Infinitely many non-radial solutions for a Choquard equation

Fashun Gao; Minbo Yang

doi:10.1515/anona-2022-0224

Open Access Published by De Gruyter March 9, 2022

Infinitely many non-radial solutions for a Choquard equation

Fashun Gao and Minbo Yang

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0224

Abstract

In this article, we consider the non-linear Choquard equation

− Δ u + V ( ∣ x ∣ ) u = ∫ R 3 ∣ u ( y ) ∣ 2 ∣ x − y ∣ d y u in R 3 ,

where V ( r ) is a positive bounded function. Under some proper assumptions on V ( r ) , we are able to establish the existence of infinitely many non-radial solutions.

Keywords: Choquard equation; Hardy-Littlewood-Sobolev inequality; infinitely many non-radial solutions

MSC 2010: 35J20; 35J60; 35A15

1 Introduction and main results

In the past two decades, many authors have devoted to the study of existence, multiplicity, and properties of the solutions of the non-linear Choquard equation (1.1),

(1.1) − Δ u + V ( x ) u = ( ∣ x ∣ − μ ∗ ∣ u ∣ 2 ) u , in R N .

In a early paper [1], Lieb proved that the ground state U of the equation

(1.2) − Δ u + u = ( ∣ x ∣ − 1 ∗ ∣ u ∣ 2 ) u in R 3 ,

is radial and unique up to translations. While Lions [2] showed the existence of a sequence of radially symmetric solutions via variational methods. In [3,4], the authors proved, if u is a ground state of equation (1.2), then u is either positive or negative and there exist x 0 ∈ R 3 and a monotone function v ∈ C ∞ ( 0 , ∞ ) such that for every x ∈ R 3 , u ( x ) = v ( ∣ x − x 0 ∣ ) . Without loss of generality, we can suppose U > 0 and x 0 = 0 , that is, U ( x ) = U ( ∣ x ∣ ) . For the non-degeneracy of the ground states, we may see [5,6,7,8]. Chen [9] proved that the ground state solution U is non-degenerate, i.e., the kernel of the linearized equation

− Δ ϕ + ϕ − ( ∣ x ∣ − μ ∗ ∣ U ∣ 2 ) ϕ − 2 ( ∣ x ∣ − μ ∗ ( U ϕ ) ) U = 0

and can be spanned by ∂ U ∂ x 1 , ∂ U ∂ x 2 , ∂ U ∂ x 3 . Moreover,

∣ U ′ ∣ ≤ C e − τ r ,

where τ is an arbitrary number in (0, 1), r = ∣ x ∣ , and C is a positive constant depending on τ . By this fact, we also have

(1.3) 0 < U ( x ) ≤ C e − β ∣ x ∣ , x ∈ R N ,

for some C , β > 0 . For more background and recent literature of the non-linear Choquard equation, we may turn to [1,2,3,4,7,8,10,11,12,13,14] and references therein.

The aim of the present article is to consider the following non-linear Choquard equation:

(1.4) − Δ u + V ( ∣ x ∣ ) u = ∫ R 3 ∣ u ( y ) ∣ 2 ∣ x − y ∣ d y u in R 3 , u ∈ H 1 ( R 3 ) ,

where potential V ( x ) satisfies the following assumptions:

( V ) There are constants a > 0 , m ≥ 3 , θ > 0 , and V 0 > 0 such that

V ( r ) = V 0 + a r m + O 1 r m + θ ,

as r → + ∞ . (Without loss of generality, we may assume that V 0 = 1 ).

To apply variational methods, we introduce the energy functional associated with equation (1.4) by

J ( u ) = 1 2 ∫ R 3 ( ∣ ∇ u ∣ 2 + V ( ∣ x ∣ ) ∣ u ∣ 2 ) d x − 1 4 ∫ R 3 ∫ R 3 ∣ u ( x ) ∣ 2 ∣ u ( y ) ∣ 2 ∣ x − y ∣ d x d y .

The Hardy-Littlewood-Sobolev inequality implies that J is well defined on H 1 ( R 3 ) and belongs to C 1 . And so u is a weak solution of (1.4) if and only if u is a critical point of the functional J .

The main result of this article is to establish the existence of infinitely many non-radial solution for (1.4) under assumption ( V ) . The result says that

Theorem 1.1

Suppose that assumption ( V ) holds. Then equation (1.4) has infinitely many non-radial solutions.

To prove the main results, we will adopt the idea introduced by Wei and Yan in [15] to use the unique ground state U of equation (1.2) to build up the approximate solutions for (1.4) with large number of bumps near the infinity. As in [15], let

(1.5) z j = r cos 2 ( j − 1 ) π k , r sin 2 ( j − 1 ) π k , 0 , j = 1 , ⋯ , k ,

and

r ∈ S k ≔ m + 1 2 β π − δ k ln k , m + 1 2 β π + δ k ln k ,

where m is the constant in the expansion for V , δ > 0 is a small constant, and β is given in (1.3). We denote

W r ( x ) ≔ ∑ j = 1 k U z j ( x ) , j = 1 , ⋯ , k ,

where U z j ( x ) = U ( x − z j ) . Set x = ( x ′ , x ″ ) , x ′ ∈ R 2 , and x ″ ∈ R . Define

H s = u ∈ H 1 ( R 3 ) , u is even in x h , h = 2 , 3 , u ( r cos θ , r sin θ , x ″ ) = u r cos θ + 2 j π k , r sin θ + 2 j π k , x ″ .

To prove Theorem 1.1 we only need to prove the following result:

Theorem 1.2

Suppose that V ( r ) satisfies ( V ) . Then there is an integer k 0 > 0 , such that for any integer k ≥ k 0 , (1.4) has a solution u k of the form

u k = W r k ( x ) + w k ,

where w k ∈ H s , r k ∈ S k , and as k → + ∞ ,

∫ R 3 ( ∣ ∇ w k ∣ 2 + w k 2 ) d x → 0 .

This article is organized as follows. In Section 2, we prove two basic estimates. In Section 3, we carry out the reduction. Then, we study the reduced finite dimensional problem and prove Theorem 1.2 in Section 4.

2 Preliminaries

Throughout this article we write ∣ ⋅ ∣ q for the L q ( R 3 ) -norm, q ∈ [ 1 , ∞ ] , always assume that condition ( V ) holds, and the norm of H 1 ( R 3 ) is defined as follows:

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

Let

Ω j = x = ( x ′ , x ″ ) ∈ R 2 × R : x ′ ∣ x ′ ∣ , z j ′ ∣ z j ′ ∣ ≥ cos π k , j = 1 , ⋯ , k .

Then, we have the following basic estimates:

Lemma 2.1

For any x ∈ Ω 1 and η ∈ ( 0 , 1 ] , there is a constant C > 0 , such that

(2.1) ∑ i = 2 k U z i ( x ) ≤ C e − β π η r k e − β ( 1 − η ) ∣ x − z 1 ∣ .

Proof

For any x ∈ Ω 1 , we have ∣ x − z i ∣ ≥ ∣ x − z 1 ∣ . If ∣ x − z 1 ∣ ≥ 1 2 ∣ z i − z 1 ∣ , then for any x ∈ Ω 1 ,

∣ x − z i ∣ ≥ 1 2 ∣ z i − z 1 ∣ .

If ∣ x − z 1 ∣ ≤ 1 2 ∣ z i − z 1 ∣ , then for any x ∈ Ω 1 ,

∣ x − z i ∣ ≥ ∣ z i − z 1 ∣ − ∣ x − z 1 ∣ ≥ 1 2 ∣ z i − z 1 ∣ .

So, for any x ∈ Ω 1 , we have

U z i ( x ) ≤ C e − β ∣ x − z i ∣ ≤ C e − β ∣ x − z 1 ∣ = C e − β η ∣ x − z 1 ∣ e − β ( 1 − η ) ∣ x − z 1 ∣ ≤ C e − 1 2 β η ∣ z i − z 1 ∣ e − β ( 1 − η ) ∣ x − z 1 ∣ .

Thus,

∑ i = 2 k U z i ( x ) ≤ C e − β ( 1 − η ) ∣ x − z 1 ∣ ∑ i = 2 k e − 1 2 β η ∣ z i − z 1 ∣ ≤ C e − β ( 1 − η ) ∣ x − z 1 ∣ ∑ i = 2 k e − β η ∣ z 1 ∣ sin i π k ≤ C e − β ( 1 − η ) ∣ x − z 1 ∣ e − β π η r k . □

Lemma 2.2

For any q ≥ 1 , there is a constant σ > 0 , such that

(2.2) W r q ( x ) = U z i q ( x ) + O 1 r σ e − 1 2 β ∣ x − z i ∣ , ∀ x ∈ Ω i .

Proof

In view of the symmetry, we only estimate the function W r q ( x ) in Ω 1 . Notice that

W r q ( x ) = U z 1 q ( x ) + O U z 1 q − 1 ( x ) ∑ i = 2 k U z i ( x ) + ∑ i = 2 k U z i ( x ) q .

By Lemma 2.1, we have

(2.3) ∣ x − z i ∣ ≥ 1 2 ∣ z i − z 1 ∣ , ∀ x ∈ Ω 1 .

Thus for any α > 0 , using (2.3), we know

∑ i = 2 k U z i α ( x ) ≤ C ∑ i = 2 k e − α β ∣ x − z i ∣ ≤ C ∑ i = 2 k e − 1 2 α β ∣ z 1 − z i ∣ ≤ C e − α β π r k ≤ C k σ , ∀ x ∈ Ω 1 .

Therefore,

U z 1 q − 1 ( x ) ∑ i = 2 k U z i ( x ) ≤ U z 1 q − 1 2 ( x ) ∑ i = 2 k U z i 1 2 ( x ) ≤ C k σ e − 1 2 β ∣ x − z 1 ∣ , ∀ x ∈ Ω 1 ,

and

∑ i = 2 k U z i ( x ) q ≤ U z 1 q 2 ( x ) ∑ i = 2 k U z i 1 2 ( x ) q ≤ C k σ e − 1 2 β ∣ x − z 1 ∣ , ∀ x ∈ Ω 1 .

Consequently, (2.2) follows.□

3 The reduction argument

Let T j = ∂ U z j ∂ r , j = 1 , ⋯ , k , where z j is given in (1.5). We denote by

E = u ∈ H s : ∫ R 3 ∫ R 3 ∣ U z j ( x ) ∣ 2 T j ( y ) u ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U z j ( x ) u ( x ) U z j ( y ) T j ( y ) ∣ x − y ∣ d x d y = 0 , j = 1 , ⋯ , k .

Applying Lemma 2.2, there exists a bounded linear operator L from E to E such that

⟨ L v 1 , v 2 ⟩ = ∫ R 3 ( ∇ v 1 ∇ v 2 + V ( ∣ x ∣ ) v 1 v 2 ) d x − ∫ R 3 ∫ R 3 ∣ W r ( x ) ∣ 2 v 1 ( y ) v 2 ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 W r ( x ) v 1 ( x ) W r ( y ) v 2 ( y ) ∣ x − y ∣ d x d y , v 1 , v 2 ∈ E .

Thus, we have

Lemma 3.1

There is a constant C ′ > 0 , independent of k , such that for any r ∈ S k ,

‖ L v ‖ ≤ C ′ ‖ v ‖ , v ∈ E .

Next, we show that L is invertible in E .

Lemma 3.2

There is a constant C ″ > 0 , independent of k , such that for any r ∈ S k ,

‖ L v ‖ ≥ C ″ ‖ v ‖ , v ∈ E .

Proof

Suppose to the contrary that there are k → + ∞ , r k ∈ S k , and v k ∈ E , with

‖ L v k ‖ = o ( 1 ) ‖ v k ‖ .

Then

(3.1) ⟨ L v k , ψ ⟩ = o ( 1 ) ‖ v k ‖ ‖ ψ ‖ , ∀ ψ ∈ E .

We may assume that ‖ v k ‖ 2 = k .

By symmetry, we see from (3.1),

(3.2) ∫ Ω 1 ( ∇ v k ∇ ψ + V ( ∣ x ∣ ) v k ψ ) d x − ∫ Ω 1 ∫ R 3 ∣ W r ( x ) ∣ 2 v k ( y ) ψ ( y ) ∣ x − y ∣ d x d y − 2 ∫ Ω 1 ∫ R 3 W r ( x ) v k ( x ) W r ( y ) ψ ( y ) ∣ x − y ∣ d x d y = 1 k ⟨ L v k , ψ ⟩ = o ( 1 k ) ‖ ψ ‖ , ∀ ψ ∈ E .

In particular,

∫ Ω 1 ( ∣ ∇ v k ∣ 2 + V ( ∣ x ∣ ) ∣ v k ∣ 2 ) d x − ∫ Ω 1 ∫ R 3 ∣ W r ( x ) ∣ 2 ∣ v k ( y ) ∣ 2 ∣ x − y ∣ d x d y − 2 ∫ Ω 1 ∫ R 3 W r ( x ) v k ( x ) W r ( y ) v k ( y ) ∣ x − y ∣ d x d y = o ( 1 )

and

(3.3) ∫ Ω 1 ( ∣ ∇ v k ∣ 2 + V ( ∣ x ∣ ) ∣ v k ∣ 2 ) d x = 1 .

Let v ¯ k ( x ) = v k ( x − z 1 ) . Then for any R > 0 , since ∣ z 2 − z 1 ∣ = r sin π k ≥ m 4 ln k , we see that B R ( z 1 ) ⊂ Ω 1 . As a result, from (3.3), we find that for any R > 0 ,

∫ B R ( z 1 ) ( ∣ ∇ v ¯ k ∣ 2 + V ( ∣ x ∣ ) ∣ v ¯ k ∣ 2 ) d x ≤ 1 .

So, we may assume that there is a v ∈ H 1 ( R 3 ) , such that as k → + ∞ ,

v ¯ k ⇀ v , weakly in H loc 1 ( R 3 ) ,

and

v ¯ k → v , strongly in L loc 2 ( R 3 ) .

Since v ¯ k is even in x h , h = 2 , 3 , it is easy to see that v is even in x h , h = 2 , 3 . On the other hand, from

∫ R 3 ∫ R 3 ∣ U z 1 ( x ) ∣ 2 T 1 ( y ) v k ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U z 1 ( x ) v k ( x ) U z 1 ( y ) T 1 ( y ) ∣ x − y ∣ d x d y = 0 ,

we obtain

∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 ∂ U ( y ) ∂ z 1 ( y ) v ¯ k ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U ( x ) v ¯ k ( x ) U ( y ) ∂ U ( y ) ∂ z 1 ( y ) ∣ x − y ∣ d x d y = 0 .

So, v satisfies

(3.4) ∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 ∂ U ( y ) ∂ z 1 ( y ) v ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U ( x ) v ( x ) U ( y ) ∂ U ( y ) ∂ z 1 ( y ) ∣ x − y ∣ d x d y = 0 .

Now, we claim that v satisfies

(3.5) − Δ v + v = ∫ R 3 ∣ U ( y ) ∣ 2 ∣ x − y ∣ d y v + 2 ∫ R 3 U ( y ) v ( y ) ∣ x − y ∣ d y U in R 3 .

Define

E ˜ = u : u ∈ H 1 ( R 3 ) : ∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 ∂ U ( y ) ∂ z 1 ( y ) u ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U ( x ) u ( x ) U ( y ) ∂ U ( y ) ∂ z 1 ( y ) ∣ x − y ∣ d x d y = 0 .

For any R > 0 , let ψ ∈ C 0 ∞ ( B R ( 0 ) ) ∩ E ˜ be any function, satisfying that ψ is even in x h , h = 2 , 3 . Then ψ k ≔ ψ ( x − z 1 ) ∈ C 0 ∞ ( B R ( 0 ) ) . With the argument in [15] we find

∫ Ω 1 ( ∇ v k ∇ ψ k + V ( ∣ x ∣ ) v k ψ k ) d x → ∫ R 3 ( ∇ v ∇ ψ + v ψ ) d x .

By Lemma 2.2, we know

∫ Ω 1 ∫ R 3 ∣ W r ( x ) ∣ 2 v k ( y ) ψ k ( y ) ∣ x − y ∣ d x d y = ∫ Ω 1 ∫ Ω 1 U z 1 2 ( x ) + O ( 1 r σ e − 1 2 β ∣ x − z 1 ∣ ) v k ( y ) ψ k ( y ) ∣ x − y ∣ d x d y → ∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 v ( y ) ψ ( y ) ∣ x − y ∣ d x d y .

Similarly,

∫ Ω 1 ∫ R 3 W r ( x ) v k ( x ) W r ( y ) ψ k ( y ) ∣ x − y ∣ d x d y → ∫ R 3 ∫ R 3 U ( x ) v ( x ) U ( y ) ψ ( y ) ∣ x − y ∣ d x d y .

Thus, we have

(3.6) ∫ R 3 ( ∇ v ∇ ψ + v ψ ) d x = ∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 v ( y ) ψ ( y ) ∣ x − y ∣ d x d y + 2 ∫ R 3 ∫ R 3 U ( x ) v ( x ) U ( y ) ψ ( y ) ∣ x − y ∣ d x d y .

On the other hand, since v is even in x h , h = 2 , 3 , (3.6) holds for any function ψ ∈ C 0 ∞ ( R 3 ) , which is odd in x h , h = 2 , 3 . Therefore, (3.6) holds for any ψ ∈ C 0 ∞ ( B R ( 0 ) ) ∩ E ˜ . By the density of C 0 ∞ ( R 3 ) in H 1 ( R 3 ) , it is easy to show that

(3.7) ∫ R 3 ( ∇ v ∇ ψ + v ψ ) d x − ∫ R 3 ∫ R 3 U ( x ) ∣ 2 v ( y ) ψ ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 U ( x ) v ( x ) U ( y ) ψ ( y ) ∣ x − y ∣ d x d y = 0 , ∀ ψ ∈ E ˜ .

But (3.7) holds for ψ = ∂ U ∂ z 1 . Thus, (3.7) is true for any ψ ∈ H 1 ( R 3 ) . So, we have proved (3.5). Since U is non-degenerate, we see v = c ∂ U ∂ z 1 that because v is even in x h , h = 2 , 3 . From (3.4), we find

v = 0 .

As a result,

∫ B R ( z 1 ) v k 2 d x = o ( 1 ) , ∀ R > 0 .

On the other hand, it follows from Lemma 2.1 that for any small η > 0 , there is a constant C > 0 , such that

(3.8) W r k ( x ) ≤ C e − ( 1 − η ) β ∣ x − z 1 ∣ , x ∈ Ω 1 .

Thus,

o ( 1 ) = ∫ Ω 1 ( ∣ ∇ v k ∣ 2 + V ( ∣ x ∣ ) ∣ v k ∣ 2 ) d x − ∫ Ω 1 ∫ R 3 ∣ W r ( x ) ∣ 2 ∣ v k ( y ) ∣ 2 ∣ x − y ∣ d x d y − 2 ∫ Ω 1 ∫ R 3 W r ( x ) v k ( x ) W r ( y ) v k ( y ) ∣ x − y ∣ d x d y

≥ ∫ Ω 1 ( ∣ ∇ v k ∣ 2 + V ( ∣ x ∣ ) ∣ v k ∣ 2 ) d x + o ( 1 ) + O ( e − ( 1 − η ) β R ) ∫ Ω 1 ∣ v k ∣ 2 d x ≥ 1 2 ∫ Ω 1 ( ∣ ∇ v k ∣ 2 + V ( ∣ x ∣ ) ∣ v k ∣ 2 ) d x + o ( 1 ) .

This is a contradiction to (3.3).□

Let

I ( φ ) = J ( W r + φ ) , φ ∈ E .

Expand I ( φ ) as follows:

I ( φ ) = I ( 0 ) + l ( φ ) + 1 2 ⟨ L φ , φ ⟩ + ℛ ( φ ) , φ ∈ E ,

where

l ( φ ) = ∑ j = 1 k ∫ R 3 ( V ( ∣ x ∣ ) − 1 ) U z j φ d x − ∫ R 3 ∫ R 3 ∣ W r ( x ) ∣ 2 W r ( y ) φ ( y ) ∣ x − y ∣ d x d y + ∑ j = 1 k ∫ R 3 ∫ R 3 ∣ U z j ( x ) ∣ 2 U z j ( y ) φ ( y ) ∣ x − y ∣ d x d y

and

ℛ ( φ ) = − ∫ R 3 ∫ R 3 W r ( x ) φ ( x ) φ 2 ( y ) ∣ x − y ∣ d x d y − 1 4 ∫ R 3 ∫ R 3 φ 2 ( x ) φ 2 ( y ) ∣ x − y ∣ d x d y .

In order to find a critical point φ ∈ E for I ( φ ) , we need to estimate each term in the expansion.

Lemma 3.3

There is a constant C > 0 , independent of k , such that for any φ ∈ H 1 ( R 3 ) ,

(3.9) ∣ ℛ ( φ ) ∣ ≤ C ( ‖ φ ‖ 3 + ‖ φ ‖ 4 ) ,

(3.10) ‖ ℛ ′ ( φ ) ‖ ≤ C ( ‖ φ ‖ 2 + ‖ φ ‖ 3 ) ,

and

(3.11) ‖ ℛ ″ ( φ ) ‖ ≤ C ( ‖ φ ‖ + ‖ φ ‖ 2 ) .

Proof

Similar to the proof of (3.1), we have that for any v , w ∈ H 1 ( R 3 )

∣ ℛ ( φ ) ∣ ≤ C ( ‖ φ ‖ 3 + ‖ φ ‖ 4 ) , ∣ ⟨ ℛ ′ ( φ ) , v ⟩ ∣ = − ∫ R 3 ∫ R 3 W r ( x ) v ( x ) φ 2 ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 W r ( x ) φ ( x ) φ ( y ) v ( y ) ∣ x − y ∣ d x d y − ∫ R 3 ∫ R 3 φ 2 ( x ) φ ( y ) v ( y ) ∣ x − y ∣ d x d y ≤ C ( ‖ φ ‖ 2 + ‖ φ ‖ 3 ) ‖ v ‖ ,

and

∣ ⟨ ℛ ″ ( φ ) v , w ⟩ ∣ = − 2 ∫ R 3 ∫ R 3 W r ( x ) v ( x ) φ ( y ) w ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 W r ( x ) w ( x ) φ ( y ) v ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 W r ( x ) φ ( x ) w ( y ) v ( y ) ∣ x − y ∣ d x d y − ∫ R 3 ∫ R 3 φ 2 ( x ) w ( y ) v ( y ) ∣ x − y ∣ d x d y − 2 ∫ R 3 ∫ R 3 φ ( x ) w ( x ) φ ( y ) v ( y ) ∣ x − y ∣ d x d y ≤ C ( ‖ φ ‖ + ‖ φ ‖ 2 ) ‖ v ‖ ‖ w ‖ .

So, (3.10) and (3.11) follow.□

Lemma 3.4

Moreover, there is a small σ > 0 , such that

(3.12) ‖ l k ‖ ≤ C k m − 3 2 + σ .

Proof

By the symmetry of the problem,

∑ j = 1 k ∫ R 3 ( V ( ∣ x ∣ ) − 1 ) U z j φ d x = k ∫ R 3 ( V ( ∣ x ∣ ) − 1 ) U z 1 φ d x = k ∫ R 3 ( V ( ∣ x − z 1 ∣ ) − 1 ) U z 1 φ ( x − z 1 ) d x ≤ k O ( 1 r m ) ‖ φ ‖ ≤ C k m − 1 2 + σ ‖ φ ‖ ,

because m > 1 .

By Lemma 2.1, we have

∑ j ≠ i U z j ≤ C e − β π r k ∀ x ∈ Ω i ,

and

∑ i ≠ j ∫ Ω i ∣ U z i U z j ∣ ∣ x − y ∣ d y ≤ C ∑ i ≠ j ∫ Ω i ∣ U z i ∣ ∣ x − y ∣ d y e − β 2 ∣ z j − z i ∣ ≤ C ∑ i ≠ j e − β ∣ z j ∣ sin i π k ∫ Ω i ∣ U z i ∣ ∣ x − y ∣ d y ≤ C e − β π r k ∑ i ≠ j ∫ Ω i ∣ U z i ∣ ∣ x − y ∣ d y .

Combining these with

W r ( x ) = U z i ( x ) + O 1 r σ e − 1 2 β ∣ x − z i ∣ , ∀ x ∈ Ω i ,

we have,

∑ j = 1 k ∫ R 3 ∫ R 3 ∣ U z j ( x ) ∣ 2 U z j ( y ) φ ( y ) ∣ x − y ∣ d x d y − ∫ R 3 ∫ R 3 ∣ W r ( x ) ∣ 2 W r ( y ) φ ( y ) ∣ x − y ∣ d x d y = ∑ i = 1 k ∫ R 3 ∫ R 3 ∣ U z i ( x ) ∣ 2 ∑ j ≠ i U z j ( y ) φ ( y ) ∣ x − y ∣ d x d y + 2 ∑ j = 1 k ∑ i ≠ j ∫ R 3 ∫ R 3 ∣ U z i ( x ) U z j ( x ) ∣ W r ( y ) φ ( y ) ∣ x − y ∣ d x d y = k ∑ i = 1 k ∫ R 3 ∫ Ω i ∣ U z i ( x ) ∣ 2 ∑ j ≠ i U z j ( y ) φ ( y ) ∣ x − y ∣ d x d y + 2 k ∑ j = 1 k ∑ i ≠ j ∫ Ω i ∫ R 3 ∣ U z i ( x ) U z j ( x ) ∣ W r ( y ) φ ( y ) ∣ x − y ∣ d x d y ≤ k e − β π r k ∑ j = 1 k ∫ R 3 ∫ Ω i ∣ U z j ( x ) ∣ 2 φ ( y ) ∣ x − y ∣ d x d y + 2 e − β π r k ∑ j = 1 k ∫ Ω i ∫ R 3 ∣ U z i ( x ) ∣ W r ( y ) φ ( y ) ∣ x − y ∣ d x d y ≤ C k 2 e − β π r k ‖ φ ‖ .

Since m ≥ 3 , we see

k 2 e − β π r k ≤ k 2 e − β π m + 1 β π − δ ln k ≤ C k m − 3 2 + σ .

Thus, we have

‖ l k ‖ ≤ C k m − 3 2 + σ . □

Proposition 3.5

There is an integer k 0 > 0 , such that for each k ≥ k 0 , there is a C 1 map from S k to H s : φ = φ ( r ) , r = ∣ x 1 ∣ satisfying φ ∈ E , and

∂ I ( φ ) ∂ φ , ψ = 0 , ∀ ψ ∈ E .

Moreover, there is a small σ > 0 , such that

(3.13) ‖ φ ‖ ≤ C k m − 3 2 + σ .

Proof

Since l ( φ ) is a bounded linear functional in E , we know that there is an l k ∈ E , such that

l ( φ ) = ⟨ l k , φ ⟩ .

Thus, finding a critical point for I ( φ ) is equivalent to solving

(3.14) l k + L ( φ ) + ℛ ′ ( φ ) = 0 .

By Lemma 3.2, L is invertible. Thus, (3.14) can be rewritten as

φ = A ( φ ) ≔ − L − 1 l k − L − 1 ℛ ′ ( φ ) .

Let

S = φ : φ ∈ E , ‖ φ ‖ ≤ 1 k m − 3 2 .

So, from Lemma 3.4,

‖ A ( φ ) ‖ ≤ C ‖ l k ‖ + C ( ‖ φ ‖ 2 + ‖ φ ‖ 3 ) ≤ C k m − 3 2 + σ + C k m − 3 + C k 3 ( m − 3 ) 2 ≤ 1 k m − 3 2 .

Thus, A maps S into S .

By (3.11), we have,

‖ A ( φ 1 ) − A ( φ 2 ) ‖ = ‖ L − 1 ℛ ′ ( φ 1 ) − L − 1 ℛ ′ ( φ 2 ) ‖ ≤ C ( ‖ φ 1 ‖ + ‖ φ 1 ‖ 2 + ‖ φ 2 ‖ + ‖ φ 2 ‖ 2 ) ‖ φ 1 − φ 2 ‖ ≤ 1 2 ‖ φ 1 − φ 2 ‖ .

So, we have proved that A is a contraction map from S to S . Therefore, the result follows from the contraction mapping theorem.□

4 Proof of Theorem 1.2

Lemma 4.1

There is a small constant σ > 0 , such that

J ( W r ) = k A k + B 1 r m − k B 2 e − 2 π r k + O 1 k 2 m − 2 + σ ,

where

A k = 1 2 − k 4 ∫ R 3 ∫ R 3 U 4 ( x ) U 4 ( y ) ∣ x − y ∣ d x d y , B 1 = a 2 ∫ R 3 U 2 d x

and B 2 > 0 is a positive constant.

Proof

Using the symmetry,

∫ R 3 ( ∣ ∇ W r ∣ 2 + W r 2 ) d x = ∑ j = 1 k ∑ i = 1 k ∫ R 3 ∫ R 3 ∣ U z j ( x ) ∣ 2 U z j ( y ) U z i ( y ) ∣ x − y ∣ d x d y = k ∫ R 3 ∫ R 3 ∣ U ( x ) ∣ 2 ∣ U ( y ) ∣ 2 ∣ x − y ∣ d x d y + k ∑ i = 2 k ∫ R 3 ∫ R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) ∣ x − y ∣ d x d y .

It follows from Lemma 2.1 that

∫ R 3 ( V ( ∣ x ∣ ) − 1 ) W r 2 d x = k ∫ Ω 1 ( V ( ∣ x ∣ ) − 1 ) W r 2 d x = k ∫ Ω 1 ( V ( ∣ x ∣ ) − 1 ) U z 1 + O e − 1 2 β ∣ z 1 ∣ π k e − 1 2 β ∣ x − z 1 ∣ 2 d x = k ∫ Ω 1 ( V ( ∣ x ∣ ) − 1 ) U z 1 2 d x + k O ∫ Ω 1 ∣ V ( ∣ x ∣ ) − 1 ∣ e − β ∣ z 1 ∣ π k e − β ∣ x − z 1 ∣ d x = k B 1 r m + O 1 k m + θ .

Using Lemma 2.1 and the fact

∑ i = 2 k e − β ∣ z i − z 1 ∣ ≤ C e − 2 π β r k ≤ C k m + 1 − τ , for any r ∈ S k ,

we obtain that if τ > 0 is small enough, for any x ∈ Ω 1 ,

∣ W r ( x ) ∣ 2 = U z 1 2 ( x ) + 2 U z 1 ( x ) ∑ i = 2 k U z i ( x ) + ∑ i = 2 k U z i ( x ) 2 = U z 1 2 ( x ) + 2 U z 1 ( x ) ∑ i = 2 k U z i ( x ) + O 1 k 2 m + σ .

Thus, we have

∫ R 3 ∫ R 3 ∣ W r ( x ) ∣ 2 ∣ W r ( y ) ∣ 2 ∣ x − y ∣ d x d y = k 2 ∫ Ω 1 ∫ Ω 1 ∣ W r ( x ) ∣ 2 ∣ W r ( y ) ∣ 2 ∣ x − y ∣ d x d y = k 2 ∫ R 3 ∫ R 3 U 2 ( x ) U 2 ( y ) ∣ x − y ∣ d x d y + 4 k 2 ∑ i = 2 k ∫ R 3 ∫ R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) ∣ x − y ∣ d x d y + 4 k 2 ∑ i = 2 k ∑ j = 2 k ∫ R 3 ∫ R 3 U z 1 ( x ) U z i ( x ) U z 1 ( y ) U z j ( y ) ∣ x − y ∣ d x d y + k 2 O 1 k 2 m + σ .

On the other hand, we have

∑ i = 2 k ∑ j = 2 k ∫ R 3 ∫ R 3 U z 1 ( x ) U z i ( x ) U z 1 ( y ) U z j ( y ) ∣ x − y ∣ d x d y = B 2 ′ ∑ i = 2 k e − β ∣ z i − z 1 ∣ ∑ j = 2 k e − β ∣ z j − z 1 ∣ = B 2 ″ e − 4 π β r k

and

∑ i = 2 k ∫ R 3 ∫ R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) ∣ x − y ∣ d x d y = B 3 ′ ∑ i = 2 k e − β ∣ z i − z 1 ∣ = B 3 ″ e − 2 β r π k .

So,

J ( W r ) = k A k + B 1 r m − k B 2 e − 2 β π r k + O 1 k 2 m − 2 + σ ,

where

A k = 1 2 − k 4 ∫ R 3 ∫ R 3 U 2 ( x ) U 2 ( y ) ∣ x − y ∣ d x d y , B 1 = a 2 ∫ R 3 U 2 d x .

and B 2 > 0 is a positive constant.□

We are ready to prove Theorem 1.2. Let φ r = φ ( r ) be the map obtained in Proposition 3.5. Define

F ( r ) = J ( W r + φ r ) , ∀ r ∈ S k .

With the same argument in [15,16], we can easily check that for k sufficiently large, if r is a critical point of F ( r ) , then W r + φ r is a solution of (1.4).

Proof of Theorem 1.2

It follows from Lemmas 3.1 and 3.3 that

‖ L φ r ‖ ≤ C ‖ φ r ‖ , ∣ ℛ ( φ ) ∣ ≤ C ( ‖ φ ‖ 3 + ‖ φ ‖ 4 ) .

So, Proposition 3.5 and Lemma 4.1 give

F ( r ) = J ( W r ) + l ( φ r ) + 1 2 ⟨ L φ r , φ r ⟩ + ℛ ( φ r ) = J ( W r ) + O ( ‖ l k ‖ ‖ φ r ‖ + ‖ φ r ‖ 2 + ‖ φ r ‖ 3 + ‖ φ r ‖ 4 ) = J ( W r ) + O 1 k m − 3 + σ ,

where

J ( W r ) = k A k + B 1 r m − k B 2 e − 2 β π r k + O 1 k 2 m − 2 + σ .

Consider

(4.1) max { F ( r ) : r ∈ S k } ,

where S k is defined in Section 1. Since the function

B 1 r m − k B 2 e − 2 β π r k

has a maximum point

r ¯ k = m + 1 2 β π + o ( 1 ) k ln k ,

which is an interior point of S k , it is easy to check that (4.1) is achieved by some r k , which is in the interior of S k . Thus, r k is a critical point of F ( r ) . As a result,

W r + φ r k

is a solution of (1.4).□

Funding information: Fashun Gao was partially supported by NSFC (11901155). Minbo Yang is the corresponding author who was partially supported by NSFC (11971436, 12011530199) and ZJNSF (LZ22A010001, LD19A010001).
Conflict of interest: Authors state no conflict of interest.

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Received: 2021-10-05

Accepted: 2022-01-04

Published Online: 2022-03-09

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

Infinitely many non-radial solutions for a Choquard equation

Abstract

1 Introduction and main results

Theorem 1.1

Theorem 1.2

2 Preliminaries

Lemma 2.1

Proof

Lemma 2.2

Proof

3 The reduction argument

Lemma 3.1

Lemma 3.2

Proof

Lemma 3.3

Proof

Lemma 3.4

Proof

Proposition 3.5

Proof

4 Proof of Theorem 1.2

Lemma 4.1

Proof

Proof of Theorem 1.2

References

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