Standing waves to upper critical Choquard equation with a local perturbation: Multiplicity, qualitative properties and stability

Xinfu Li

doi:10.1515/anona-2022-0230

Open Access Published by De Gruyter March 9, 2022

Standing waves to upper critical Choquard equation with a local perturbation: Multiplicity, qualitative properties and stability

Xinfu Li

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0230

Abstract

In this article, we consider the upper critical Choquard equation with a local perturbation

− Δ u = λ u + ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − 2 u + μ ∣ u ∣ q − 2 u , x ∈ R N , u ∈ H 1 ( R N ) , ∫ R N ∣ u ∣ 2 = a ,

where N ≥ 3 , μ > 0 , a > 0 , λ ∈ R , α ∈ ( 0 , N ) , p = p ¯ ≔ N + α N − 2 , q ∈ 2 , 2 + 4 N , and I α = C ∣ x ∣ N − α with C > 0 . When μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) with γ q = N 2 − N q and K being some positive constant, we prove

Existence and orbital stability of the ground states.
Existence, positivity, radial symmetry, exponential decay, and orbital instability of the “second class” solutions.

This article generalized and improved parts of the results obtained for the Schrödinger equation.

Keywords: normalized solutions; multiplicity; symmetry; stability; upper critical exponent

MSC 2010: 35J20; 35B06; 35B33; 35B35

1 Introduction and main results

In this article, we study standing waves of prescribed mass to the Choquard equation with a local perturbation

(1.1) i ∂ t ψ + Δ ψ + ( I α ∗ ∣ ψ ∣ p ) ∣ ψ ∣ p − 2 ψ + μ ∣ ψ ∣ q − 2 ψ = 0 , ( t , x ) ∈ R × R N ,

where N ≥ 3 , ψ : R × R N → C , μ > 0 , α ∈ ( 0 , N ) , I α is the Riesz potential defined for every x ∈ R N ⧹ { 0 } by

(1.2) I α ( x ) ≔ A α ( N ) ∣ x ∣ N − α , A α ( N ) ≔ Γ N − α 2 Γ α 2 π N / 2 2 α

with Γ denoting the Gamma function (see [1], p. 19), p and q will be defined later.

The equation (1.1) has several physical origins. When N = 3 , p = 2 , α = 2 , and μ = 0 , (1.1) was investigated by Pekar in [2] to study the quantum theory of a polaron at rest. In [3], Choquard applied it as an approximation to Hartree-Fock theory of one component plasma. It also arises in multiple particle systems [4] and quantum mechanics [5]. When p = 2 , equation (1.1) reduces to the well-known Hartree equation. The Choquard equation (1.1) with or without a local perturbation has attracted much attention nowadays, see [6,7,8, 9,10,11] for the local existence, global existence, blow up, and more in general dynamical properties.

Standing waves to (1.1) are solutions of the form ψ ( t , x ) = e − i λ t u ( x ) , where λ ∈ R and u : R N → C . Then u satisfies the equation

(1.3) − Δ u = λ u + ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − 2 u + μ ∣ u ∣ q − 2 u , x ∈ R N .

When looking for solutions to (1.3) one choice is to fix λ < 0 and to search for solutions to (1.3) as critical points of the action functional

J ( u ) ≔ ∫ R N 1 2 ∣ ∇ u ∣ 2 − λ 2 ∣ u ∣ 2 − 1 2 p ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − μ q ∣ u ∣ q d x ,

see for example [12,13, 14,15,16] and references therein. Another choice is to fix the L 2 -norm of the unknown u , that is, to consider the problem

(1.4) − Δ u = λ u + ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − 2 u + μ ∣ u ∣ q − 2 u , x ∈ R N , u ∈ H 1 ( R N ) , ∫ R N ∣ u ∣ 2 = a

with fixed a > 0 and unknown λ ∈ R . In this direction, define on H 1 ( R N ) the energy functional

E ( u ) ≔ 1 2 ∫ R N ∣ ∇ u ∣ 2 − 1 2 p ∫ R N ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − μ q ∫ R N ∣ u ∣ q .

It is standard to check that E ∈ C 1 under some assumptions on p and q , and a critical point of E constrained to

S a ≔ u ∈ H 1 ( R N ) : ∫ R N ∣ u ∣ 2 = a

gives rise to a solution to (1.4). Such solution is usually called a normalized solution of (1.3) on S a , which is the aim of this article.

For future reference, we recall.

Definition 1.1

We say that u is a normalized ground state to (1.3) on S a if

E ( u ) = c a g ≔ inf { E ( v ) : v ∈ S a , ( E ∣ S a ) ′ ( v ) = 0 } .

The set of the normalized ground states will be denoted by G a .

Definition 1.2

G a is orbitally stable if for every ε > 0 there exists δ > 0 such that, for any ψ 0 ∈ H 1 ( R N ) with inf v ∈ G a ‖ ψ 0 − v ‖ H 1 < δ , we have

inf v ∈ G a ‖ ψ ( t , ⋅ ) − v ‖ H 1 < ε for any t > 0 ,

where ψ ( t , x ) denotes the solution to (1.1) with initial value ψ 0 .

A standing wave e − i λ t u is strongly unstable if for every ε > 0 there exists ψ 0 ∈ H 1 ( R N ) such that ‖ ψ 0 − u ‖ H 1 < ε and ψ ( t , x ) blows up in finite time.

When studying normalized solutions to the Choquard equation

(1.5) − Δ u = λ u + ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p − 2 u , x ∈ R N ,

the L 2 -critical exponent p ∗ ≔ 1 + 2 + α N , the Hardy-Littlewood-Sobolev upper critical exponent p ¯ ≔ N + α N − 2 , and lower critical exponent p ̲ ≔ N + α N play an important role. For p ̲ < p < p ∗ , the existence of normalized ground state to (1.5) was studied by Cazenave and Lions [17] and Ye [18] by considering the minimizer of E constrained on S a . Cazenave and Lions [17] also studied the orbital stability of the normalized ground states set by using the concentration compactness principle. For p ∗ < p < p ¯ , the functional E is no longer bounded from below on S a . By considering the minimizer of E constrained on the Pohožaev set, Luo [19] obtained the existence and instability of normalized ground state to (1.5). For p = p ∗ , by scaling invariance, the result is delicate, see [17] and [18] for details. See [20,21,22] for studies of Choquard equation with general nonlinearity. For (1.5) with p = p ¯ , Moroz and Van Schaftingen [23] showed that (1.5) has no solutions in H 1 ( R N ) for fixed λ < 0 . While Gao and Yang [24] obtained the solution to the equation

(1.6) − Δ u = ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u , x ∈ R N

in D 1 , 2 ( R N ) . So it is interesting to study the normalized solutions to (1.5) with p = p ¯ under a local perturbation μ ∣ u ∣ q − 2 u , namely equation (1.4). In a recent article, Li [25] considered the existence and symmetry of solutions to (1.4) with p = p ¯ and 2 + 4 N ≤ q < 2 ∗ ≔ 2 N N − 2 . Note that 2 + 4 N is the L 2 -critical exponent in studying normalized solutions to the Schrödinger equation

− Δ u = λ u + ∣ u ∣ q − 2 u , x ∈ R N .

In this article, we consider (1.4) with p = p ¯ and 2 < q < 2 + 4 N . This article is motivated by [26,27, 28,29], which considered normalized solutions to the Schrödinger equation with mixed nonlinearities

(1.7) − Δ u = λ u + ∣ u ∣ 2 ∗ − 2 u + μ ∣ u ∣ q − 2 u , x ∈ R N .

We should point out that Liu and Shi [10] studied the existence and orbital stability of ground states to (1.4) with p = p ∗ and 2 < q < 2 + 4 N .

Before stating the main results of this article, we make some notations. In the following, we assume p = p ¯ ≔ N + α N − 2 in (1.4). Set

(1.8) S α ≔ inf u ∈ D 1 , 2 ( R N ) ⧹ { 0 } ∫ R N ∣ ∇ u ∣ 2 ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ 1 / p ¯ ,

(1.9) K ≔ 2 p ¯ − q γ q 2 p ¯ ( 2 − q γ q ) S α p ¯ p ¯ ( 2 − q γ q ) C N , q q S α p ¯ q ( p ¯ − 1 ) 2 p ¯ − 2 2 p ¯ − q γ q

with γ q ≔ N 2 − N q and C N , q defined in Lemma 2.1,

(1.10) ρ 0 ≔ p ¯ ( 2 − q γ q ) S α p ¯ 2 p ¯ − q γ q 1 p ¯ − 1 ,

B ρ 0 ≔ { u ∈ H 1 ( R N ) : ‖ ∇ u ‖ 2 2 < ρ 0 } , V a ≔ S a ∩ B ρ 0 , m a ≔ inf u ∈ V a E ( u ) .

Now we state the first two main results of this article.

Theorem 1.3

Let N ≥ 3 , α ∈ ( 0 , N ) , 2 < q < 2 + 4 N , p = p ¯ , μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then

E ∣ S a has a critical point u ˜ at negative level m a < 0 , which is an interior local minimizer of E on the set V a .
m a = c a g (that is, u ˜ is a ground state to (1.4)), and any other ground state to (1.4) is a local minimizer of E on V a .
G a is compact, up to translation.
c a g is reached by a positive and radially symmetric non-increasing function.
For any u ∈ G a , there exists λ < 0 such that u satisfies (1.4).

Theorem 1.4

Let the assumptions in Theorem 1.3 hold, α ≥ N − 4 (i.e., p ¯ ≥ 2 ) and α < N − 2 . Then the set G a is orbitally stable.

To prove Theorems 1.3 and 1.4, we follow the strategy of [26]. In the proofs, a special role will be played by the Pohožaev set

P a ≔ { u ∈ S a : P ( u ) = 0 } ,

where

P ( u ) ≔ ∫ R N ∣ ∇ u ∣ 2 − ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ γ q ∫ R N ∣ u ∣ q .

The set P a is quite related to the fiber map

(1.11) Ψ u ( τ ) ≔ E ( u τ ) = 1 2 τ 2 ‖ ∇ u ‖ 2 2 − 1 2 p ¯ τ 2 p ¯ ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ q τ q γ q ‖ u ‖ q q ,

where

(1.12) u τ ( x ) ≔ τ N 2 u ( τ x ) , x ∈ R N , τ > 0 .

The fiber map Ψ u ( τ ) is introduced by Jeanjean in [30] for the Schrödinger equation and is well studied by Soave in [31]. According to Ψ u ( τ ) , P a = P a , + ∪ P a , − , where

P a , + ≔ { u ∈ P a : E ( u ) < 0 } , P a , − ≔ { u ∈ P a : E ( u ) > 0 }

if μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , see Lemmas 3.3 and 4.4.

The ground state u + obtained in Theorem 1.3 lies on P a , + and can be characterized by

E ( u + ) = inf u ∈ P a , + E ( u ) = inf u ∈ V a E ( u ) = m a .

The critical point u − which will be obtained in the following theorem lies on P a , − and can be characterized by E ( u − ) = inf u ∈ P a , − E ( u ) . Precisely,

Theorem 1.5

Let N ≥ 3 , α ∈ ( 0 , N ) , 2 < q < 2 + 4 N , p = p ¯ , μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then there exists a second solution u − to (1.4) which satisfies

0 < E ( u − ) = inf u ∈ P a , − E ( u ) < m a + 2 + α 2 ( N + α ) S α N + α 2 + α .

In particular, u − is not a ground state.

Remark 1.6

Note that the result is new in the case μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . There is not corresponding result even to the Schrödinger equation (1.7). During the proof, the lower bound of inf u ∈ P a , − E ( u ) obtained in Lemma 4.4 plays an important role. The proof of Lemma 4.4 is interesting. Maybe it gives us some insights to consider the case μ a q ( 1 − γ q ) 2 > ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) .

We combine the methods used in [27] and [29] to prove Theorem 1.5. Precisely, we first use the mountain pass lemma to obtain a Palais-Smale sequence { u n } of E on S a ∩ H r 1 ( R N ) with P ( u n ) → 0 and E ( u n ) → M r ( a ) as n → ∞ , see Lemma 4.1. Second, by using the Pohožaev constraint method and the Schwartz rearrangement, we can show that

M r ( a ) = M ( a ) = inf u ∈ P a , − E ( u ) = inf P a , − ∩ H r 1 ( R N ) E ( u ) ,

see Lemma 4.2. Third, by using the radial symmetry of { u n } and the bounds of inf u ∈ P a , − E ( u ) , we can show that { u n } converges to a solution to (1.4). In the proof, to obtain the upper bound of inf u ∈ P a , − E ( u ) is a difficult task. When N ≥ 5 and p ¯ < 2 , the methods used in [29] can not threat the nonlocal term ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u directly, see Lemma 4.5. Inspired by [27], by using the radially non-increasing of u + and by calculating the nonlocal term carefully (see (4.34)), we can choose { y ε } satisfying (4.30) and (4.31). Based on which, we can give the upper bound of inf u ∈ P a , − E ( u ) when N ≥ 5 and p ¯ < 2 , see Lemma 4.6.

The following result is about the positivity, radial symmetry, and exponential decay of the “second class” solution.

Theorem 1.7

Assume the conditions in Theorem 1.5 hold. Let u be a solution to (1.4) with E ( u ) = inf v ∈ P a , − E ( v ) , then

∣ u ∣ > 0 ;
There exist x 0 ∈ R N and a non-increasing positive function v : ( 0 , ∞ ) → R such that ∣ u ( x ) ∣ = v ( ∣ x − x 0 ∣ ) for almost every x ∈ R N ;
If α ≥ N − 4 (i.e., p ¯ ≥ 2 ), then ∣ u ∣ has exponential decay at infinity:
∣ u ( x ) ∣ ≤ C e − δ ∣ x ∣ , ∣ x ∣ ≥ r 0 ,
for some C > 0 , δ > 0 , and r 0 > 0 .

The positivity is obtained by using the properties of Ψ u ( τ ) and inf v ∈ P a , − E ( v ) . The symmetry is obtained by using the theories of polarization and the fact that inf v ∈ P a , − E ( v ) is a mountain pass level value. This method is motivated by [32]. The exponential decay follows the radial symmetry, the estimate of ( I α ∗ ∣ u ∣ p ¯ ) , and the exponential decay studied in [33] to the Schrödinger equation. Theorem 1.7 plays an important role in proving the following result.

Theorem 1.8 is about the instability of the “second class” solution, which is very new in the existing research. As we know, most existing results are about the instability of a solution but not all solutions.

Theorem 1.8

Assume the conditions in Theorem 1.4 hold. Let u be a solution to (1.4) with E ( u ) = inf v ∈ P a , − E ( v ) , then λ < 0 and the associated standing wave e − i λ t u is strongly unstable.

Remark 1.9

The conditions α ≥ N − 4 (i.e., p ¯ ≥ 2 ) and α < N − 2 in Theorems 1.4 and 1.8 are added for obtaining the local existence of solution to (1.1), see Lemma 6.4. The condition α ≥ N − 4 in Theorem 1.8 is also needed to prove the exponential decay of u (see (3) in Theorem 1.7), which is used to show that ∣ x ∣ u ∈ L 2 ( R N ) .

The condition p ¯ ≥ 2 is added since the nonlinearity ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u is singular when p ¯ < 2 . We do not know whether it can be removed. While, the condition α < N − 2 is added for technical reason, and we guess it can be removed.

This article is organized as follows. In Section 2, we cite some preliminaries. Sections 3–5 are devoted to the proofs of Theorems 1.3, 1.5, and 1.7, respectively. In Section 6, we first give a local existence result, and then prove Theorems 1.4 and 1.8.

Notation: In this article, it is understood that all functions, unless otherwise stated, are complex valued, but for simplicity we write L r ( R N ) , W 1 , r ( R N ) , H 1 ( R N ) D 1 , 2 ( R N ) , … . For 1 ≤ r < ∞ , L r ( R N ) is the usual Lebesgue space endowed with the norm ‖ u ‖ r r ≔ ∫ R N ∣ u ∣ r , W 1 , r ( R N ) is the usual Sobolev space endowed with the norm ‖ u ‖ W 1 , r r ≔ ‖ ∇ u ‖ r r + ‖ u ‖ r r , H 1 ( R N ) = W 1 , 2 ( R N ) and ‖ u ‖ H 1 2 ≔ ‖ u ‖ W 1 , 2 2 , D 1 , 2 ( R N ) ≔ { u ∈ L 2 ∗ ( R N ) : ∣ ∇ u ∣ ∈ L 2 ( R N ) } . H r 1 ( R N ) denotes the subspace of functions in H 1 ( R N ) which are radially symmetric with respect to zero. S a , r ≔ S a ∩ H r 1 ( R N ) . C , C 1 , C 2 , … denote positive constants, whose values can change from line to line. The notation A ≲ B means that A ≤ C B for some constant C > 0 . If A ≲ B ≲ A , we write A ≈ B .

2 Preliminaries

The following Gagliardo-Nirenberg inequality can be found in [34].

Lemma 2.1

Let N ≥ 1 and 2 < p < 2 ∗ , then the following sharp Gagliardo-Nirenberg inequality

‖ u ‖ p ≤ C N , p ‖ u ‖ 2 1 − γ p ‖ ∇ u ‖ 2 γ p

holds for any u ∈ H 1 ( R N ) , where the sharp constant C N , p is

C N , p p = 2 p 2 N + ( 2 − N ) p 2 N + ( 2 − N ) p N ( p − 2 ) N ( p − 2 ) 4 1 ‖ Q p ‖ 2 p − 2

and Q p is the unique positive radial solution of equation

− Δ Q + Q = ∣ Q ∣ p − 2 Q .

The following well-known Hardy-Littlewood-Sobolev inequality can be found in [35].

Lemma 2.2

Let N ≥ 1 , p , r > 1 , and 0 < β < N with 1 / p + ( N − β ) / N + 1 / r = 2 . Let u ∈ L p ( R N ) and v ∈ L r ( R N ) . Then there exists a sharp constant C ( N , β , p ) , independent of u and v , such that

∫ R N ∫ R N u ( x ) v ( y ) ∣ x − y ∣ N − β d x d y ≤ C ( N , β , p ) ‖ u ‖ p ‖ v ‖ r .

If p = r = 2 N N + β , then

C ( N , β , p ) = C β ( N ) = π N − β 2 Γ β 2 Γ N + β 2 Γ N 2 Γ ( N ) − β N .

Remark 2.3

(1). By the Hardy-Littlewood-Sobolev inequality above, for any v ∈ L s ( R N ) with s ∈ ( 1 , N / α ) , I α ∗ v ∈ L N s N − α s ( R N ) and

‖ I α ∗ v ‖ L N s N − α s ≤ C ‖ v ‖ L s ,

where C > 0 is a constant depending only on N , α , and s .

(2). By the Hardy-Littlewood-Sobolev inequality above and the Sobolev embedding theorem, we obtain

(2.1) ∫ R N ( I β ∗ ∣ u ∣ p ) ∣ u ∣ p ≤ C ∫ R N ∣ u ∣ 2 N p N + β 1 + β / N ≤ C ‖ u ‖ H 1 ( R N ) 2 p

for any p ∈ [ 1 + β / N , ( N + β ) / ( N − 2 ) ] if N ≥ 3 and p ∈ [ 1 + β / N , + ∞ ) if N = 1 , 2 , where C > 0 is a constant depending only on N , β , and p .

The following fact is used in this article (see [36]).

Lemma 2.4

Let N ≥ 3 , α ∈ ( 0 , N ) , and p ∈ N + α N , N + α N − 2 . Assume that { w n } n = 1 ∞ ⊂ H 1 ( R N ) satisfying w n ⇀ w weakly in H 1 ( R N ) as n → ∞ , then

( I α ∗ ∣ w n ∣ p ) ∣ w n ∣ p − 2 w n ⇀ ( I α ∗ ∣ w ∣ p ) ∣ w ∣ p − 2 w weakly in H − 1 ( R N ) as n → ∞ .

The following lemma is used in this article, see [33] for its proof.

Lemma 2.5

Let N ≥ 3 and 1 ≤ t < + ∞ . If u ∈ L t ( R N ) is a radial non-increasing function (i.e., 0 ≤ u ( x ) ≤ u ( y ) if ∣ x ∣ ≥ ∣ y ∣ ), then one has

∣ u ( x ) ∣ ≤ ∣ x ∣ − N / t N ∣ S N − 1 ∣ 1 / t ‖ u ‖ t , x ≠ 0 ,

where ∣ S N − 1 ∣ is the area of the unit sphere in R N .

The following Pohožaev identity is cited from [13], where the proof is given for λ > 0 but it clearly extends to λ ∈ R .

Lemma 2.6

Let N ≥ 3 , α ∈ ( 0 , N ) , λ ∈ R , μ ∈ R , p ∈ N + α N , N + α N − 2 , and q ∈ [ 2 , 2 ∗ ] . If u ∈ H 1 ( R N ) is a solution to (1.3), then u satisfies the Pohožaev identity

N − 2 2 ∫ R N ∣ ∇ u ∣ 2 = N λ 2 ∫ R N ∣ u ∣ 2 + N + α 2 p ∫ R N ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p + μ N q ∫ R N ∣ u ∣ q .

Lemma 2.7

Assume the conditions in Lemma 2.6 hold. If u ∈ H 1 ( R N ) is a solution to (1.3), then P ( u ) = 0 .

Proof

Multiplying (1.3) by u and integrating over R N , we derive

∫ R N ∣ ∇ u ∣ 2 = λ ∫ R N ∣ u ∣ 2 + ∫ R N ( I α ∗ ∣ u ∣ p ) ∣ u ∣ p + μ ∫ R N ∣ u ∣ q ,

which combined with the Pohožaev identity from Lemma 2.6 gives that P ( u ) = 0 .□

3 Existence of normalized ground state standing waves

In this section, we prove Theorem 1.3. We first study the lower bound of E ( u ) . By (1.8) and Lemma 2.1, we obtain, for any u ∈ S a ,

(3.1) E ( u ) ≥ 1 2 ‖ ∇ u ‖ 2 2 − 1 2 p ¯ S α − p ¯ ‖ ∇ u ‖ 2 2 p ¯ − μ q C N , q q a q ( 1 − γ q ) / 2 ‖ ∇ u ‖ 2 q γ q = ‖ ∇ u ‖ 2 2 f μ , a ( ‖ ∇ u ‖ 2 2 )

with

(3.2) f μ , a ( ρ ) ≔ 1 2 − 1 2 p ¯ S α − p ¯ ρ p ¯ − 1 − μ q C N , q q a q ( 1 − γ q ) 2 ρ q γ q − 2 2 , ρ ∈ ( 0 , ∞ ) .

Next we study the properties of f μ , a ( ρ ) .

Lemma 3.1

Let N ≥ 3 , α ∈ ( 0 , N ) , μ > 0 , a > 0 , p = p ¯ , q ∈ 2 , 2 + 4 N , and K be defined in (1.9). Then

(3.3) max ρ > 0 f μ , a ( ρ ) > 0 , i f μ a q ( 1 − γ q ) 2 < ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , = 0 , i f μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , < 0 , i f μ a q ( 1 − γ q ) 2 > ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) .

Proof

By the definition of f μ , a ( ρ ) , we have that

f μ , a ′ ( ρ ) = − p ¯ − 1 2 p ¯ S α − p ¯ ρ p ¯ − 2 − μ q q γ q − 2 2 C N , q q a q ( 1 − γ q ) 2 ρ q γ q − 2 2 − 1 .

Hence, the equation f μ , a ′ ( ρ ) = 0 has a unique solution given by

(3.4) ρ μ , a = p ¯ μ ( 2 − q γ q ) q ( p ¯ − 1 ) C N , q q a q ( 1 − γ q ) 2 S α p ¯ 2 2 p ¯ − q γ q .

Taking into account that f μ , a ( ρ ) → − ∞ as ρ → 0 + and f μ , a ( ρ ) → − ∞ as ρ → + ∞ , we obtain that ρ μ , a is the unique global maximum point of f μ , a ( ρ ) and the maximum value is

max ρ > 0 f μ , a ( ρ ) = f μ , a ( ρ μ , a ) = 1 2 − K μ a q ( 1 − γ q ) 2 2 ( p ¯ − 1 ) 2 p ¯ − q γ q ,

which implies that (3.3) holds.□

Lemma 3.2

Let N ≥ 3 , α ∈ ( 0 , N ) , μ > 0 , p = p ¯ , and q ∈ 2 , 2 + 4 N . If a 1 > 0 and ρ 1 > 0 are such that f μ , a 1 ( ρ 1 ) ≥ 0 , then for any a 2 ∈ ( 0 , a 1 ) , we have

(3.5) f μ , a 2 ( ρ 2 ) > 0 for ρ 2 ∈ a 2 a 1 ρ 1 , ρ 1 .

Proof

It is obvious that f μ , a 2 ( ρ 1 ) > f μ , a 1 ( ρ 1 ) ≥ 0 , and by direct calculation,

f μ , a 2 a 2 a 1 ρ 1 = 1 2 − 1 2 p ¯ S α − p ¯ a 2 a 1 p ¯ − 1 ρ 1 p ¯ − 1 − μ q C N , q q a 2 a 1 q 2 − 1 a 1 q ( 1 − γ q ) 2 ρ 1 q γ q − 2 2 > 1 2 − 1 2 p ¯ S α − p ¯ ρ 1 p ¯ − 1 − μ q C N , q q a 1 q ( 1 − γ q ) 2 ρ 1 q γ q − 2 2 = f μ , a 1 ( ρ 1 ) ≥ 0 .

It follows from the properties of f μ , a 2 ( ρ ) studied in Lemma 3.1 that (3.5) holds.□

By Lemma 3.1, the domain { ( μ , a ) ∈ R 2 : μ > 0 , a > 0 } is divided into three parts Ω 1 , Ω 2 , and Ω 3 by the curve μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) with

Ω 1 = ( μ , a ) ∈ R 2 : μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 < ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) ,

Ω 2 = ( μ , a ) ∈ R 2 : μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) ,

and

Ω 3 = ( μ , a ) ∈ R 2 : μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 > ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) .

In this article, we will consider the domain Ω 1 ∪ Ω 2 . For fixed μ > 0 , define a 0 such that

(3.6) μ a 0 q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) .

Then Ω 1 ∪ Ω 2 = { ( μ , a ) ∈ R 2 , μ > 0 , 0 < a ≤ a 0 } . Note that ρ 0 defined in (1.10) is ρ μ , a 0 , and by Lemmas 3.1 and 3.2, f μ , a 0 ( ρ 0 ) = 0 and f μ , a ( ρ 0 ) > 0 for a ∈ ( 0 , a 0 ) . Hence, inf u ∈ ∂ V a E ( u ) ≥ 0 . Moreover, V a is a potential well, see Lemma 3.4.

For future use, we study the properties of Ψ u ( τ ) defined in (1.11).

Lemma 3.3

Let N ≥ 3 , α ∈ ( 0 , N ) , μ > 0 , p = p ¯ , q ∈ 2 , 2 + 4 N , and a ∈ ( 0 , a 0 ] . Then for every u ∈ S a , the function Ψ u ( τ ) has exactly two critical points τ u + and τ u − with 0 < τ u + < τ u − . Moreover,

τ u + is a local minimum point for Ψ u ( τ ) , E ( u τ u + ) < 0 , and u τ u + ∈ V a .
τ u − is a global maximum point for Ψ u ( τ ) , Ψ u ′ ( τ ) < 0 for τ > τ u − , and
E ( u τ u − ) ≥ inf u ∈ ∂ V a E ( u ) ≥ 0 .
In particular, if a ∈ ( 0 , a 0 ) , then inf u ∈ ∂ V a E ( u ) > 0 .
Ψ u ″ ( τ u − ) < 0 and the maps u ∈ S a ↦ τ u − ∈ R is of class C 1 .

Proof

The proof can be obtained by modifying the proof of ([27], Lemma 2.4) in a trivial way. So we omit it.□

Lemma 3.4

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , and a ∈ ( 0 , a 0 ] . Then

m a = inf u ∈ V a E ( u ) < 0 ≤ inf u ∈ ∂ V a E ( u ) .
If m a is reached, then any ground state to (1.4) is contained in V a .

Proof

In view of Lemma 3.3, we just need to prove inf u ∈ V a E ( u ) < 0 . For any fixed u ∈ S a , let u τ ( x ) and Ψ u ( τ ) be defined in (1.12) and (1.11), respectively. It is obvious that ‖ ∇ u τ ‖ 2 2 → 0 and E ( u τ ) = Ψ u ( τ ) → 0 − as τ → 0 + . Hence, we can choose τ 0 > 0 sufficiently small such that u τ 0 ∈ V a and E ( u τ 0 ) < 0 .
Let u ∈ V a be such that E ( u ) = m a . By (1), u is a solution to (1.4). Let v be any ground state to (1.4). Then E ( v ) ≤ E ( u ) = m a < 0 , and by Lemma 2.7, P ( v ) = 0 . Consequently, by Lemma 3.3, τ v + = 1 and v = v τ v + ∈ V a .□

Lemma 3.5

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , and μ > 0 . Then

a ∈ ( 0 , a 0 ] ↦ m a is a continuous mapping.
Let a ∈ ( 0 , a 0 ] . We have for every a 1 ∈ ( 0 , a ) : m a ≤ m a 1 + m a − a 1 , and if m a 1 or m a − a 1 is reached, then the inequality is strict.

Proof

The proof can be done by modifying the proof of ([26], Lemma 2.6) in a trivial way. So we omit it.□

The following result will both imply the existence of a ground state to (1.4) and will be a crucial step to derive the orbital stability of the set G a .

Proposition 3.6

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , and a ∈ ( 0 , a 0 ] . If { u n } ⊂ B ρ 0 is such that ‖ u n ‖ 2 2 → a and E ( u n ) → m a , then, up to translation, u n converges to u ∈ V a strongly in H 1 ( R N ) .

Proof

Since { u n } ⊂ B ρ 0 and ‖ u n ‖ 2 2 → a , we obtain that { u n } is bounded in H 1 ( R N ) . We claim that

(3.7) liminf n → ∞ sup y ∈ R N ∫ B 1 ( y ) ∣ u n ( x ) ∣ 2 d x > 0 .

If it is false, ‖ u n ‖ q → 0 as n → ∞ by Lions’ vanishing lemma, see ([37], Lemma 1.21). By using (1.8) and { u n } ⊂ B ρ 0 , we obtain that

E ( u n ) = 1 2 ‖ ∇ u n ‖ 2 2 − 1 2 p ¯ ∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ + o n ( 1 ) ≥ 1 2 ‖ ∇ u n ‖ 2 2 − 1 2 p ¯ S α − p ¯ ‖ ∇ u n ‖ 2 2 p ¯ + o n ( 1 ) = ‖ ∇ u n ‖ 2 2 1 2 − 1 2 p ¯ S α − p ¯ ‖ ∇ u n ‖ 2 2 p ¯ − 2 + o n ( 1 ) ≥ ‖ ∇ u n ‖ 2 2 1 2 − 1 2 p ¯ S α − p ¯ ρ 0 p ¯ − 1 + o n ( 1 ) .

Since f μ , a 0 ( ρ 0 ) = 0 , we have that

1 2 − 1 2 p ¯ S α − p ¯ ρ 0 p ¯ − 1 = μ q C N , q q a 0 q ( 1 − γ q ) 2 ρ 0 q γ q − 2 2 > 0 .

Consequently, E ( u n ) ≥ o n ( 1 ) , which contradicts E ( u n ) → m a < 0 .

So (3.7) holds. Going if necessary to a subsequence, there exists a sequence { y n } ⊂ R N such that, u ˜ n ( x ) ≔ u n ( x − y n ) ⇀ u ∈ H 1 ( R N ) ⧹ { 0 } weakly in H 1 ( R N ) . Set v n = u ˜ n − u . Then by the weak convergence and the Brezis-Lieb lemma, we know

‖ u n ‖ 2 2 = ‖ u ˜ n ‖ 2 2 = ‖ u ‖ 2 2 + ‖ v n ‖ 2 2 + o n ( 1 ) , ‖ ∇ u n ‖ 2 2 = ‖ ∇ u ˜ n ‖ 2 2 = ‖ ∇ u ‖ 2 2 + ‖ ∇ v n ‖ 2 2 + o n ( 1 ) , ‖ u n ‖ q q = ‖ u ˜ n ‖ q q = ‖ u ‖ q q + ‖ v n ‖ q q + o n ( 1 ) ,

and

∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ = ∫ R N ( I α ∗ ∣ u ˜ n ∣ p ¯ ) ∣ u ˜ n ∣ p ¯ = ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ + ∫ R N ( I α ∗ ∣ v n ∣ p ¯ ) ∣ v n ∣ p ¯ + o n ( 1 ) .

Consequently,

E ( u n ) = E ( u ˜ n ) = E ( u ) + E ( v n ) + o n ( 1 ) .

Next, by repeating word by word the proof of Theorem 2.5 in [26], we can show that ‖ v n ‖ 2 2 → 0 and ‖ ∇ v n ‖ 2 2 → 0 as n → ∞ . Thus, u ˜ n → u ∈ V a strongly in H 1 ( R N ) .□

Proof of Theorem 1.3

(1), (2), and (3) follow from Proposition 3.6 and Lemma 3.4. To prove (4), we let ∣ u ˜ ∣ ∗ denote the Schwartz rearrangement of ∣ u ˜ ∣ . Then

‖ ∣ u ˜ ∣ ∗ ‖ 2 2 = ‖ u ˜ ‖ 2 2 = a , ‖ ∇ ∣ u ˜ ∣ ∗ ‖ 2 2 ≤ ‖ ∇ ∣ u ˜ ∣ ‖ 2 2 ≤ ‖ ∇ u ˜ ‖ 2 2 < ρ 0 , ‖ ∣ u ˜ ∣ ∗ ‖ q q = ‖ u ˜ ‖ q q , ∫ R N ( I α ∗ ∣ ∣ u ˜ ∣ ∗ ∣ p ¯ ) ∣ ∣ u ˜ ∣ ∗ ∣ p ¯ ≥ ∫ R N ( I α ∗ ∣ u ˜ ∣ p ¯ ) ∣ u ˜ ∣ p ¯ .

These imply that ∣ u ˜ ∣ ∗ ∈ V a and E ( ∣ u ˜ ∣ ∗ ) ≤ E ( u ˜ ) = m a . By the definition of m a , we know m a is attained by the positive and radially symmetric non-increasing function ∣ u ˜ ∣ ∗ . Finally, we prove (5). By using the equation (1.4), P ( u ) = 0 , 0 < γ q < 1 , and μ > 0 , we obtain

(3.8) λ a = ‖ ∇ u ‖ 2 2 − ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ ‖ u ‖ q q = μ ( γ q − 1 ) ‖ u ‖ q q < 0 ,

which implies λ < 0 . The proof is complete.□

4 Existence of mountain pass-type normalized standing waves

In this section, we prove Theorem 1.5. First, we use the mountain pass lemma to obtain a special Palais-Smale sequence. Now we set

M r ( a ) ≔ inf g ∈ Γ r ( a ) max t ∈ [ 0 , ∞ ) E ( g ( t ) ) ,

where

Γ r ( a ) ≔ { g ∈ C ( [ 0 , ∞ ) , S a , r ) : g ( 0 ) ∈ P a , + , ∃ t g s.t. g ( t ) ∈ E 2 m a for t ≥ t g }

with E c ≔ { u ∈ H 1 ( R N ) : E ( u ) < c } . Then we have

Lemma 4.1

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then there exists a Palais-Smale sequence { u n } ⊂ S a , r for E ∣ S a at level M r ( a ) , with P ( u n ) → 0 as n → ∞ .

Proof

We follow the strategy introduced in [30] and consider the functional E ˜ : R + × H 1 ( R N ) → R defined by

E ˜ ( τ , u ) ≔ E ( u τ ) = Ψ u ( τ ) .

Define

M ˜ r ( a ) ≔ inf g ˜ ∈ Γ ˜ r ( a ) max t ∈ [ 0 , ∞ ) E ˜ ( g ˜ ( t ) ) ,

where

Γ ˜ r ( a ) ≔ { g ˜ ∈ C ( [ 0 , ∞ ) , R + × S a , r ) : g ˜ ( 0 ) ∈ ( 1 , P a , + ) , ∃ t g ˜ s.t. g ˜ ( t ) ∈ ( 1 , E 2 m a ) , t ≥ t g ˜ } .

Similar to Lemma 3.3 in [27], we can show that M ˜ r ( a ) = M r ( a ) and

M ˜ r ( a ) = inf g ˜ ∈ Γ ˜ r ( a ) max t ∈ [ 0 , ∞ ) E ˜ ( g ˜ ( t ) ) ≥ 0 > max { E ˜ ( g ˜ ( 0 ) ) , E ˜ ( g ˜ ( t g ˜ ) ) } .

Then repeating word by word the proof of Proposition 1.10 in [27], we can obtain a Palais-Smale sequence { u n } ⊂ S a , r for E ∣ S a at level M r ( a ) , with P ( u n ) → 0 as n → ∞ .□

Next we study the value of M r ( a ) . For this aim, we set

M ( a ) ≔ inf g ∈ Γ ( a ) max t ∈ [ 0 , ∞ ) E ( g ( t ) ) ,

where

(4.1) Γ ( a ) ≔ { g ∈ C ( [ 0 , ∞ ) , S a ) : g ( 0 ) ∈ V a ∩ E 0 , ∃ t g s.t. g ( t ) ∈ E 2 m a , t ≥ t g } .

Lemma 4.2

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then

M r ( a ) = M ( a ) = inf P a , − E ( u ) = inf P a , − ∩ H r 1 ( R N ) E ( u ) .

Proof

Obviously, M r ( a ) ≥ M ( a ) .

For any g ( t ) ∈ Γ ( a ) , since g ( 0 ) ∈ V a , E ( g ( 0 ) ) < 0 , and E ( g ( t g ) ) < 2 m a < m a , by Lemma 3.3, we have τ g ( 0 ) − > 1 and τ g ( t g ) − < 1 . So by the continuity of g ( t ) and of u ↦ τ u − , we know that there exists t 0 such that τ g ( t 0 ) − = 1 , i.e., g ( t 0 ) ∈ P a , − . Thus,

M ( a ) = inf g ∈ Γ ( a ) max t ∈ [ 0 , ∞ ) E ( g ( t ) ) ≥ inf g ∈ Γ ( a ) E ( g ( t 0 ) ) ≥ inf u ∈ P a , − E ( u ) .

For any u ∈ P a , − , let ∣ u ∣ ∗ be the Schwartz rearrangement of ∣ u ∣ . Since ‖ ∣ u ∣ ∗ ‖ t = ‖ u ‖ t with t ∈ [ 1 , ∞ ) , ‖ ∇ ( ∣ u ∣ ∗ ) ‖ 2 ≤ ‖ ∇ u ‖ 2 and

∫ R N ( I α ∗ ( ∣ u ∣ ∗ ) p ¯ ) ( ∣ u ∣ ∗ ) p ¯ ≥ ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ ,

we obtain that Ψ ∣ u ∣ ∗ ( τ ) ≤ Ψ u ( τ ) for any τ ∈ [ 0 , ∞ ) . Let τ u − be defined by Lemma 3.3 such that P ( u τ u − ) = 0 . Then

E ( u ) = Ψ u ( 1 ) = Ψ u ( τ u − ) ≥ Ψ u ( τ ∣ u ∣ ∗ − ) ≥ Ψ ∣ u ∣ ∗ ( τ ∣ u ∣ ∗ − ) .

Since ∣ u ∣ τ ∣ u ∣ ∗ − ∗ ∈ P a , − ∩ H r 1 ( R N ) , we have that

E ( u ) ≥ inf P a , − ∩ H r 1 ( R N ) E ( u ) .

By the arbitrariness of u , we obtain that

(4.2) inf P a , − E ( u ) ≥ inf P a , − ∩ H r 1 ( R N ) E ( u ) .

For any u ∈ P a , − ∩ H r 1 ( R N ) , define

g u ( t ) ≔ u t + τ u + ,

where τ u + is defined by Lemma 3.3. Then g u ( t ) ∈ Γ r ( a ) and

E ( u ) = max t ∈ [ 0 , ∞ ) E ( g u ( t ) ) ≥ M r ( a ) ,

which implies that inf P a , − ∩ H r 1 ( R N ) E ( u ) ≥ M r ( a ) . The proof is complete.□

When μ a q ( 1 − γ q ) 2 < ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , the lower bound of inf u ∈ P a , − E ( u ) is already studied in Lemma 3.3. For clarity, we restate it in the following lemma.

Lemma 4.3

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , 2 < q < 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 < ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then inf u ∈ P a , − E ( u ) > 0 .

When μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , by Lemma 3.3, inf u ∈ P a , − E ( u ) ≥ 0 . Now we prove that the strict inequality holds.

Lemma 4.4

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , 2 < q < 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 = ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then inf u ∈ P a , − E ( u ) > 0 .

Proof

Suppose by contradiction that inf u ∈ P a , − E ( u ) = 0 . Since by Lemma 4.2, inf P a , − E ( u ) = inf P a , − ∩ H r 1 ( R N ) E ( u ) , there exists { u n } ⊂ S a ∩ H r 1 ( R N ) such that P ( u n ) = 0 and E ( u n ) = A n , where A n → 0 + as n → ∞ . By using E ( u n ) = A n , P ( u n ) = 0 , ‖ u n ‖ 2 2 = a , (1.8), and Lemma 2.1, we obtain that

(4.3) ‖ ∇ u n ‖ 2 2 = 2 p ¯ − q γ q q ( p ¯ − 1 ) μ ‖ u n ‖ q q + C 1 A n ≤ 2 p ¯ − q γ q q ( p ¯ − 1 ) μ C N , q q a q 2 ( 1 − γ q ) ‖ ∇ u n ‖ 2 q γ q + C 1 A n , ‖ ∇ u n ‖ 2 2 = 2 p ¯ − q γ q p ¯ ( 2 − q γ q ) ∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ − C 2 A n ≤ 2 p ¯ − q γ q p ¯ ( 2 − q γ q ) S α − p ¯ ‖ ∇ u n ‖ 2 2 p ¯ − C 2 A n ,

where C 1 and C 2 are some positive constants. Consequently, liminf n → ∞ ‖ ∇ u n ‖ 2 2 > 0 and

‖ ∇ u n ‖ 2 2 ( 1 − q γ q / 2 ) ≤ 2 p ¯ − q γ q p ¯ ( 2 − q γ q ) S α p ¯ q γ q − 2 2 ( p ¯ − 1 ) + o n ( 1 ) , ‖ ∇ u n ‖ 2 2 ( p ¯ − 1 ) ≥ p ¯ ( 2 − q γ q ) S α p ¯ 2 p ¯ − q γ q + o n ( 1 ) ,

which implies that

‖ ∇ u n ‖ 2 2 ≤ ρ 0 + o n ( 1 ) , ‖ ∇ u n ‖ 2 2 ≥ ρ 0 + o n ( 1 ) .

Hence, ‖ ∇ u n ‖ 2 2 → ρ 0 as n → ∞ , which combined with (4.3) gives that

(4.4) ‖ u n ‖ q q → C N , q q ‖ u n ‖ 2 q ( 1 − γ q ) ‖ ∇ u n ‖ 2 q γ q ∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ → S α − p ¯ ‖ ∇ u n ‖ 2 2 p ¯

as n → ∞ . That is, { u n } ⊂ H r 1 ( R N ) is a minimizing sequence of

(4.5) 1 C N , q q ≔ inf u ∈ H 1 ( R N ) ⧹ { 0 } ‖ ∇ u ‖ 2 q γ q ‖ u ‖ 2 q ( 1 − γ q ) ‖ u ‖ q q

and

(4.6) S α ≔ inf u ∈ D 1 , 2 ( R N ) ⧹ { 0 } ‖ ∇ u ‖ 2 2 ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ 1 / p ¯ .

Since { u n } ⊂ H r 1 ( R N ) is bounded, there exists u 0 ∈ H r 1 ( R N ) such that u n ⇀ u 0 weakly in H 1 ( R N ) , u n → u 0 strongly in L t ( R N ) with t ∈ ( 2 , 2 ∗ ) , and u n → u 0 a.e. in R N . By the weak convergence, we have ‖ u 0 ‖ 2 2 ≤ ‖ u n ‖ 2 2 and ‖ ∇ u 0 ‖ 2 2 ≤ ‖ ∇ u n ‖ 2 2 . Consequently, u 0 is a minimizer of (4.5) and u n → u 0 strongly in H 1 ( R N ) . By Theorem B in [34], u 0 is the ground state of the equation

(4.7) ( q − 2 ) N 4 Δ u − 1 + ( q − 2 ) ( 2 − N ) 4 u + ∣ u ∣ q − 2 u = 0 .

By using (4.6) and u n → u 0 strongly in H 1 ( R N ) , we obtain that u 0 is a minimizer of S α . So u 0 is of the form

(4.8) u 0 = C b b 2 + ∣ x ∣ 2 N − 2 2 ,

where C > 0 is a fixed constant and b ∈ ( 0 , ∞ ) is a parameter, see [24]. (4.8) contradicts to (4.7). Thus, inf u ∈ P a , − E ( u ) > 0 .□

The next two lemmas are about the upper bound of inf u ∈ P a , − E ( u ) .

Lemma 4.5

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . If N ≥ 5 , we further assume that p ¯ ≥ 2 (i.e., α ≥ N − 4 ). Then

inf u ∈ P a , − E ( u ) < m a + 2 + α 2 ( N + α ) S α N + α 2 + α .

Proof

For any ε > 0 , we define

(4.9) u ε ( x ) = φ ( x ) U ε ( x ) ,

where φ ( x ) ∈ C c ∞ ( R N ) is a cut off function satisfying: (a) 0 ≤ φ ( x ) ≤ 1 for any x ∈ R N ; (b) φ ( x ) ≡ 1 in B 1 ; (c) φ ( x ) ≡ 0 in R N ⧹ B 2 ¯ . Here, B s denotes the ball in R N of center at origin and radius s .

U ε ( x ) = ( N ( N − 2 ) ε 2 ) N − 2 4 ( ε 2 + ∣ x ∣ 2 ) N − 2 2 ,

where U 1 ( x ) is the extremal function of the minimizing problem (1.8). In [24], they proved that S α = S ( A α ( N ) C α ( N ) ) 1 / p ¯ , where A α ( N ) is defined in (1.2), C α ( N ) is in Lemma 2.2, and

S ≔ inf u ∈ D 1 , 2 ( R N ) ⧹ { 0 } ∫ R N ∣ ∇ u ∣ 2 ∫ R N ∣ u ∣ 2 N N − 2 N − 2 N .

By [38] (see also [37]), we have the following estimates.

(4.10) ∫ R N ∣ ∇ u ε ∣ 2 = S N 2 + O ( ε N − 2 ) , N ≥ 3 ,

and

(4.11) ∫ R N ∣ u ε ∣ 2 = K 2 ε 2 + O ( ε N − 2 ) , N ≥ 5 , K 2 ε 2 ∣ ln ε ∣ + O ( ε 2 ) , N = 4 , K 2 ε + O ( ε 2 ) , N = 3 ,

where K 2 > 0 . By direct calculation, for t ∈ ( 2 , 2 ∗ ) , there exists K 1 > 0 such that

(4.12) ∫ R N ∣ u ε ∣ t ≥ ( N ( N − 2 ) ) N − 2 4 t ε N − N − 2 2 t ∫ B 1 ε ( 0 ) 1 ( 1 + ∣ x ∣ 2 ) N − 2 2 t d x ≥ K 1 ε N − N − 2 2 t , ( N − 2 ) t > N , K 1 ε N − N − 2 2 t ∣ ln ε ∣ , ( N − 2 ) t = N , K 1 ε N − 2 2 t , ( N − 2 ) t < N .

Moreover, similar to that [39] and [24], by direct computation, we have

(4.13) ∫ R N ( I α ∗ ∣ u ε ∣ p ¯ ) ∣ u ε ∣ p ¯ ≥ ( A α ( N ) C α ( N ) ) N 2 S α N + α 2 + O ( ε N + α 2 ) .

Let u + be a positive and radially symmetric non-increasing ground state to (1.4). For t ≥ 0 , we define

(4.14) u ˆ ε , t = u + + t u ε and u ¯ ε , t = a − 1 2 ‖ u ˆ ε , t ‖ 2 N − 2 2 u ˆ ε , t a − 1 2 ‖ u ˆ ε , t ‖ 2 x .

Then

∫ R N ∣ u ¯ ε , t ∣ 2 = a , ∫ R N ∣ ∇ u ¯ ε , t ∣ 2 = ∫ R N ∣ ∇ u ˆ ε , t ∣ 2 ,

∫ R N ( I α ∗ ∣ u ¯ ε , t ∣ p ¯ ) ∣ u ¯ ε , t ∣ p ¯ = ∫ R N ( I α ∗ ∣ u ˆ ε , t ∣ p ¯ ) ∣ u ˆ ε , t ∣ p ¯ ,

∫ R N ∣ u ¯ ε , t ∣ q = a − 1 2 ‖ u ˆ ε , t ‖ 2 q γ q − q ∫ R N ∣ u ˆ ε , t ∣ q .

Since u ¯ ε , t ∈ S a , by Lemma 3.3, there exists a unique τ ε , t − > 0 such that ( u ¯ ε , t ) τ ε , t − ∈ P a , − , which implies that

(4.15) ( τ ε , t − ) 2 − q γ q ‖ ∇ u ¯ ε , t ‖ 2 2 = ( τ ε , t − ) 2 p ¯ − q γ q ∫ R N ( I α ∗ ∣ u ¯ ε , t ∣ p ¯ ) ∣ u ¯ ε , t ∣ p ¯ + μ γ q ‖ u ¯ ε , t ‖ q q .

Since u ¯ ε , 0 = u + ∈ P a , + , by Lemma 3.3, τ ε , 0 − > 1 . By (4.10), (4.13), and (4.15), τ ε , t − → 0 as t → + ∞ uniformly for ε > 0 sufficiently small. Since τ ε , t − is unique by Lemma 3.3, it is standard to show that τ ε , t − is continuous for t ≥ 0 , which implies that there exists t ε > 0 such that τ ε , t ε − = 1 . Consequently, inf u ∈ P a , − E ( u ) ≤ sup t ≥ 0 E ( u ¯ ε , t ) for any ε small enough. By (4.10)–(4.13), and the expression

(4.16) E ( u ¯ ε , t ) = 1 2 ‖ ∇ u ˆ ε , t ‖ 2 2 − 1 2 p ¯ ∫ R N ( I α ∗ ∣ u ˆ ε , t ∣ p ¯ ) ∣ u ˆ ε , t ∣ p ¯ − μ q a − 1 2 ‖ u ˆ ε , t ‖ 2 q γ q − q ‖ u ˆ ε , t ‖ q q ,

we have E ( u ¯ ε , t ) → m a as t → 0 , and

E ( u ¯ ε , t ) ≤ t 2 ‖ ∇ u ε ‖ 2 2 − 1 2 p ¯ t 2 p ¯ ∫ R N ( I α ∗ ∣ u ε ∣ p ¯ ) ∣ u ε ∣ p ¯ → − ∞

as t → + ∞ uniformly for ε > 0 sufficiently small. Hence, there exists t 0 > 0 large enough and ε 0 > 0 small enough such that

E ( u ¯ ε , t ) < m a + 2 + α 2 ( N + α ) S α N + α 2 + α

for t < 1 t 0 and t > t 0 uniformly for 0 < ε < ε 0 .

Next we estimate E ( u ¯ ε , t ) for 1 t 0 < t < t 0 . By using the inequalities

( a + b ) r ≥ a r + r a r − 1 b + b r , a > 0 , b > 0 , r ≥ 2 ,

and

( a + b ) r ≥ a r + r a r − 1 b + r a b r − 1 + b r , a > 0 , b > 0 , r ≥ 3 ,

we obtain that

(4.17) ‖ ∇ u ˆ ε , t ‖ 2 2 = ‖ ∇ u + ‖ 2 2 + 2 t ∫ R N ∇ u + ∇ u ε + ‖ ∇ ( t u ε ) ‖ 2 2 ,

(4.18) ‖ u ˆ ε , t ‖ q q ≥ ‖ u + ‖ q q + ‖ t u ε ‖ q q + q t ∫ R N ∣ u + ∣ q − 1 u ε ,

‖ u ˆ ε , t ‖ 2 2 = ‖ u + ‖ 2 2 + 2 t ∫ R N u + u ε + ‖ t u ε ‖ 2 2 ,

(4.19) a − 1 2 ‖ u ˆ ε , t ‖ 2 2 = 1 + 2 t a ∫ R N u + u ε + t 2 a ‖ u ε ‖ 2 2 ,

(4.20) ∫ R N ( I α ∗ ∣ u ˆ ε , t ∣ p ¯ ) ∣ u ˆ ε , t ∣ p ¯ ≥ ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ + 2 p ¯ t ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − 1 u ε + 2 p ¯ ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − 1 u + + ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯

for N = 3 , and

(4.21) ∫ R N ( I α ∗ ∣ u ˆ ε , t ∣ p ¯ ) ∣ u ˆ ε , t ∣ p ¯ ≥ ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ + ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ + 2 p ¯ t ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − 1 u ε

for N ≥ 4 , and N ≥ 5 , p ¯ ≥ 2 .

By the positivity of u + , we have

(4.22) ∫ R N u + u ε ≈ ∫ B 1 φ ( x ) U ε ( x ) ≈ ε N + 2 2 ∫ 0 1 ε 1 ( 1 + r 2 ) N − 2 2 r N − 1 d r ≈ ε N + 2 2 1 ε 2 ≈ ε N − 2 2 .

By (4.11), (4.19), (4.22), and the inequality ( 1 + t ) a ≥ 1 + a t for t ≥ 0 and a < 0 , we obtain that

(4.23) a − 1 2 ‖ u ˆ ε , t ‖ 2 q γ q − q = 1 + 2 t a ∫ R N u + u ε + t 2 a ‖ u ε ‖ 2 2 q γ q − q 2 ≥ 1 + q γ q − q 2 2 t a ∫ R N u + u ε + t 2 a ‖ u ε ‖ 2 2 .

Case N = 3 . Noting that u + satisfies the equation

− Δ u + = λ u + + ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − 2 u + + μ ∣ u + ∣ q − 2 u +

with λ < 0 and λ a = μ ( γ q − 1 ) ‖ u + ‖ q q (see (3.8)), and by using (4.16), (4.17), (4.18), (4.20), and (4.23), we obtain that

(4.24) E ( u ¯ ε , t ) ≤ 1 2 ‖ ∇ u + ‖ 2 2 + t ∫ R N ∇ u + ∇ u ε + 1 2 ‖ ∇ ( t u ε ) ‖ 2 2 − 1 2 p ¯ ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − t ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − 1 u ε − ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − 1 u + − 1 2 p ¯ ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − μ q ‖ u + ‖ q q − μ q ‖ t u ε ‖ q q − μ t ∫ R N ∣ u + ∣ q − 1 u ε − μ ( γ q − 1 ) 2 2 t a ∫ R N u + u ε + t 2 a ‖ u ε ‖ 2 2 ‖ u ˆ ε , t ‖ q q = E ( u + ) + E ( t u ε ) + t λ ∫ R N u + u ε − μ t ( γ q − 1 ) a ‖ u ˆ ε , t ‖ q q ∫ R N u + u ε − μ ( γ q − 1 ) t 2 2 a ‖ u ε ‖ 2 2 ‖ u ˆ ε , t ‖ q q − ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − 1 u + = m a + E ( t u ε ) + μ t ( 1 − γ q ) a ( ‖ u ˆ ε , t ‖ q q − ‖ u + ‖ q q ) ∫ R N u + u ε + μ ( 1 − γ q ) t 2 2 a ‖ u ε ‖ 2 2 ‖ u ˆ ε , t ‖ q q − ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − 1 u + .

By direct calculation, we have

(4.25) ∫ R N ( I α ∗ ∣ u ε ∣ p ¯ ) ∣ u ε ∣ p ¯ − 1 u + ≳ ∫ R N ( I α ∗ ∣ u ε ∣ p ¯ ) ∣ u ε ∣ p ¯ − 1 ≳ ∫ B 1 ∫ B 1 ∣ U ε ( x ) ∣ p ¯ ∣ U ε ( y ) ∣ p ¯ − 1 ∣ x − y ∣ N − α d x d y = ε N − 2 2 ∫ B 1 ε ∫ B 1 ε 1 ( 1 + ∣ x ∣ 2 ) N − 2 2 p ¯ ∣ x − y ∣ N − α ( 1 + ∣ y ∣ 2 ) N − 2 2 ( p ¯ − 1 ) d x d y ≳ ε N − 2 2 ,

(4.26) ‖ u ˆ ε , t ‖ q q − ‖ u + ‖ q q = ‖ u + + t u ε ‖ q q − ‖ u + ‖ q q ≲ ∫ R N ∣ u + ∣ q − 1 t u ε + ‖ t u ε ‖ q q ,

and similar to (4.22),

(4.27) ∫ R N u + q − 1 u ε ≲ ∫ B 2 U ε ( x ) ≲ ε N − 2 2 .

By (4.10), (4.11), (4.13), (4.22), (4.24), (4.25), (4.26), and (4.27), we obtain

E ( u ¯ ε , t ) ≤ m a + t 2 2 S N 2 + O ( ε N − 2 ) − t 2 p ¯ 2 p ¯ ( A α ( N ) C α ( N ) ) N 2 S α N + α 2 + O ( ε N + α 2 ) − μ q t q ‖ u ε ‖ q q + O ( ε N − 2 ) + O ( ε N − 2 2 ) ‖ u ε ‖ q q + O ( ‖ u ε ‖ 2 2 ) − C ε N − 2 2 < m a + t 2 2 S N 2 − t 2 p ¯ 2 p ¯ ( A α ( N ) C α ( N ) ) N 2 S α N + α 2 ≤ m a + 2 + α 2 ( N + α ) S α N + α 2 + α

for 1 t 0 < t < t 0 uniformly for ε ∈ ( 0 , ε 0 ) small enough.

Case N ≥ 4 . Similar to case N = 3 , by using (4.16), (4.17), (4.18), (4.21), and (4.23), we have

(4.28) E ( u ¯ ε , t ) ≤ m a + E ( t u ε ) + μ t ( 1 − γ q ) a ( ‖ u ˆ ε , t ‖ q q − ‖ u + ‖ q q ) ∫ R N u + u ε + μ ( 1 − γ q ) t 2 2 a ‖ u ε ‖ 2 2 ‖ u ˆ ε , t ‖ q q .

Thus, by using (4.10), (4.11), (4.12), (4.13), (4.22), (4.26), (4.27), and (4.28), we obtain

(4.29) E ( u ¯ ε , t ) ≤ m a + t 2 2 S N 2 + O ( ε N − 2 ) − t 2 p ¯ 2 p ¯ ( A α ( N ) C α ( N ) ) N 2 S α N + α 2 + O ( ε N + α 2 ) − μ q t q ‖ u ε ‖ q q + O ( ε N − 2 ) + O ( ε N − 2 2 ) ‖ u ε ‖ q q + O ( ‖ u ε ‖ 2 2 ) < m a + t 2 2 S N 2 − t 2 p ¯ 2 p ¯ ( A α ( N ) C α ( N ) ) N 2 S α N + α 2 ≤ m a + 2 + α 2 ( N + α ) S α N + α 2 + α

for 1 t 0 < t < t 0 uniformly for ε ∈ ( 0 , ε 0 ) small enough. The proof is complete.□

Lemma 4.6

Let N ≥ 4 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Then

inf u ∈ P a , − E ( u ) < m a + 2 + α 2 ( N + α ) S α N + α 2 + α .

Proof

Step 1. Let u + and u ε be defined in Lemma 4.5. We claim that for any ε > 0 , there exists y ε ∈ R N such that

(4.30) ∫ R N u + ( x − y ε ) u ε ( x ) d x ≤ ‖ u ε ‖ 2 2

and

(4.31) ∫ R N ∇ u + ( x − y ε ) ⋅ ∇ u ε ( x ) d x ≤ ‖ u ε ‖ 2 2 .

Indeed, since u + is radial and non-increasing, by Lemma 2.5, we obtain that

∫ R N u + ( x − y ) u ε ( x ) d x ≤ N ∣ S N − 1 ∣ 1 / 2 a ∫ R N ∣ x − y ∣ − N / 2 u ε ( x ) d x .

Noting that supp ( u ε ) ⊂ B 2 , and by using the Hölder inequality, we have, for ∣ y ∣ > 10 ,

∫ R N u + ( x − y ) u ε ( x ) d x ≲ ∫ B 2 ∣ y ∣ 2 − N / 2 u ε ( x ) d x ≲ ∣ y ∣ 2 − N / 2 ∣ B 2 ∣ 1 / 2 ‖ u ε ‖ 2 ,

which combined with (4.11) implies that (4.30) holds for y ε large enough.

Noting that for any y ∈ R N , u + ( x − y ) is a solution to the equation

− Δ u = λ u + ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u + μ ∣ u ∣ q − 2 u

with some λ < 0 , we obtain that

(4.32) ∫ R N ∇ u + ( x − y ) ⋅ ∇ u ε ( x ) d x ≤ ∫ R N ( I α ∗ ∣ u + ( x − y ) ∣ p ¯ ) ∣ u + ( x − y ) ∣ p ¯ − 2 u + ( x − y ) u ε ( x ) d x + μ ∫ R N ∣ u + ( x − y ) ∣ q − 2 u + ( x − y ) u ε ( x ) d x .

Similar to the proof of (4.30), we have

(4.33) μ ∫ R N ∣ u + ( x − y ) ∣ q − 2 u + ( x − y ) u ε ( x ) d x ≤ 1 2 ‖ u ε ‖ 2 2

for y large enough. Now, for x ∈ B 2 and ∣ y ∣ > 100 , we calculate

(4.34) ∫ R N 1 ∣ x − z ∣ N − α ∣ u + ( z − y ) ∣ p ¯ d z = ∫ R N \ B 4 ∣ y ∣ ( y ) + ∫ B 4 ∣ y ∣ ( y ) \ B ∣ y ∣ 2 ( y ) + ∫ B ∣ y ∣ 2 ( y ) 1 ∣ x − z ∣ N − α ∣ u + ( z − y ) ∣ p ¯ d z ≔ I 1 + I 2 + I 3 .

It follows from x ∈ B 2 , ∣ y ∣ > 100 , and z ∈ R N \ B 4 ∣ y ∣ ( y ) that

∣ x − z ∣ ≥ ∣ z ∣ − ∣ x ∣ ≥ 1 2 ∣ z ∣ ≥ 1 2 ∣ y ∣

and

∣ x − z ∣ ≥ 1 2 ∣ z ∣ ≥ 1 4 ( ∣ z ∣ + ∣ y ∣ ) ≥ 1 4 ∣ z − y ∣ .

By using Lemma 2.5 with t = 2 , for any δ ∈ ( 0 , min { N − α , N p ¯ / 2 − α } ) , we have

(4.35) I 1 ≲ ∫ R N \ B 4 ∣ y ∣ ( y ) 1 ∣ y ∣ δ ∣ z − y ∣ N − α − δ ∣ z − y ∣ − N p ¯ 2 ‖ u + ( z − y ) ‖ 2 p ¯ d z ≲ 1 ∣ y ∣ δ ∫ 4 ∣ y ∣ + ∞ 1 r N − α − δ r − N p ¯ 2 r N − 1 d r ≲ 1 ∣ y ∣ δ .

For x ∈ B 2 , ∣ y ∣ > 100 , and z ∈ B 4 ∣ y ∣ ( y ) \ B ∣ y ∣ 2 ( y ) , we obtain that ∣ z − x ∣ ≤ 8 ∣ y ∣ and ∣ z − y ∣ ≥ ∣ y ∣ 2 . By using Lemma 2.5 with t = 2 , we have

(4.36) I 2 ≲ ∫ B 4 ∣ y ∣ ( y ) \ B ∣ y ∣ 2 ( y ) 1 ∣ x − z ∣ N − α ∣ y ∣ 2 − N p ¯ 2 ‖ u + ( z − y ) ‖ 2 p ¯ d z ≲ ∣ y ∣ − N p ¯ 2 ∫ 0 8 ∣ y ∣ 1 r N − α r N − 1 d r ≲ ∣ y ∣ α − N p ¯ 2 .

For x ∈ B 2 , ∣ y ∣ > 100 , and z ∈ B ∣ y ∣ 2 ( y ) , we obtain that ∣ z − x ∣ ≥ ∣ y ∣ 3 and ∣ z − y ∣ ≤ ∣ y ∣ 2 . Then by using Lemma 2.5 with t = 2 ∗ and ‖ u + ‖ 2 ∗ ≤ C , we obtain that

(4.37) I 3 ≲ ∫ B ∣ y ∣ 2 ( y ) 1 ∣ x − z ∣ N − α ∣ z − y ∣ − N p ¯ / 2 ∗ d z ≲ ∫ B ∣ y ∣ 2 ( 0 ) 1 ∣ y ∣ 3 N − α ∣ z ∣ − N p ¯ / 2 ∗ d z ≲ 1 ∣ y ∣ N − α ∫ 0 ∣ y ∣ 2 r − N p ¯ / 2 ∗ r N − 1 d r ≲ ∣ y ∣ α − N 2 .

By using (4.34), (4.35), (4.36), and (4.37), similar to the proof of (4.30), we obtain

(4.38) ∫ R N ( I α ∗ ∣ u + ( x − y ) ∣ p ¯ ) ∣ u + ( x − y ) ∣ p ¯ − 2 u + ( x − y ) u ε ( x ) d x ≤ 1 2 ‖ u ε ‖ 2 2

for y large enough. In view of (4.32), (4.33), and (4.38), we complete the proof of (4.31).

Step 2. Let y ε be given in Step 1 such that (4.30) and (4.31) hold. As in (4.14), we define

u ˆ ε , t = u + ( x − y ε ) + t u ε ( x ) and u ¯ ε , t = a − 1 2 ‖ u ˆ ε , t ‖ 2 N − 2 2 u ˆ ε , t a − 1 2 ‖ u ˆ ε , t ‖ 2 x .

Similar to the proof of Lemma 4.5, there exists t 0 > 0 large enough and ε 0 > 0 small enough such that

(4.39) E ( u ¯ ε , t ) < m a + 2 + α 2 ( N + α ) S α N + α 2 + α

for t < 1 t 0 and t > t 0 uniformly for 0 < ε < ε 0 .

By using the inequality

( a + b ) r ≥ a r + b r , a > 0 , b > 0 , r ≥ 1 ,

(4.10), (4.11), (4.12), (4.13), (4.30), and (4.31), similar to (4.29), we obtain

(4.40) E ( u ¯ ε , t ) ≤ 1 2 ‖ ∇ u + ‖ 2 2 + t ∫ R N ∇ u + ( x − y ε ) ∇ u ε ( x ) + 1 2 ‖ ∇ ( t u ε ) ‖ 2 2 − 1 2 p ¯ ∫ R N ( I α ∗ ∣ u + ∣ p ¯ ) ∣ u + ∣ p ¯ − 1 2 p ¯ ∫ R N ( I α ∗ ∣ t u ε ∣ p ¯ ) ∣ t u ε ∣ p ¯ − μ q ‖ u + ‖ q q − μ q ‖ t u ε ‖ q q − μ ( γ q − 1 ) 2 2 t a ∫ R N u + ( x − y ε ) u ε ( x ) + t 2 a ‖ u ε ‖ 2 2 ‖ u ˆ ε , t ‖ q q ≤ E ( u + ) + E ( t u ε ) + O ( ‖ u ε ‖ 2 2 ) + O ( ‖ u ε ‖ 2 2 ) ‖ u ˆ ε , t ‖ q q < m a + 2 + α 2 ( N + α ) S α N + α 2 + α

for 1 t 0 < t < t 0 uniformly for ε ∈ ( 0 , ε 0 ) small enough. In view of (4.39) and (4.40), we complete the proof.□

The next lemma is about the convergence of the Palais-Smale sequence.

Lemma 4.7

Assume N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . Let { u n } ⊂ S a , r be a Palais-Smale sequence for E ∣ S a at level c , with P ( u n ) → 0 as n → ∞ . If

0 < c < m a + 2 + α 2 ( N + α ) S α N + α 2 + α ,

then up a subsequence, u n → u strongly in H 1 ( R N ) , and u is a radial solution to (1.4) with E ( u ) = c and some λ < 0 .

Proof

The proof is divided into four steps.

Step 1. We show { u n } is bounded in H 1 ( R N ) . It follows from P ( u n ) = o n ( 1 ) and E ( u n ) = c + o n ( 1 ) that

E ( u n ) = 1 2 − 1 2 p ¯ ‖ ∇ u n ‖ 2 2 + γ q 2 p ¯ − 1 q μ ‖ u n ‖ q q + o n ( 1 ) .

Since q γ q < 2 < 2 p ¯ , by using the Gagliardo-Nirenberg inequality, we obtain that

1 2 − 1 2 p ¯ ‖ ∇ u n ‖ 2 2 ≤ c + 1 q − γ q 2 p ¯ μ ‖ u n ‖ q q + o n ( 1 ) ≤ c + 1 q − γ q 2 p ¯ μ C N , q q a q 2 ( 1 − γ q ) ‖ ∇ u n ‖ 2 q γ q + o n ( 1 ) ,

which implies that { ‖ ∇ u n ‖ 2 2 } is bounded. Since { u n } ⊂ S a , we obtain that { u n } is bounded in H 1 ( R N ) .

There exists u ∈ H r 1 ( R N ) such that, up to a subsequence, u n ⇀ u weakly in H 1 ( R N ) , u n → u strongly in L t ( R N ) with t ∈ ( 2 , 2 ∗ ) and u n → u a.e. in R N .

Step 2. We claim that u ≢ 0 . Suppose by contradiction that u ≡ 0 . By using E ( u n ) = c + o n ( 1 ) , P ( u n ) = o n ( 1 ) , ‖ u n ‖ q q = o n ( 1 ) , and (1.8), we obtain that

E ( u n ) = 1 2 − 1 2 p ¯ ‖ ∇ u n ‖ 2 2 + o n ( 1 )

and

(4.41) ‖ ∇ u n ‖ 2 2 = ∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ + o n ( 1 ) ≤ ( S α − 1 ‖ ∇ u n ‖ 2 2 ) p ¯ + o n ( 1 ) .

Since c > 0 , we obtain liminf n → ∞ ‖ ∇ u n ‖ 2 2 > 0 and hence

limsup n → ∞ ‖ ∇ u n ‖ 2 2 ≥ S α N + α 2 + α .

Consequently,

c = lim n → ∞ 1 2 − 1 2 p ¯ ‖ ∇ u n ‖ 2 2 + o n ( 1 ) ≥ 2 + α 2 ( N + α ) S α N + α 2 + α ,

which contradicts to

c < m a + 2 + α 2 ( N + α ) S α N + α 2 + α

and m a < 0 . So u ≢ 0 .

Step 3. We show u is a solution to (1.3) with some λ < 0 . Since { u n } is a Palais-Smale sequence of E ∣ S a , by the Lagrange multipliers rule, there exists λ n such that

(4.42) ∫ R N ( ∇ u n ⋅ ∇ φ − λ n u n φ − ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ − 2 u n φ − μ ∣ u n ∣ q − 2 u n φ ) = o n ( 1 ) ‖ φ ‖ H 1

for every φ ∈ H 1 ( R N ) . The choice φ = u n provides

(4.43) λ n a = ‖ ∇ u n ‖ 2 2 − ∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ − μ ‖ u n ‖ q q + o n ( 1 )

and the boundedness of { u n } in H 1 ( R N ) implies that λ n is bounded as well; thus, up to a subsequence λ n → λ ∈ R . Furthermore, by using P ( u n ) = o n ( 1 ) , (4.43), μ > 0 , γ q ∈ ( 0 , 1 ) , and u n ⇀ u weakly in H 1 ( R N ) , we obtain that

− λ n a = μ ( 1 − γ q ) ‖ u n ‖ q q + o n ( 1 )

and then

− λ a ≥ μ ( 1 − γ q ) ‖ u ‖ q q > 0 ,

which implies that λ < 0 . By using (4.42) and Lemma 2.4, we obtain that

(4.44) ∫ R N ( ∇ u ⋅ ∇ φ − λ u φ − ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u φ − μ ∣ u ∣ q − 2 u φ ) = lim n → ∞ ∫ R N ( ∇ u n ⋅ ∇ φ − λ n u n φ − ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ − 2 u n φ − μ ∣ u n ∣ q − 2 u n φ ) = lim n → ∞ o n ( 1 ) ‖ φ ‖ H 1 = 0 ,

which implies that u satisfies the equation

(4.45) − Δ u = λ u + ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u + μ ∣ u ∣ q − 2 u .

Thus, P ( u ) = 0 by Lemma 2.7.

Step 4. We show u n → u strongly in H 1 ( R N ) . Set v n ≔ u n − u . Then we have

(4.46) ‖ u n ‖ 2 2 = ‖ u ‖ 2 2 + ‖ v n ‖ 2 2 + o n ( 1 ) , ‖ ∇ u n ‖ 2 2 = ‖ ∇ u ‖ 2 2 + ‖ ∇ v n ‖ 2 2 + o n ( 1 ) ,

(4.47) ‖ u n ‖ q q = ‖ u ‖ q q + ‖ v n ‖ q q + o n ( 1 ) = ‖ u ‖ q q + o n ( 1 )

and

∫ R N ( I α ∗ ∣ u n ∣ p ¯ ) ∣ u n ∣ p ¯ = ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ + ∫ R N ( I α ∗ ∣ v n ∣ p ¯ ) ∣ v n ∣ p ¯ + o n ( 1 ) ,

which combined with P ( u n ) = o n ( 1 ) and P ( u ) = 0 gives that

(4.48) ‖ ∇ v n ‖ 2 2 = ∫ R N ( I α ∗ ∣ v n ∣ p ¯ ) ∣ v n ∣ p ¯ + o n ( 1 ) .

Similarly to (4.41), we infer that

limsup n → ∞ ‖ ∇ v n ‖ 2 2 ≥ S α N + α 2 + α or liminf n → ∞ ‖ ∇ v n ‖ 2 2 = 0 .

If limsup n → ∞ ‖ ∇ v n ‖ 2 2 ≥ S α N + α 2 + α , then by using the fact that u satisfies (4.45), ‖ u ‖ 2 2 ≤ a , (4.46), (4.47), and Lemma 3.5, we obtain that

E ( u n ) = 1 2 − 1 2 p ¯ ‖ ∇ u n ‖ 2 2 + γ q 2 p ¯ − 1 q μ ‖ u n ‖ q q + o n ( 1 ) = 1 2 − 1 2 p ¯ ‖ ∇ u ‖ 2 2 + γ q 2 p ¯ − 1 q μ ‖ u ‖ q + 1 2 − 1 2 p ¯ ‖ ∇ v n ‖ 2 2 + o n ( 1 ) ≥ m ‖ u ‖ 2 2 + 2 + α 2 ( N + α ) S α N + α 2 + α + o n ( 1 ) ≥ m a + 2 + α 2 ( N + α ) S α N + α 2 + α + o n ( 1 ) ,

which contradicts E ( u n ) = c + o n ( 1 ) and c < m a + 2 + α 2 ( N + α ) S α N + α 2 + α . Thus,

liminf n → ∞ ‖ ∇ v n ‖ 2 2 = 0

holds. So up to a subsequence, ∇ u n → ∇ u in L 2 ( R N ) . Choosing φ = u n − u in (4.42) and (4.44), and subtracting, we obtain that

∫ R N ( ∣ ∇ ( u n − u ) ∣ 2 − λ ∣ u n − u ∣ 2 ) → 0 .

Since λ < 0 , we get that u n → u strongly in H 1 ( R N ) . The proof is complete.□

Proof of Theorem 1.5

It is a direct result of Lemmas 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, and 4.7.□

5 Positivity, symmetry, and exponential decay of solution u to (1.4) with E ( u ) = inf P a , − E ( v )

In this section, we prove Theorem 1.7. For future use, we first give the following result.

Lemma 5.1

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . If u ∈ P a , − such that E ( u ) = inf P a , − E ( v ) , then u satisfies the equation (1.4) with some λ < 0 .

Proof

By the Lagrange multiplier rule, there exist λ and η such that u satisfies

(5.1) − Δ u − ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u − μ ∣ u ∣ q − 2 u = λ u + η [ − 2 Δ u − 2 p ¯ ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u − μ q γ q ∣ u ∣ q − 2 u ] ,

or equivalently,

− ( 1 − 2 η ) Δ u = λ u + ( 1 − η 2 p ¯ ) ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u + μ ( 1 − η q γ q ) ∣ u ∣ q − 2 u .

Next we show η = 0 . Similar to the definition of P ( u ) (see Lemma 2.7), we obtain

( 1 − 2 η ) ‖ ∇ u ‖ 2 2 − ( 1 − η 2 p ¯ ) ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ ( 1 − η q γ q ) γ q ‖ u ‖ q q = 0 ,

which combined with P ( u ) = 0 gives that

η 2 ‖ ∇ u ‖ 2 2 − 2 p ¯ ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ q γ q 2 ‖ u ‖ q q = 0 .

If η ≠ 0 , then

2 ‖ ∇ u ‖ 2 2 − 2 p ¯ ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ q γ q 2 ‖ u ‖ q q = 0 ,

which combined with P ( u ) = 0 gives that

μ γ q ( 2 p ¯ − q γ q ) ‖ u ‖ q q = ( 2 p ¯ − 2 ) ‖ ∇ u ‖ 2 2 , ( q γ q − 2 p ¯ ) ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ = ( q γ q − 2 ) ‖ ∇ u ‖ 2 2 .

Hence,

E ( u ) = ( p ¯ − 1 ) ( q γ q − 2 ) 2 p ¯ q γ q ‖ ∇ u ‖ 2 2 < 0 ,

which contradicts to inf P a , − E ( v ) ≥ 0 , see Lemma 3.3. So η = 0 .

It follows from (5.1) with η = 0 , P ( u ) = 0 , 0 < γ q < 1 , and μ > 0 that

λ a = ‖ ∇ u ‖ 2 2 − ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − μ ‖ u ‖ q q = μ ( γ q − 1 ) ‖ u ‖ q q < 0 ,

which implies λ < 0 . The proof is complete.□

Next, we study the positivity of the solution u to (1.4) with E ( u ) = inf P a , − E ( v ) . By Lemma 5.1, it is enough to prove the following result.

Proposition 5.2

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . If u ∈ P a , − such that E ( u ) = inf P a , − E ( v ) , then ∣ u ∣ τ ∣ u ∣ − ∈ P a , − and E ( ∣ u ∣ τ ∣ u ∣ − ) = inf P a , − E ( v ) . Moreover, ∣ u ∣ τ ∣ u ∣ − > 0 in R N .

Proof

It follows from ‖ ∇ ∣ u ∣ ‖ 2 2 ≤ ‖ ∇ u ‖ 2 2 that Ψ ∣ u ∣ ( τ ) ≤ Ψ u ( τ ) for any τ > 0 . By Lemma 3.3, we have

E ( ∣ u ∣ τ ∣ u ∣ − ) = Ψ ∣ u ∣ ( τ ∣ u ∣ − ) ≤ Ψ u ( τ ∣ u ∣ − ) ≤ Ψ u ( τ u − ) = E ( u ) .

Since ∣ u ∣ τ ∣ u ∣ − ∈ P a , − , we obtain that E ( ∣ u ∣ τ ∣ u ∣ − ) = inf P a , − E ( v ) . By Lemma 5.1, there exists λ < 0 such that ∣ u ∣ τ ∣ u ∣ − satisfies the equation

− Δ u = λ u + ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u + μ ∣ u ∣ q − 2 u .

Since ∣ u ∣ τ ∣ u ∣ − is continuous by Theorem 2.1 in [13], the strong maximum principle implies that ∣ u ∣ τ ∣ u ∣ − > 0 in R N .□

Next, we study the radial symmetry of the solution u to (1.4) with E ( u ) = inf P a , − E ( v ) . We follow the arguments of [32], which relies on polarization. So we first recall some theories of polarization [40,23,41].

Assume that H ⊂ R N is a closed half-space and that σ H is the reflection with respect to ∂ H . The polarization u H : R N → R of u : R N → R is defined for x ∈ R N by

u H ( x ) = max { u ( x ) , u ( σ H ( x ) ) } , if x ∈ H , min { u ( x ) , u ( σ H ( x ) ) } , if x ∉ H .

Lemma 5.3

(Polarization and Dirichlet integrals, Lemma 5.3 in [40]). Let H ⊂ R N be a closed half-space. If u ∈ H 1 ( R N ) , then u H ∈ H 1 ( R N ) , and

∫ R N ∣ ∇ u H ∣ 2 = ∫ R N ∣ ∇ u ∣ 2 .

Lemma 5.4

(Polarization and nonlocal integrals, Lemma 5.3 in [23]). Let α ∈ ( 0 , N ) , u ∈ L 2 N N + α ( R N ) , and H ⊂ R N be a closed half-space. If u ≥ 0 , then

∫ R N ∫ R N u ( x ) u ( y ) ∣ x − y ∣ N − α d x d y ≤ ∫ R N ∫ R N u H ( x ) u H ( y ) ∣ x − y ∣ N − α d x d y ,

with equality if and only if either u H = u or u H = u ∘ σ H .

Lemma 5.5

(Symmetry and polarization, Proposition 3.15 in [41], Lemma 5.4 in [23]). Assume that u ∈ L 2 ( R N ) is non-negative. There exist x 0 ∈ R N and a non-increasing function v : ( 0 , ∞ ) → R such that for almost every x ∈ R N , u ( x ) = v ( ∣ x − x 0 ∣ ) if and only if for every closed half-space H ⊂ R N , u H = u or u H = u ∘ σ H .

Now we are ready to prove the radial symmetry result.

Proposition 5.6

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , and μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) . If u is a positive solution to (1.4) with E ( u ) = inf P a , − E ( v ) , then there exist x 0 ∈ R N and a non-increasing positive function v : ( 0 , ∞ ) → R such that u ( x ) = v ( ∣ x − x 0 ∣ ) for almost every x ∈ R N .

Proof

By Lemmas 2.7 and 3.3, u ∈ P a , − . Let Γ ( a ) be defined in (4.1), τ 1 ≥ 0 be small enough such that u τ 1 ∈ V a and E ( u τ 1 ) < 0 . Then g u ( t ) = u t + τ 1 ∈ Γ ( a ) , g u ( τ u − − τ 1 ) = u τ u − , g u ( t ) ≥ 0 for every t ≥ 0 , E ( g u ( t ) ) < E ( u τ u − ) = E ( u ) = inf v ∈ P a , − E ( v ) for any t ∈ ( [ 0 , ∞ ) ⧹ { τ u − − τ 1 } ) .

For every closed half-space H define the path g u H : [ 0 , ∞ ) → S a by g u H ( t ) = ( g u ( t ) ) H . By Lemma 5.3 and ‖ u H ‖ r = ‖ u ‖ r with r ∈ [ 1 , ∞ ) , we have g u H ∈ C ( [ 0 , ∞ ) , S a ) . By Lemmas 5.3 and 5.4, we obtain that g u H ( 0 ) ∈ V a and E ( g u H ( t ) ) ≤ E ( g u ( t ) ) for every t ∈ [ 0 , ∞ ) and thus g u H ∈ Γ ( a ) . Hence,

max t ∈ [ 0 , ∞ ) E ( g u H ( t ) ) ≥ inf v ∈ P a , − E ( v ) .

Since for every t ∈ ( [ 0 , ∞ ) ⧹ { τ u − − τ 1 } ) ,

E ( g u H ( t ) ) ≤ E ( g u ( t ) ) < E ( u ) = inf v ∈ P a , − E ( v ) ,

we deduce that

E ( g u H ( τ u − − τ 1 ) ) = E ( u H ) = inf v ∈ P a , − E ( v ) .

Hence E ( u H ) = E ( u ) , which implies that

∫ R N ( I α ∗ ∣ u H ∣ p ¯ ) ∣ u H ∣ p ¯ = ∫ R N ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ .

By Lemma 5.4, we have u H = u or u H = u ∘ σ H . By Lemma 5.5, we complete the proof.□

Proposition 5.7

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , μ > 0 , a > 0 , μ a q ( 1 − γ q ) 2 ≤ ( 2 K ) q γ q − 2 p ¯ 2 ( p ¯ − 1 ) , and α ≥ N − 4 (i.e., p ¯ ≥ 2 ). If u is a positive solution to (1.4) with E ( u ) = inf P a , − E ( v ) , then u has exponential decay at infinity:

u ( x ) ≤ C e − δ ∣ x ∣ , ∣ x ∣ ≥ r 0 ,

for some C > 0 , δ > 0 , and r 0 > 0 .

Proof

By Lemmas 2.7 and 5.1, there exists λ < 0 such that u satisfies the equation

(5.2) − Δ u = λ u + ( I α ∗ ∣ u ∣ p ¯ ) ∣ u ∣ p ¯ − 2 u + μ ∣ u ∣ q − 2 u .

By Proposition 5.6, there exist x 0 ∈ R N and a non-increasing positive function v : ( 0 , ∞ ) → R such that u ( x ) = v ( ∣ x − x 0 ∣ ) for almost every x ∈ R N . Hence, w ≔ u ( x + x 0 ) is a positive and radially non-increasing solution to (5.2). Similar to the estimate of (4.34), there exists r 0 > 0 such that

( I α ∗ ∣ w ∣ p ¯ ) ( x ) = C ∫ R N ∣ w ( x − z ) ∣ p ¯ ∣ z ∣ N − α d z ≤ − λ 2

for ∣ x ∣ > r 0 . Hence, if p ¯ > 2 , there exists C > 0 such that w satisfies

− Δ w ≤ λ w + C w p ¯ − 2 w + μ w q − 2 w , ∣ x ∣ ≥ r 0 ,

and if p ¯ = 2 , w satisfies

− Δ w ≤ λ 2 w + μ w q − 2 w , ∣ x ∣ ≥ r 0 .

Now, repeating word by word the proof of Lemma 2 in [33], we can show that w decays exponentially at infinity. The proof is complete.□

Proof of Theorem 1.7

By Proposition 5.2, w ≔ ∣ u ∣ τ ∣ u ∣ − ∈ P a , − is a positive solution to (1.4) with E ( w ) = inf v ∈ P a , − E ( v ) . Hence, w has exponential decay at infinity by Proposition 5.7, and by Proposition 5.6, there exist x 0 ∈ R N and a non-increasing positive function v : ( 0 , ∞ ) → R such that w = v ( ∣ x − x 0 ∣ ) for almost every x ∈ R N . The proof is complete by using the fact ∣ u ∣ ( x ) = ( τ ∣ u ∣ − ) − N / 2 w x τ ∣ u ∣ − .□

6 Dynamical studies to the equation (1.1)

In this section, we first study the local existence, global existence, and the finite time blow up to the Cauchy problem (1.1), and then study the stability and instability of the standing waves obtained in Sections 3 and 4.

6.1 Local existence

In this subsection, we consider the local existence to the Cauchy problem

(6.1) i ∂ t φ + Δ φ + ( I α ∗ ∣ φ ∣ p ¯ ) ∣ φ ∣ p ¯ − 2 φ + μ ∣ φ ∣ q − 2 φ , ( t , x ) ∈ R × R N , φ ( 0 , x ) = φ 0 ( x ) ∈ H 1 ( R N ) , x ∈ R N .

Definition 6.1

Let N ≥ 3 . The pair ( p , r ) is said to be Schrödinger admissible, for short ( p , r ) ∈ S , if

2 p + N r = N 2 , p , r ∈ [ 2 , ∞ ] .

Define

(6.2) ( p 1 , r 1 ) ≔ 2 ( N + α ) N − 2 = 2 p ¯ , 2 N ( N + α ) N α + 4 + N 2 − 2 N

and

(6.3) ( p 2 , r 2 ) ≔ 4 q ( q − 2 ) ( N − 2 ) , N q q + N − 2 .

Then ( p 1 , r 1 ) , ( p 2 , r 2 ) ∈ S by direct calculation. For such defined admissible pairs, we define the spaces Y T ≔ Y p 1 , r 1 , T ∩ Y p 2 , r 2 , T and X T ≔ X p 1 , r 1 , T ∩ X p 2 , r 2 , T equipped with the following norms:

(6.4) ‖ ψ ‖ Y T = ‖ ψ ‖ Y p 1 , r 1 , T + ‖ ψ ‖ Y p 2 , r 2 , T and ‖ ψ ‖ X T = ‖ ψ ‖ X p 1 , r 1 , T + ‖ ψ ‖ X p 2 , r 2 , T ,

where, for any p , r ∈ ( 1 , ∞ ) ,

‖ ψ ( t , x ) ‖ Y p , r , T ≔ ∫ 0 T ‖ ψ ( t , ⋅ ) ‖ r p d t 1 / p

and

‖ ψ ( t , x ) ‖ X p , r , T ≔ ∫ 0 T ‖ ψ ( t , ⋅ ) ‖ W 1 , r p d t 1 / p .

Definition 6.2

Let T > 0 . We say that φ ( t , x ) is an integral solution of the Cauchy problem (6.1) on the time interval [ 0 , T ] if φ ∈ C ( [ 0 , T ] , H 1 ( R N ) ) ∩ X T , and φ ( t , x ) = e i t Δ φ 0 ( x ) − i ∫ 0 t e i ( t − s ) Δ g ( φ ( s , x ) ) d s for all t ∈ ( 0 , T ) , where g ( φ ) ≔ g 1 ( φ ) + g 2 ( φ ) , g 1 ( φ ) ≔ ( I α ∗ ∣ φ ∣ p ¯ ) ∣ φ ∣ p ¯ − 2 φ , and g 2 ( φ ) ≔ μ ∣ φ ∣ q − 2 φ .

Let us recall Strichartz’s estimates that will be useful in the sequel (see, e.g., ([7], Theorem 2.3.3 and Remark 2.3.8) and [42] for the endpoint estimates).

Lemma 6.3

Let N ≥ 3 , ( p , r ) , and ( p ˜ , r ˜ ) ∈ S . Then there exists a constant C > 0 such that for any T > 0 , the following properties hold:

For any u ∈ L 2 ( R N ) , the function t ↦ e i t Δ u belongs to Y p , r , T ∩ C ( [ 0 , T ] , L 2 ( R N ) ) and ‖ e i t Δ u ‖ Y p , r , T ≤ C ‖ u ‖ 2 .
Let F ( t , x ) ∈ Y p ˜ ′ , r ˜ ′ , T , where we use a prime to denote conjugate indices. Then the function
t ↦ Φ F ( t , x ) ≔ ∫ 0 t e i ( t − s ) Δ F ( s , x ) d s
belongs to Y p , r , T ∩ C ( [ 0 , T ] , L 2 ( R N ) ) and ‖ Φ F ‖ Y p , r , T ≤ C ‖ F ‖ Y p ˜ ′ , r ˜ ′ , T .
For every u ∈ H 1 ( R N ) , the function t ↦ e i t Δ u belongs to X p , r , T ∩ C ( [ 0 , T ] , H 1 ( R N ) ) and ‖ e i t Δ u ‖ X p , r , T ≤ C ‖ u ‖ H 1 .

Lemma 6.4

Let N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , α ≥ N − 4 (i.e., p ¯ ≥ 2 ), α < N − 2 , ( p 1 , r 1 ) be defined in (6.2) and g 1 ( φ ) be defined in Definition 6.2. Then for every ( p ˜ , r ˜ ) ∈ S there exists a constant C > 0 such that for every T > 0 ,

(6.5) ∫ 0 t e i ( t − s ) Δ [ ∇ g 1 ( φ ( s ) ) ] d s Y p ˜ , r ˜ , T ≤ C ‖ ∇ φ ‖ Y p 1 , r 1 , T 2 p ¯ − 1

and

(6.6) ∫ 0 t e i ( t − s ) Δ [ g 1 ( φ ( s ) ) − g 1 ( ψ ( s ) ) ] d s Y p ˜ , r ˜ , T ≤ C ( ‖ ∇ φ ‖ Y p 1 , r 1 , T 2 p ¯ − 2 + ‖ ∇ ψ ‖ Y p 1 , r 1 , T 2 p ¯ − 2 ) ‖ φ − ψ ‖ Y p 1 , r 1 , T .

Proof

By using

∣ ∇ ( ∣ φ ∣ p ¯ ) ∣ ≲ ∣ φ ∣ p ¯ − 1 ∣ ∇ φ ∣ and ∣ ∇ ( ∣ φ ∣ p ¯ − 2 φ ) ∣ ≲ ∣ φ ∣ p ¯ − 2 ∣ ∇ φ ∣ ,

we obtain that

(6.7) ∣ ∇ g 1 ( φ ) ∣ ≲ ∣ ( I α ∗ ∣ φ ∣ p ¯ ) ∣ φ ∣ p ¯ − 2 ∇ φ ∣ + ∣ ( I α ∗ [ ∣ φ ∣ p ¯ − 1 ∇ φ ] ) ∣ φ ∣ p ¯ − 2 φ ∣ ≔ I 1 + I 2 .

By using

∣ ∣ φ ∣ p ¯ − ∣ ψ ∣ p ¯ ∣ ≲ ( ∣ φ ∣ + ∣ ψ ∣ ) p ¯ − 1 ∣ φ − ψ ∣

and

∣ ∣ φ ∣ p ¯ − 2 φ − ∣ ψ ∣ p ¯ − 2 ψ ∣ ≲ ( ∣ φ ∣ p ¯ − 2 + ∣ ψ ∣ p ¯ − 2 ) ∣ φ − ψ ∣ ,

we obtain that

∣ g 1 ( φ ) − g 1 ( ψ ) ∣ ≤ ∣ ( I α ∗ ∣ φ ∣ p ¯ ) ( ∣ φ ∣ p ¯ − 2 φ − ∣ ψ ∣ p ¯ − 2 ψ ) ∣ + ∣ ( I α ∗ ∣ φ ∣ p ¯ − I α ∗ ∣ ψ ∣ p ¯ ) ∣ ψ ∣ p ¯ − 2 ψ ∣ ≲ ∣ ( I α ∗ ∣ φ ∣ p ¯ ) [ ( ∣ φ ∣ p ¯ − 2 + ∣ ψ ∣ p ¯ − 2 ) ∣ φ − ψ ∣ ] ∣ + ∣ ( I α ∗ [ ( ∣ φ ∣ + ∣ ψ ∣ ) p ¯ − 1 ∣ φ − ψ ∣ ] ) ∣ ψ ∣ p ¯ − 2 ψ ∣ ≔ I 3 + I 4 .

Case p ¯ > 2 . Set

a 1 = 2 N N − 2 − α and q 1 = 2 N ( N + α ) ( α + 4 − N ) ( N − 2 + α ) ,

then

N a 1 N + α a 1 ∈ 1 , N α , N a 1 N + α a 1 p ¯ = ( p ¯ − 2 ) q 1 = r 1 ∗ ≔ N r 1 N − r 1 , 1 r 1 ′ = 1 a 1 + 1 q 1 + 1 r 1 .

By using the Hölder inequality, the Hardy-Littlewood-Sobolev inequality, and the Sobolev embedding W 1 , r 1 ( R N ) ↪ L r 1 ∗ ( R N ) , we have

(6.8) ‖ I 1 ‖ r 1 ′ ≤ ‖ I α ∗ ∣ φ ∣ p ¯ ‖ a 1 ‖ ∣ φ ∣ p ¯ − 2 ‖ q 1 ‖ ∇ φ ‖ r 1 ≲ ‖ ∣ φ ∣ p ¯ ‖ N a 1 N + α a 1 ‖ φ ‖ ( p ¯ − 2 ) q 1 p ¯ − 2 ‖ ∇ φ ‖ r 1 = ‖ φ ‖ N a 1 N + α a 1 p ¯ p ¯ ‖ φ ‖ ( p ¯ − 2 ) q 1 p ¯ − 2 ‖ ∇ φ ‖ r 1 ≲ ‖ ∇ φ ‖ r 1 2 p ¯ − 1 .

Set

a 2 = 2 N N − α and q 2 = 2 N ( N + α ) ( α + 2 ) ( N − 2 + α ) ,

then

1 r 1 ′ = 1 a 2 + 1 q 2 , N a 2 N + α a 2 ∈ 1 , N α , 1 N a 2 N + α a 2 = 1 r 1 + 1 q 2 , and ( p ¯ − 1 ) q 2 = r 1 ∗ .

By using the Hölder inequality, the Hardy-Littlewood-Sobolev inequality, and the Sobolev embedding W 1 , r 1 ( R N ) ↪ L r 1 ∗ ( R N ) , we have

(6.9) ‖ I 2 ‖ r 1 ′ ≤ ‖ I α ∗ ( ∣ φ ∣ p ¯ − 1 ∇ φ ) ‖ a 2 ‖ ∣ φ ∣ p ¯ − 2 φ ‖ q 2 ≲ ‖ ∣ φ ∣ p ¯ − 1 ∇ φ ‖ N a 2 N + α a 2 ‖ φ ‖ ( p ¯ − 1 ) q 2 p ¯ − 1 ≲ ‖ ∣ φ ∣ p ¯ − 1 ‖ q 2 ‖ ∇ φ ‖ r 1 ‖ φ ‖ ( p ¯ − 1 ) q 2 p ¯ − 1 = ‖ φ ‖ ( p ¯ − 1 ) q 2 p ¯ − 1 ‖ φ ‖ ( p ¯ − 1 ) q 2 p ¯ − 1 ‖ ∇ φ ‖ r 1 ≲ ‖ ∇ φ ‖ r 1 2 p ¯ − 1 .

By using (6.7), (6.8), and (6.9), we have

‖ ∇ g 1 ( φ ) ‖ Y p 1 ′ , r 1 ′ , T = ∫ 0 T ‖ ∇ g 1 ( φ ( t ) ) ‖ r 1 ′ p 1 ′ d t 1 p 1 ′ ≲ ∫ 0 T ‖ ∇ φ ( t ) ‖ r 1 ( 2 p ¯ − 1 ) p 1 ′ d t 1 p 1 ′ = ‖ ∇ φ ‖ Y p 1 , r 1 , T 2 p ¯ − 1 .

Hence, by Lemma 6.3(ii), we obtain that

∫ 0 t e i ( t − s ) Δ [ ∇ g 1 ( φ ( s ) ) ] d s Y p ˜ , r ˜ , T ≲ ‖ ∇ g 1 ( φ ) ‖ Y p 1 ′ , r 1 ′ , T ≲ ‖ ∇ φ ‖ Y p 1 , r 1 , T 2 p ¯ − 1 ;

that is, (6.5) holds.

Similar to (6.8) and (6.9), we obtain that

(6.10) ‖ I 3 ‖ r 1 ′ ≲ ( ‖ φ ‖ r 1 ∗ + ‖ ψ ‖ r 1 ∗ ) 2 p ¯ − 2 ‖ φ − ψ ‖ r 1

and

(6.11) ‖ I 4 ‖ r 1 ′ ≲ ( ‖ φ ‖ r 1 ∗ + ‖ ψ ‖ r 1 ∗ ) 2 p ¯ − 2 ‖ φ − ψ ‖ r 1 .

By using the Hölder inequality, we have

‖ I 3 ‖ Y p 1 ′ , r 1 ′ , T ≲ ∫ 0 T ( ‖ φ ( t ) ‖ r 1 ∗ + ‖ ψ ( t ) ‖ r 1 ∗ ) ( 2 p ¯ − 2 ) p 1 ′ ‖ φ ( t ) − ψ ( t ) ‖ r 1 p 1 ′ d t 1 p 1 ′ ≲ ∫ 0 T ( ‖ φ ( t ) ‖ r 1 ∗ + ‖ ψ ( t ) ‖ r 1 ∗ ) p 1 d t ( 2 p ¯ − 2 ) p 1 ′ p 1 1 p 1 ′ ∫ 0 T ‖ φ ( t ) − ψ ( t ) ‖ r 1 p 1 d t p 1 ′ p 1 1 p 1 ′ ≲ ( ‖ ∇ φ ‖ Y p 1 , r 1 , T + ‖ ∇ ψ ‖ Y p 1 , r 1 , T ) 2 p ¯ − 2 ‖ φ ( t ) − ψ ( t ) ‖ Y p 1 , r 1 , T .

Hence, by Lemma 6.3 (ii), we obtain (6.6) holds.

Case p ¯ = 2 . Similar to case p ¯ > 2 , just in the estimate of ‖ I 1 ‖ r 1 ′ and ‖ I 3 ‖ r 1 ′ by choosing q 1 = ∞ and a 1 = 2 N N − 2 − α , we have (6.8), (6.9), (6.10), and (6.11) hold and then (6.5) and (6.6) hold. The proof is complete.□

The following lemma is cited from [26].

Lemma 6.5

Let N ≥ 3 , q ∈ ( 2 , 2 ∗ ) , ( p 2 , r 2 ) be defined in (6.3) and g 2 ( φ ) be defined in Definition 6.2. Then for every ( p ˜ , r ˜ ) ∈ S there exists a constant C > 0 such that for every T > 0 ,

∫ 0 t e i ( t − s ) Δ [ ∇ g 2 ( φ ( s ) ) ] d s Y p ˜ , r ˜ , T ≤ C T ( N − 2 ) ( 2 ∗ − q ) 4 ‖ ∇ φ ‖ Y p 2 , r 2 , T q − 1

and

∫ 0 t e i ( t − s ) Δ [ g 2 ( φ ( s ) ) − g 2 ( ψ ( s ) ) ] d s Y p ˜ , r ˜ , T ≤ C T ( N − 2 ) ( 2 ∗ − q ) 4 ( ‖ ∇ φ ‖ Y p 2 , r 2 , T q − 2 + ‖ ∇ ψ ‖ Y p 2 , r 2 , T q − 2 ) ‖ φ − ψ ‖ Y p 2 , r 2 , T .

Similar to the proof of Lemma 3.7 in [26], we have the following result.

Lemma 6.6

For all R , T > 0 the metric space ( B R , T , d ) is complete, where

B R , T ≔ { u ∈ X T : ‖ u ‖ X T ≤ R } and d ( u , v ) ≔ ‖ u − v ‖ Y T .

Now, we are ready to prove the following local existence result.

Proposition 6.7

There exists γ 0 > 0 such that if φ 0 ∈ H 1 ( R N ) and T ∈ ( 0 , 1 ] satisfy

(6.12) ‖ e i t Δ φ 0 ‖ X T ≤ γ 0 ,

then there exists a unique integral solution φ ( t , x ) to (6.1) on the time interval [ 0 , T ]. Moreover, φ ( t , x ) ∈ X p , r , T for every ( p , r ) ∈ S and satisfies the following conservation laws:

(6.13) E ( φ ( t ) ) = E ( φ 0 ) , ‖ φ ( t ) ‖ 2 = ‖ φ 0 ‖ 2 , for all t ∈ [ 0 , T ] .

Proof

By modifying the proof of Proposition 3.3 in [26], we can show that there exists a unique integral solution φ ( t , x ) to (6.1) on the time interval [ 0 , T ] and φ ( t , x ) ∈ X p , r , T for every ( p , r ) ∈ S . The proofs of the conservation laws (6.13) follow the proofs of Propositions 1 and 2 in [43], which can be repeated mutatis mutandis in the context of (6.1).□

6.2 Orbital stability

Now we prove Theorem 1.4.

Proof of Theorem 1.4

Since in the context of (6.1), we have the local existence result (Proposition 6.7), the proof of Theorem 1.4 can be done by repeating word by word Section 4 in [26] and we omit it.□

6.3 Orbital instability

In this subsection, we prove Theorem 1.8. For this aim, we first give the following result.

Lemma 6.8

Assume N ≥ 3 , α ∈ ( 0 , N ) , p = p ¯ , q ∈ 2 , 2 + 4 N , α ≥ N − 4 (i.e., p ¯ ≥ 2 ), and α < N − 2 . Let u ∈ S a be such that E ( u ) < inf P a , − E ( v ) and let τ u − be the unique global maximum point of Ψ u ( τ ) determined in Lemma 3.3. If τ u − < 1 , and ∣ x ∣ u ∈ L 2 ( R N ) , then the solution ψ ( t , x ) of (1.1) with initial value u blows up in finite time.

Proof

We claim that

(6.14) if u ∈ S a and τ u − ∈ ( 0 , 1 ) , then P ( u ) ≤ E ( u ) − inf P a , − E ( v ) .

Indeed, by using the equality

Ψ u ( τ u − ) = Ψ u ( 1 ) + Ψ u ′ ( 1 ) ( τ u − − 1 ) + Ψ u ″ ( ξ ) ( τ u − − 1 ) 2 , for some ξ ∈ ( τ u − , 1 ) ,

and noting that P ( u ) < P ( u τ u − ) = 0 , Ψ u ″ ( ξ ) < 0 for ξ > τ u − , Ψ u ′ ( 1 ) = P ( u ) , and Ψ u ( 1 ) = E ( u ) , we obtain that

inf P a , − E ( v ) ≤ Ψ u ( τ u − ) ≤ E ( u ) − P ( u ) ,

which implies that (6.14) holds.

Now, let us consider the solution ψ ( t , x ) with initial value u . By Proposition 6.7, ψ ( t , x ) ∈ C ( [ 0 , T max ) , H 1 ( R N ) ) , where T max ∈ ( 0 , + ∞ ] is the maximal lifespan of ψ ( t , x ) . Since by assumption τ u − < 1 , and the map u ↦ τ u − is continuous, we deduce that τ ψ ( t ) − < 1 as well for t small, say t ∈ [ 0 , t 1 ) . By (6.14), the assumption E ( u ) < inf P a , − E ( v ) , and the conservation laws of mass and energy, we obtain that for t ∈ [ 0 , t 1 ) ,

P ( ψ ( t ) ) ≤ E ( ψ ( t ) ) − inf P a , − E ( v ) = E ( u ) − inf P a , − E ( v ) < − δ .

Hence, P ( ψ ( t 1 ) ) ≤ − δ and then τ ψ ( t 1 ) − < 1 . Hence, by continuity, the above argument yields

P ( ψ ( t ) ) ≤ − δ , for any t ∈ [ 0 , T max ) .

To obtain a contradiction we recall that, since ∣ x ∣ u ∈ L 2 ( R N ) by assumption, by the virial identity (see Proposition 6.5.1 in [7]), the function

Φ ( t ) ≔ ∫ R N ∣ x ∣ 2 ∣ ψ ( t , x ) ∣ 2 d x

is of class C 2 , with Φ ″ ( t ) = 8 P ( ψ ( t ) ) ≤ − 8 δ for every t ∈ [ 0 , T max ) . Therefore,

0 ≤ Φ ( t ) ≤ Φ ( 0 ) + Φ ′ ( 0 ) t − 4 δ t 2 for every t ∈ [ 0 , T max ) .

Since the right hand side becomes negative for t large, this yields an upper bound on T max , which in turn implies finite time blow up.□

Proof of Theorem 1.8

By Lemmas 2.7 and 5.1, u satisfies (1.4) with some λ < 0 . Next, we prove the strong instability of e − i λ t u ( x ) . For s > 1 , let u s ≔ s N / 2 u ( s x ) and ψ s ( t , x ) be the solution to (1.1) with initial value u s . We have u s → u strongly in H 1 ( R N ) as s → 1 + , and hence it is sufficient to prove that ψ s blows up in finite time. Let τ u s − be defined by Lemma 3.3. Clearly τ u s − = s − 1 < 1 , and by the definition of τ u s − ,

E ( u s ) < E ( ( u s ) τ u s − ) = E ( u ) = inf P a , − E ( v ) .

By Theorem 1.7 (3), ∣ x ∣ u s ∈ L 2 ( R N ) . Hence, by Lemma 6.8, ψ s blows up in finite time. The proof is complete.□

Acknowledgements

This work was supported by the National Natural Science Foundation of China (No. 12001403).

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

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Received: 2021-05-26

Revised: 2021-12-31

Accepted: 2022-01-06

Published Online: 2022-03-09

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