < + ∞ , and there exists α 0 > 0 such that

(1.1) α 0 ≔ inf u ∈ H 1 ( R 3 ) ⧹ { 0 } ∫ R 3 ( ∣ ∇ u ∣ 2 + V ( x ) ∣ u ∣ 2 ) d x ∫ R 3 ∣ u ∣ 2 d x > 0 ;

( K 0 ) K ∈ L 2 ( R 3 ) , there exist γ > 0 and C K > 0 such that

0 ≤ K ( x ) ≤ C K e − γ ∣ x ∣ , for all x ∈ R 3 , K ( x ) ≢ 0 ;

f ∈ C ( R , R ) and lim t → 0 f ( t ) t = 0 ;
lim ∣ t ∣ → ∞ f ( t ) ∣ t ∣ 4 t = 0 ;
there exists ζ > 0 such that F ( ± ζ ) > 0 , where F ( ζ ) = ∫ 0 ζ f ( s ) d s ;
f ( t ) ∣ t ∣ 3 is a non-decreasing function on t ∈ ( − ∞ , 0 ) ∪ ( 0 , ∞ ) .

System ( SP ) or the more general one

(1.2) − Δ u + V ( x ) u + K ( x ) ϕ u = f ( x , u ) , x ∈ R 3 , − Δ ϕ = K ( x ) u 2 , x ∈ R 3 ,

has been studied extensively, where V , K ∈ C ( R 3 , R ) and f ∈ C 1 ( R 3 × R , R ) . The Schrödinger-Poisson system, of the form similar to system (1.2), also known as the Schrödinger-Maxwell system, was first proposed by Benci and Fortunato [3] to describe the interaction of a charged particle with the electrostatic field in quantum mechanics. For more detailed mathematical and physical background of the Schrödinger-Poisson system, we refer the readers to [4] and references therein.

When K ( x ) ≡ 0 , system (1.2) is reduced to the following well-known Schrödinger equation:

(1.3) − Δ u + V ( x ) u = f ( x , u ) , x ∈ R 3 ,

which has received increasingly more attention under various assumptions on the potential and the nonlinearities. For more results about problem (1.3), we also refer the readers to [12,17,20] and references therein.

In recent years, there has been increasing attention to system (1.2), especially on the existence of positive solutions, least energy (or ground state) solutions, multiple solutions, sign-changing solutions, fractional problems, and so on, see [2,3,8, 9,10,11,13,22,23, 24,25,26, 27,29,30] and references therein. Since ∫ R 3 ϕ u u 2 d x is homogeneous of degree 4, the greatest part of the literature focuses on system (1.2) assume that f ( x , u ) either satisfies the 3-superlinear condition:

( SF ) lim ∣ t ∣ → ∞ F ( x , t ) ∣ t ∣ 4 = + ∞ , where F ( x , t ) = ∫ 0 t f ( x , s ) d s ,

or satisfies the following well-known Ambrosetti-Rabinowitz (AR) condition:

( AR ) there exists μ > 4 such that 0 < μ F ( x , t ) ≤ f ( x , t ) t , for all t ∈ R 3 \ { 0 } .

When inf x ∈ R 3 V ( x ) > 0 , the authors always assumed that V ∈ C ( R 3 , R ) is radial or satisfies suitable assumptions to ensure the compactness of the embedding

H ≔ u ∈ H 1 ( R 3 ) : ∫ R 3 V ( x ) u 2 d x < + ∞ ↪ L q ( R 3 )

for all q ∈ [ 2 , 6 ) . Recently, via the variational methods and the Brouwer degree, Wang and Zhou [27] proved that system (1.2) when f ( x , u ) replaced by ∣ u ∣ p − 1 u ( 3 < p < 5 ) has a least energy sign-changing solution with V ∈ C ( R 3 , R + ) satisfying lim ∣ x ∣ → ∞ V ( x ) = ∞ . Afterward, combining variational methods with the quantitative deformation lemma, Shuai and Wang [22] proved that system (1.2) has a least energy sign-changing solution when f ( x , u ) ≔ f ( u ) satisfies ( f 1 ), ( f 2 ), (SF), and the following monotonicity condition:

( NE ) f ( t ) ∣ t ∣ 3 is an increasing function on t ∈ ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ,

and the potential V ∈ C ( R 3 , R + ) satisfies some assumptions that guarantee the compactness of the Sobolev embedding. More results about the existence of sign-changing solutions with the compactness of the Sobolev embedding, we refer the readers to [13] and references therein.

When V ∈ C ( R 3 , R ) satisfies lim ∣ x ∣ → ∞ V ( x ) = V ∞ ≔ sup x ∈ R 3 V ( x ) ≥ inf x ∈ R 3 V ( x ) > 0 , based on variational methods in association with the quantitative deformation lemma, Alves et al. [2] proved that system ( SP ) possesses a least energy sign-changing solution, when the nonlinearity f ∈ C 1 ( R , R ) satisfies ( f 1 ), ( f 2 ), (NE), and (SF), and the potential V : R 3 → R is locally Hölder continuous and satisfies the following assumption:

( V 1 ) there exist R 0 > 0 and ρ : ( R 0 , ∞ ) → ( 0 , ∞ ) a non-increasing function such that

lim r → ∞ ρ ( r ) e ϱ r = ∞ , for any ϱ > 0 , V ( x ) ≤ V ∞ − ρ ( ∣ x ∣ ) , for any ∣ x ∣ ≥ R 0 .

We especially refer to [9], where Chen and Tang studied the following Schrödinger-Poisson system:

(1.4) − Δ u + V ( x ) u + ϕ u = a ( x ) f ( u ) , x ∈ R 3 , − Δ ϕ = u 2 , x ∈ R 3 ,

when V ∈ C ( R 3 , R ) satisfies ( V 0 ) , ( V 1 ) and a ∈ C ( R 3 , R ) satisfies the following condition:

( a 0 ) inf x ∈ R 3 a ( x ) = a ∞ ≔ lim ∣ x ∣ → ∞ a ( x ) > 0 ,

they proved that system (1.4) possesses a least energy sign-changing solution with exactly two nodal domains, where f satisfies ( f 1 ), ( f 2 ), (NE), and the following super-quadratic condition:

( CF ) lim ∣ t ∣ → ∞ F ( t ) ∣ t ∣ 3 = + ∞ , where F ( t ) = ∫ 0 t f ( s ) d s .

Under the same assumptions on f , when V satisfies ( V 0 ) and the following assumption:

( V 2 ) there exist μ > 0 and C V > 0 such that

V ( x ) ≤ V ∞ − C V e − μ ∣ x ∣ , for all x ∈ R 3 .

Chen and Tang [9] also obtained the existence of least energy sign-changing solutions for system (1.4). The key point in [2,9] is to overcome the lack of compactness of the Sobolev embedding, and authors chose a class of least energy sign-changing critical points { u R } of the corresponding functional in B R ( 0 ) as a minimizing sequence for the nodal level, then proved that u R converges toward a sign-changing critical point of ℐ as R → ∞ , provided that m < c + c ∞ . Here we must point out that the methods to estimate energy used in [2,9] heavily rely on that ϕ w , ϕ v have the same decay as w , v (where w , v denote the positive least energy solutions of system (1.4) and the limit problem, respectively). To the best of our knowledge, even if w has compact support, ϕ w decays at infinity as 1 ∣ x ∣ , the exponential decay of ϕ w , ϕ v cannot hold. More results about the existence of sign-changing solutions of system ( SP ) or similar systems, we refer the readers to [10,11,26,30] for the critical case, [29] for 3-linear case, [13] for the quadratic, and sub-quadratic case, and so on.

Motivated by the aforementioned works, we will further study the existence of least energy sign-changing solutions for system ( SP ) with lack of compactness and improve the results received in [2,9] by using some much weaker conditions ( f 3 ) and ( f 4 ). As we know, there are no works concerning the existence of least energy sign-changing solutions for system ( SP ) with ( f 3 ) until now.

Now, we are ready to state our main results.

Theorem 1.1

Assume that ( V 0 ) , ( V 1 ) , ( K 0 ) , and ( f 1 )–( f 4 ) hold. If γ < 2 V ∞ , then system ( SP ) possesses a least energy sign-changing solution which has exactly two nodal domains, and its energy is strictly larger than twice that of least energy solutions.

Theorem 1.2

Assume that ( V 0 ) , ( V 2 ) , ( K 0 ) , and ( f 1 )–( f 4 ) hold. If μ < min { γ , V ∞ ∕ 2 } , and γ < 2 V ∞ , then all the conclusions of Theorem 1.1 remain true.

Remark 1.3

Compared with [1,2,9], we prove the existence of least energy sign-changing solutions for system ( SP ) without compactness by using a much weaker growth condition ( f 3 ). Note that ( f 4 ) is weaker than the usual monotonicity condition (NE).

Remark 1.4

Conditions ( V 1 ) , ( V 2 ) , and ( K 0 ) are key to estimate energy (see Lemmas 3.2 and 3.3). Observe that by ( K 0 ) , for any x ∈ R 3 ,

0 ≤ K ( x ) ≤ C K , lim ∣ x ∣ → ∞ K ( x ) = 0 .

There are many functions satisfying ( V 0 ) and ( V 1 ) , such as V ( x ) = V ∞ − e − ∣ x ∣ θ with V ∞ > 1 and 0 < θ < 1 , and a simple example of V ( x ) satisfying ( V 0 ) and ( V 2 ) is given by V ( x ) = V ∞ − e − μ ∣ x ∣ with V ∞ > 1 and μ > 0 . Let f ( t ) = − t 3 1 + ∣ t ∣ for t ∈ R , then for some constant C > 0 ,

F ( t ) = − 1 3 t 3 + 1 2 t 2 − t + ln ∣ 1 + t ∣ + C , if t ≥ 0 , 1 3 t 3 + 1 2 t 2 + t + ln ∣ 1 − t ∣ + C , if t < 0 .

It is evident to prove that f satisfies conditions ( f 1 ) − ( f 4 ) , but not satisfy ( SF ) or ( CF ) .

The proof is based on variational methods in association with the global compactness lemma. The main difficulties lie in three aspects. The first difficulty is to prove that the sign-changing Nehari manifold ℳ is non-empty. In this article, we assume that the nonlinearity f satisfies ( f 3 ) ((AR) or (SF) may be not hold), which means that there would be a competition between the non-local term and the nonlinearity. In view of this, we construct a non-empty set S (defined by (2.16)) and apply some new techniques to deal with this difficulty, see Lemma 2.1.

The second difficulty is to construct a sign-changing Palais-Smale (PS) sequence of ℐ at the nodal level m . Here we borrow an idea from [6] to overcome it, see Lemma 3.4.

The third difficulty is to obtain the compactness of bounded (PS) sequence of energy functional ℐ , we use the concentration-compactness principle of Lions (see [28, Lemma 1.21]) to apply the global compactness lemma for the functional ℐ , see Lemma 3.5, which contains a relationship among least energy nodal level m of ℐ , least energy level c of ℐ , and least energy level c ∞ of the limiting functional ℐ ∞ , see Lemmas 3.2 and 3.3. Such a relationship can help us to obtain compactness of bounded (PS) sequence.

This article is organized as follows. In Section 2, we present the variational framework and prove some preliminary lemmas. In Section 3, we are devoted to proving Theorems 1.1 and 1.2.

2 Preliminaries

In this section, we present some notations, work space stuff, and some preliminary lemmas which are important for proving our main results. We begin this section by giving some notations.

H 1 ( R 3 ) is the usual Sobolev space endowed with the norm
‖ u ‖ H 1 = ∫ R 3 ( ∣ ∇ u ∣ 2 + u 2 ) d x 1 2 .
D 1 , 2 ( R 3 ) is the completion of C 0 ∞ ( R 3 ) with respect to the norm
‖ u ‖ D 1 , 2 = ∫ R 3 ∣ ∇ u ∣ 2 d x 1 2 .
L q ( R 3 ) is the usual Lebesgue space equipped with the norm
∣ u ∣ q = ∫ R 3 ∣ u ∣ q d x 1 q , for all q ∈ [ 1 , + ∞ ) .
“ ⇀ ” and “ → ” denote the weak and strong convergence, respectively.
o ( 1 ) denotes a quantity which goes to zero as n → ∞ .
B r ( x ) ≔ { y ∈ R 3 : ∣ x − y ∣ < r } .
C , C i ( i = 1 , 2 , … ) denote positive constants which may change from line to line.
u + ( x ) ≔ max { u ( x ) , 0 } and u − ( x ) ≔ min { u ( x ) , 0 } .
S denotes the best Sobolev constant for the embedding of D 1 , 2 ( R 3 ) in L 6 ( R 3 ) , that is,
(2.1) S ≔ inf u ∈ D 1 , 2 ( R 3 ) ⧹ { 0 } ∫ R 3 ∣ ∇ u ∣ 2 d x ∫ R 3 ∣ u ∣ 6 d x 1 3 .

2.1 Work space stuff

Throughout this article, we define

H ≔ u ∈ H 1 ( R 3 ) : ∫ R 3 V ( x ) u 2 d x < + ∞

equipped with the inner product

⟨ u , v ⟩ ≔ ∫ R 3 ( ∇ u ⋅ ∇ v + V ( x ) u v ) d x

and the corresponding norm

‖ u ‖ ≔ ∫ R 3 ( ∣ ∇ u ∣ 2 + V ( x ) u 2 ) d x 1 2 .

Since ‖ ⋅ ‖ is equivalent to the usual norm of H 1 ( R 3 ) , the embedding H ↪ L q ( R 3 ) is continuous for any q ∈ [ 2 , 6 ] , thus there exists a constant S q > 0 (only depend on q ) such that

(2.2) ∣ u ∣ q ≤ S q ‖ u ‖ .

For u ∈ H 1 ( R 3 ) , the Lax-Milgram theorem implies that there is a unique ϕ u ∈ D 1 , 2 ( R 3 ) such that

(2.3) ∫ R 3 ∇ ϕ ⋅ ∇ v d x = ∫ R 3 K ( x ) u 2 v d x ,

for v ∈ D 1 , 2 ( R 3 ) , that is, ϕ u is a weak solution of − Δ ϕ = K ( x ) u 2 and the representation formula

(2.4) ϕ u ( x ) = 1 4 π ∫ R 3 K ( y ) u 2 ( y ) ∣ x − y ∣ d y

holds. Hölder’s inequality and (2.1) imply that

(2.5) ∫ R 3 K ( x ) u 2 v d x ≤ ∣ K ∣ 2 ∣ u 2 ∣ 3 ∣ v ∣ 6 = ∣ K ∣ 2 ∣ u ∣ 6 2 ∣ v ∣ 6 ≤ S − 1 2 ∣ K ∣ 2 ∣ u ∣ 6 2 ‖ v ‖ D 1 , 2 .

Moreover, by (2.2), (2.3), and (2.5), there hold

(2.6) ‖ ϕ u ‖ D 1 , 2 ≤ S − 1 2 ∣ K ∣ 2 ∣ u ∣ 6 2 ≤ S − 1 2 S 6 2 ∣ K ∣ 2 ‖ u ‖ 2

and

(2.7) L ϕ u ( u ) ≔ ∫ R 3 K ( x ) ϕ u u 2 d x = 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) u 2 ( x ) u 2 ( y ) ∣ x − y ∣ d x d y ≤ S − 1 S 6 4 ∣ K ∣ 2 2 ‖ u ‖ 4 .

It can derive from (2.7) and the Fubini theorem that

L ϕ u + ( u − ) = ∫ R 3 K ( x ) ϕ u + ( u − ) 2 d x = ∫ R 3 K ( x ) ϕ u − ( u + ) 2 d x = L ϕ u − ( u + ) .

Substituting (2.4) into system ( SP ), we rewrite system ( SP ) into the following equation:

− Δ u + V ( x ) u + K ( x ) ϕ u u = f ( u ) , x ∈ R 3 .

Now, we define the energy functional ℐ : H → R by

ℐ ( u ) = 1 2 ∫ R 3 ( ∣ ∇ u ∣ 2 + V ( x ) u 2 ) d x + 1 4 ∫ R 3 K ( x ) ϕ u u 2 d x − ∫ R 3 F ( u ) d x .

Obviously, ℐ is well defined in H and of class C 1 . Moreover, for any v ∈ H ,

⟨ ℐ ′ ( u ) , v ⟩ = ∫ R 3 ( ∇ u ⋅ ∇ v + V ( x ) u v ) d x + ∫ R 3 K ( x ) ϕ u u v d x − ∫ R 3 f ( u ) v d x .

Recall that u is a weak solution of system ( SP ) if u ∈ H such that ⟨ ℐ ′ ( u ) , v ⟩ = 0 for any v ∈ H . Furthermore, if u ∈ H is a weak solution of system ( SP ) with u ± ≠ 0 , then u is a sign-changing solution of system ( SP ).

Our main purpose in this article is to prove the existence of least energy sign-changing solutions of system ( SP ), in view of this, we will consider the following minimization problem:

m ≔ inf u ∈ ℳ ℐ ( u ) ,

where

ℳ ≔ { u ∈ H : u ± ≠ 0 , ⟨ ℐ ′ ( u ) , u + ⟩ = ⟨ ℐ ′ ( u ) , u − ⟩ = 0 } .

Obviously, the set ℳ contains all of the sign-changing solutions of system ( SP ).

Another purpose of this article is to show the relationship between the energy of least energy sign-changing solutions of system ( SP ) and that of least energy solutions. We can seek the least energy solution of system ( SP ) in the following Nehari manifold:

N ≔ { u ∈ H ⧹ { 0 } : ⟨ ℐ ′ ( u ) , u ⟩ = 0 } ,

and define the following minimization problem

c ≔ inf u ∈ N ℐ ( u ) .

Using a global compactness lemma which is related to the functional ℐ and its limit functional ℐ ∞ , we can deduce that ( PS ) m condition holds. Before applying the global compactness lemma, it is crucial to consider the “limiting problem” associated with system ( SP ), which is defined by

− Δ u + V ∞ u = f ( u ) , x ∈ R 3 , ( SP ∞ )

with the corresponding energy functional

(2.8) ℐ ∞ ( u ) = 1 2 ‖ u ‖ ∞ 2 − ∫ R 3 F ( u ) d x , where ∣ u ‖ ∞ 2 ≔ ∫ R 3 ( ∣ ∇ u ∣ 2 + V ∞ u 2 ) d x .

By a standard argument (see [5, Theorem A.VI]), we can show that ℐ ∞ ∈ C 1 ( H , R ) with

⟨ ℐ ∞ ′ ( u ) , v ⟩ = ∫ R 3 ( ∇ u ⋅ ∇ v + V ∞ u v ) d x − ∫ R 3 f ( u ) v d x , for any v ∈ H .

Analogously, define the Nehari manifold and the least energy corresponding to ( SP ∞ ) as follows:

N ∞ ≔ { u ∈ H ⧹ { 0 } : ⟨ ℐ ∞ ′ ( u ) , u ⟩ = 0 } , c ∞ ≔ inf u ∈ N ∞ ℐ ∞ ( u ) .

2.2 Preliminary lemmas

In this subsection, we present some technical lemmas. We first give some properties on the nonlinearity f and its primitive F ( t ) = ∫ 0 t f ( s ) d s . Combining ( f 1 ) with ( f 2 ) , for any ε > 0 and p ∈ ( 2 , 6 ) , there exists C ε > 0 such that, for any t ∈ R ,

(2.9) ∣ f ( t ) ∣ ≤ ε ∣ t ∣ + C ε ∣ t ∣ 5 ,

(2.10) ∣ f ( t ) ∣ ≤ ε ( ∣ t ∣ + ∣ t ∣ 5 ) + C ε ∣ t ∣ p − 1 .

Condition ( f 4 ) implies that, for any t ∈ R ,

(2.11) ℋ ( t ) ≔ 1 4 f ( t ) t − F ( t ) ≥ 0 ,

and ℋ ( t ) is non-decreasing when t > 0 and non-increasing when t < 0 , which implies that

(2.12) f ′ ( t ) t 2 − 3 f ( t ) t ≥ 0 .

Using ( f 4 ) again, one can easily check that for any t ≥ 1 and τ ∈ R ,

(2.13) f ( t τ ) t τ ≥ t 4 f ( τ ) τ ,

which means that for any t ≥ 1 and τ ∈ R ,

(2.14) L ϕ t τ ( t τ ) − ∫ R 3 f ( t τ ) t τ d x = t 4 L ϕ τ ( τ ) − ∫ R 3 f ( t τ ) t τ d x ≤ t 4 L ϕ τ ( τ ) − ∫ R 3 f ( τ ) τ d x .

Different from the 3-superlinear case, in order to show ℳ ≠ ∅ and N ≠ ∅ , we have to overcome the competing effect of the non-local term. Now, we define sets S 0 and S as follows:

(2.15) S 0 ≔ u ∈ H : L ϕ u ( u ) − ∫ R 3 f ( u ) u d x < 0 ,

(2.16) S ≔ u ∈ H : L ϕ u ( u ± ) − ∫ R 3 f ( u ± ) u ± d x < 0 .

Lemma 2.1

Assume that ( V 0 ) , ( K 0 ) , and ( f 1 )–( f 4 ) hold. Then S ≠ ∅ and S 0 ≠ ∅ . Moreover, ℳ ⊂ S and N ⊂ S 0 .

Proof

We first claim that there exists z ∈ H such that

(2.17) ∫ R 3 F ( z ± ) d x > 0 .

Borrowing the method in [5, Proof of Theorem 2], for R > 0 , we define

w ( R , x ) ≔ ζ , for ∣ x ∣ ≤ R , ζ ( R + 1 − r ) , for r = ∣ x ∣ ∈ [ R , R + 1 ] , 0 , for ∣ x ∣ ≥ R + 1 .

Then, w ( R , x ) ∈ H . It is not difficult to check that

∫ R 3 F ( ± w ( R , x ) ) d x ≥ F ( ± ζ ) meas ( B R ( 0 ) ) − meas ( B R + 1 ( 0 ) − B R ( 0 ) ) ( max γ ∈ [ 0 , ζ ] ∣ F ( ± γ ) ∣ ) ,

where meas ( ⋅ ) denotes the Lebesgue measure. Hence, there exist constants C 1 , C 2 > 0 such that

∫ R 3 F ( ± w ( R , x ) ) d x ≥ C 1 R 3 − C 2 R 2 .

So there exists R ˜ > 1 sufficiently large such that ∫ R 3 F ( ± w ( R ˜ , x ) ) d x > 0 . Let z ± ≔ ± w ( R ˜ , x ) , then (2.17) holds.

For any fixed u ∈ H , let u t ( x ) ≔ u ( t x ) for t ≥ 1 , then (2.11) and (2.13) imply that

(2.18) L ϕ t u t ( t u t ± ) − ∫ R 3 f ( t u t ± ) t u t ± d x ≤ C K 2 ∫ R 3 ∫ R 3 ( t u t ( y ) ) 2 ( t u t ± ( x ) ) 2 4 π ∣ x − y ∣ d x d y − ∫ R 3 f ( t u t ± ) t u t ± d x = C K 2 t − 1 ∫ R 3 ∫ R 3 u 2 ( y ) ( u ± ( x ) ) 2 4 π ∣ x − y ∣ d x d y − t − 3 ∫ R 3 f ( t u ± ) t u ± d x ≤ C K 2 t − 1 ∫ R 3 ∫ R 3 u 2 ( y ) ( u ± ( x ) ) 2 4 π ∣ x − y ∣ d x d y − t ∫ R 3 f ( u ± ) u ± d x ≤ C K 2 t − 1 ∫ R 3 ∫ R 3 u 2 ( y ) ( u ± ( x ) ) 2 4 π ∣ x − y ∣ d x d y − 4 t ∫ R 3 F ( u ± ) d x .

Note that there exists z ∈ H such that ∫ R 3 F ( z ± ) d x > 0 , then (2.17) and (2.18) imply that

L ϕ t z t ( t z t ± ) − ∫ R 3 f ( t z t ± ) t z t ± d x ≤ C K 2 t − 1 ∫ R 3 ∫ R 3 z 2 ( y ) ( z ± ( x ) ) 2 4 π ∣ x − y ∣ d x d y − 4 t ∫ R 3 F ( z ± ) d x → − ∞ ,

as t → + ∞ . Thus, taking v = t ∞ z t ∞ with t ∞ sufficiently large such that v ∈ S , which implies that S ≠ ∅ . Obviously, S ⊂ S 0 , then S 0 ≠ ∅ . Moreover, by the definition of ℳ and N , it is easy to deduce that ℳ ⊂ S and N ⊂ S 0 . Thus, we complete the proof of Lemma 2.1.□

With the help of Lemma 2.1, we will prove that ℳ ≠ ∅ and N ≠ ∅ .

Lemma 2.2

Assume that ( V 0 ) , ( K 0 ) , and ( f 1 )–( f 4 ) hold. If u ∈ S , then there exists a unique pair ( s u , t u ) of positive numbers such that s u u + + t u u − ∈ ℳ . If u ∈ S 0 , then there exists a unique t ( u ) > 0 such that t ( u ) u ∈ N .

Proof

First, we prove the existence of ( s u , t u ) . For s , t ≥ 0 , let ψ 1 ( s , t ) ≔ ⟨ ℐ ′ ( s u + + t u − ) , s u + ⟩ and ψ 2 ( s , t ) ≔ ⟨ ℐ ′ ( s u + + t u − ) , t u − ⟩ , that is,

(2.19) ψ 1 ( s , t ) = s 2 ‖ u + ‖ 2 + s 4 L ϕ u + ( u + ) − ∫ R 3 f ( s u + ) s u + d x + s 2 t 2 L ϕ u − ( u + ) ,

(2.20) ψ 2 ( s , t ) = t 2 ‖ u − ‖ 2 + t 4 L ϕ u − ( u − ) − ∫ R 3 f ( t u − ) t u − d x + s 2 t 2 L ϕ u + ( u − ) .

From (2.14) and (2.19), for any s ≥ 1 and t ≥ 0 , it follows that

(2.21) ψ 1 ( s , t ) ≤ s 2 ‖ u + ‖ 2 + s 4 L ϕ u + ( u + ) − ∫ R 3 f ( u + ) u + d x + s 2 t 2 L ϕ u − ( u + ) = s 2 ‖ u + ‖ 2 + s 4 L ϕ u ( u + ) − ∫ R 3 f ( u + ) u + d x + s 2 ( t 2 − s 2 ) L ϕ u − ( u + ) .

For any fixed t ≥ 0 , using ( f 1 ) , ( f 2 ) , and (2.21), it is easy to verify that ψ 1 ( 0 , t ) = 0 , ψ 1 ( s , t ) > 0 for s > 0 sufficiently small and ψ 1 ( s , t ) < 0 for s > 0 sufficiently large due to u ∈ S . Thus, there exist 0 < r < R such that

(2.22) ψ 1 ( r , r ) > 0 , ψ 2 ( r , r ) > 0 ; ψ 1 ( R , R ) < 0 , ψ 2 ( R , R ) < 0 .

By (2.19) and (2.20), we obtain that the functions ψ 1 ( s , ⋅ ) and ψ 2 ( ⋅ , t ) are increasing in R + for any fixed s > 0 and t > 0 , respectively. Then, we can conclude from (2.22) that

ψ 1 ( r , t ) > 0 , ψ 1 ( R , t ) < 0 , for any t ∈ [ r , R ] ; ψ 2 ( s , r ) > 0 , ψ 2 ( s , R ) < 0 , for any s ∈ [ r , R ] .

Consequently, ( ψ 1 , ψ 2 ) ≠ ( 0 , 0 ) on the boundary of ( r , R ) × ( r , R ) . Applying Miranda’s theorem [19], there exists ( s u , t u ) ∈ ( r , R ) × ( r , R ) such that ψ 1 ( s u , t u ) = 0 and ψ 2 ( s u , t u ) = 0 . That is, s u u + + t u u − ∈ ℳ , which implies that ℳ ≠ ∅ .

Second, we prove the uniqueness of the pair ( s u , t u ) . The proof will be divided into two cases.

Case 1: u ∈ ℳ .

If u ∈ ℳ , which implies that

(2.23) ‖ u ± ‖ 2 + L ϕ u ( u ± ) = ∫ R 3 f ( u ± ) u ± d x .

Now, we prove ( s u , t u ) = ( 1 , 1 ) is the unique pair of positive numbers such that s u u + + t u u − ∈ ℳ . We can assume ( s 0 , t 0 ) is another pair of positive numbers such that s 0 u + + t 0 u − ∈ ℳ , without loss of generality, we may suppose that 0 < s 0 ≤ t 0 . Hence, we obtain

(2.24) s 0 2 ‖ u + ‖ 2 + s 0 4 L ϕ u ( u + ) ≤ s 0 2 ‖ u + ‖ 2 + s 0 4 L ϕ u + ( u + ) + s 0 2 t 0 2 L ϕ u − ( u + ) = ∫ R 3 f ( s 0 u + ) s 0 u + d x ,

(2.25) t 0 2 ‖ u − ‖ 2 + t 0 4 L ϕ u ( u − ) ≥ t 0 2 ‖ u − ‖ 2 + t 0 4 L ϕ u − ( u − ) + s 0 2 t 0 2 L ϕ u + ( u − ) = ∫ R 3 f ( t 0 u − ) t 0 u − d x .

Combining (2.23) with (2.24), we deduce that

(2.26) 1 − 1 s 0 2 ‖ u + ‖ 2 ≥ ∫ R 3 f ( u + ) ( u + ) 3 − f ( s 0 u + ) ( s 0 u + ) 3 ∣ u + ∣ 4 d x .

If s 0 < 1 , the left side of (2.26) is negative, which is absurd because the right side is non-negative by ( f 4 ) . So, we obtain 1 ≤ s 0 ≤ t 0 . Meanwhile, by (2.23) and (2.25), one has

(2.27) 1 − 1 t 0 2 ‖ u − ‖ 2 ≤ ∫ R 3 f ( u − ) ( u − ) 3 − f ( t 0 u − ) ( t 0 u − ) 3 ∣ u − ∣ 4 d x .

If t 0 > 1 , the left side of (2.27) is positive, which is a contradiction because the right side is non-positive by ( f 4 ) . So, we obtain 0 < s 0 ≤ t 0 ≤ 1 . Consequently, s 0 = t 0 = 1 .

Case 2: u ∈ S ⧹ ℳ .

If u ∈ S ⧹ ℳ , we know there exists ( s u , t u ) of positive numbers such that s u u + + t u u − ∈ ℳ . We can assume there exists another pair ( s ′ , t ′ ) of positive numbers such that s ′ u + + t ′ u − ∈ ℳ . Let ν ≔ s u u + + t u u − and ν ′ ≔ s ′ u u + + t ′ u u − , then

ν ′ = s ′ u s u s u u + + t ′ u t u t u u − = s ′ s u ν + + t ′ t u ν − ∈ ℳ .

Thanks to ν ∈ ℳ , we have s ′ u s u = t ′ u t u = 1 , which yields to s u = s ′ u , t u = t ′ . Thus, we assert that ( s u , t u ) is the unique pair of positive numbers such that s u u + + t u u − ∈ ℳ .

Similarly, we obtain that for all u ∈ S 0 , there exists a unique t ( u ) > 0 such that t ( u ) u ∈ N .□

Lemma 2.3

Assume that ( V 0 ) , ( K 0 ) , and ( f 1 )–( f 4 ) hold. Let u ∈ H with u ± ≠ 0 , then

if ⟨ ℐ ′ ( u ) , u ± ⟩ ≤ 0 , then 0 < s u , t u ≤ 1 , where ( s u , t u ) is obtained by Lemma 2.2;
if { u n } ⊂ ℳ , lim n → ∞ ℐ ( u n ) = m , then m > 0 and C 1 ≤ ‖ u n ± ‖ ≤ C 2 for some C 1 , C 2 > 0 .

Proof

(1) Since s u u + + t u u − ∈ ℳ , without loss of generality, suppose that 0 < t u ≤ s u , then

(2.28) s u 2 ‖ u + ‖ 2 + s u 4 L ϕ u ( u + ) ≥ ∫ R 3 f ( s u u + ) s u u + d x .

The assumption ⟨ ℐ ′ ( u ) , u + ⟩ ≤ 0 yields that

(2.29) ‖ u + ‖ 2 + L ϕ u ( u + ) ≤ ∫ R 3 f ( u + ) u + d x .

Combining (2.28) with (2.29), we obtain

(2.30) 1 − 1 s u 2 ‖ u + ‖ 2 ≤ ∫ R 3 f ( u + ) ( u + ) 3 − f ( s u u + ) ( s u u + ) 3 ∣ u + ∣ 4 d x .

If s u > 1 , the left side of (2.30) is positive, which is absurd because the right side is non-positive by ( f 4 ) . Thus, s u ≤ 1 . Hence, 0 < t u ≤ s u ≤ 1 , and finishes the proof of (1) of Lemma 2.3.

( 2 ) On one hand, by using (2.2), (2.9), and the fact that { u n } ⊂ ℳ , we can arrive at

‖ u n ± ‖ 2 ≤ ∫ R 3 f ( u n ± ) u n ± d x ≤ ε ∫ R 3 ∣ u n ± ∣ 2 d x + C ε ∫ R 3 ∣ u n ± ∣ 6 d x ≤ ε S 2 2 ‖ u n ± ‖ 2 + C ε S 6 6 ‖ u n ± ‖ 6 .

Choosing ε > 0 such that ( 1 − ε S 2 2 ) > 0 , which implies that ‖ u n ± ‖ ≥ C 1 for some C 1 > 0 .

On the other hand, for any { u n } ⊂ ℳ ⊂ N , we obtain ⟨ ℐ ′ ( u n ) , u n ⟩ = 0 . By (2.11), one has

m + o ( 1 ) = ℐ ( u n ) = ℐ ( u n ) − 1 4 ⟨ ℐ ′ ( u n ) , u n ⟩ = 1 4 ‖ u n ‖ 2 + ∫ R 3 ℋ ( u n ) d x ≥ 1 4 ‖ u n ‖ 2 ≥ 1 4 ‖ u n ± ‖ 2 > 0 ,

which means that m > 0 and there is C 2 > 0 such that ‖ u n ± ‖ ≤ C 2 . Thus, C 1 ≤ ‖ u n ± ‖ ≤ C 2 .□

3 Proof of main results

In this section, we shall prove Theorems 1.1 and 1.2. This section is divided into three parts. The first part is to verify m < c + c ∞ and c < c ∞ . The second part is to construct a sign-changing ( PS ) m sequence for the functional ℐ . The third part is to prove that the ( PS ) m condition holds via a global compactness lemma.

3.1 Estimate energy on c and m

In this subsection, we are devoted to verifying m < c + c ∞ and c < c ∞ . Let w , v ∈ H be a positive least energy solution of system ( SP ) and problem ( SP ∞ ), respectively, that is,

ℐ ( w ) = c , ℐ ′ ( w ) = 0 ; ℐ ∞ ( v ) = c ∞ , ℐ ∞ ′ ( v ) = 0 .

The exponential decay of w ( x ) , v ( x ) at infinity will be shown in the following lemma.

Lemma 3.1

For any 0 < θ < V ∞ , there exists C = C ( θ ) > 0 such that

(3.1) 0 < v ( x ) + w ( x ) ≤ C e − θ ∣ x ∣ .

Proof

By applying a similar argument to [16, Theorem 1.11], it can deduce that v ∈ L ∞ ( R 3 ) and v ( x ) → 0 as ∣ x ∣ → ∞ . Then, for any 0 < θ < V ∞ , there exists R = R ( θ ) > 0 such that

V ∞ v − f ( v ) ≥ V ∞ v − ε ∣ v ∣ − C ε ∣ v ∣ 5 ≥ θ 2 v , for all ∣ x ∣ ≥ R .

Thus, we have − Δ v + θ 2 v ≤ 0 for all ∣ x ∣ ≥ R and there exists C = C ( θ ) > 0 such that v ( x ) ≤ C for ∣ x ∣ = R . Let v ¯ ( x ) = C e − θ ( ∣ x ∣ − R ) , a direct calculation derives that − Δ v ¯ + θ 2 v ¯ ≥ 0 for x ≠ 0 . The maximum principle implies that v ( x ) ≤ C e − θ ( ∣ x ∣ − R ) for ∣ x ∣ ≥ R . Thus, v ( x ) ≤ C e − θ ∣ x ∣ .

We can proceed as above (or see [15, Theorem 3.1]) to conclude that for any 0 < θ < V ∞ , there exists C = C ( θ ) > 0 such that w ( x ) ≤ C e − θ ∣ x ∣ . Thus, the proof is completed.□

Setting v n ( x ) ≔ v ( x + n e 1 ) for each n ∈ N , where e 1 = ( 1 , 0 , 0 ) is the fixed unit vector in R 3 .

Lemma 3.2

Under assumptions ( V 0 ) , ( K 0 ) , and ( f 1 )–( f 4 ), suppose that ( V 1 ) or ( V 2 ) holds. Then, 0 < c < c ∞ .

Proof

By ⟨ ℐ ∞ ′ ( v ) , v ⟩ = 0 and ( f 4 ) , there exist constants α > 0 sufficiently small and β sufficiently large such that

(3.2) ⟨ ℐ ∞ ′ ( α v ) , α v ⟩ > 0 , ⟨ ℐ ∞ ′ ( β v ) , β v ⟩ < 0 .

Observe that

∫ R 3 ( V ( x ) − V ∞ ) v n 2 d x + ∫ R 3 K ( x ) ϕ v n v n 2 d x = o ( 1 ) ,

then (3.2) implies that

(3.3) ⟨ ℐ ′ ( α v n ) , α v n ⟩ → ⟨ ℐ ∞ ′ ( α v ) , α v ⟩ > 0 , ⟨ ℐ ′ ( β v n ) , β v n ⟩ → ⟨ ℐ ∞ ′ ( β v ) , β v ⟩ < 0 ,

as n → ∞ . In view of (3.3), we deduce that there exist n ∈ N sufficiently large and t ¯ ∈ ( α , β ) such that ⟨ ℐ ′ ( t ¯ v n ) , t ¯ v n ⟩ = 0 , which implies that t ¯ v n ∈ N and ℐ ( t ¯ v n ) ≥ c .

By a direct calculation and the definition of c , we can obtain that

(3.4) c ≤ ℐ ( t ¯ v n ) = ℐ ∞ ( t ¯ v n ) + 1 2 ∫ R 3 ( V ( x ) − V ∞ ) ( t ¯ v n ) 2 d x + 1 4 L ϕ t ¯ v n ( t ¯ v n ) = ℐ ∞ ( t ¯ v n ) + T 1 + T 2 .

Claim: For n sufficiently large, we have

(3.5) T 1 + T 2 < 0 .

If the above claim comes true, combining with (3.4), there holds

(3.6) c < ℐ ∞ ( t ¯ v n ) = ℐ ∞ ( t ¯ v ) ≤ ℐ ∞ ( v ) = c ∞ ,

which completes the proof of Lemma 3.2.

It suffices to verify inequality (3.5). Since γ < 2 V ∞ , we can choose 0 < θ < V ∞ such that γ < 2 θ . Then, it derives from (3.1), ( K 0 ), Hölder’s inequality, and ∣ x − n e 1 ∣ ≥ n − ∣ x ∣ that

(3.7) T 2 ≤ C ∣ ϕ v n ∣ 6 ∫ R 3 ∣ e − γ ∣ x − n e 1 ∣ v 2 ∣ 6 5 d x 5 6 ≤ C e − γ n ∫ R 3 ∣ e ( γ − 2 θ ) ∣ x ∣ ∣ 6 5 d x 5 6 ≤ C 1 e − γ n .

If V satisfies ( V 1 ) , for R 0 > 0 and ρ : a non-increasing function, we deduce that

(3.8) T 1 ≤ 1 2 ∫ ∣ x − n e 1 ∣ ≤ 1 ( V ( x ) − V ∞ ) ( t ¯ v n ) 2 d x ≤ − ρ ( n + 1 ) 2 ∫ ∣ x − n e 1 ∣ ≤ 1 ( t ¯ v n ) 2 d x ,

for any n ≥ R 0 + 1 . (3.7), (3.8), and the boundedness of t ¯ imply that, for n sufficiently large,

(3.9) T 1 + T 2 ≤ e − γ n C 1 − ρ ( n + 1 ) 2 e γ n t ¯ 2 ∫ B 1 ( 0 ) v 2 d x < 0 .

If V satisfies ( V 2 ) , we can derive from ∣ x − n e 1 ∣ ≤ n + ∣ x ∣ that

(3.10) T 1 ≤ − C V ∫ R 3 e − μ ∣ x − n e 1 ∣ v 2 d x ≤ − C V e − μ n ∫ R 3 e − μ ∣ x ∣ v 2 d x ≤ − C 2 e − μ n .

Since μ < γ , it derives from (3.7), (3.10), and the boundedness of t ¯ that

(3.11) T 1 + T 2 ≤ C 1 e − γ n − C 2 e − μ n = e − μ n ( C 1 e ( μ − γ ) n − C 2 ) < 0 ,

for n sufficiently large. Thus, when ( V 1 ) or ( V 2 ) holds, it follows from (3.9) or (3.11) that (3.5) is satisfied for n sufficiently large. This completes the proof of Lemma 3.2.□

Lemma 3.3

Under assumptions ( V 0 ) , ( K 0 ) , and ( f 1 )–( f 4 ), suppose that ( V 1 ) or ( V 2 ) holds. Then, 0 < m < c + c ∞ .

Proof

Let T ≔ [ α , β ] × [ α , β ] ( α , β are defined by Lemma 3.2). We first prove that there exists ( ξ 0 , κ 0 ) ∈ T such that ξ 0 w − κ 0 v n ∈ ℳ for some large n ∈ N . Using w ∈ N and ( f 4 ) , we obtain

(3.12) ⟨ ℐ ′ ( α w ) , α w ⟩ > 0 , ⟨ ℐ ′ ( β w ) , β w ⟩ < 0 .

For any ξ , κ > 0 , let us define

Ψ ( ξ , κ ) ≔ ( ⟨ ℐ ′ ( ξ w − κ v n ) , ( ξ w − κ v n ) + ⟩ , ⟨ ℐ ′ ( ξ w − κ v n ) , ( ξ w − κ v n ) − ⟩ ) .

Since v ( x ) → 0 as ∣ x ∣ → ∞ , it follows from (3.3) and (3.12) that there exists n 0 > 0 such that

(3.13) ⟨ ℐ ′ ( α w − κ v n ) , ( α w − κ v n ) + ⟩ > 0 , ⟨ ℐ ′ ( β w − κ v n ) , ( β w − κ v n ) + ⟩ < 0 ;

(3.14) ⟨ ℐ ′ ( ξ w − α v n ) , ( ξ w − α v n ) − ⟩ > 0 , ⟨ ℐ ′ ( ξ w − β v n ) , ( ξ w − β v n ) − ⟩ < 0 ,

for any n ≥ n 0 and ξ , κ ∈ [ α , β ] . Note that the function Ψ is continuous in T and by (3.13) and (3.14), we can apply Miranda’s theorem [19] to conclude that there exists ( ξ 0 , κ 0 ) ∈ T such that Ψ ( ξ 0 , κ 0 ) = ( 0 , 0 ) , which means that ξ 0 w − κ 0 v n ∈ ℳ for any n ≥ n 0 . Consequently,

m ≤ ℐ ( ξ 0 w − κ 0 v n ) ≤ sup ( s , t ) ∈ T ℐ ( s w − t v n ) .

In view of this, it suffices to show that

sup ( s , t ) ∈ T ℐ ( s w − t v n ) < c + c ∞ ,

for n sufficiently large. Indeed, by a direct calculation, we can obtain that, for any ( s , t ) ∈ T ,

(3.15) ℐ ( s w − t v n ) = 1 2 ‖ s w − t v n ‖ 2 + 1 4 L ϕ ( s w − t v n ) ( s w − t v n ) − ∫ R 3 F ( s w − t v n ) d x = ℐ ( s w ) + ℐ ∞ ( − t v n ) + Π 1 + Π 2 + Π 3 + Π 4 ,

where

Π 1 = − s t ∫ R 3 ( ∇ w ⋅ ∇ v n + V ( x ) w v n ) d x , Π 2 = 1 4 L ϕ ( s w − t v n ) ( s w − t v n ) − 1 4 L ϕ s w ( s w ) ,

Π 3 = ∫ R 3 F ( s w ) + F ( − t v n ) − F ( s w − t v n ) d x , Π 4 = 1 2 ∫ R 3 ( V ( x ) − V ∞ ) ( t v n ) 2 d x .

Claim: For n sufficiently large, we have

(3.16) Π 1 + Π 2 + Π 3 + Π 4 < 0 .

If the above claim comes true, combining with (3.15), there holds

sup ( s , t ) ∈ T ℐ ( s w − t v n ) < ℐ ( s w ) + ℐ ∞ ( t v n ) ≤ ℐ ( w ) + ℐ ∞ ( v ) = c + c ∞ ,

which completes the proof of Lemma 3.3.

It suffices to prove inequality (3.16). Let us define Θ ≔ B n ∕ ( q + 1 ) ( 0 ) . For any 1 ≤ q ≤ 5 , by Hölder’s inequality and (3.1), we can obtain that

∫ Θ w q v n d x ≤ ∫ R 3 w q + 1 d x q q + 1 ∫ Θ v n q + 1 d x 1 q + 1 ≤ C ∫ Θ e − θ ( q + 1 ) ∣ x + n e 1 ∣ d x 1 q + 1 .

This, together with ∣ x + n e 1 ∣ ≥ ∣ n e 1 ∣ − ∣ x ∣ = n − ∣ x ∣ , we obtain that

(3.17) ∫ Θ w q v n d x ≤ C e − θ n ∫ Θ e θ ( q + 1 ) ∣ x ∣ d x 1 q + 1 = C e − θ n ∫ 0 n q + 1 e θ ( q + 1 ) r r 2 d r 1 q + 1 .

We now recall that, for any fixed t > 0 and N ∈ N ∗ , we have

∫ e t r r N − 1 d r = e t r P ( r ) ,

where

P ( r ) ≔ r N − 1 t − ( N − 1 ) t 2 r N − 2 + ( N − 1 ) ( N − 2 ) t 3 r N − 3 + ⋯ + ( − 1 ) N + 1 ( N − 1 ) ! t N .

In view of this, by taking t ≔ θ ( q + 1 ) and N ≔ 3 , then there holds

∫ 0 n q + 1 e θ ( q + 1 ) r r 2 d r = e θ n P n q + 1 = C 3 e θ n + C 4 ,

where C 3 , C 4 > 0 depend only on θ . The above estimate and (3.17) imply that

(3.18) ∫ Θ w q v n d x ≤ C e − θ n e θ n q + 1 + e − θ n ≤ C e − θ n q q + 1 .

Analogously, by (3.1) and Hölder’s inequality, we have

(3.19) ∫ R 3 ⧹ Θ w q v n d x ≤ ∫ R 3 ⧹ Θ w q + 1 d x q q + 1 ∫ R 3 ∣ v ( x + n e 1 ) ∣ q + 1 d x 1 q + 1 ≤ C ∫ R 3 ⧹ Θ e − θ ( q + 1 ) ∣ x ∣ d x q q + 1 = C ∫ n q + 1 ∞ e − θ ( q + 1 ) r r 2 d r q q + 1 ≤ C e − θ n q q + 1 .

Then, (3.18) and (3.19) imply that

(3.20) ∫ R 3 w q v n d x ≤ C e − θ n q q + 1 .

Meanwhile, we can proceed as above to obtain that

(3.21) ∫ R 3 w v n q d x ≤ C e − θ n q q + 1 .

Since w is a positive least energy solution of system ( SP ), one has

(3.22) Π 1 = − s t ∫ R 3 ( ∇ w ⋅ ∇ v n + V ( x ) w v n ) d x = s t ∫ R 3 K ( x ) ϕ w w v n d x − ∫ R 3 f ( w ) v n d x .

Then, (3.1), ( K 0 ) , and Hölder’s inequality, together with ∣ x + n e 1 ∣ ≥ n − ∣ x ∣ imply that

(3.23) ∫ R 3 K ( x ) ϕ w w v n d x ≤ ∣ ϕ w ∣ 6 ∫ R 3 ∣ K ( x ) w v n ∣ 6 5 d x 5 6 ≤ C e − θ n ∫ R 3 ∣ e − γ ∣ x ∣ ∣ 6 5 d x 5 6 ≤ C e − θ n .

According to (2.9), (3.20), (3.22), (3.23), and the boundedness of s , t , we can deduce that

(3.24) Π 1 ≤ s t ∫ R 3 K ( x ) ϕ w w v n d x + ε s t ∫ R 3 w v n d x + C ε s t ∫ R 3 w 5 v n d x ≤ C 5 e − θ n 2 .

Indeed, by using (2.7), a simple calculation shows that

L ϕ ( s w − t v n ) ( s w − t v n ) = 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) ( s w ( x ) − t v n ( x ) ) 2 ( s w ( y ) − t v n ( y ) ) 2 ∣ x − y ∣ d x d y ≤ 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) ( s 2 w 2 ( x ) + t 2 v n 2 ( x ) ) ( s 2 w 2 ( y ) + t 2 v n 2 ( y ) ) ∣ x − y ∣ d x d y = 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) s 4 w 2 ( x ) w 2 ( y ) ∣ x − y ∣ d x d y + 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) s 2 t 2 ( w 2 ( x ) v n 2 ( y ) + v n 2 ( x ) w 2 ( y ) ) ∣ x − y ∣ d x d y + 1 4 π ∫ R 3 ∫ R 3 K ( x ) K ( y ) t 4 v n 2 ( x ) v n 2 ( y ) ∣ x − y ∣ d x d y = L ϕ s w ( s w ) + 2 L ϕ s w ( t v n ) + L ϕ t v n ( t v n ) .

Then, it follows from the above inequality that

(3.25) Π 2 ≤ 1 2 L ϕ s w ( t v n ) + 1 4 L ϕ t v n ( t v n ) = s 2 t 2 2 L ϕ w ( v n ) + t 4 4 L ϕ v n ( v n ) .

Since γ < 2 V ∞ , choosing 0 < θ < V ∞ satisfying γ < 2 θ . Thus, it derives from (3.1), ( K 0 ) , Hölder’s inequality, and ∣ x − n e 1 ∣ ≥ n − ∣ x ∣ that

(3.26) L ϕ w ( v n ) ≤ ∣ ϕ w ∣ 6 ∫ R 3 ∣ e − γ ∣ x − n e 1 ∣ v 2 ∣ 6 5 d x 5 6 ≤ C e − γ n ∫ R 3 ∣ e ( γ − 2 θ ) ∣ x ∣ ∣ 6 5 d x 5 6 ≤ C e − γ n ,

(3.27) L ϕ v n ( v n ) ≤ ∣ ϕ v n ∣ 6 ∫ R 3 ∣ e − γ ∣ x − n e 1 ∣ v 2 ∣ 6 5 d x 5 6 ≤ C e − γ n ∫ R 3 ∣ e ( γ − 2 θ ) ∣ x ∣ ∣ 6 5 d x 5 6 ≤ C e − γ n .

Thus, we can deduce from the boundedness of s , t , and (3.25)–(3.27) that

(3.28) Π 2 ≤ C 6 e − γ n .

In view of (2.9), (3.20), (3.21), and [10, Lemma 3.2], we conclude that

(3.29) ∣ Π 3 ∣ ≤ 4 ∫ R 3 ( ∣ f ( s w ) t v n ∣ + ∣ f ( − s w ) t v n ∣ + ∣ f ( t v n ) s w ∣ + ∣ f ( − t v n ) s w ∣ ) d x ≤ C ∫ R 3 ( ∣ s w ∣ ∣ t v n ∣ + ∣ s w ∣ 5 ∣ t v n ∣ + ∣ s w ∣ ∣ t v n ∣ 5 ) d x ≤ C 7 e − θ n 2 .

If V satisfies ( V 1 ) , for R 0 > 0 and ρ : a non-increasing function given by ( V 1 ) , we deduce that

(3.30) Π 4 ≤ 1 2 ∫ ∣ x − n e 1 ∣ ≤ 1 ( V ( x ) − V ∞ ) ( t v n ) 2 d x ≤ − ρ ( n + 1 ) 2 ∫ ∣ x − n e 1 ∣ ≤ 1 ( t v n ) 2 d x ,

for any n ≥ R 0 + 1 . Taking λ > 0 such that λ ≔ min { γ , θ ∕ 2 } , it follows from (3.24), (3.28)–(3.30) that, for n sufficiently large,

(3.31) Π 1 + Π 2 + Π 3 + Π 4 ≤ C 5 e − θ n 2 + C 6 e − γ n + C 7 e − θ n 2 − ρ ( n + 1 ) 2 ∫ ∣ x − n e 1 ∣ ≤ 1 ( t v n ) 2 d x ≤ e − λ n C 6 e ( λ − γ ) n + C 8 e λ − θ 2 n − ρ ( n + 1 ) 2 e λ n t 2 ∫ B 1 ( 0 ) v 2 d x < 0 .

If V satisfies ( V 2 ) , we can derive from ∣ x − n e 1 ∣ ≤ n + ∣ x ∣ that

(3.32) Π 4 ≤ − C V ∫ R 3 e − μ ∣ x − n e 1 ∣ v 2 d x ≤ − C V e − μ n ∫ R 3 e − μ ∣ x ∣ v 2 d x ≤ − C 9 e − μ n .

Since μ < V ∞ 2 , choosing 0 < θ < V ∞ such that 2 μ < θ , it derives from μ < γ , (3.24), (3.28), (3.29), and (3.32) that

(3.33) Π 1 + Π 2 + Π 3 + Π 4 ≤ C 5 e − θ n 2 + C 6 e − γ n + C 7 e − θ n 2 − C 9 e − μ n = e − μ n C 6 e ( μ − γ ) n + C 10 e ( 2 μ − θ ) n 2 − C 9 < 0 ,

for n sufficiently large. Thus, when ( V 1 ) or ( V 2 ) holds, it follows from (3.31) or (3.33) that (3.16) is satisfied for n sufficiently large. This completes the proof of Lemma 3.3.□

3.2 Construction of the sign-changing ( PS ) m sequence

In this subsection, we are devoted to constructing a sign-changing ( PS ) m sequence, we borrow an idea from [6] to construct it. Let P be the cone of nonnegative functions in H , Q ∈ [ 0 , 1 ] × [ 0 , 1 ] and Σ be the set of continuous maps σ such that, for any s , t ∈ [ 0 , 1 ] ,

σ ( s , 0 ) = 0 , σ ( 0 , t ) ∈ P and σ ( 1 , t ) ∈ − P ;
( ℐ ∘ σ ) ( s , 1 ) ≤ 0 , ∫ R 3 f ( σ ( s , 1 ) ) ( σ ( s , 1 ) ) d x ‖ σ ( s , 1 ) ‖ 2 + L ϕ σ ( s , 1 ) ( σ ( s , 1 ) ) ≥ 2 .

For each u ∈ H with u ± ≠ 0 , let σ ( s , t ) = k t ( 1 − s ) u + + k t s u − , where k > 0 , s , t ∈ [ 0 , 1 ] . It is easy to check that σ ( s , t ) ∈ Σ for k > 0 sufficiently large, which means that Σ ≠ ∅ . Define

l ( u , v ) ≔ ∫ R 3 f ( u ) u d x ‖ u ‖ 2 + L ϕ u ( u ) + L ϕ v ( u ) , if u ≠ 0 ; 0 , if u = 0 .

And u ∈ ℳ if and only if l ( u + , u − ) = l ( u − , u + ) = 1 . Let us define

U ≔ u ∈ H : 1 2 < l ( u + , u − ) < 3 2 , 1 2 < l ( u − , u + ) < 3 2 .

Lemma 3.4

There exists a sequence { u n } ⊂ U such that ℐ ( u n ) → m and ℐ ′ ( u n ) → 0 .

Proof

We divide the proof into three claims.

Claim 1: inf σ ∈ Σ sup u ∈ σ ( Q ) ℐ ( u ) = inf u ∈ ℳ ℐ ( u ) = m .

In fact, for any u ∈ ℳ , there exists σ ( s , t ) = k t ( 1 − s ) u + + k t s u − ∈ Σ for k > 0 sufficiently large. Then, from Lemma 2.2, we can arrive at

ℐ ( u ) = max s , t ≥ 0 ℐ ( s u + + t u − ) ≥ sup u ∈ σ ( Q ) ℐ ( u ) ≥ inf σ ∈ Σ sup u ∈ σ ( Q ) ℐ ( u ) ,

and then

inf u ∈ ℳ ℐ ( u ) ≥ inf σ ∈ Σ sup u ∈ σ ( Q ) ℐ ( u ) .

To complete Claim 1, it remains to prove that for each σ ∈ Σ , there is u σ ∈ σ ( Q ) ∩ ℳ such that

sup u ∈ σ ( Q ) ℐ ( u ) ≥ ℐ ( u σ ) ≥ inf u ∈ ℳ ℐ ( u ) ,

which implies that

inf σ ∈ Σ sup u ∈ σ ( Q ) ℐ ( u ) ≥ inf u ∈ ℳ ℐ ( u ) .

Indeed, for each σ ∈ Σ and t ∈ [ 0 , 1 ] , it is easy to verify that σ ( 0 , t ) ∈ P and σ ( 1 , t ) ∈ − P , thus

(3.34) l ( σ + ( 0 , t ) , σ − ( 0 , t ) ) − l ( σ − ( 0 , t ) , σ + ( 0 , t ) ) = l ( σ + ( 0 , t ) , σ − ( 0 , t ) ) ≥ 0 ,

(3.35) l ( σ + ( 1 , t ) , σ − ( 1 , t ) ) − l ( σ − ( 1 , t ) , σ + ( 1 , t ) ) = − l ( σ − ( 1 , t ) , σ + ( 1 , t ) ) ≤ 0 .

Meanwhile, from the definition of Σ , for any σ ∈ Σ and s ∈ [ 0 , 1 ] , using the elementary inequality b a + d c ≥ b + d a + c for any a , b , c , d > 0 , we can deduce that

l ( σ + ( s , 1 ) , σ − ( s , 1 ) ) + l ( σ − ( s , 1 ) , σ + ( s , 1 ) ) ≥ ∫ R 3 f ( σ ( s , 1 ) ) ( σ ( s , 1 ) ) d x ‖ σ ( s , 1 ) ‖ 2 + L ϕ σ ( s , 1 ) ( σ ( s , 1 ) ) ≥ 2 .

Thus,

(3.36) l ( σ + ( s , 1 ) , σ − ( s , 1 ) ) + l ( σ − ( s , 1 ) , σ + ( s , 1 ) ) − 2 ≥ 0 ,

(3.37) l ( σ + ( s , 0 ) , σ − ( s , 0 ) ) + l ( σ − ( s , 0 ) , σ + ( s , 0 ) ) − 2 = − 2 < 0 .

Combining Miranda’s theorem [19] with (3.34)–(3.37), there exists ( s σ , t σ ) ∈ Q such that

0 = l ( σ + ( s σ , t σ ) , σ − ( s σ , t σ ) ) − l ( σ − ( s σ , t σ ) , σ + ( s σ , t σ ) ) = l ( σ + ( s σ , t σ ) , σ − ( s σ , t σ ) ) + l ( σ − ( s σ , t σ ) , σ + ( s σ , t σ ) ) − 2 ,

then

l ( σ + ( s σ , t σ ) , σ − ( s σ , t σ ) ) = l ( σ − ( s σ , t σ ) , σ + ( s σ , t σ ) ) = 1 .

That is, for any σ ∈ Σ , there exists u σ = σ ( s σ , t σ ) ∈ σ ( Q ) ∩ ℳ . Hence,

inf σ ∈ Σ sup u ∈ σ ( Q ) ℐ ( u ) = inf u ∈ ℳ ℐ ( u ) = m .

Claim 2: There exists a ( PS ) m sequence { u n } ⊂ H for the functional ℐ .

Consider a minimizing sequence { w n } ⊂ ℳ and σ n ( s , t ) = k t ( 1 − s ) w n + + k t s w n − ∈ Σ , then

lim n → ∞ max w ∈ σ n ( Q ) ℐ ( w ) = lim n → ∞ ℐ ( w n ) .

Using a variant form of the classical deformation lemma [21] due to Hofer [14], we claim that there exists { u n } ⊂ H such that, as n → ∞ ,

(3.38) ℐ ( u n ) → m , ℐ ′ ( u n ) → 0 , dist ( u n , σ n ( Q ) ) → 0 .

Suppose by contradiction, there is δ > 0 such that σ n ( Q ) ∩ V δ = ∅ for n sufficiently large, where

V δ = { u ∈ H : ∃ v ∈ H , s . t . ∣ v − u ‖ ≤ δ , ∣ ℐ ′ ( v ) ‖ ≤ δ , ∣ ℐ ( v ) − m ∣ ≤ δ } .

By [14], there exists a map η ∈ C ( [ 0 , 1 ] × H , H ) satisfying for some ε ∈ ( 0 , m 2 ) and all t ∈ [ 0 , 1 ] ,

η ( 0 , u ) = u , η ( t , − u ) = − η ( t , u ) ;
η ( t , u ) = u , for any u ∈ ℐ m − ε ∪ ( H ⧹ ℐ m + ε ) ;
η ( 1 , ℐ m + ε 2 ⧹ V δ ) ⊂ ℐ m − ε 2 ;
η ( 1 , ( ℐ m + ε 2 ∩ P ) ⧹ V δ ) ⊂ ℐ m − ε 2 ∩ P , where ℐ d = { u ∈ H : ℐ ( u ) ≤ d } .

Since lim n → ∞ max w ∈ σ n ( Q ) ℐ ( w ) = m , we can choose n sufficiently large such that

(3.39) σ n ( Q ) ⊂ ℐ m + ε 2 , σ n ( Q ) ∩ V δ = ∅ .

Let us define σ ˜ n ( s , t ) ≔ η ( 1 , σ n ( s , t ) ) for all ( s , t ) ∈ Q . Then we declare that σ ˜ n ∈ Σ , it derives from (3.39) and property (iii) of η that σ ˜ n ( Q ) ⊂ ℐ m − ε 2 , which leads to a contradiction with

m = inf σ ∈ Σ sup w ∈ σ ( Q ) ℐ ( w ) ≤ max w ∈ σ ˜ n ( Q ) ℐ ( w ) ≤ m − ε 2 .

Indeed, property (ii) of η and σ n ∈ Σ imply σ ˜ n ( s , 0 ) = η ( 1 , σ n ( s , 0 ) ) = η ( 1 , 0 ) = 0 . On one hand, it follows from σ n ( 0 , t ) ∈ P , (3.39), and property (iv) of η that σ ˜ n ( 0 , t ) ∈ P . On the other hand, thanks to σ n ( 1 , t ) ∈ − P and (3.39), we see that − σ n ( 1 , t ) ∈ ( ℐ m + ε 2 ∩ P ) ⧹ V δ , which implies that σ ˜ n ( 1 , t ) = − η ( 1 , − σ n ( 1 , t ) ) ∈ − P in view of properties (i) and (iv) of η . Then, σ ˜ n satisfies property ( I ) . Moreover, using the fact that ( ℐ ∘ σ n ) ( s , 1 ) ≤ 0 and property (ii) of η , we have σ ˜ n ( s , 1 ) = η ( 1 , σ n ( s , 1 ) ) = σ n ( s , 1 ) , then σ ˜ n satisfies property ( ii ) . Therefore, it can deduce from the continuity of η and σ n that σ ˜ n ∈ Σ .

Claim 3: The sequence { u n } obtained in Claim 2 satisfies { u n } ⊂ U for n sufficiently large.

Since ℐ ′ ( u n ) → 0 , we have ⟨ ℐ ′ ( u n ) , u n ± ⟩ = o ( 1 ) . Then it suffices to prove that u n ± ≠ 0 , which means that l ( u n + , u n − ) → 1 , l ( u n − , u n + ) → 1 , and thus { u n } ⊂ U for n sufficiently large. By (3.38), there exists a sequence { ν n } such that

(3.40) ν n = s n w n + + t n w n − ∈ σ n ( Q ) , ‖ ν n − u n ‖ → 0 .

Hence, in order to obtain u n ± ≠ 0 , we show that s n w n + ≠ 0 and t n w n − ≠ 0 for n sufficiently large. By { w n } ⊂ ℳ and Lemma 2.3(2), we have C 1 ≤ ‖ w n ± ‖ ≤ C 2 for some C 1 , C 2 > 0 . Now, we only need to show that s n ↛ 0 and t n ↛ 0 for n sufficiently large. Suppose by contradiction that s n → 0 , by the continuity of ℐ and (3.40), we obtain

m = lim n → ∞ ℐ ( ν n ) = lim n → ∞ ℐ ( s n w n + + t n w n − ) = lim n → ∞ ℐ ( t n w n − ) .

Then, by (2.9), Lemma 2.2, and C 1 ≤ ‖ w n ± ‖ ≤ C 2 , choosing ε > 0 such that ( 1 − ε S 2 2 ) > 0 , thus

m = lim n → ∞ ℐ ( w n ) = lim n → ∞ max s , t > 0 ℐ ( s w n + + t w n − ) ≥ lim n → ∞ max s > 0 ℐ ( s w n + + t n w n − ) = lim n → ∞ max s > 0 s 2 2 ‖ w n + ‖ 2 + s 4 4 L ϕ w n + ( w n + ) − ∫ R 3 F ( s w n + ) d x + s 2 t n 2 2 L ϕ w n − ( w n + ) + ℐ ( t n w n − ) ≥ lim n → ∞ max s > 0 s 2 2 ‖ w n + ‖ 2 − ∫ R 3 F ( s w n + ) d x + lim n → ∞ ℐ ( t n w n − ) ≥ lim n → ∞ max s > 0 ( 1 − ε S 2 2 ) s 2 2 ‖ w n + ‖ 2 − C ε S 6 6 s 6 6 ‖ w n + ‖ 6 + m ≥ max s > 0 { C 3 s 2 − C 4 s 6 } + m > m ,

which leads to a contradiction for some C 3 , C 4 > 0 . Then { u n } ⊂ U for n sufficiently large.□

3.3 The compactness condition

In this subsection, we are ready to establish a global compactness lemma inspired by [25, Lemma 4.2] (see also [24, Lemma 4.3]), which is a key point to prove our main results, then prove the ( PS ) m condition holds.

Lemma 3.5

[Global compactness lemma] Assume that ( V 0 ) , ( K 0 ) , and ( f 1 ) − ( f 4 ) hold. Let { u n } ⊂ U be a bounded ( PS ) m sequence for the functional ℐ , then there exists u 0 ∈ H such that u n ⇀ u 0 with ℐ ′ ( u 0 ) = 0 , and integer l ∈ N ∪ { 0 } , sequence { y n k } ⊂ R 3 , non-trivial points w k ∈ H for k = 1 , 2 , … , l such that

∣ y n k ∣ → + ∞ and ∣ y n k − y n k ′ ∣ → + ∞ for k ≠ k ′ ;
w k ≠ 0 and ℐ ∞ ′ ( w k ) = 0 for k = 1 , 2 , … , l ;
u n − u 0 − ∑ k = 1 l w k ( ⋅ − y n k ) → 0 ;
ℐ ( u n ) → ℐ ( u 0 ) + ∑ k = 1 l ℐ ∞ ( w k ) .

Moreover, if l = 0 , then u n → u 0 in H without w k and { y n k } .

Proof

Let us prove that ℐ ′ ( u 0 ) = 0 . Since { u n } is bounded in H , we may assume that, up to a subsequence, u n ⇀ u 0 in H , u n → u 0 in L loc q ( R 3 ) for q ∈ [ 1 , 6 ) , and u n ( x ) → u 0 ( x ) a . e . in R 3 . In order to prove that ℐ ′ ( u 0 ) = 0 . It is sufficient to check that ⟨ ℐ ′ ( u 0 ) , φ ⟩ = 0 for all φ ∈ C 0 ∞ ( R 3 ) . Observe that

(3.41) ⟨ ℐ ′ ( u n ) , φ ⟩ − ⟨ ℐ ′ ( u 0 ) , φ ⟩ = ⟨ u n − u 0 , φ ⟩ + ∫ R 3 K ( x ) ( ϕ u n u n − ϕ u 0 u 0 ) φ d x − ∫ R 3 ( f ( u n ) − f ( u 0 ) ) φ d x .

Recalling that u n ⇀ u 0 in H , then ⟨ u n − u 0 , φ ⟩ → 0 .

By using (2.1), (2.6), [24, Lemma 2.4], and Hölder’s inequality, we can deduce that

∫ R 3 K ( x ) ( ϕ u n u n − ϕ u 0 u 0 ) φ d x = ∫ R 3 K ( x ) ϕ u n − u 0 ( u n − u 0 ) φ d x + o ( 1 ) ≤ C K ∣ φ ∣ ∞ ∫ R 3 ∣ ϕ u n − u 0 ∣ 6 d x 1 6 ∫ supp φ ∣ u n − u 0 ∣ 6 5 d x 5 6 + o ( 1 ) ≤ C K S − 1 2 ∣ φ ∣ ∞ ‖ ϕ u n − u 0 ‖ D 1 , 2 ∫ supp φ ∣ u n − u 0 ∣ 6 5 d x 5 6 + o ( 1 ) ≤ C K S − 1 S 6 2 ∣ φ ∣ ∞ ∣ K ∣ 2 ‖ u n − u 0 ‖ 2 ∫ supp φ ∣ u n − u 0 ∣ 6 5 d x 5 6 + o ( 1 ) ≤ C ∫ supp φ ∣ u n − u 0 ∣ 6 5 d x 5 6 → 0 .

In view of this, it remains to prove that

(3.42) ∫ R 3 ( f ( u n ) − f ( u 0 ) ) φ d x → 0 .

By (2.9) and Young’s inequality, we conclude that

∣ ( f ( u n ) − f ( u 0 ) ) φ ∣ ≤ ∣ u n ∣ ∣ φ ∣ + ∣ u 0 ∣ ∣ φ ∣ + C 1 ∣ u n ∣ 5 ∣ φ ∣ + C 2 ∣ u 0 ∣ 5 ∣ φ ∣ ≤ ∣ u n − u 0 ∣ ∣ φ ∣ + 2 ∣ u 0 ∣ ∣ φ ∣ + C 1 ∣ u n − u 0 ∣ 5 ∣ φ ∣ + C 3 ∣ u 0 ∣ 5 ∣ φ ∣ ≤ ε ∣ u n − u 0 ∣ 2 + C ε ∣ φ ∣ 2 + 2 ∣ u 0 ∣ ∣ φ ∣ + ε ∣ u n − u 0 ∣ 6 + C 1 C ε ∣ φ ∣ 6 + C 3 ∣ u 0 ∣ 5 ∣ φ ∣ .

Let us define G ε , n ( x ) ≔ max { ∣ ( f ( u n ) − f ( u 0 ) ) φ ∣ − ε ∣ u n − u 0 ∣ 2 − ε ∣ u n − u 0 ∣ 6 , 0 } , then

0 ≤ G ε , n ( x ) ≤ C ε ∣ φ ∣ 2 + 2 ∣ u 0 ∣ ∣ φ ∣ + C 1 C ε ∣ φ ∣ 6 + C 3 ∣ u 0 ∣ 5 ∣ φ ∣ ∈ L 1 ( R 3 ) ,

and G ε , n ( x ) → 0 a.e. in R 3 . By the Lebesgue dominated convergence theorem, one has

∫ R 3 G ε , n ( x ) d x → 0 , as n → ∞ .

Therefore,

limsup n → ∞ ∫ R 3 ( f ( u n ) − f ( u 0 ) ) φ d x ≤ limsup n → ∞ ∫ R 3 G ε , n ( x ) d x + ε limsup n → ∞ ∫ R 3 ∣ u n − u 0 ∣ 2 d x + ε limsup n → ∞ ∫ R 3 ∣ u n − u 0 ∣ 6 d x ≤ C 4 ε .

It follows from the arbitrariness of ε that (3.42) is satisfied. Thus, recalling that ℐ ′ ( u n ) → 0 , we deduce from (3.41) that ℐ ′ ( u 0 ) = 0 . Now, we prove that ℐ ( u 0 ) ≥ 0 . Indeed, by (2.11), we have

(3.43) ℐ ( u 0 ) = ℐ ( u 0 ) − 1 4 ⟨ ℐ ′ ( u 0 ) , u 0 ⟩ = 1 4 ‖ u 0 ‖ 2 + ∫ R 3 ℋ ( u 0 ) d x ≥ 1 4 ‖ u 0 ‖ 2 ≥ 0 .

The remaining proof will be divided into three steps.

Step 1. Set z n 1 ≔ u n − u 0 , then we have z n 1 ⇀ 0 in H . Let us define

δ ≔ limsup n → + ∞ sup y ∈ R 3 ∫ B 1 ( y ) ∣ z n 1 ∣ 2 d x .

Case 1 (Vanishing): δ = 0 . That is,

sup y ∈ R 3 ∫ B 1 ( y ) ∣ z n 1 ∣ 2 d x → 0 .

Using the concentration-compactness principle of Lions [28, Lemma 1.21], we obtain that z n 1 → 0 in L q ( R 3 ) for any q ∈ [ 2 , 6 ) . Since ℐ ′ ( u 0 ) = 0 , we have

‖ z n 1 ‖ 2 = ⟨ ℐ ′ ( u n ) , z n 1 ⟩ − ⟨ ℐ ′ ( u 0 ) , z n 1 ⟩ + ∫ R 3 K ( x ) ( ϕ u 0 u 0 − ϕ u n u n ) z n 1 d x + ∫ R 3 ( f ( u n ) − f ( u 0 ) ) z n 1 d x = ∫ R 3 K ( x ) ( ϕ u 0 u 0 − ϕ u n u n ) z n 1 d x + ∫ R 3 ( f ( u n ) − f ( u 0 ) ) z n 1 d x + o ( 1 ) .

By using (2.1), (2.6), [24, Lemma 2.4] and Hölder’s inequality, we can deduce that

∫ R 3 K ( x ) ( ϕ u n u n − ϕ u 0 u 0 ) z n 1 d x = ∫ R 3 K ( x ) ϕ z n 1 ( z n 1 ) 2 d x + o ( 1 ) ≤ C K ∫ R 3 ∣ ϕ z n 1 ∣ 6 d x 1 6 ∫ R 3 ∣ z n 1 ∣ 12 5 d x 5 6 + o ( 1 ) ≤ C K S − 1 2 ‖ ϕ z n 1 ‖ D 1 , 2 ∫ R 3 ∣ z n 1 ∣ 12 5 d x 5 6 + o ( 1 ) ≤ C ∫ R 3 ∣ z n 1 ∣ 12 5 d x 5 6 → 0 .

Using (2.10) and Hölder’s inequality, we deduce that

∫ R 3 ( f ( u n ) − f ( u 0 ) ) z n 1 d x ≤ ∫ R 3 [ ε ( ∣ u n ∣ + ∣ u 0 ∣ + ∣ u n ∣ 5 + ∣ u 0 ∣ 5 ) + C ε ( ∣ u n ∣ p − 1 + ∣ u 0 ∣ p − 1 ) ] ∣ z n 1 ∣ d x ≤ ε C 5 + C ε ( ∣ u n ∣ p p − 1 + ∣ u 0 ∣ p p − 1 ) ∣ z n 1 ∣ p ≤ ε C 5 + C 6 C ε ∣ z n 1 ∣ p ,

where C 5 , C 6 are independent of ε and n . By the arbitrariness of ε and ∣ z n 1 ∣ p → 0 , we obtain

∫ R 3 ( f ( u n ) − f ( u 0 ) ) z n 1 d x → 0 , as n → ∞ .

Therefore, ‖ z n 1 ‖ → 0 in H . Thus, z n 1 → 0 in H , and Lemma 3.5 holds with l = 0 .

Case 2 (Non-vanishing): δ > 0 . Assume that there exists a sequence { y n 1 } ⊂ R 3 such that

∫ B 1 ( y n 1 ) ∣ z n 1 ∣ 2 d x > δ 2 > 0 .

Then, along a subsequence if necessary, we have, for some w 1 ∈ H ,

∣ y n 1 ∣ → + ∞ , z n 1 ( ⋅ + y n 1 ) ⇀ w 1 , ℐ ∞ ′ ( w 1 ) = 0 .

Let us define z n 1 ˜ ( ⋅ ) ≔ z n 1 ( ⋅ + y n 1 ) . Then, z n 1 ˜ is bounded in H and we may assume that z n 1 ˜ ⇀ w 1 in H , z n 1 ˜ → w 1 in L loc q ( R 3 ) for q ∈ [ 2 , 6 ) , and z n 1 ˜ ( x ) → w 1 ( x ) a.e. in R 3 . Since

∫ B 1 ( 0 ) ∣ z n 1 ˜ ∣ 2 d x > δ 2 ,

then

∫ B 1 ( 0 ) ∣ w 1 ∣ 2 d x > δ 2 ,

and then w 1 ≠ 0 . But it can derive from z n 1 ⇀ 0 in H that { y n 1 } must be unbounded. Passing to a subsequence, we assume that ∣ y n 1 ∣ → + ∞ .

Now, it remains to prove that ℐ ∞ ′ ( w 1 ) = 0 . Indeed, similar to the proof of (3.41), we see that

⟨ ℐ ∞ ′ ( z n 1 ˜ ) , φ ⟩ − ⟨ ℐ ∞ ′ ( w 1 ) , φ ⟩ → 0 , for any fixed φ ∈ C 0 ∞ ( R 3 ) .

Therefore, it suffices to prove that ⟨ ℐ ∞ ′ ( z n 1 ˜ ) , φ ⟩ → 0 for any fixed φ ∈ C 0 ∞ ( R 3 ) . We conclude that

(3.44) ⟨ ℐ ′ ( z n 1 ) , φ ( ⋅ − y n 1 ) ⟩ = ∫ R 3 ∇ z n 1 ⋅ ∇ φ ( ⋅ − y n 1 ) d x + ∫ R 3 V ( x ) z n 1 φ ( ⋅ − y n 1 ) d x + ∫ R 3 K ( x ) ϕ z n 1 z n 1 φ ( ⋅ − y n 1 ) d x − ∫ R 3 f ( z n 1 ) φ ( ⋅ − y n 1 ) d x = ∫ R 3 ∇ z n 1 ˜ ⋅ ∇ φ d x + ∫ R 3 V ( x + y n 1 ) z n 1 ˜ φ d x + ∫ R 3 K ( x + y n 1 ) ϕ z n 1 ˜ z n 1 ˜ φ d x − ∫ R 3 f ( z n 1 ˜ ) φ d x .

Since z n 1 ⇀ 0 in H , similar to the proof of (3.41), we obtain that

⟨ ℐ ′ ( z n 1 ) , φ ( ⋅ − y n 1 ) ⟩ − ⟨ ℐ ′ ( 0 ) , φ ( ⋅ − y n 1 ) ⟩ → 0 ,

which implies that

⟨ ℐ ′ ( z n 1 ) , φ ( ⋅ − y n 1 ) ⟩ → 0 .

Then, it deduces from (3.44) that

(3.45) ∫ R 3 ∇ z n 1 ˜ ⋅ ∇ φ d x + ∫ R 3 V ( x + y n 1 ) z n 1 ˜ φ d x + ∫ R 3 K ( x + y n 1 ) ϕ z n 1 ˜ z n 1 ˜ φ d x − ∫ R 3 f ( z n 1 ˜ ) φ d x → 0 .

Since ∣ y n 1 ∣ → + ∞ as n → ∞ and for any φ ∈ C 0 ∞ ( R 3 ) , by ( V 0 ) and ( K 0 ) , one can prove that

(3.46) ∫ R 3 ( V ( x + y n 1 ) − V ∞ ) z n 1 ˜ φ d x → 0 , as n → ∞ ,

and

(3.47) ∫ R 3 K ( x + y n 1 ) ϕ z n 1 ˜ z n 1 ˜ φ d x → 0 , as n → ∞ .

Thus, we use (3.45) minus ⟨ ℐ ∞ ′ ( z n 1 ˜ ) , φ ⟩ , by (3.46) and (3.47), we deduce that

⟨ ℐ ∞ ′ ( z n 1 ˜ ) , φ ⟩ → 0 .

Therefore, ℐ ∞ ′ ( w 1 ) = 0 .

Step 2. In the following, we are devoted to showing that

(3.48) ℐ ( z n 1 ) = m − ℐ ( u 0 ) + o ( 1 ) , ℐ ∞ ( z n 1 ) = ℐ ( u n ) − ℐ ( u 0 ) + o ( 1 ) .

Indeed, it follows from the Brézis-Lieb lemma [7] that

(3.49) ‖ z n 1 ‖ 2 = ‖ u n ‖ 2 − ‖ u 0 ‖ 2 + o ( 1 ) .

We can deduce from Lemma 2.4 in [24] that

(3.50) L ϕ z n 1 ( z n 1 ) = L ϕ u n ( u n ) − L ϕ u 0 ( u 0 ) + o ( 1 ) .

Lemma 3.2 in [18] implies that

(3.51) ∫ R 3 F ( z n 1 ) d x = ∫ R 3 F ( u n ) d x − ∫ R 3 F ( u 0 ) d x + o ( 1 ) ,

(3.52) ∫ R 3 f ( z n 1 ) z n 1 d x = ∫ R 3 f ( u n ) u n d x − ∫ R 3 f ( u 0 ) u 0 d x + o ( 1 ) .

Then, using (3.49)–(3.52), it is easy to verify that

ℐ ( z n 1 ) = m − ℐ ( u 0 ) + o ( 1 ) .

Through direct calculations, we deduce that

(3.53) ℐ ∞ ( z n 1 ) = ℐ ( u n ) − ⟨ u n , u 0 ⟩ + 1 2 ‖ u 0 ‖ 2 − 1 2 ∫ R 3 ( V ( x ) − V ∞ ) ( u n − u 0 ) 2 d x − 1 4 L ϕ u n ( u n ) + ∫ R 3 ( F ( u n ) − F ( z n 1 ) ) d x = ℐ ( u n ) − ℐ ( u 0 ) − ⟨ u n − u 0 , u 0 ⟩ − 1 2 ∫ R 3 ( V ( x ) − V ∞ ) ( u n − u 0 ) 2 d x − 1 4 ( L ϕ u n ( u n ) − L ϕ u 0 ( u 0 ) ) + ∫ R 3 F ( u n ) − F ( z n 1 ) − F ( u 0 ) d x .

In view of ( V 1 ) and the locally compactness of Sobolev embedding, we see that

(3.54) ∫ R 3 ( V ( x ) − V ∞ ) ( u n − u 0 ) 2 d x = o ( 1 ) .

By ( K 0 ) , we have lim ∣ x ∣ → ∞ K ( x ) = 0 , then for any ε > 0 , there exists R ( ε ) > 0 such that

∫ ∣ x ∣ ≥ R ( ε ) K ( x ) ϕ z n 1 ( z n 1 ) 2 d x ≤ ε .

We also have

lim n → ∞ ∫ ∣ x ∣ ≤ R ( ε ) K ( x ) ϕ z n 1 ( z n 1 ) 2 d x ≤ C lim n → ∞ ∫ ∣ x ∣ ≤ R ( ε ) ∣ ϕ z n 1 ∣ 6 d x 1 6 ∫ ∣ x ∣ ≤ R ( ε ) ∣ z n 1 ∣ 12 5 d x 5 6 = 0 .

Combining the aforementioned two inequalities with (3.50), we can deduce that

(3.55) L ϕ u n ( u n ) − L ϕ u 0 ( u 0 ) = o ( 1 ) .

Substituting (3.51), (3.54), and (3.55) into (3.53), there holds

ℐ ∞ ( z n 1 ) = ℐ ( u n ) − ℐ ( u 0 ) + o ( 1 ) .

Step 3. Set z n 2 ( ⋅ ) ≔ z n 1 ( ⋅ ) − w 1 ( ⋅ − y n 1 ) , then z n 2 ⇀ 0 in H . We can similarly check that

(3.56) ℐ ( z n 2 ) = ℐ ( u n ) − ℐ ( u 0 ) − ℐ ∞ ( w 1 ) + o ( 1 ) ,

(3.57) ℐ ∞ ( z n 2 ) = ℐ ( z n 1 ) − ℐ ∞ ( w 1 ) + o ( 1 ) ,

(3.58) ⟨ ℐ ′ ( z n 2 ) , φ ⟩ = ⟨ ℐ ′ ( u n ) , φ ⟩ − ⟨ ℐ ′ ( u 0 ) , φ ⟩ − ⟨ ℐ ∞ ′ ( w 1 ) , φ ⟩ + o ( 1 ) = o ( 1 ) .

Hence, using (3.48), we see that

ℐ ( u n ) = ℐ ( u 0 ) + ℐ ∞ ( z n 1 ) + o ( 1 ) = ℐ ( u 0 ) + ℐ ∞ ( w 1 ) + ℐ ∞ ( z n 2 ) + o ( 1 ) .

Then, by ℐ ∞ ′ ( w 1 ) = 0 , similar to (3.43), we obtain ℐ ∞ ( w 1 ) ≥ 0 , together with (3.43) imply that

ℐ ( z n 2 ) = ℐ ( u n ) − ℐ ( u 0 ) − ℐ ∞ ( w 1 ) + o ( 1 ) ≤ m .

Similar to the argument in Step 1, let us define

δ 1 ≔ limsup n → + ∞ sup y ∈ R 3 ∫ B 1 ( y ) ∣ z n 2 ∣ 2 d x .

If vanishing case occurs, then ‖ z n 2 ‖ → 0 , i.e., ‖ u n − u 0 − w 1 ( ⋅ − y n 1 ) ‖ → 0 , and thus Lemma 3.5 holds with k = 1 . If non-vanishing case occurs, then there exists a sequence { y n 2 } ⊂ R 3 and w 2 ∈ H such that z n 2 ˜ ( x ) = z n 2 ( x + y n 2 ) ⇀ w 2 in H . By using (3.58), we can similarly check that ℐ ∞ ′ ( w 2 ) = 0 . Furthermore, z n 2 ⇀ 0 in H implies that ∣ y n 2 ∣ → + ∞ and ∣ y n 1 − y n 2 ∣ → + ∞ . By iterating this procedure, we obtain sequences of points { y n k } ⊂ R 3 such that ∣ y n k ∣ → + ∞ and ∣ y n k − y n k ′ ∣ → + ∞ for k ≠ k ′ and z n k = z n k − 1 − w k − 1 ( ⋅ − y n k − 1 ) with k ≥ 2 such that

z n k ⇀ 0 in H , ℐ ∞ ′ ( w k ) = 0

and

(3.59) ‖ u n ‖ 2 − ‖ u 0 ‖ 2 − ∑ j = 1 k − 1 ∥ w j ( ⋅ − y n j ) ∥ 2 = u n − u 0 − ∑ j = 1 k − 1 w j ( ⋅ − y n j ) 2 + o ( 1 ) ,

(3.60) ℐ ( u n ) − ℐ ( u 0 ) − ∑ j = 1 k − 1 ℐ ∞ ( w j ) − ℐ ∞ ( z n k ) = o ( 1 ) .

Since { u n } is bounded, (3.59) and (3.60) imply that the iteration stops at some finite l + 1 ∈ N . Therefore, z n l + 1 → 0 in H , by (3.59) and (3.60), it is easy to verify that (3) and (4) hold.□

Based on Lemma 3.5, we can prove that the ( PS ) m condition for the functional ℐ holds.

Lemma 3.6

Under assumptions ( V 0 ) , ( K 0 ) , and ( f 1 ) − ( f 4 ) , suppose that ( V 1 ) or ( V 2 ) holds. Then any sequence { u n } ⊂ U satisfying

ℐ ( u n ) → m ∈ ( 0 , c + c ∞ ) , ℐ ′ ( u n ) → 0

contains a convergent subsequence in H.

Proof

Since { u n } is bounded in H , passing to a subsequence, we may assume that there exists u 0 ∈ H such that u n ⇀ u 0 in H . In view of Lemma 3.5, there exists l ∈ N ∪ { 0 } and { y n k } ⊂ R 3 with ∣ y n k ∣ → + ∞ for k = 1 , 2 , … , l , and u 0 ∈ H , w k ∈ H such that

ℐ ′ ( u 0 ) = 0 , u n − u 0 − ∑ k = 1 l w k ( ⋅ − y n k ) → 0 and ℐ ( u n ) → ℐ ( u 0 ) + ∑ k = 1 l ℐ ∞ ( w k ) ,

where w k are non-trivial critical points of ℐ ∞ for k = 1 , 2 , … , l . Since c < c ∞ (see Lemma 3.2), it follows from (4) of Lemma 3.5 that l ≤ 1 , that is, l = 0 or l = 1 .

Suppose that u 0 = 0 . In this case, since m > 0 , we have that l = 1 , and therefore

(3.61) ∥ u n − w 1 ( ⋅ − y n 1 ) ∥ → 0 .

From { u n } ⊂ U and the definition of U , we obtain 1 2 < l ( u n ± , u n ∓ ) < 3 2 , then (2.2) and (2.9) imply that

1 2 ‖ u n ± ‖ < ∫ R 3 f ( u n ± ) u n ± d x ≤ ε ∫ R 3 ∣ u n ± ∣ 2 d x + C ε ∫ R 3 ∣ u n ± ∣ 6 d x ≤ ε S 2 2 ‖ u n ± ‖ 2 + C ε S 6 6 ‖ u n ± ‖ 6 ,

choosing ε > 0 such that 1 2 − ε S 2 2 > 0 , which implies that there exists ρ > 0 such that

(3.62) ‖ u n ± ‖ ≥ ρ > 0 .

Since ∣ y n 1 ∣ → + ∞ and { u n } ⊂ U , (3.61) and (3.62) imply that ( w 1 ) ± ∈ N ∞ . Hence,

c + c ∞ > m = ℐ ∞ ( w 1 ) = ℐ ∞ ( ( w 1 ) + ) + ℐ ∞ ( ( w 1 ) − ) ≥ 2 c ∞ ,

contradicting to c < c ∞ . So, u 0 ≠ 0 and we can use ℐ ′ ( u 0 ) = 0 to obtain that u 0 ∈ N and ℐ ( u 0 ) ≥ c .

If l = 1 , we obtain

m = lim n → ∞ ℐ ( u n ) = ℐ ( u 0 ) + ∑ k = 1 l ℐ ∞ ( w k ) = ℐ ( u 0 ) + ℐ ∞ ( w 1 ) ≥ c + c ∞ ,

which leads to a contradiction with m < c + c ∞ . Thus, l = 0 , that is, u n → u 0 in H .□

Proof of Theorems 1.1 and 1.2

In view of Lemma 3.4, there exists a sequence { u n } ⊂ U such that ℐ ( u n ) → m and ℐ ′ ( u n ) → 0 as n → ∞ . Lemmas 3.5 and 3.6 imply that { u n } admits a convergent subsequence in H . Thus, under assumptions of Theorem 1.1 or Theorem 1.2, there exists u 0 ∈ H such that u n → u 0 in H , then ℐ ( u 0 ) = m and ℐ ′ ( u 0 ) = 0 . Furthermore, from { u n } ⊂ U and the definition of U , we see that 1 2 < l ( u n ± , u n ∓ ) < 3 2 , similar to the proof of (3.62), we can obtain that

‖ u 0 ± ‖ = lim n → ∞ ‖ u n ± ‖ ≥ ρ > 0 ,

that is, u 0 ± ≠ 0 . Therefore, u 0 is a least energy sign-changing solution of system ( SP ).

Next, we prove m > 2 c . By Lemma 2.2, there exist s ¯ , t ¯ > 0 such that s ¯ u 0 + , t ¯ u 0 − ∈ N , then

m = ℐ ( u 0 ) ≥ ℐ ( s ¯ u 0 + + t ¯ u 0 − ) = ℐ ( s ¯ u 0 + ) + ℐ ( t ¯ u 0 − ) + s ¯ 2 t ¯ 2 2 L ϕ u 0 − ( u 0 + ) > ℐ ( s ¯ u 0 + ) + ℐ ( t ¯ u 0 − ) ≥ 2 c .

The proof is completed by showing that u 0 has exactly two nodal domains. By contradiction, we assume that u 0 has at least three nodal domains such that

u 0 = u 1 + u 2 + u 3

with u i ≠ 0 , u 1 ≥ 0 , u 2 ≤ 0 , and supp ( u i ) ∩ supp ( u j ) = ∅ for i ≠ j ( i , j = 1 , 2 , 3 ) and

⟨ ℐ ′ ( u 0 ) , u i ⟩ = 0 , for i = 1 , 2 , 3 .

Setting v ≔ u 1 + u 2 , we deduce that v + = u 1 and v − = u 2 , i.e., v ± ≠ 0 . In view of Lemma 2.2, there exists a unique pair ( s v , t v ) of positive numbers such that s v v + + t v v − ∈ ℳ , then

ℐ ( s v v + + t v v − ) ≥ m .

Moreover, using the fact that ℐ ′ ( u 0 ) = 0 , it follows that ⟨ ℐ ′ ( v ) , v ± ⟩ < 0 , then by Lemma 2.3(1), we have that ( s v , t v ) ∈ ( 0 , 1 ] × ( 0 , 1 ] . On the other hand, by (2.11), we conclude that

(3.63) 0 = 1 4 ⟨ ℐ ′ ( u 0 ) , u 3 ⟩ = 1 4 ‖ u 3 ‖ 2 + 1 4 L ϕ u 1 ( u 3 ) + 1 4 L ϕ u 2 ( u 3 ) + 1 4 L ϕ u 3 ( u 3 ) − 1 4 ∫ R 3 f ( u 3 ) u 3 d x ≤ 1 4 ‖ u 3 ‖ 2 + 1 4 L ϕ u 1 ( u 3 ) + 1 4 L ϕ u 2 ( u 3 ) + 1 4 L ϕ u 3 ( u 3 ) − ∫ R 3 F ( u 3 ) d x < ℐ ( u 3 ) + 1 4 L ϕ u 1 ( u 3 ) + 1 4 L ϕ u 2 ( u 3 ) .

Then, by (3.63), we can deduce that

m ≤ ℐ ( s v u 1 + t v u 2 ) = ℐ ( s v u 1 + t v u 2 ) − 1 4 ⟨ ℐ ′ ( s v u 1 + t v u 2 ) , s v u 1 + t v u 2 ⟩ = s v 2 4 ‖ u 1 ‖ 2 + t v 2 4 ‖ u 2 ‖ 2 + ∫ R 3 ℋ ( s v u 1 ) d x + ∫ R 3 ℋ ( t v u 2 ) d x ≤ 1 4 ‖ u 1 ‖ 2 + 1 4 ‖ u 2 ‖ 2 + ∫ R 3 ℋ ( u 1 ) d x + ∫ R 3 ℋ ( u 2 ) d x = ℐ ( u 1 ) + ℐ ( u 2 ) + 1 2 L ϕ u 1 ( u 2 ) + 1 4 L ϕ u 1 ( u 3 ) + 1 4 L ϕ u 2 ( u 3 ) < ℐ ( u 1 ) + ℐ ( u 2 ) + ℐ ( u 3 ) + 1 2 L ϕ u 1 ( u 2 ) + 1 2 L ϕ u 1 ( u 3 ) + 1 2 L ϕ u 2 ( u 3 ) = ℐ ( u 0 ) = m ,

which is a contradiction. That is, u 3 = 0 , and u 0 has exactly two nodal domains.□

Acknowledgments

The authors would like to thank editors and the anonymous referees for a careful reading of the results and for making suggestions to improve the original manuscript.

Funding information: This study was supported by National Natural Science Foundation of China (No. 11971393).
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2021-12-05

Revised: 2022-07-08

Accepted: 2022-07-30

Published Online: 2022-08-25

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https://doi.org/10.1515/ans-2022-0021

Keywords for this article

Schrödinger-Poisson system; sign-changing solutions; variational methods

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