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Choquard equations with recurrent potentials

  • Hui-Sheng Ding , Quan Liu , Wei Long EMAIL logo and Lan Zhong
Published/Copyright: December 6, 2024

Abstract

In this article, we are concerned with the existence of nontrivial solutions to the Choquard equation

Δ u + α ( x ) u = ( x μ u q ) u q 2 u in R N ( N 2 ) ,

with recurrent potential α , where 0 < μ < N and 2 N μ N < q < 2 N μ N 2 . Our results include some classical cases where α is constant and α is periodic, as well as some new cases, such as α being almost periodic and α being only bounded and uniformly continuous.

MSC 2010: 35J20; 35J60; 47G20

1 Introduction

We consider the problem

(1.1) Δ u + α ( x ) u = ( x μ u q ) u q 2 u in R N ( N 2 ) ,

where 0 < μ < N and 2 N μ N < q < 2 N μ N 2 . Equation (1.1) is usually called the nonlinear Choquard (or Choquard-Pekar) equation, and it has a strong physical background. Equation (1.1) originates from the model of Fröhlich and Pekar for polaron (cf. [7,21]). In the approximation developed by the Hartree-Fock theory of one component plasma, Choquard also used equation (1.1) to describe an electron trapped in its own hole (cf. [13]). In addition, the Choquard equation is also known as the Schrödinger-Newton equation in models coupling the Schrödinger equation of quantum physics with nonrelativistic Newtonian gravity (cf. [11,22]). Moreover, the equation can be derived from the Einstein-Klein-Gordon and Einstein-Dirac system (cf. [9]). Such a model was also proposed for boson stars and for the collapse of galaxy fluctuations of scalar field dark matter (cf. [10,26]).

When the potential α 1 , problem (1.1) becomes

(1.2) Δ u + u = ( x μ u q ) u q 2 u in R N ( N 2 ) ,

and it is called the autonomous homogeneous Choquard equation. Some classical results about existence, symmetry and decay of ground state solutions for equation (1.2) have been established (cf. [13,15,16,18,19,28] and the references therein). For some recent relevant contributions to some variants of equation (1.2), we refer the reader to [3,17]. When the potential α is nonconstant, problem (1.1) becomes more tricky, and most known results focus on the cases of α being a perturbation of a constant or periodic. For example, Lions [15] studied the existence of solutions to problem (1.1) with p = 2 when α is a perturbation of a constant potential, and recently in [2,4], minimizers for the energy function have been constructed when α is a positive periodic function. Moreover, Ackermann and Qin et al. [1,24] considered the existence of solutions for problem (1.1) with a sign changing periodic potential.

In this article, we are interested in problem (1.1) with a more general positive potential α . As one will see, our recurrent assumptions on α include many special cases such as α being constant, α being periodic, α being almost periodic, and even α being only bounded and uniformly continuous. More specifically, we assume that the potential function α satisfies the following conditions:

  • ( α 1 ) α is bounded and continuous on R N .

  • ( α 2 ) inf x R N α ( x ) > 0 .

  • ( α 3 ) for any sequence { y n } R N , there exist a subsequence { y n k } and a function α h ( x ) : R N R such that

    lim k α ( x + y n k ) = α h ( x ) , for all x R N .

    Hereafter, we denote Hull ( α ) as the set of all these α h ( x ) .

  • ( α 3 ) α ( x ) = α ( x 1 , , x N ) is T i -periodic in x i for i = 2 , , N and almost periodic in x 1 uniformly with respect to ( x 2 , x 3 , , x N ) [ 0 , T 2 ] × [ 0 , T 3 ] × × [ 0 , T N ] (c.f. Definition A.1 in Appendix).

Remark 1.1

We note that ( α 3 ) is a pretty weak assumption, which includes many special cases. In fact, ( α 3 ) automatically holds if α is constant, continuous and periodic, almost periodic, or even bounded and uniformly continuous. Moreover, if α = β + γ , where β satisfies ( α 3 ) and γ is continuous on R N with

lim x γ ( x ) = 0 ,

then α still satisfies ( α 3 ) . Moreover, ( α 3 ) is weaker than the recurrent assumptions in [20], where Pankov considered a semilinear uniformly elliptic partial differential equation with nonstabilizing coefficients.

Since the function α h in ( α 3 ) is not necessarily a translation of α , which is the biggest difference of ( α 3 ) from the corresponding assumptions in earlier results, we need consider the following shell equation of (1.1)

(1.3) Δ u + α h ( x ) u = ( x μ u q ) u q 2 u in R N ,

where α h ( x ) Hull ( α ) . Denote

A α = { β C ( R N , R ) inf x R N α ( x ) β ( x ) sup x R N α ( x ) for all x R N } .

It is easy to see that Hull ( α ) A α . For every β A α and u H 1 ( R N ) , let

J ( β , u ) = 1 2 R N ( u 2 + β ( x ) u 2 ) d x 1 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y ,

and

J ( β , u ) , φ = R N ( u φ + β ( x ) u φ ) d x R N R N u ( x ) q u ( y ) q 2 u ( y ) φ ( y ) x y μ d x d y , φ H 1 ( R N ) .

Hereafter, for simplicity, we denote J ( α , u ) = J ( u ) , J ( α , u ) = J ( u ) , J ( α h , u ) = J h ( u ) , and J ( α h , u ) = J h ( u ) if there is no confusion.

The Hardy-Littlewood-Sobolev inequality (cf. [14]) implies that R N R N u ( x ) q u ( y ) q x y μ d x d y is well defined for u H 1 ( R N ) if 2 N μ N q 2 N μ N 2 . Usually, 2 N μ N is called the lower critical exponent and 2 N μ N 2 is the upper critical exponent of the Choquard equation. The upper critical exponent plays a similar role as the Sobolev critical exponent in local semilinear equations, while the lower critical exponent is related to the bubbling at infinity phenomenon. Several existence and nonexistence results of solutions have been established for various of q . For instance, in view of the Pohozaev identity, the autonomous Choquard equation (1.1) (if α 1 ) does not have nontrivial solutions if either q 2 N μ N or q 2 N μ N 2 . We refer to Li and Ma [12] for details.

We know that solutions of problems (1.1) and (1.3) are critical points of the energy functionals J ( α , u ) and J ( α h , u ) , respectively. By a classical argument, the energy functional J ( β , u ) is of class C 1 on H 1 ( R N ) for all β A α . Moreover, J ( β , u ) is a C 2 -functional on H 1 ( R N ) for all β A α if q 2 .

Consider the following constrained minimization problems:

I = inf { J ( u ) u H 1 ( R N ) , J ( u ) , u = 0 , u 0 }

and

I h = inf { J h ( u ) u H 1 ( R N ) , J h ( u ) , u = 0 , u 0 } .

Theorem 1.2

Assume that ( α 1 ) ( α 3 ) hold and 2 N μ N < q < 2 N μ N 2 . Then, problem (1.3) has a nontrivial solution for some α h ( x ) Hull ( α ) . Moreover, if I < I h for every α h ( x ) Hull ( α ) , then there exists a nontrivial solution of (1.1).

Remark 1.3

The assumptions ( α 1 ) ( α 3 ) include several classical cases such as α being constant and α being periodic. If α is constant then α α h , and if α is periodic, then α h is a translation of α . So in both these two cases, the original equation (1.1) admits a nontrivial solution. This means that Theorem 1.2 recover some classical results in [2] and [16].

Corollary 1.4

Let q be as in Theorem 1.2and α ( x ) = α ¯ ( x ) + α ˜ ( x ) with

  1. α ¯ satisfying ( α 1 ) ( α 3 ) ;

  2. α ˜ C ( R N , R ) , α ˜ ( x ) 0 for every x R N with α ˜ ( x ) 0 , and lim x α ˜ ( x ) = 0 .

If

(1.4) Δ u + α ¯ ( x ) u = ( x μ u q ) u q 2 u , x R N

has a ground state solution, then I < I h for every α h ( x ) Hull ( α ) . Moreover, equation (1.1) has a nontrivial solution.

Remark 1.5

Corollary 1.4 means that any result for the ground state solution to equation (1.4) would lead to a corresponding result for equation (1.1). Moreover, by Corollary 1.4, we know that the condition of I < I h for every α h ( x ) Hull ( α ) does occur in many cases. In particular, for the case of α ¯ being a constant, then by Corollary 1.4, problem (1.1) has a nontrivial solution, which recovers a classical result in Part III of [15].

Although Theorem 1.2 establishes the existence of nontrivial solutions to equation (1.1) in many cases (Remarks 1.3 and 1.5), one cannot use Theorem 1.2 to obtain directly the existence of nontrivial solution to equation (1.1) for some cases. In fact, from the proof of Theorem 1.2 and Lemma 2.4, we will know that I I h if α is an almost periodic function, that is,

lim k sup x R N α ( x + y n k ) α h ( x ) = 0 in ( α 3 ) .

Moreover, in this case, one cannot expect that α h is always a translation of α . Thus, for a general almost periodic potential, Theorem 1.2 can only show that some shell equation (1.3) (not necessarily the original equation) has a nontrivial solution. In fact, the problem on nontrivial solution to equation (1.1) with almost periodic potentials appears to be difficult to answer completely. Fortunately, by utilizing an idea in [27], we can solve partially this problem.

Theorem 1.6

If α satisfies ( α 1 ) , ( α 2 ) , ( α 3 ) , and 2 < q < 2 N μ N 2 , then problem (1.1) has a nontrivial solution.

This article is organized as follows. In Section 2, we present some estimates for the constrained minimizers and the proof of Theorem 1.2. The proof of Theorem 1.6 is given in Section 3. In the final part of this article, we state some auxiliary results in the Appendix.

2 Constrained minimizers and proof of Theorem 1.2

In this section, we give some basic estimates of energy functionals, which are used in the sequel.

Denote

l ( u ) = J ( u ) 1 2 J ( u ) , u = 1 2 1 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y , I a = inf { l ( u ) u H 1 ( R N ) , J ( u ) , u = a } , a R .

Let N = { u H 1 ( R N ) J ( u ) , u = 0 , u 0 } and I = inf u N J ( u ) .

Lemma 2.1

We have I > 0 and I a 0 . Moreover, I a > I if a < 0 .

Proof

We know

I = inf u N J ( u ) = inf u N l ( u ) .

For every u N , we have

J ( u ) = l ( u ) = J ( u ) 1 2 J ( u ) , u = 1 2 1 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y = 1 2 1 2 q R N ( u 2 + α ( x ) u 2 ) d x d y 1 2 1 2 q min { 1 , α ̲ } u 2 ,

where α ̲ = inf x R N α ( x ) . There exists r > 0 such that u > r for every u N . It follows that I 1 2 1 2 q min { 1 , α ̲ } r 2 > 0 .

Moreover, for every u { u H 1 ( R N ) J ( u ) , u = a } , we obtain l ( u ) 0 . So

I a 0 .

We will prove that I a > I if a < 0 , and this will be done in four steps.

Step 1: We claim that

I a = inf { l ( u ) u H 1 ( R N ) , J ( u ) , u = a , J ( u ) c 0 } ,

where c 0 = I a 1 2 a + 1 > 0 .

Denote

I M 1 = inf { l ( u ) u M 1 } , M 1 = { u H 1 ( R N ) J ( u ) , u = a }

and

I M 2 = inf { l ( u ) u M } , M = { u H 1 ( R N ) J ( u ) , u = a , J ( u ) c 0 } .

Since M M 1 , we obtain I M 1 I M 2 . On the other hand, for 0 < ε < 1 , there exists v M 1 such that

I M 1 + ε > J ( v ) 1 2 J ( v ) , v = J ( v ) 1 2 a ,

which yields

J ( v ) < I M 1 + ε + 1 2 a I M 1 1 2 a + 1 = c 0 .

Thus, v M and I M 1 + ε > J ( v ) 1 2 J ( v ) , v I M 2 . According to the arbitrariness of ε , we have I M 1 I M 2 . So

I M 1 = I M 2 .

Step 2: For u M , there exists a 0 < t = t ( u ) < 1 such that t ( u ) u N .

We define

G ( t ) = J ( t u ) , t u = t 2 R N ( u 2 + α ( x ) u 2 ) d x t 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y , t > 0 , u M .

By

R N R N u ( x ) q u ( y ) q x y μ d x d y c u 2 q ,

we have

G ( t ) min { 1 , α ̲ } t 2 u 2 c t 2 q u 2 q = t u 2 ( min { 1 , α ̲ } c t u ( 2 q 2 ) ) .

We can choose r > 0 such that G ( t ) = J ( t u ) , t u > 0 when 0 < t < r u . In addition, G ( 1 ) = J ( u ) , u = a < 0 and G ( t ) C ( R ) . Therefore, there exists t = t ( u ) ( 0 , 1 ) such that G ( t ( u ) ) = J ( t ( u ) u ) , t ( u ) u = 0 , that is, t ( u ) u N .

Step 3: We claim that t ( u ) s < 1 , where s ( 0 , 1 ) and u M .

We prove this by contradiction. Suppose that there are s n = 1 1 n < 1 , { v n } M and t n = t ( v n ) > s n . Since s n < t n < 1 and lim n s n = 1 , we obtain lim n t n = 1 .

According to t n v n N , we obtain

J ( t n v n ) , t n v n = t n 2 R N ( v n 2 + α ( x ) v n 2 ) d x t n 2 q R N R N v n ( x ) q v n ( y ) q x y μ d x d y = 0 ,

that is,

R N R N v n ( x ) q v n ( y ) q x y μ d x d y = t n 2 2 q R N ( v n 2 + α ( x ) v n 2 ) d x .

By v n M , one obtains

(2.1) ( 1 t n 2 2 q ) R N ( v n 2 + α ( x ) v n 2 ) d x = a .

On the one hand, according to v n M and choosing γ = 1 2 1 2 + 1 2 q , 1 2 q < γ < 1 2 , we have

I a 1 2 a + 1 γ a J ( v n ) γ J ( v n ) , v n = 1 2 γ min { 1 , α ̲ } v n 2 + γ 1 2 q R N R N v n ( x ) q v n ( y ) q x y μ d x d y 1 2 γ min { 1 , α ̲ } v n 2 > 0 .

Thus, { v n } is bounded. Taking the limit on both sides of form (2.1), then a = 0 , which contradicts with a < 0 .

Step 4: I a > I if a < 0 .

Let

g ( t ) = l ( t u ) = 1 2 1 2 q t 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y .

Thus,

g ( 1 ) g ( t ( u ) ) = 1 2 1 2 q [ 1 ( t ( u ) ) 2 q ] R N R N u ( x ) q u ( y ) q x y μ d x d y = 1 2 1 2 q [ 1 ( t ( u ) ) 2 q ] R N ( u 2 + α ( x ) u 2 ) d x a 1 2 1 2 q ( 1 s 2 q ) ( a ) .

Moreover,

l ( u ) 1 2 1 2 q ( 1 s 2 q ) ( a ) + l ( t ( u ) u ) 1 2 1 2 q ( 1 s 2 q ) ( a ) + I .

So

I a 1 2 1 2 q ( 1 s 2 q ) ( a ) + I > I ,

which completes the proof.□

Lemma 2.2

{ u n } H 1 ( R N ) is bounded. There exists a constant k such that

R N R N u n ( x ) q u n ( y ) q x y μ d x d y > k > 0 .

Then there is a subsequence { t n } ( a 1 , a 2 ) ( a 1 , a 2 > 0 ) , such that J ( t n u n ) , t n u n = 0 .

Proof

By R N R N u n ( x ) q u n ( y ) q x y μ d x d y > k > 0 , we know for every fixed n ,

J ( t n u n ) , t n u n = t n 2 R N ( u n 2 + α ( x ) u n 2 ) d x t n 2 q R N R N u n ( x ) q u n ( y ) q x y μ d x d y = 0

has a solution

t n = R N ( u n 2 + α ( x ) u n 2 ) d x R N R N u n ( x ) q u n ( y ) q x y μ d x d y 1 2 q 2 .

If t n 0 as n , we obtain

t n u n 2 = t n 2 u n 2 0 .

Moreover, t n u n 0 and t n u n N , which is a contradiction. Thus, t n 0 . On the other hand, by R N ( u n 2 + α ( x ) u n 2 ) d x < c , we have

0 = J ( t n u n ) , t n u n t n 2 c t n 2 q k .

This is a contradiction. Thus, the result follows.□

Remark 2.3

As the aforementioned discussion, we have the following results:

  1. By the similar way with N , l ( u ) , J ( u ) , I and I a , we define N h , l h ( u ) , J h ( u ) , I h , and I a h , respectively. Thus, we also have I h > 0 , I a h 0 and I a h > I h when a < 0 .

  2. Similar to Lemma 2.2, if { u n } H 1 ( R N ) is bounded, there exists a constant k such that

    R N R N u n ( x ) q u n ( y ) q x y μ d x d y > k > 0 .

    Then there is a subsequence { t n } ( a 1 , a 2 ) ( a 1 , a 2 > 0 ) and J h ( t n u n ) , t n u n = 0 .

Lemma 2.4

For every α h ( x ) Hull ( α ) , we have I I h .

Proof

By the definition of Hull ( α ) , for every α h ( x ) Hull ( α ) , there exists { y n } R N such that

(2.2) lim n α ( x + y n ) α h ( x ) = 0 , for x R N .

By the definition of I h , for all ε > 0 , there exists w N h such that I h + ε > J h ( w ) . Let w n ( x ) = w ( x y n ) , then we have

R N R N w n ( x ) q w n ( y ) q x y μ d x d y = R N R N w ( x y n ) q w ( y y n ) q x y μ d x d y = R N R N w ( x ) q w ( y ) q x y μ d x d y > 0 .

According to Lemma 2.2, going if necessary to a subsequence, there exists t n t 0 as n such that

0 = J ( t n w n ) , t n w n = t n 2 R N ( w n 2 + α ( x ) w n 2 ) d x t n 2 q R N R N w n ( x ) q w n ( y ) q x y μ d x d y = t n 2 R N ( w 2 + α ( x + y n ) w 2 ) d x t n 2 q R N R N w ( x ) q w ( y ) q x y μ d x d y .

We obtain

(2.3) ( t n 2 t 0 2 ) R N w 2 d x t n 2 t 0 2 R N w 2 d x 0 ,

(2.4) ( t n 2 q t 0 2 q ) R N R N w ( x ) q w ( y ) q x y μ d x d y t n 2 q t 0 2 q R N R N w ( x ) q w ( y ) q x y μ d x d y 0

and

(2.5) t n 2 R N α ( x + y n ) w 2 d x t 0 2 R N α h ( x ) w 2 d x t n 2 R N α ( x + y n ) w 2 d x t 0 2 R N α ( x + y n ) w 2 d x + t 0 2 R N α ( x + y n ) w 2 d x t 0 2 R N α h w 2 d x t n 2 t 0 2 α ¯ R N w 2 d x + t 0 2 R N lim n α ( x + y n ) α h w 2 d x .

By combining with (2.2)–(2.5), we have

t 0 2 R N ( w 2 + α h ( x ) w 2 ) d x t 0 2 q R N R N w ( x ) q w ( y ) q x y μ d x d y = 0 .

That is, J h ( t 0 w ) , t 0 w = 0 . Since J h ( w ) , w = 0 , we obtain t 0 = 1 . By the direct computations, we obtain

lim n J ( t n w n ) = lim n t n 2 2 R N ( w n 2 + α ( x ) w n 2 ) d x t n 2 q 2 q R N R N w n ( x ) q w n ( y ) q x y μ d x d y = 1 2 R N ( w 2 + α h ( x ) w 2 ) d x 1 2 q R N R N w ( x ) q w ( y ) q x y μ d x d y = J h ( w ) .

Thus,

I h + ε > J h ( w ) = lim n J ( t n w n ) ,

that is,

I h + ε J ( t n w n ) .

Since t n w n N , J ( t n w n ) I . Thus,

I h I .

The proof is now complete.□

2.1 Proof of Theorem 1.2

Proof

We choose a minimizing sequence { u n } H 1 ( R N ) of I . Then, going if necessary to a subsequence, { u n } is bounded in H 1 ( R N ) . We can assume that there exists λ 0 such that

lim n R N u n 2 d x = λ .

If λ = 0 , that is, lim n R N u n 2 d x = 0 , setting γ = 2 N 2 N μ > 0 , there exists 0 < θ < 1 such that 2 < q γ < 2 * and q γ = θ 2 + ( 1 θ ) 2 satisfying

(2.6) lim n R N u n q γ d x = lim n R N u n θ 2 + ( 1 θ ) 2 d x = lim n R N u n θ 2 u n ( 1 θ ) 2 d x lim n R N u n 2 d x θ R N u n 2 d x 1 θ = 0 .

By Hardy-Littlewood-Sobolev inequality, we obtain

R N R N u n ( x ) q u n ( y ) q x y μ d x d y c u n L q γ ( R N ) 2 q 0 .

Since { u n } N , we have

lim n R N ( u n 2 + α ( x ) u n 2 ) d x = 0 .

Thus,

lim n J ( u n ) = lim n 1 2 R N ( u n 2 + α ( x ) u n 2 ) d x 1 2 q R N R N u n ( x ) q u n ( y ) q x y μ d x d y = 0 ,

that is, I = 0 . This contradicts I > 0 , so lim n R N u n 2 d x = λ > 0 . The vanishing Lemma [29] implies that

lim R lim n sup y R N B R ( y ) u n 2 d x = τ ( 0 , λ ] .

If τ ( 0 , λ ) , we have

τ = lim R lim n sup y R N B R ( y ) u n 2 d x ,

then for all 0 < ε < min { 3 ( λ τ ) 8 , τ 4 } , there exists R 0 > 0 , n 0 > 0 and { y n } . When n > n 0 , we have

(2.7) B R 0 ( y n ) u n 2 d x ( τ ε , τ + ε ) .

Setting u ¯ n ( x ) = u n ( x + y n ) , we know u ¯ n ( x ) 0 is bounded in H 1 ( R N ) . We may assume, going if necessary to a subsequence

u ¯ n u ¯ in H 1 ( R N ) ,

u ¯ n u ¯ in L loc 2 ( R N ) .

On the one hand, if lim n y n = , we can rewrite constraint condition

(2.8) 0 = ξ ( u n ) = J ( u n ) , u n = R N ( u n 2 + α ( x ) u n 2 ) d x R N R N u n ( x ) q u n ( y ) q x y μ d x d y = R N ( u ¯ n 2 + α ( x + y n ) u ¯ n 2 ) d x R N R N u ¯ n ( x ) q u ¯ n ( y ) q x y μ d x d y .

By ( α 3 ) , there exists α h 1 ( x ) Hull ( α ) such that

lim n R N α ( x + y n ) u ¯ 2 d x = R N α h 1 ( x ) u ¯ 2 d x .

Then

(2.9) lim n R N u ¯ n 2 + α ( x + y n ) u ¯ n 2 d x = lim n R N ( u ¯ n ( x y n ) u ¯ ( x y n ) ) 2 + α ( x ) u ¯ n ( x y n ) u ¯ ( x y n ) 2 d x + R N u ¯ 2 + α h 1 ( x ) u ¯ 2 d x .

On the other hand,

(2.10) lim n R N ( x μ u ¯ n q ) u ¯ n q d x = lim n R N ( x μ u ¯ n ( x y n ) u ¯ ( x y n ) q ) u ¯ n ( x y n ) u ¯ ( x y n ) q d x + R N ( x μ u ¯ q ) u ¯ q d x .

Combining to (2.8)–(2.10), we have

(2.11) 0 = lim n ξ ( u n ) = lim n ξ ( u ¯ n ( x y n ) u ¯ ( x y n ) ) + ξ h 1 ( u ¯ ) .

Similar to (2.11), we have

(2.12) I = lim n J ( u n ) = lim n J ( u n ) 1 2 J ( u n ) , u n = lim n l ( u n ) = 1 2 1 2 q lim n R N ( x μ u n q ) u n q d x = 1 2 1 2 q lim n R N ( x μ u ¯ n ( x y n ) u ¯ ( x y n ) q ) u ¯ n ( x y n ) u ¯ ( x y n ) q d x + 1 2 1 2 q R N ( x μ u ¯ q ) u ¯ q d x = lim n l ( u ¯ n ( x y n ) u ¯ ( x y n ) ) + l h 1 ( u ¯ ) .

Set lim n ξ n = lim n ξ ( u ¯ n ( x y n ) u ¯ ( x y n ) ) = ξ , ξ h 1 ( u ¯ ) = a ˜ and lim n l ( u ¯ n ( x y n ) u ¯ ( x y n ) ) = l ˜ . We obtain

0 = ξ + a ˜ , I = l ˜ + l h 1 ( u ¯ ) .

1 0 . If a ˜ < 0 , for all n N + , l ( u ¯ n ( x y n ) u ¯ ( x y n ) ) 0 , then l ˜ 0 . By definition I a ˜ h 1 , we have

I = l ˜ + l h 1 ( u ¯ ) l h 1 ( u ¯ ) I a ˜ h 1 > I h 1 .

But by Lemma 2.4, we have I I h 1 . This is a contradiction.

2 0 . If a ˜ > 0 , then ξ < 0 . Hence, when n enough large, we have ξ ( u ¯ n ( x y n ) u ¯ ( x y n ) ) < 0 . By definition I ξ n and Lemma 2.1,

l ( u ¯ n ( x y n ) u ¯ ( x y n ) ) I ξ n > I .

So,

l ˜ I .

But u ¯ 0 , we have l h 1 ( u ¯ ) > 0 , and thus

I = l ˜ + l h 1 ( u ¯ ) > l ˜ I .

This is a contradiction.

3 0 . If a ˜ = 0 , then ξ = 0 and

0 = ξ = lim n ξ ( u ¯ n ( x y n ) u ¯ ( x y n ) ) = lim n R N ( u ¯ n ( x y n ) u ¯ ( x y n ) ) 2 + α ( x ) u ¯ n ( x y n ) u ¯ ( x y n ) 2 d x lim n R N ( x μ u ¯ n ( x y n ) u ¯ ( x y n ) q ) u ¯ n ( x y n ) u ¯ ( x y n ) q d x .

Moreover,

l ˜ = lim n l ( u ¯ n ( x y n ) u ¯ ( x y n ) ) = 1 2 1 2 q lim n R N ( x μ u ¯ n ( x y n ) u ¯ ( x y n ) q ) u ¯ n ( x y n ) u ¯ ( x y n ) q d x 1 2 1 2 q lim n R N α ( x + y n ) u ¯ n u ¯ 2 d x 1 2 1 2 q α ̲ lim n R N u ¯ n 2 d x R N u ¯ 2 d x .

By lim R B R u ¯ 2 d x = R N u ¯ 2 d x , we obtain that for any 0 < ε < min { 3 ( λ τ ) 8 , τ 4 } , there exists R 1 > R 0 > 0 such that R N u ¯ 2 d x < B R u ¯ 2 d x + ε when R R 1 . So

R N u ¯ 2 d x < B R 1 u ¯ 2 d x + ε = lim n B R 1 u ¯ n 2 d x + ε = lim n B R 1 ( y n ) u n 2 d x + ε lim n sup y R N B R 1 ( y ) u n 2 d x + ε τ + 4 ε 3 τ + λ τ 2 .

Since lim n R N u ¯ n 2 d x = lim n R N u n 2 d x = λ , we have

l ˜ 1 2 1 2 q α ̲ lim n R N u ¯ n 2 d x R N u ¯ 2 d x 1 2 1 2 q α ̲ λ τ 2 > 0 .

Moreover,

I = l ˜ + l h 1 ( u ¯ ) > l h 1 ( u ¯ ) I h 1 .

This is a contradiction with I I h 1 .

On the other hand, if y n is bounded, then going if necessary to a subsequence { y n } , we have lim n y n = y 0 , that is, it follows from (2.7), and there exists R 2 > 0 , and we have B R 0 ( y n ) B R 2 ,

0 < τ ε < B R 0 ( y n ) u n 2 d x < B R 2 u n 2 d x .

Similar to y n , by the definition of I , one can obtain a contradiction. Thus, the dichotomy cannot happen. Thereby, we have

lim R lim n sup y R N B R ( y ) u n 2 d x = λ .

For any 0 < ε < λ 4 , there exists r > 0 , n 1 > 0 , { y ˜ n } R N , we obtain

(2.13) 0 < B r ( y ˜ n ) u n 2 d x ( λ ε , λ + ε )

when n > n 1 . If lim n y ˜ n = , set u ˜ n ( x ) = u n ( x + y ˜ n ) . It is verified through the compactness that

(2.14) u ˜ n u ˜ in L 2 ( R N ) ,

where u ˜ 0 . Hence, by (2.14) and ( α 3 ) , there exists α h 2 Hull ( α ) such that

(2.15) lim n + R N ( u ˜ n 2 + α ( x + y ˜ n ) u ˜ n 2 ) d x R N ( u ˜ 2 + α h 2 ( x ) u ˜ 2 ) d x .

Moreover,

0 = lim n ξ ( u n ) ξ h 2 ( u ˜ ) ,

and

I = lim n J ( u n ) = lim n J ( u n ) 1 2 J ( u n ) , u n = lim n l ( u n ) = l h 2 ( u ˜ ) .

If ξ h 2 ( u ˜ ) < 0 , by definition of I ξ h 2 ( u ˜ ) h 2 and Remark 2.3, we have

I = l h 2 ( u ˜ ) I ξ h 2 ( u ˜ ) h 2 > I h 2 .

This is a contradiction to Lemma 2.4. If ξ h 2 ( u ˜ ) = 0 , by definition of I h 2 ,

I = l h 2 ( u ˜ ) I h 2 .

By I I h 2 , we obtain I = I h 2 = l h 2 ( u ˜ ) , ξ h 2 ( u ˜ ) = 0 , that is, u ˜ is a minimax function of I h 2 . Thus, u ˜ is a nontrivial solution of

Δ u + α h 2 ( x ) u = ( x μ u q ) u q 2 u in R N .

If y ˜ n is bounded, by (2.13), there exists { y ˜ n } , lim n y ˜ n = y ˜ 0 , for enough large r 0 > 0 such that

0 < λ ε < B r ( y ˜ n ) u n 2 d x < B r 0 u n 2 d x ,

when B r ( y ˜ n ) B r 0 . By { u n } is bounded in H 1 ( R N ) , there exists u ˆ H 1 ( R N ) such that

u n u ˆ in H 1 ( R N ) .

Analogous to y ˜ n , u ˆ is a nontrivial solution of (1.1).

Particularly, if I < I h for every α h ( x ) Hull ( α ) , then lim n y ˜ n = never happens. Hence, lim n y ˜ n is necessary, which means that equation (1.1) has a nontrivial solution.□

2.2 Proof of Corollary 1.4

Proof

Set

J 1 ( u ) = 1 2 R N ( u 2 + α ¯ ( x ) u 2 ) d x 1 2 q R N R N u ( x ) q u ( y ) q x y μ d x d y

and

I 1 = inf { J 1 ( u ) u H 1 ( R N ) , J 1 ( u ) , u = 0 , u 0 } .

We have I < I 1 . Suppose I 1 is attained by u 0 , then there holds

a J ( u 0 ) , u 0 = R N ( u 0 2 + α ( x ) u 0 2 ) d x R N R N u 0 ( x ) q u 0 ( y ) q x y μ d x d y < R N ( u 0 2 + α ¯ ( x ) u 0 2 ) d x R N R N u 0 ( x ) q u 0 ( y ) q x y μ d x d y = 0

and

(2.16) I 1 = J 1 ( u 0 ) = J 1 ( u 0 ) 1 2 J 1 ( u 0 ) , u 0 = 1 2 1 2 q R N R N u 0 ( x ) q u 0 ( y ) q x y μ d x d y = l ( u 0 ) inf { l ( u ) u H 1 ( R N ) , J ( u ) , u = a } I a .

By Lemma 2.1, we obtain I a > I . So I < I 1 .

On the other hand, for all α h ( x ) Hull ( α ) , by definition of Hull ( α ) , we know there exists y n such that

lim n α ( x + y n ) = α h ( x ) , for any x R N .

Since α ¯ ( x ) satisfies ( α 3 ) , there is α ¯ h ¯ C ( R N ) such that

lim n α ¯ ( x + y n ) = α ¯ h ¯ ( x ) , for any x R N .

By lim n α ˜ ( x + y n ) = 0 , we have α h ( x ) = α ¯ h ¯ ( x ) , that is, I h = I 1 h ¯ . Thus, I 1 I h by Lemma 2.4. Hence,

I < I h .

By Theorem 1.2, Equation (1.1) has a nontrivial solution.□

3 Proof of Theorem 1.6

Set

K { v H 1 ( R N ) v 0 , β A α , J ( β , v ) = 0 } .

For every given l ¯ N + , define

Q Φ H 1 ( R N ) Φ = i = 1 l ̲ v i ( x + z i ) , v i K , z i R N , i = 1 , , l ̲ , l ̲ = 1 , , l ¯ .

By Lemma A.1 in [5], we have

Theorem 3.1

If v K , then v C 2 ( R N ) and v is a solution of the following equation:

Δ v + β ( x ) v = ( x μ v q ) v q 2 v , lim x v ( x ) = 0 .

Hence, we can assume that v ( x ) > 0 and v C 2 ( R N ) for all v K . So if Φ Q , we have

  1. Φ C 2 ( R N ) ;

  2. lim x Φ ( x ) = 0 ;

  3. Φ ( x ) > 0 .

The next result contains an uniform qualitative property of the elements in Q .

Lemma 3.2

There exists δ > 0 such that Δ Φ ( x ) > α ̲ 2 Φ ( x ) when Φ ( x ) < δ .

Proof

For v K , we have

R N v ( y ) q x y μ d y = B R ( x ) v ( y ) q x y μ d y + R N \ B R ( x ) v ( y ) q x y μ d y max y B R ( x ) v ( y ) q c ( N ) 1 N μ R N μ + c R μ v q ,

that is, R N v ( y ) q x y μ d y is bounded. Set h i ( x ) = R N v i ( y ) q x y μ d y , i = 1 , , l ̲ , h 0 = max { h i ( x + z i ) , i = 1 , , l ̲ } . By q > 2 , for every ε > 0 , δ > 0 , we have h 0 v ( x ) q 2 v ( x ) v ( x ) < ε when v ( x ) < δ . Choosing ε = α ̲ 2 , when Φ ( x ) < δ , for every v i , i = 1 , , l ̲ , we have v i ( x + z i ) < δ and

i = 1 l ̲ h i ( x + z i ) v i ( x + z i ) q 2 v i ( x + z i ) < i = 1 l ̲ α ̲ 2 v i ( x + z i ) = α ̲ 2 i = 1 l ̲ v i ( x + z i ) .

For v i K , it follows from Theorem 3.1 that v i satisfy

Δ v i + β i ( x ) v i = ( x μ v i q ) v i q 2 v i = h i ( x ) v i q 2 v i .

Thus,

Δ v i ( x + z i ) + β i ( x + z i ) v i ( x + z i ) = h i ( x + z i ) v i ( x + z i ) q 2 v i ( x + z i ) .

Moreover,

Δ Φ = i = 1 l ̲ Δ v i ( x + z i ) = i = 1 l ̲ β i ( x + z i ) v i ( x + z i ) i = 1 l ̲ h i ( x + z i ) v i ( x + z i ) q 2 v i ( x + z i ) > α ̲ i = 1 l ̲ v i ( x + z i ) α ̲ 2 i = 1 l ̲ v i ( x + z i ) = α ̲ 2 i = 1 l ̲ v i ( x + z i ) = α ̲ 2 Φ .

The proof is now complete.□

Similar to regularity arguments ([8] or [25]), using α 1 , we obtain that for all v K and for every open ball B R N , there results

v C 2 , γ ( B ) C v ,

where C > 0 depends on N , α ¯ , γ ( 0 , 1 ) and diam( B ). Then it follows that for all ϕ Q , we have

ϕ C 2 , γ ( B ) C 1 v ,

where C 1 > 0 depends on N , α ¯ , γ ( 0 , 1 ) and diam( B ). Consider a bounded subset Q in Q . It follows from the aforementioned estimates that

ϕ L ( R N ) C

for every ϕ Q , where C > 0 does not depend on ϕ Q . Therefore, we have the following:

Theorem 3.3

If Q is a bounded set in Q . Then for every Φ Q , there exists k ( Q ) > 0 , such that

  1. Φ ( x ) Φ ( y ) k x y , for all x , y R N ;

  2. for every ε > 0 , there exists r = r ( ε , Q ) > 0 such that Φ ( x ) Φ ( y ) ε when x y < r .

By using the properties, we can give an estimate of the L distance between elements in Q .

Lemma 3.4

For every bounded subset Q in Q , there exists a constant c = c ( Q , N ) > 0 such that

Φ Ψ L ( R N ) c Φ Ψ 2 N + 2 , for a l l Φ , Ψ Q .

Proof

For every x 0 R N , by Theorem 3.3, there exists k = k ( Q ) > 0 , for all x R N , such that

Φ ( x ) Φ ( x 0 ) k x x 0 ; Ψ ( x ) Ψ ( x 0 ) k x x 0 , Φ , Ψ Q .

Without loss of generality, we suppose Ψ ( x 0 ) > Φ ( x 0 ) and set r = Ψ ( x 0 ) Φ ( x 0 ) 4 k . Then, for every x R N , we have

Φ ( x ) Φ ( x 0 ) k x x 0 k r ; Ψ ( x ) Ψ ( x 0 ) k x x 0 k r ,

when x x 0 r , that is,

Φ ( x ) Φ ( x 0 ) + k r ; Ψ ( x 0 ) k r Ψ ( x ) .

It follows that

Ψ ( x ) Φ ( x ) Ψ ( x 0 ) k r Φ ( x 0 ) k r = 1 2 ( Ψ ( x 0 ) Φ ( x 0 ) ) .

Moreover, we have

Ψ Φ 2 B r ( x 0 ) Ψ Φ 2 d x 1 4 ( Ψ ( x 0 ) Φ ( x 0 ) ) 2 c ( N ) r N = c ( k , N ) ( Ψ ( x 0 ) Φ ( x 0 ) ) N + 2 .

Hence, Ψ ( x 0 ) Φ ( x 0 ) c Ψ Φ 2 N + 2 . By arbitrariness of x 0 , we obtain

Φ Ψ L ( R N ) c Φ Ψ 2 N + 2 ,

which completes the proof.□

Define a mapping X 1 : Q H 1 ( R N ) R by

X 1 ( Φ ) = max { x 1 R Φ ( x 1 , x ) = δ for some x R N 1 } , Φ Q ,

where δ > 0 is the real number given in Lemma 3.2.

Lemma 3.5

The function X 1 : Q H 1 ( R N ) R is uniformly continuous on bounded subsets of Q .

Proof

Set Ω = { x R N x 1 > X 1 ( Φ ) } . We will prove Φ ( x ) < δ , x Ω . In fact, suppose x 0 Ω such that Φ ( x 0 ) δ . If Φ ( x 0 ) = δ , then x 0 1 X 1 ( Φ ) . This contradicts x 0 Ω ( x 0 1 > X 1 ( Φ ) ) . If Φ ( x 0 ) > δ . By lim x Φ ( x ) = 0 , there exists y 0 Ω and y 0 1 > x 0 1 such that Φ ( y 0 ) < δ . By Φ C 2 ( R N ) , there exists y 1 in line x 0 y 0 such that Φ ( y 1 ) = δ . Then y 1 1 X 1 ( Φ ) . By y 1 in line x 0 y 0 , then y 1 1 > x 0 1 > X 1 ( Φ ) . This is a contradiction. Similarly, we can prove Φ ( x ) δ , x Ω .

Let v ( x ) = δ e ω ( x 1 X 1 ( Φ ) ) , where ω ( 0 , ( α ̲ 2 ) 1 2 ) and

L ( v ( x ) Φ ( x ) ) = Δ ( v ( x ) Φ ( x ) ) + α ̲ 2 ( v ( x ) Φ ( x ) ) , x Ω .

By Lemma 3.2, we know L ( v ( x ) Φ ( x ) ) 0 , x Ω and v ( x ) Φ ( x ) 0 , x Ω . By liminf x ( v ( x ) Φ ( x ) ) = liminf x v ( x ) 0 and maximum principle applied to the unbounded domain (see [23]), we know v ( x ) Φ ( x ) 0 , x Ω . So, ( x 1 , x ) R N , x 1 X 1 ( Φ ) , we have

(3.1) Φ ( x 1 , x ) v ( x 1 , x ) = δ e ω ( x 1 X 1 ( Φ ) ) .

We prove the continuity of X 1 by the aforementioned estimate. Taking Φ , Ψ Q , there exists x ( Φ ) = ( X 1 ( Φ ) , x ( Φ ) ) , x ( Ψ ) = ( X 1 ( Ψ ) , x ( Ψ ) ) R N such that Φ ( X 1 ( Φ ) , x ( Φ ) ) = δ and Ψ ( X 1 ( Ψ ) , x ( Ψ ) ) = δ . Without losing of generality, we assume X 1 ( Ψ ) > X 1 ( Φ ) . By (3.1),

Φ ( X 1 ( Ψ ) , x ( Ψ ) ) δ e ω ( X 1 ( Ψ ) X 1 ( Φ ) ) ,

that is,

Φ ( X 1 ( Ψ ) , x ( Ψ ) ) δ e ω ( X 1 ( Ψ ) X 1 ( Φ ) ) .

By Ψ ( x ( Ψ ) ) = δ , we have

Φ Ψ L ( R N ) Ψ ( x ( Ψ ) ) Φ ( x ( Ψ ) ) δ δ e ω ( X 1 ( Ψ ) X 1 ( Φ ) ) .

By Lemma 3.4, we obtain

c Φ Ψ 2 N + 2 Φ Ψ L ( R N ) δ δ e ω ( X 1 ( Ψ ) X 1 ( Φ ) ) .

Thus,

0 < X 1 ( Ψ ) X 1 ( Φ ) ln δ ln ( δ c Φ Ψ 2 N + 2 ) ω 0 ( Φ Ψ 0 ) .

This completes the proof.□

Theorem 3.6

Set { u m } H 1 ( R N ) be a ( P S ) sequence for J at level c ˜ , u m 0 in H 1 ( R N ) and lim m u m u m 1 = 0 . Then there exists { x m } R N such that

  1. lim m x m 1 x m 1 1 = 0 , where x m = ( x m 1 , , x m N ) ;

  2. there exists R > 0 such that lim m B R ( x m ) u m 2 d x > 0 ;

  3. lim m x m = .

Proof

By Theorem A.9, going if necessary to a subsequence { u m } , there exists β i ( x ) A α and v i ( 0 ) H 1 ( R N ) , i = 1 , , l such that

  1. u m i = 1 l v i ( x + y m ( i ) ) 0 in H 1 ( R N ) ;

  2. J ( β i , v i ) = 0 , i = 1 , , l ;

  3. c ˜ = i = 1 l J ( β i , v i ) ;

  4. y m ( i ) , i = 1 , , l .

Set Φ m ( x ) = i = 1 l v i ( x + y m ( i ) ) , then { Φ m ( x ) } Q and lim m u m Φ m = 0 . Moreover, lim m Φ m Φ m 1 = 0 . There exists x ( Φ m ) = ( X 1 ( Φ m ) , x ( Φ m ) ) R N such that Φ m ( X 1 ( Φ m ) , x ( Φ m ) ) = δ . Let x m = x ( Φ m ) . By { u m } is bounded in H 1 ( R N ) , { Φ m } is bounded in Q . By Lemma 3.5, we have lim m x m 1 x m 1 1 = 0 . ( 1 0 ) is proved.

By Theorem 3.3 (ii) and Φ m ( x m ) = δ , choosing ε = δ 2 , there exists R > 0 such that Φ m ( x ) δ 2 for all x B R ( x m ) . By lim m u m Φ m = 0 , we know

lim m B R ( x m ) u m 2 d x = lim m B R ( x m ) Φ m 2 d x δ 2 2 B R ( x m ) > 0 .

Thus, ( 2 0 ) is proved.

Next, we will use the contradiction method to prove ( 3 0 ) . Going if necessary to a subsequence, suppose that { x m } and x m < R ˜ . Then

lim m B R + R ˜ u m 2 d x lim m B R ( x m ) u m 2 d x > 0 .

This contradicts with u m 0 in L loc 2 ( R N ) . Thus, ( 3 0 ) is proved.□

Proof of Theorem 1.6

Since J satisfies the geometric assumption of the mountain pass theorem, using Theorem 1.2 in [27], we can find a Palais-Smale sequence { u m } for J at some level c ˜ > 0 such that

(3.2) lim m u m u m 1 = 0 .

We have { u m } admits a subsequence converging weakly to a critical point u . If u 0 , one find a solution of (1.1).

Next, if u = 0 , we will find a nontrivial solution of (1.1). By (3.2) and Theorem 3.6, there exists { x m } R N satisfying

  1. lim m x m 1 x m 1 1 = 0 , where x m = ( x m 1 , , x m N ) ,

  2. there exists R > 0 such that lim m B R ( x m ) u m 2 d x > 0 ,

  3. lim m x m = .

Take τ m ( x m 1 , k m 2 T 2 , , k m N T N ) , where k m i = min { k N + k T i x m i T i } , i = 2 , , N . So

τ m x m = i = 2 N k m i T i x m i 2 1 2 i = 2 N T i 2 1 2 T < + .

Denote v m ( x ) = u m ( x + τ m ) . We know there exists v 0 H 1 ( R N ) such that v m v 0 in H 1 ( R N ) and v m v 0 in L loc 2 ( R N ) . Moreover,

B R + T ( 0 ) v m 2 d x = B R + T ( 0 ) u m ( x + τ m ) 2 d x = B R + T ( τ m ) u m ( x ) 2 d x B R ( x m ) u m ( x ) 2 d x .

By ( 2 0 ) , we have lim m B R + T v m 2 d x > 0 . Hence, v 0 0 .

Case I: lim m x m 1 where x m 1 is the first component of τ m , There exists a subsequence { x m 1 } such that lim m x m 1 = x 0 1 . By ( α 3 ) , we have

α ( x + τ m ) α ( x 1 + x 0 1 , x ) = α ( x 1 + x m 1 , x 2 + k m 2 T 2 , , x N + k m N T N ) α ( x 1 + x 0 1 , x ) = α ( x 1 + x m 1 , x ) α ( x 1 + x 0 1 , x ) ,

where x = ( x 2 , , x N ) U = [ 0 , T 2 ] × × [ 0 , T N ] . By Theorem A.2, we have

sup x R N α ( x + τ m ) α ( x 1 + x 0 1 , x ) = sup x R N α ( x 1 + x m 1 , x ) α ( x 1 + x 0 1 , x ) = sup x R × U α ( x 1 + x m 1 , x ) α ( x 1 + x 0 1 , x ) 0 .

By Lemma A.5,

J ( α ( x + τ m ) , v m ) J ( α ( x 1 + x 0 1 , x ) , v m ) H 1 ( R N ) c α ( x + τ m ) α ( x 1 + x 0 1 , x ) L ( R N ) c sup x R N α ( x + τ m ) α ( x 1 + x 0 1 , x ) 0 .

So for all φ H 1 ( R N ) , denoting φ ˜ ( x ) = φ ( x 1 + x 0 1 , x ) H 1 ( R N ) , we have

lim m J ( α ( x 1 + x 0 1 , x ) , v m ) , φ ˜ = lim m J ( α ( x + τ m ) , v m ) , φ ˜ = lim m J ( α ( x ) , v m ( x τ m ) ) , φ ˜ ( x τ m ) = lim m J ( u m ( x ) ) , φ ˜ ( x τ m ) = 0 .

By v m v 0 , we have J ( α ( x 1 + x 0 1 , x ) , v 0 ) , φ ˜ = lim m J ( α ( x 1 + x 0 1 , x ) , v m ) , φ ˜ = 0 , we obtain

R N v 0 ( x ) φ ˜ ( x ) + α ( x 1 + x 0 1 , x ) v 0 ( x ) φ ˜ ( x ) d x R N R N v 0 ( x ) q v 0 ( y ) q 2 v 0 ( y ) φ ˜ ( y ) x y μ d x d y = 0 .

Moreover,

R N v 0 ( x 1 x 0 1 , x ) φ ( x ) + α ( x ) v 0 ( x 1 x 0 1 , x ) φ ( x ) d x R N R N v 0 ( x 1 x 0 1 , x ) q v 0 ( y 1 x 0 1 , y ) q 2 v 0 ( y 1 x 0 1 , y ) φ ( y ) x y μ d x d y = 0 .

Letting v ¯ 0 = v 0 ( x 1 x 0 1 , x ) , we have v ¯ 0 is a nontrivial solution of (1.1).

Case II: lim m x m 1 = . Because α ( x ) satisfies ( α 3 ), there exists { σ m } R and lim m σ m = , such that

lim m sup x R N α ( x 1 + σ m , x ) α ( x 1 , x ) = lim m sup x R × U α ( x 1 + σ m , x ) α ( x 1 , x ) = 0 .

By lim m x m 1 = , lim m x m 1 x m 1 1 = 0 , we can choose { x m k 1 } { x m 1 } such that

lim k x m k 1 σ k = 0 .

Moreover,

sup x R N α ( x + τ m k ) α ( x ) = sup x R N α ( x 1 + x m k 1 , x 2 + k m k 2 T 2 , , x N + k m k N T N ) α ( x ) = sup x R N α ( x 1 + x m k 1 , x 2 , , x N ) α ( x ) sup x R N α ( x 1 + x m k 1 , x ) α ( x 1 + σ k , x ) + sup x R N α ( x 1 + σ k , x ) α ( x ) = sup x R × U α ( x 1 + x m k 1 , x ) α ( x 1 + σ k , x ) + sup x R × U α ( x 1 + σ k , x ) α ( x ) 0 .

By Lemma A.5, we know

J ( α ( x + τ m k ) , v m k ) J ( α ( x ) , v m k ) H 1 ( R N ) c α ( x + τ m k ) α ( x ) L ( R N ) c sup x R N α ( x + τ m k ) α ( x ) 0 .

By v m k v 0 , for all φ H 1 ( R N ) , we have

J ( α ( x ) , v 0 ) , φ = lim k J ( α ( x ) , v m k ) , φ = lim k J ( α ( x + τ m k ) , v m k ) , φ = lim k J ( α ( x ) , v m k ( x τ m k ) ) , φ ( x τ m k ) = lim k J ( α ( x ) , u m k ( x ) ) , φ ( x τ m k ) = 0 ,

that is, J ( v 0 ) = 0 . So v 0 is a nontrivial solution of (1.1).

By Theorem 3.1, the nontrivial weak solution of (1.1) is a classical solution. Thus, we end the proof.□

  1. Funding information: Hui-Sheng Ding acknowledges support from the NSFC (12361023), Two Thousand Talents Program of Jiangxi Province (jxsq2019201001), and the Key Project of Jiangxi Provincial NSF (20242BAB26001). Wei Long acknowledges support from the NSFC (12271223), and the Key Project of Jiangxi Provincial NSF (20212ACB201003).

  2. Author contributions: The all authors contribute equally.

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

Appendix

In this section, we give some preliminaries. First, we recall some definitions and properties concerning almost periodic functions depending on a parameter (cf. [6,30]).

Definition A.1

Let Ω R N 1 be a compact subset. A continuous function α ( x 1 , x ) is called almost periodic in x 1 R , uniformly with respect to x Ω , if for any ε > 0 , there corresponds a number l ( ε ) > 0 such that any interval of the real line of length l ( ε ) contains at least a real σ for which

sup ( x 1 , x ) R × Ω α ( x 1 + σ , x ) α ( x 1 , x ) ε .

The number σ is called an ε -period for α . The uniform dependence on parameters follows from the fact that l ( ε ) and σ are independent of x .

The following two theorems contain classical results on almost periodic functions depending on parameters.

Theorem A.2

Let Ω R N 1 be a compact subset. Then an almost periodic function α ( x 1 , x ) in x 1 uniformly with respect to x Ω is uniformly continuous and bounded on R × Ω .

Theorem A.3

Let Ω R N 1 be a compact subset. A necessary and sufficient condition for a function α ( x 1 , x ) to be almost periodic in x 1 uniformly with respect to x Ω , is that the family of its translates { α ( x 1 + τ , x ) , τ R } is (uniformly) pre-compact in C ( R × Ω , R ) .

Next, we will give the property of the Choquard terms. Since the proof is simple or can be found directly, we omit the proof.

Lemma A.4

For β A α , let K β = { u H 1 ( R N ) J ( β , u ) = 0 , u 0 } . Then we have

  1. inf β A α inf u K β u > 0 ;

  2. inf β A α inf u K β J ( β , u ) > 0 .

Lemma A.5

Assume that Λ is a bounded subset of H 1 ( R N ) . Then, for β 1 , β 2 L ( R N ) and all u Λ , there exists a constant c ( Λ ) > 0 such that

  1. J ( β 1 , u ) J ( β 2 , u ) c β 1 β 2 L ( R N ) ;

  2. J ( β 1 , u ) J ( β 2 , u ) H 1 ( R N ) c β 1 β 2 L ( R N ) .

Proposition A.6

If u n u in H 1 ( R N ) , then for every φ H 1 ( R N ) , we have

(A1) lim n R N ( x μ u n u q ) u n u q 2 ( u n u ) φ d x = 0 ;

(A2) lim n R N [ ( x μ u n q ) u n q 2 u n φ ( x μ u q ) u q 2 u φ ] d x = 0 .

Moreover, we have

Lemma A.7

(Brezis-Lieb lemma)  [18, Lemma 2.4] Let N 3 , p [ 1 , 2 N 2 N μ ] and 0 < μ < N . If { u n } is a bounded sequence in L 2 N p N μ ( R N ) such that u n u a.e. in R N , then

lim n R N ( x μ u n p ) u n p d x R N ( x μ u n u p ) u n u p d x = R N ( x μ u p ) u p d x ,

as n .

By the Brezis-Lieb lemma of the Choquard terms, we obtain

Lemma A.8

If u m u 0 in H 1 ( R N ) , then

  1. sup β A α J ( β , u m u 0 ) J ( β , u m ) + J ( β , u 0 ) 0 ;

  2. sup β A α J ( β , u m u 0 ) J ( β , u m ) + J ( β , u 0 ) H 1 ( R N ) 0 .

Through Lemma A.8, we can prove the following representation theorem. Although the proof is standard [15], there exist some differences, which result from the almost periodic potential. To ensure the integrity of the article, we give the proof here.

Theorem A.9

(Representation theorem) Suppose that α satisfies ( α 1 ) ( α 3 ) . Let { u m } be Palais-Smale sequence with the level c ˜ . Then there exist u 0 H 1 ( R N ) , l = l ( c ˜ ) N , β i Hull ( α ) , v i H 1 ( R N ) , v i 0 , { y m ( i ) } R N , i = 1 , 2 , , l , such that

  1. u m u 0 i = 1 l v i ( x + y m ( i ) ) 0 in H 1 ( R N ) ;

  2. J ( u 0 ) = 0 and J ( β i , v i ) = 0 , i = 1 , , l ;

  3. c ˜ = J ( u 0 ) + i = 1 l J ( β i , v i ) ;

  4. y m ( i ) , i = 1 , , l .

Proof

By the boundedness of { u m } , there exists u 0 H 1 ( R N ) , such that w m = u m u 0 0 in H 1 ( R N ) . By Lemma A.8, we have

J ( u 0 ) = 0 , lim m J ( w m ) = 0 and J ( w m ) = J ( u m ) J ( u 0 ) + o m ( 1 ) .

By vanishing theorem [29], we know lim m R N w m 2 d x = λ 1 > 0 . Hence, we obtain dichotomy or compactness can occur, and hence, there exists R ˜ > 0 such that lim m sup y R N B R ˜ ( y ) w m 2 d x > 0 . Thus, for every m N + , there exists y m ( 1 ) R N such that

sup y R N B R ˜ ( y ) w m 2 d x 1 m < B R ˜ ( y m ( 1 ) ) w m 2 d x ,

that is,

lim m B R ˜ ( y m ( 1 ) ) w m 2 d x lim m sup y R N B R ˜ ( y ) w m 2 d x > 0 .

Hence, when m is enough large,

B R ˜ ( y m ( 1 ) ) w m 2 d x 1 2 lim m B R ˜ ( y m ( 1 ) ) w m 2 d x .

Moreover, we know lim m y m ( 1 ) = . In fact, going if necessary to subsequence, we assume y m ( 1 ) < c ˜ 1 . Then

B c ˜ 1 + 2 R ˜ w m 2 d x > B R ˜ ( y m ( 1 ) ) w m 2 d x 1 2 lim m B R ˜ ( y m ( 1 ) ) w m 2 d x > 0 .

This contradicts with w m 0 .

So, we have lim m y m ( 1 ) = and lim m B R ˜ ( y m ( 1 ) ) w m 2 d x > 0 . We may assume that there exists v 1 H 1 ( R N ) , such that

w m ( x + y m ( 1 ) ) v 1 in H 1 ( R N ) ,

w m ( x + y m ( 1 ) ) v 1 in L loc 2 ( R N ) .

By

lim m B R ˜ w m ( x + y m ( 1 ) ) 2 d x = lim m B R ˜ ( y m ( 1 ) ) w m 2 d x > 0 ,

we have v 1 0 .

For each m N and α ( x + y m ( 1 ) ) , we choose k m ( 1 ) , i N , i = 2 , , N , ( z m ( 1 ) , 2 , z m ( 1 ) , 3 , , z m ( 1 ) , N ) [ 0 , T 2 ] × × [ 0 , T N ] such that

α ( x + y m ( 1 ) ) = α ( x 1 + y m ( 1 ) , 1 , z m ( 1 ) , 2 + k m ( 1 ) , 2 T 2 , , z m ( 1 ) , N + k m ( 1 ) , N T N ) = α ( x 1 + y m ( 1 ) , 1 , z m ( 1 ) , 2 , , z m ( 1 ) , N )

since the aforementioned equality is from ( α 3 ) . By Theorem A.3, we know there exists β 1 A α = { β C ( R N , R ) α ̲ β ( x ) α ¯ , x R N } such that lim m sup x R N α ( x + y m ( 1 ) ) β 1 ( x ) = 0 .

Since { w m ( x + y m ( 1 ) ) } is bounded in H 1 ( R N ) and Lemma A.5, we have

J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) J ( β 1 , w m ( x + y m ( 1 ) ) ) c α ( x + y m ( 1 ) ) β 1 ( x ) L ( R N ) c sup x R N α ( x + y m ( 1 ) ) β 1 ( x )

and

J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) J ( β 1 , w m ( x + y m ( 1 ) ) ) H 1 ( R N ) c α ( x + y m ( 1 ) ) β 1 ( x ) L ( R N ) c sup x R N α ( x + y m ( 1 ) ) β 1 ( x ) .

So

lim m J ( β 1 , w m ( x + y m ( 1 ) ) ) = lim m J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) = lim m J ( w m ) = c ˜ > 0

and

lim m J ( β 1 , w m ( x + y m ( 1 ) ) ) H 1 ( R N ) = lim m J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) H 1 ( R N ) = lim m J ( w m ) H 1 ( R N ) = 0 .

By w m ( x + y m ( 1 ) ) v 1 , we know

(A3) J ( β 1 , v 1 ) = 0 .

By Lemmas A.4A.8, the energy function

J ( w m v 1 ( x y m ( 1 ) ) ) J ( w m ) + J ( β 1 , v 1 ) = J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) v 1 ) J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) + J ( β 1 , v 1 ) J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) v 1 ) J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) + J ( α ( x + y m ( 1 ) ) , v 1 ) + J ( α ( x + y m ( 1 ) ) , v 1 ) J ( β 1 , v 1 ) sup β A α J ( β , w m ( x + y m ( 1 ) ) v 1 ) J ( β , w m ( x + y m ( 1 ) ) ) + J ( β , v 1 ) + c sup x R N α ( x + y m ( 1 ) ) β 1 .

So

(A4) lim m J ( w m v 1 ( x y m ( 1 ) ) ) = c ˜ J ( β 1 , v 1 ) .

By Lemma A.4, we know c ˜ J ( β 1 , v 1 ) 0 .

For every φ H 1 ( R N ) , we have

(A5) J ( w m v 1 ( x y m ( 1 ) ) ) , φ J ( w m v 1 ( x y m ( 1 ) ) ) , φ J ( w m ) , φ + J ( v 1 ( x y m ( 1 ) ) ) , φ + J ( w m ) , φ J ( v 1 ( x y m ( 1 ) ) ) , φ J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) v 1 ) J ( α ( x + y m ( 1 ) ) , w m ( x + y m ( 1 ) ) ) + J ( α ( x + y m ( 1 ) ) , v 1 ) H 1 ( R N ) φ + J ( w m ) , φ J ( v 1 ( x y m ( 1 ) ) ) , φ .

On the other hand, we know

(A6) J ( w m ) , φ J ( v 1 ( x y m ( 1 ) ) ) , φ J ( w m ) H 1 ( R N ) φ + J ( α ( x + y m ( 1 ) ) , v 1 ) H 1 ( R N ) φ J ( w m ) H 1 ( R N ) φ + J ( α ( x + y m ( 1 ) ) , v 1 ) J ( β 1 , v 1 ) H 1 ( R N ) φ + J ( β 1 , v 1 ) H 1 ( R N ) φ .

By Lemmas A.5 and A.8, combining (A3) and (A5), with (A6), we have

(A7) lim m J ( w m v 1 ( x y m ( 1 ) ) ) H 1 ( R N ) = 0 .

Combining to (A3), (A4) and (A7), as m , we have

J ( β 1 , v 1 ) = 0 in H 1 ( R N ) , J ( w m v 1 ( x y m ( 1 ) ) ) c ˜ J ( β 1 , v 1 ) 0 , J ( w m v 1 ( x y m ( 1 ) ) ) 0 .

( 1 0 ) If c ˜ J ( β 1 , v 1 ) = 0 , that is,

J ( w m v 1 ( x y m ( 1 ) ) ) 0 , J ( w m v 1 ( x y m ( 1 ) ) ) 0 ,

then

lim m { J ( w m v 1 ( x y m ( 1 ) ) ) 1 2 q J ( w m v 1 ( x y m ( 1 ) ) ) , w m v 1 ( x y m ( 1 ) ) } = lim m 1 2 1 2 q R N ( ( w m v 1 ( x y m ( 1 ) ) ) 2 + α ( x ) w m v 1 ( x y m ( 1 ) ) 2 ) d x = 0 .

So, lim m w m v 1 ( x y m ( 1 ) ) = 0 and c ˜ = J ( β 1 , v 1 ) . Thus, the theorem is proved with l = 1 .

( 2 0 ) If c ˜ J ( β 1 , v 1 ) > 0 , that is, J ( β 1 , v 1 ) = c ˜ 1 < c ˜ , setting w m ( 1 ) ( x ) = w m ( x ) v 1 ( x y m ( 1 ) ) , then

J ( w m ( 1 ) ( x ) ) c ˜ c ˜ 1 > 0 , J ( w m ( 1 ) ( x ) ) 0 .

We iterate as in the aforementioned steps.

To prove that this procedure ends, it is enough to show that for some l N , we obtain J ( β l , v l ) = c l = c c 1 c 2 c l 1 . By Lemma A.4, we obtain c c 1 c 2 c l = 0 after most for some l .□

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Received: 2024-05-13
Revised: 2024-08-19
Accepted: 2024-11-07
Published Online: 2024-12-06

© 2024 the author(s), published by De Gruyter

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

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