On fractional logarithmic Schrödinger equations

Qi Li; Shuangjie Peng; Wei Shuai

doi:10.1515/ans-2022-0002

Article Open Access

On fractional logarithmic Schrödinger equations

Qi Li , Shuangjie Peng and Wei Shuai

Published/Copyright: March 7, 2022

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From the journal Advanced Nonlinear Studies Volume 22 Issue 1

Abstract

We study the following fractional logarithmic Schrödinger equation:

( − Δ ) s u + V ( x ) u = u log u 2 , x ∈ R N ,

where N ≥ 1 , ( − Δ ) s denotes the fractional Laplace operator, 0 < s < 1 and V ( x ) ∈ C ( R N ) . Under different assumptions on the potential V ( x ) , we prove the existence of positive ground state solution and least energy sign-changing solution for the equation. It is known that the corresponding variational functional is not well defined in H s ( R N ) , and inspired by Cazenave (Stable solutions of the logarithmic Schrödinger equation, Nonlinear Anal. 7 (1983), 1127–1140), we first prove that the variational functional is well defined in a subspace of H s ( R N ) . Then, by using minimization method and Lions’ concentration-compactness principle, we prove that the existence results.

Keywords: fractional logarithmic Schrödinger equations; positive ground state solutions; sign-changing solutions; variational methods

MSC 2010: 35R11 (35B38 35Q55 47J30)

1 Introduction

In this article, we study the following fractional logarithmic Schrödinger equation:

(1.1) ( − Δ ) s u + V ( x ) u = u log u 2 , x ∈ R N ,

where N ≥ 1 , 0 < s < 1 , V ( x ) ∈ C ( R N ) and V 0 ≔ inf x V ( x ) > 0 . The fractional Laplace operator ( − Δ ) s is defined by

( − Δ ) s u ( x ) = C N , s P.V. ∫ R N u ( x ) − u ( y ) ∣ x − y ∣ N + 2 s d y ,

where P.V. stands for the principle value and C N , s is a normalization constant, see for instance [1] and references therein.

Equation (1.1) arises from looking for standing waves Φ ( t , x ) = e − i ω t u ( x ) of the following fractional logarithmic Schrödinger equation:

(1.2) i ∂ Φ ∂ t = ( − Δ ) s Φ + ( V ( x ) + ω ) Φ − Φ log ∣ Φ ∣ 2 , ( t , x ) ∈ R + × R N ,

which is a generalization of the following logarithmic Schrödinger equation:

(1.3) i ∂ Φ ∂ t = − Δ Φ + ( V ( x ) + λ ) Φ − Φ log ∣ Φ ∣ 2 , ( t , x ) ∈ R + × R N .

For physical and mathematical background of equation (1.3), we refer the reader to [2,3, 4,5,6, 7,8] and references therein.

There are few results on fractional logarithmic Schrödinger equation. For example, Ardila [9] recently studied the existence and stability of standing waves for nonlinear fractional Schrödinger equation (1.2). In [10], d’Avenia et al. employed non-smooth critical point theory and obtained infinitely many standing wave solutions to equation (1.2).

Equation (1.1) formally associated with the energy functional I : H s ( R N ) → R ∪ { + ∞ } is defined by

I ( u ) = 1 2 ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + 1 2 ∫ R N ( V ( x ) + 1 ) u 2 d x − 1 2 ∫ R N u 2 log u 2 d x ,

where

H s ( R N ) ≔ u ∈ L 2 ( R N ) ∣ ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y < ∞ ,

with the norm

‖ u ‖ H s ≔ ( [ u ] s 2 + ‖ u ‖ L 2 ( R N ) 2 ) 1 2 , [ u ] s ≔ ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y 1 2 .

By the fractional logarithmic Sobolev inequality (see [11]), for each u ∈ H s ( R N ) , then

(1.4) ∫ R N u 2 log u 2 ∣ ∣ u ∣ ∣ L 2 2 d x + N + N s log α + log s Γ N 2 Γ N 2 s ‖ u ‖ L 2 2 ≤ α 2 π s [ u ] s 2 , for any α > 0 .

It is easy to see that ∫ R N u 2 log u 2 < + ∞ , but the functional fails to be finite since the logarithm is singular at origin. Indeed, let u be a smooth function that satisfies

(1.5) u ( x ) = ( ∣ x ∣ N / 2 log ∣ x ∣ ) − 1 , ∣ x ∣ ≥ 3 , 0 , ∣ x ∣ ≤ 2 .

One can verify directly that u ∈ H s ( R N ) but ∫ R N u 2 log u 2 d x = − ∞ . Thus, I fails to be C 1 on H s ( R N ) .

Due to loss of smoothness, the classical critical point theory cannot be applied for I . The same difficulty also occurs for s = 1 . In order to investigate the following logarithmic Schrödinger equation:

(1.6) − Δ u + V ( x ) u = u log u 2 , x ∈ R N ,

several approaches developed so far in the literature. Cazenave [5] worked in a suitable Banach space endowed with a Luxemburg-type norm, in which the corresponding energy functional is well defined. Squassina and Szulkin [12] applied non-smooth critical point theory for lower semi-continuous functionals, see also [13,14,15]. Tanaka and Zhang [16] used penalization technique. Recently, via direction derivative and constrained minimization method, Shuai [17] proved the existence of ground state solutions and sign-changing solutions for equation (1.6). Recently, Wang and Zhang [18] proved that the positive ground state solution of the power-law equations

(1.7) − Δ u + λ u = ∣ u ∣ p − 2 u , x ∈ R N , lim ∣ x ∣ → ∞ u ( x ) = 0

converges to the Gaussian

U ( x ) ≔ e N 2 e − λ 4 ∣ x ∣ ,

up to translations, which is the unique positive solution of the logarithmic equation

− Δ u = λ u log u 2 , x ∈ R N , lim ∣ x ∣ → ∞ u ( x ) = 0 .

They also proved that the same result holds for bound state solutions.

Inspired by [5], we first find a suitable Banach space, in which the energy functional I is well defined. Then, we study the existence of positive ground state solution and least energy sign-changing solution for equation (1.1).

Throughout this article, we assume V ( x ) ∈ C 0 , γ ( R N ) for some γ ∈ ( 0 , 1 ) and bounded from below. The following different types of potential are considered:

lim ∣ x ∣ → ∞ V ( x ) = + ∞ .
V ( x ) is radially symmetric, i.e., V ( x ) = V ( ∣ x ∣ ) .
V ( x ) is 1-periodic in each variable of x 1 , x 2 , … , x N .
For almost every x ∈ R N , V ( x ) ≤ lim inf ∣ x ∣ → ∞ V ( x ) ≔ V ∞ < + ∞ , and the inequality is strict in a subset of positive Lebesgue measure.
There exist positive constants M , C 0 and m ∈ ( 0 , 2 s ) such that
V ( x ) ≤ V ∞ − C 0 1 + ∣ x ∣ m , ∀ ∣ x ∣ ≥ M .

Next, we denote

H ≔ u ∈ H s ( R N ) ∣ ∫ R N V ( x ) u 2 d x < ∞

with the norm

‖ u ‖ H ≔ ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) u 2 d x 1 2 .

The Orlicz space L A ( R N ) is defined by

L A ( R N ) = { u ∈ L loc 1 ( R N ) ∣ A ( ∣ u ∣ ) ∈ L 1 ( R N ) }

which is equipped with the Luxemburg norm

‖ u ‖ L A ≔ inf k > 0 ∣ ∫ R N A ( k − 1 ∣ u ( x ) ∣ ) d x ≤ 1 ,

where A is given by

A ( s ) ≔ − s 2 log s 2 , 0 ≤ s ≤ e − 3 , 3 s 2 + 4 e − 3 s − e − 6 , s ≥ e − 3 .

We also denote

F ( s ) = s 2 log s 2 , B ( s ) = F ( s ) + A ( s ) , ∀ s ≥ 0 .

It is easy to check that A , B are nonnegative convex and increasing function on [ 0 , + ∞ ) , and A ∈ C 1 ( [ 0 , + ∞ ) ) ∩ C 2 ( ( 0 , + ∞ ) ) .

Now, define

W s ( R N ) ≔ H ∩ L A ( R N ) ,

which is equipped with the norm

‖ u ‖ = ‖ u ‖ H + ‖ u ‖ L A ,

By Lemma 2.2 in [9], we know that

W s ( R N ) = u ∈ H s ( R N ) : ∫ R N u 2 ∣ log u 2 ∣ < + ∞ .

Now, we give the definition of weak solutions for equation (1.1).

Definition 1.1

We say u ∈ W s ( R N ) is a weak solution to equation (1.1) if u satisfies

∫ R N ( − Δ ) s 2 u ( − Δ ) s 2 φ + V ( x ) u φ d x = ∫ R N u φ log u 2 d x , for each φ ∈ W s ( R N ) .

It follows from Proposition 2.5 in [9] that the energy functional I is well defined on W s ( R N ) and belongs to C 1 ( W s ( R N ) , R ) . Hence, the critical points of I are corresponding to weak solutions of equation (1.1).

Remark 1.1

We remark that, for λ ≠ 0 , if v is a solution of

( − Δ ) s v + ( V ( x ) − log λ 2 ) v = v log v 2 , x ∈ R N ,

then λ v solves equation (1.1). Thus, one can choose λ > 0 small enough such that V ( x ) − log λ 2 > 0 for each x ∈ R N . Since V ( x ) is bounded from below, we always assume V 0 ≔ inf x V ( x ) > 0 in the sequent.

Before stating our main results, we give some notations. Denote

N ≔ { u ∈ W s ( R N ) \ { 0 } ∣ J ( u ) = 0 } , ℳ ≔ { u ± ≠ 0 ∣ ⟨ I ( u ) , u + ⟩ = ⟨ I ( u ) , u − ⟩ = 0 } ,

where u + ≔ max { u , 0 } , u − ≔ min { u , 0 } and J is given by

J ( u ) ≔ ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) u 2 d x − ∫ R N u 2 log u 2 d x .

By Lemma 2.1, we know that N , ℳ are nonempty. Then we define

c ≔ inf N I ( u ) , m = inf ℳ I ( u ) .

Our first result can be stated as follows.

Theorem 1.1

If u ∈ N with I ( u ) = c , then u is a positive solution of equation (1.1). If u ∈ ℳ with I ( u ) = m , then u is a sign-changing solution to equation (1.1). Moreover, if c and m are achieved, then m > 2 c .

If u is a sign-changing solution to equation (1.1), then u ∈ N , and we can easily check that

(1.8) I ( u ) = I ( u + ) + I ( u − ) − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y ,

(1.9) ⟨ I ′ ( u ) , u + ⟩ = ⟨ I ′ ( u + ) , u + ⟩ − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y ,

(1.10) ⟨ I ′ ( u ) , u − ⟩ = ⟨ I ′ ( u − ) , u − ⟩ − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y .

Note that

∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y < 0 .

Hence, by (1.8)–(1.10), one obtains

I ( u ) > I ( u + ) + I ( u − ) , and u + , u − ∉ N ,

which is totally different from the case s = 1 , because equation (1.1) is a nonlocal problem. And we cannot directly infer that I ( u ) > 2 c . This property is called energy doubling by Weth [19], which is crucial in overcoming the difficulty of lack of compactness.

Now, we focus on whether c or m is achieved.

Theorem 1.2

If one of the following four conditions hold: ( V 1 ) ; ( V 2 ) and N ≥ 2 ; ( V 3 ) ; ( V 4 ) , then c is achieved. If one of the following three conditions hold: ( V 1 ) ; ( V 2 ) and N ≥ 2 ; ( V 4 ) and ( V 5 ) , then m is achieved. In particular, if m is not achieved provided ( V 3 ) holds.

The proof of Theorem 1.2 is based on the concentration-compactness principle [20]. However, as we will see, the nonlocal operator ( − Δ ) s and the logarithmic nonlinearity cause some obstacles, which need some new technique and subtle analysis.

Remark 1.2

We point out that ( i ) the variational framework also works for equation (1.6), while the analogous results for equation (1.6) are obtained by using direction derivative and constrained minimization method; ( i i ) Theorems 1.1 and 1.2 also hold for the following fractional Schrödinger equation:

(1.11) ( − Δ ) s u + V ( x ) u = ∣ u ∣ p − 2 u , x ∈ R N ,

where 2 < p < 2 s ∗ . Even for equation (1.11), the results on the existence of sign-changing solutions are new.

In the rest of the article, we shall first prove some preliminary results and prove Theorem 1.1 in Section 2, and then we prove Theorem 1.2 in Section 3. We will use C to denote different positive constants from line to line.

2 Proof of Theorem 1.1

In this section, we first prove some technical lemmas, which is crucial for proving our main results.

Lemma 2.1

Let u ∈ W s ( R N ) ⧹ { 0 } , then there exists a unique t u > 0 such that t u u ∈ N ;
Let u ∈ W s ( R N ) with u ± ≠ 0 , then there exists a unique pair ( α u , β u ) ∈ R + × R + such that α u u + + β u u − ∈ ℳ .

Proof

Note that, by (1.14) in [21], we know that u ± ∈ H s ( R N ) if u ∈ H s ( R N ) , thus, u ± ∈ W s ( R N ) if u ∈ W s ( R N ) . We only prove ( i i ), and ( i ) can be proved by a similar argument. First, we prove the existence of ( α u , β u ) . By a direct calculation, we have

(2.1) ⟨ I ′ ( α u + + β u − ) , α u + ⟩ = α 2 [ u + ] s 2 − α β ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + α 2 ∫ R N V ( x ) ( u + ) 2 d x − α 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x − α 2 log α 2 ∫ R N ∣ u + ∣ 2 d x

and

(2.2) ⟨ I ′ ( α u + + β u − ) , β u − ⟩ = β 2 [ u − ] s 2 − α β ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + β 2 ∫ R N V ( x ) ∣ u − ∣ 2 d x − β 2 ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x − β 2 log β 2 ∫ R N ∣ u − ∣ 2 d x .

Define

F ( α , β ) = ( ⟨ I ′ ( α u + + β u − ) , α u + ⟩ , ⟨ I ′ ( α u + + β u − ) , β u − ⟩ ) .

It follows from (2.1) and (2.2) that there exists R > r > 0 such that

⟨ I ′ ( r u + + β u − ) , r u + ⟩ > 0 and ⟨ I ′ ( R u + + β u − ) , R u + ⟩ < 0 , ∀ β ∈ [ r , R ] , ⟨ I ′ ( α u + + r u − ) , r u − ⟩ > 0 and ⟨ I ′ ( α u + + R u − ) , R u − ⟩ < 0 , ∀ α ∈ [ r , R ] .

By Miranda theorem, there exists ( α u , β u ) ∈ ( r , R ) × ( r , R ) such that F ( α u , β u ) = 0 , that is, α u u + + β u u − ∈ ℳ .

Next, we prove the uniqueness of ( α u , β u ) on R + × R + . If u ∈ ℳ , then

(2.3) [ u + ] s 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) ∣ u + ∣ 2 d x = ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x

and

(2.4) [ u − ] s 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) ∣ u − ∣ 2 d x = ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x .

We claim that ( α u , β u ) = ( 1 , 1 ) is the unique pair of positive numbers such that α u u + + β u u − ∈ ℳ .

Indeed, let ( α u , β u ) satisfy α u u + + β u u − ∈ ℳ . By direct computation, we have

(2.5) α u 2 [ u + ] s 2 − α u β u ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + α u 2 ∫ R N V ( x ) ( u + ) 2 d x = α u 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x + α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x

and

(2.6) β u 2 [ u − ] s 2 − α u β u ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + β u 2 ∫ R N V ( x ) ( u − ) 2 d x = β u 2 ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x + β u 2 log β u 2 ∫ R N ∣ u − ∣ 2 d x .

Without loss of generality, we assume that 0 < α u ≤ β u . Then by (2.5),

(2.7) α u 2 [ u + ] s 2 − α u 2 ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + α u 2 ∫ R N V ( x ) ( u + ) 2 d x ≤ α u 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x + α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x .

Combining (2.3) and (2.7), we deduce that

α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x ≥ 0 ,

which implies that α u ≥ 1 . Thus, 1 ≤ α u ≤ β u . On the other hand, by using a similar argument to (2.4) and (2.6), we obtain that

β u 2 log β u 2 ∫ R N ∣ u − ∣ 2 d x ≤ 0 ,

which implies β u ≤ 1 . Therefore, α u = β u = 1 .

If u ∉ ℳ , then there exists ( α u , β u ) ∈ R + × R + such that α u u + + β u u − ∈ ℳ . If there exists another pair ( α ¯ u , β ¯ u ) ∈ R + × R + such that α ¯ u u + + β ¯ u u − ∈ ℳ . Denote v = α u u + + β u u − and v ¯ = α ¯ u u + + β ¯ u u − , then we have

α ¯ u α u v + + β ¯ u β u v − = v ¯ ∈ ℳ .

Hence, by the above analysis, we conclude that

α ¯ u α u = β ¯ u β u = 1 .

Therefore, ( α u , β u ) is unique.□

Lemma 2.2

Let u ∈ W s ( R N ) with u ± ≠ 0 such that ⟨ I ′ ( u ) , u + ⟩ ≤ 0 and ⟨ I ′ ( u ) , u − ⟩ ≤ 0 . Then the unique pair ( α u , β u ) obtained in Lemma 2.1 satisfies 0 < α u ≤ 1 and 0 < β u ≤ 1 .

Proof

Without loss of generality, we assume that 0 < β u ≤ α u . Since α u u + + β u u − ∈ ℳ , then

(2.8) α u 2 [ u + ] s 2 − α u β u ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + α u 2 ∫ R N V ( x ) ∣ u + ∣ 2 d x = α u 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x + α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x

and

(2.9) β u 2 [ u − ] s 2 − α u β u ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + β u 2 ∫ R N V ( x ) ∣ u − ∣ 2 d x = β u 2 ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x + β u 2 log β u 2 ∫ R N ∣ u − ∣ 2 d x .

Note that

∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y < 0 .

Thus, it follows from (2.8) that

(2.10) α u 2 [ u + ] s 2 − α u 2 ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + α u 2 ∫ R N V ( x ) ∣ u + ∣ 2 d x ≥ α u 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x + α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x .

By using the fact ⟨ I ′ ( u ) , u + ⟩ ≤ 0 , we have

(2.11) [ u + ] s 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) ∣ u + ∣ 2 d x ≤ ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x .

Combining (2.10) with (2.11), we deduce that

α u 2 log α u 2 ∫ R N ∣ u + ∣ 2 d x ≤ 0 ,

which implies that 0 < α u ≤ 1 . Therefore, 0 < β u ≤ α u ≤ 1 .□

Lemma 2.3

For fixed u ∈ W s ( R N ) with u ± ≠ 0 , define Φ ( α , β ) ≔ I ( α u + + β u − ) . Then ( α u , β u ) obtained in Lemma 2.1 is the unique maximum point of Φ ( α , β ) on R + × R + .

Proof

By the proof of Lemma 2.1, we deduce that ( α u , β u ) is the unique critical point of Φ ( α , β ) on R + × R + . By direct calculation, we conclude that Φ ( α , β ) → − ∞ as ∣ ( α , β ) ∣ → + ∞ . So it is sufficient to check that the maximum is not achieved on the boundary of R + × R + . Without loss of generality, we may suppose by contradiction that ( 0 , β 0 ) is a maximum point of Φ . Then we can deduce that this is impossible, since

(2.12) h ( α ) ≔ α 2 2 [ u + ] s 2 − α β 0 ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y + β 0 2 2 [ u − ] s 2 + α 2 2 ∫ R N ( V ( x ) + 1 ) ∣ u + ∣ 2 d x + β 0 2 2 ∫ R N ( V ( x ) + 1 ) ∣ u − ∣ 2 d x − α 2 2 ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x − 1 2 α 2 log α 2 ∫ R N ∣ u + ∣ 2 d x − β 0 2 2 ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x − 1 2 β 0 2 log β 0 2 ∫ R N ∣ u − ∣ 2 d x

is strictly increasing in α for α > 0 small enough.□

Lemma 2.4

Let u ∈ W s ( R N ) be a non-negative solution of (1.1), then u ∈ L ∞ ( R N ) .

Proof

The proof is inspired by Proposition 2.2 in [22]. Let us define, for β ≥ 1 and T > 0 large enough,

φ ( s ) ≔ φ β , T ( s ) = 0 , s ≤ 0 , s β , 0 < s < T , β T β − 1 ( s − T ) + T β , s ≥ T .

Obviously, φ is convex and differential function. Then, for u ∈ W s ( R N ) , one has

(2.13) ( − Δ ) s φ ( u ) ≤ φ ′ ( u ) ( − Δ ) s u .

Since φ ( u ) grows linearly and φ is Lipschitz with constant K = β T β − 1 , therefore,

‖ φ ( u ) ‖ H ≔ ∫ R N ∫ R N ∣ φ ( u ( x ) ) − φ ( u ( y ) ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) ∣ φ ( u ( x ) ) ∣ 2 d x 1 2 ≤ C ‖ u ‖ H

and

∫ R N ∣ ∣ φ ( u ( x ) ) ∣ 2 log ∣ φ ( u ( x ) ) ∣ 2 ∣ d x ≤ C 1 ∫ R N u 2 d x + C 2 ∫ R N ∣ u 2 log u 2 ∣ d x .

Thus, φ ( u ( x ) ) ∈ W s ( R N ) if u ∈ W s ( R N ) . Observe that by the Sobolev embedding theorem, we have the following inequality:

(2.14) [ φ ( u ) ] s 2 = ‖ ( − Δ ) s 2 φ ( u ) ‖ L 2 ( R N ) 2 = ∫ R N φ ( u ) ( − Δ ) s φ ( u ) d x ≥ S ( N , s ) ‖ φ ( u ) ‖ L 2 s ∗ ( R N ) 2 ,

where S ( N , s ) is the best fractional Sobolev constant defined by

S ( N , s ) ≔ inf { [ u ] s 2 ∣ u ∈ H s ( R N ) and ‖ u ‖ L 2 s ∗ ( R N ) = 1 } .

On the other hand, since φ ( u ) φ ′ ( u ) ∈ W s ( R N ) , thus

(2.15) ∫ R N φ ( u ) ( − Δ ) s φ ( u ) d x ≤ ∫ R N φ ( u ) φ ′ ( u ) ( − Δ ) s u d x ≤ C ∫ R N φ ( u ) φ ′ ( u ) ∣ u ∣ 2 s ∗ − 1 d x .

Since u φ ′ ( u ) ≤ β φ ( u ) , the above estimate (2.14)–(2.15) becomes

(2.16) ∫ R N ( φ ( u ( x ) ) ) 2 s ∗ d x 2 2 s ∗ ≤ C β ∫ R N ( φ ( u ( x ) ) ) 2 ∣ u ∣ 2 s ∗ − 2 d x .

We point out that since φ ( u ) grows linearly, both sides of (2.16) are finite.

Claim: Let β 1 be such that 2 β 1 = 2 s ∗ , then u ∈ L β 1 2 s ∗ ( R N ) .

To prove this, we take R large to be determined later. Then, Hölder’s inequality gives

∫ R N ( φ ( u ( x ) ) ) 2 ∣ u ∣ 2 s ∗ − 2 d x = ∫ { u ≤ R } ( φ ( u ( x ) ) ) 2 ∣ u ∣ 2 s ∗ − 2 d x + ∫ { u > R } ( φ ( u ( x ) ) ) 2 ∣ u ∣ 2 s ∗ − 2 d x ≤ ∫ { u ≤ R } ∣ φ ( u ( x ) ) ∣ 2 R 2 s ∗ − 2 d x + ∫ R N ∣ φ ( u ( x ) ) ∣ 2 s ∗ d x 2 2 s ∗ ∫ { u > R } ∣ ( u ( x ) ) ∣ 2 s ∗ d x 2 s ∗ − 2 2 s ∗ .

By the Monotone convergence theorem, we may take R so that

∫ { u > R } ∣ ( u ( x ) ) ∣ 2 s ∗ d x 2 s ∗ − 2 2 s ∗ ≤ 1 2 C β 1 .

In this way, the second term above is absorbed by the left-hand side of (2.16) to obtain

(2.17) ∫ R N ( φ ( u ( x ) ) ) 2 s ∗ d x 2 2 s ∗ ≤ 2 C β 1 ∫ { u ≤ R } ∣ φ ( u ( x ) ) ∣ 2 R 2 s ∗ − 2 d x .

Using that φ β 1 , T ( u ) ≤ u β 1 in the right-hand side of (2.17) and letting T → + ∞ in the left-hand side, since 2 β 1 = 2 s ∗ , we obtain

∫ R N ∣ u ( x ) ∣ 2 s ∗ β 1 d x 2 2 s ∗ ≤ 2 C β 1 ∫ R N ∣ u ( x ) ∣ 2 s ∗ R 2 s ∗ − 2 d x < + ∞ .

This proves the claim.

We now go back to inequality (2.16) and we use as before φ β 1 , T ( u ) ≤ u β in the right-hand side and then we take T → + ∞ in the left-hand side

(2.18) ∫ R N ∣ u ( x ) ∣ 2 s ∗ β d x 2 2 s ∗ ≤ C β ∫ R N ∣ u ( x ) ∣ 2 β + 2 s ∗ − 2 d x .

Then, by standard bootstrap technique, one can prove that u ∈ L ∞ ( R N ) . We omit the details here.□

Proposition 2.5

Let u ∈ W s ( R N ) be a non-negative solution of equation (1.1), then u is a classical solution of equation (1.1), i.e., ( − Δ ) s u can be written as

( − Δ ) s u = C N , s ∫ R N u ( x ) − u ( y ) ∣ x − y ∣ N + 2 s d y ,

and equation (1.1) is satisfied pointwise in all R N .

Proof

Let u ∈ W s ( R N ) be a non-negative solution of (1.1), we prove the conclusion by two cases.

Case 1. If V ( x ) is bounded, by Theorem 3.4 in [23], we can deduce that u ∈ C 0 , μ ( R N ) for some μ ∈ ( 0 , 1 ) . The function h ( x ) ≔ u ( x ) − V ( x ) u ( x ) + ∣ u ( x ) ∣ 2 log ∣ u ( x ) ∣ 2 belongs to C 0 , σ ( R N ) for certain σ > 0 . Let η 1 be a non-negative, smooth function with support in B 1 ( 0 ) such that η 1 = 1 in B 1 / 2 ( 0 ) . Let u 1 ∈ H s ( R N ) be the solution of the equation

( − Δ ) s u 1 + u 1 = η 1 h ( x ) , x ∈ R N .

It follows from Lemma 2.4 that η 1 h ( x ) ∈ L q ( R N ) for all q > 2 , then u 1 ∈ W 2 s , q ( R N ) and thus u 1 ∈ C 0 , σ 0 ( R N ) for some σ 0 ∈ ( 0 , σ ) .

Now we look at the equation

− Δ w = − u 1 + η 1 h ∈ C 0 , σ 0 ( R N ) .

By Hölder’s regularity theory for the Laplacian, we find w ∈ C 2 , σ 0 ( R N ) , so that if 2 s + σ 0 > 1 , then ( − Δ ) 1 − s w ∈ C 1 , 2 s + σ 0 − 1 ( R N ) , while if 2 s + σ 0 ≤ 1 , then ( − Δ ) 1 − s w ∈ C 0 , 2 s + σ 0 ( R N ) . Then, since

( − Δ ) s ( u 1 − ( − Δ ) 1 − s w ) = 0 ,

the function u 1 − ( − Δ ) 1 − s w is s -harmonic, we find that u 1 has the same regularity as ( − Δ ) 1 − s w . By the similar arguments as the proof of Theorems 1.3 and 3.4 in [23], one obtains that u has the same regularity as u 1 . Thus, we conclude that u ∈ C 1 , 2 s + σ 0 − 1 ( R N ) if 2 s + σ 0 > 1 , while u ∈ C 0 , 2 s + σ 0 ( R N ) if 2 s + σ 0 ≤ 1 . Note that the conclusion holds locally, but the corresponding Hölder norms depend only on η 1 , s , N , ‖ u ‖ H s and ‖ V ( x ) ‖ C 0 , σ 0 ( R N ) , so these estimates are global in R N .

Case 2. If V ( x ) is coercive, i.e., lim ∣ x ∣ → + ∞ V ( x ) = + ∞ . Similar to the proof of Case 1, we can deduce that u ∈ C loc 1 , 2 s + σ 0 − 1 ( R N ) if 2 s + σ 0 > 1 , while u ∈ C loc 0 , 2 s + σ 0 ( R N ) if 2 s + σ 0 ≤ 1 .

For each point x ∈ R N , since u ∈ L ∞ ( R N ) and u ∈ C 1 , 2 s + σ 0 − 1 ( B 1 ( x ) ) if 2 s + σ 0 > 1 , while u ∈ C 0 , 2 s + σ 0 ( B 1 ( x ) ) if 2 s + σ 0 ≤ 1 , thus

∫ B 1 ( x ) u ( x ) − u ( y ) ∣ x − y ∣ N + 2 s d y < + ∞ , ∫ R N ⧹ B 1 ( x ) u ( x ) − u ( y ) ∣ x − y ∣ N + 2 s d y < + ∞ .

Therefore,

( − Δ ) s u = C N , s ∫ R N u ( x ) − u ( y ) ∣ x − y ∣ N + 2 s d y ,

and thus equation (1.1) is satisfied pointwise in all R N .□

Now, we prove the minimizers of c or m are indeed solutions to equation (1.1) by quantitative deformation lemma and degree theory.

Proof of Theorem 1.1

( i ) Suppose that u ∈ N and I ( u ) = c , by Lagrange multiplier theorem, there exists μ ∈ R such that

I ′ ( u ) = μ J ′ ( u ) .

Thus, we have

0 = ⟨ I ′ ( u ) , u ⟩ = μ ⟨ J ′ ( u ) , u ⟩ .

Noting that

⟨ J ′ ( u ) , u ⟩ = − 2 ∫ R N u 2 d x < 0 .

Hence, μ = 0 , that is, I ′ ( u ) = 0 . By standard arguments, u ≥ 0 or u ≤ 0 . Without loss of generality, assume u ≥ 0 . If there exists x 0 ∈ R N such that u ( x 0 ) = 0 , then

(2.19) ( − Δ ) s u ( x 0 ) = C N , s ∫ R N u ( x 0 ) − u ( y ) ∣ x − y ∣ N + 2 s d y = − C N , s ∫ R N u ( y ) ∣ x − y ∣ N + 2 s d y < 0 .

On the other hand, it holds

( − Δ ) s u ( x 0 ) = − V ∞ u ( x 0 ) + u ( x 0 ) log ∣ u ( x 0 ) ∣ 2 = 0 ,

which contradicts with (2.19). Therefore, u ( x ) > 0 for each x ∈ R N . Consequently, u is a positive ground state solution to equation (1.1).

( i i ) Now, suppose that u ∈ ℳ and I ( u ) = m , then u + , u − ≠ 0 and

⟨ I ′ ( u ) , u + ⟩ = ⟨ I ′ ( u ) , u − ⟩ = 0 .

It follows from Lemma 2.3 that, for any ( α , β ) ∈ R + × R + \ ( 1 , 1 ) ,

I ( α u + + β u − ) < I ( u + + u − ) = m .

If I ′ ( u ) ≠ 0 , then there exist δ > 0 and θ > 0 such that

‖ I ′ ( v ) ‖ ≥ θ , for all ‖ v − u ‖ ≤ 3 δ .

Define D ≔ 1 2 , 3 2 × 1 2 , 3 2 and g ( α , β ) ≔ α u + + β u − . Then

m ¯ ≔ max ∂ D I ( g ( α , β ) ) < m .

For S δ ≔ B δ ( u ) and ε = min m − m ¯ 2 , θ δ 8 , it follows from deformation lemma (see Lemma 2.3, [24]) that there exists a deformation η ∈ C ( [ 0 , 1 ] × W s ( R N ) , W s ( R N ) ) such that

η ( 1 , u ) = u if u ∉ I − 1 ( [ m − 2 ε , m + 2 ε ] ) ∩ S 2 δ ,
η ( 1 , I m + ε ∩ S δ ) ⊂ I m − ε ,
I ( η ( 1 , u ) ) ≤ I ( u ) for all u ∈ W s ( R N ) ,

where S δ = { w : ‖ w − v ‖ ≤ 2 δ , v ∈ S } and I c = { u : I ( u ) ≤ c } .

It is easy to check that

(2.20) max ( α , β ) ∈ D ¯ I ( η ( 1 , g ( α , β ) ) ) < m .

Now, we prove that

η ( 1 , g ( D ) ) ∩ ℳ ≠ ∅ ,

which is contradiction to the definition of m . Setting h ( α , β ) = η ( 1 , g ( α , β ) ) and

Ψ 0 ( α , β ) = ( ⟨ I ′ ( g ( α , β ) , u + ⟩ , ⟨ I ′ ( g ( α , β ) , u − ⟩ ) , Ψ 1 ( α , β ) = 1 α ⟨ I ′ ( h ( α , β ) , h + ( α , β ) ⟩ , 1 β ⟨ I ′ ( h ( α , β ) , h − ( α , β ) ⟩ .

By Lemma 2.1 and topological degree theory, we know that deg ( Ψ 0 , D , 0 ) = 1 . Moreover, (2.20) and (a) imply that h = g on ∂ D . It follows from the property of homotopy invariance of topological degree that

deg ( Ψ 1 , D , 0 ) = deg ( Ψ 0 , D , 0 ) = 1 .

Therefore, there exists ( α 0 , β 0 ) ∈ D such that

η ( 1 , g ( α 0 , β 0 ) ) = h ( α 0 , β 0 ) ∈ ℳ ,

which yields a contradiction with (2.20). Hence, I ′ ( u ) = 0 . As a direct result, u is a least energy sign-changing solution to equation (1.1).

( i i i ) Suppose u ∈ ℳ such that I ( u ) = m , then u + , u − ≠ 0 , similar to the proof of Lemmas 2.1–2.2, we conclude that there exists a unique α u + ∈ ( 0 , 1 ) such that α u + u + ∈ N , and a unique β u − ∈ ( 0 , 1 ) such that β u − u − ∈ N .

On the other hand, one has

(2.21) I ( α u + u + + β u − u − ) = I ( α u + u + ) + I ( β u − u − ) − α u + β u − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y

and

∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y < 0 .

Thus, by Lemma 2.3, we conclude that

m = I ( u + + u − ) > I ( α u + u + + β u − u − ) > I ( α u + u + ) + I ( β u − u − ) ≥ 2 c . □

3 Existence and nonexistence of minimizer

In this section, we study whether c or m is achieved under different types of potential. The following Br é zis-Lieb-type lemma for u 2 log u 2 is crucial.

Lemma 3.1

(Lemma 2.3, [9]) Let { u n } be a bounded sequence in W s ( R N ) such that u n → u a.e. in R N . Then u ∈ W s ( R N ) and

lim n → ∞ ∫ R N [ u n 2 log u n 2 − ( u n − u ) 2 log ( u n − u ) 2 ] d x = ∫ R N u 2 log u 2 d x .

3.1 Compact potential

Lemma 3.2

Assume ( V 1 ) or ( V 2 ) , then both c > 0 and m > 0 are achieved.

Proof

We only prove the case for m , one can argue similarly to verify that c > 0 is achieved. Noting that ℳ ⊂ N , we have J ( v ) = 0 for any v ∈ ℳ . Take α = π s 2 in the logarithm Sobolev inequality (1.4), then

(3.1) ∫ R N v 2 log v 2 ‖ v ‖ L 2 2 d x + N + N log π + log s Γ N 2 Γ N 2 s ‖ v ‖ L 2 2 ≤ [ v ] s 2 .

Combining J ( v ) = 0 with (3.1)), we deduce that

∫ R N V ( x ) v 2 d x + N + N log π + log s Γ N 2 Γ N 2 s ‖ v ‖ L 2 2 ≤ ‖ v ‖ L 2 2 log ‖ v ‖ L 2 2 ,

which implies that

‖ v ‖ L 2 2 ≥ e N π N 2 s Γ N 2 Γ N 2 s > 0 .

Therefore, we can derive

I ( v ) = I ( v ) − 1 2 J ( v ) = 1 2 ‖ v ‖ L 2 2 ≥ 1 2 e N π N 2 s Γ N 2 Γ N 2 s > 0 ,

which implies that m > 0 .

Let { u n } ⊂ ℳ be a minimizing sequence of m , then

lim n → ∞ I ( u n ) = lim n → ∞ I ( u n ) − 1 2 J ( u n ) = lim n → ∞ 1 2 ∫ R N u n 2 d x = m .

Hence, { u n } is bounded in L 2 ( R N ) . By (1.4) and J ( u n ) = 0 , we can derive

(3.2) 1 − α 2 π s [ u n ] s 2 + ∫ R N V ( x ) u n 2 d x ≤ ‖ u n ‖ L 2 2 log ‖ u n ‖ L 2 2 e − N Γ N 2 s s α N 2 Γ N 2 .

Taking α > 0 is small enough in (3.2), we deduce that { u n } is bounded in H . Thus, up to a subsequence, there exists u ∈ H such that

(3.3) u n ⇀ u in H , u n → u in L p ( R N ) , 2 < p < 2 s ∗ , u n → u a.e. on R N .

For any 2 < p < 2 s ∗ , we have

0 < C ≤ ‖ u + ‖ H 2 ≤ ∫ R N ( ∣ u n + ∣ 2 log ∣ u n + ∣ 2 ) + d x ≤ C p ∫ R N ∣ u n + ∣ p d x

and

0 < C ≤ ‖ u − ‖ H 2 ≤ ∫ R N ( ∣ u n − ∣ 2 log ∣ u n − ∣ 2 ) + d x ≤ C p ∫ R N ∣ u n − ∣ p d x .

Hence, by using (3.3), we obtain

(3.4) ∫ R N ( ∣ u + ∣ 2 log ∣ u + ∣ 2 ) + d x = lim n → ∞ ∫ R N ( ∣ u n + ∣ 2 log ∣ u n + ∣ 2 ) + d x ≥ C > 0

and

(3.5) ∫ R N ( ∣ u − ∣ 2 log ∣ u − ∣ 2 ) + d x = lim n → ∞ ∫ R N ( ∣ u n − ∣ 2 log ∣ u n − ∣ 2 ) + d x ≥ C > 0 .

Now, by the fact that ⟨ I ′ ( u n ) , u n + ⟩ = 0 , (3.4) and Fatou lemma, one has

‖ u + ‖ H 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y − ∫ R N ( ∣ u + ∣ 2 log ∣ u + ∣ 2 ) − d x ≤ lim inf n → ∞ ‖ u n + ‖ H 2 − ∫ R N ∫ R N u n + ( x ) u n − ( y ) + u n − ( x ) u n + ( y ) ∣ x − y ∣ N + 2 s d x d y − ∫ R N ( ∣ u n + ∣ 2 log ∣ u n + ∣ 2 ) − d x = lim inf n → ∞ ∫ R N ( ∣ u n + ∣ 2 log ∣ u n + ∣ 2 ) + d x = ∫ R N ( ∣ u + ∣ 2 log ∣ u + ∣ 2 ) + d x ,

which implies

(3.6) ‖ u + ‖ H 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y − ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x ≤ 0 .

Similarly, one can prove

(3.7) ‖ u − ‖ H 2 − ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y − ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x ≤ 0 .

Thus, Lemma 2.1 yields a unique ( α u , β u ) ∈ R + × R + such that u ¯ ≔ α u u + + β u u − ∈ ℳ . It follows from Lemma 2.2, (3.6) and (3.7) that 0 < α u ≤ 1 and 0 < β u ≤ 1 . Thus, by direct computation, we have

m ≤ I ( u ¯ ) = I ( u ¯ ) − 1 2 J ( u ¯ ) = 1 2 ∫ R N u ¯ 2 d x ≤ α u 2 ∫ R N ∣ u + ∣ 2 d x + β u 2 ∫ R N ∣ u − ∣ 2 d x ≤ 1 2 ∫ R N ∣ u ∣ 2 d x ≤ lim inf n → ∞ 1 2 ∫ R N ∣ u n ∣ 2 d x = m ,

which implies α u = β u = 1 . As a result, u ¯ = u and I ( u ) = m .□

3.2 Periodic potential

Lemma 3.3

If potential V ( x ) satisfies ( V 3 ) , then c can be achieved.

Proof

Let { u n } be a minimizer sequence of c , then

lim n → ∞ I ( u n ) = lim n → ∞ I ( u n ) − 1 2 J ( u n ) = lim n → ∞ 1 2 ∫ R N u n 2 d x = c ,

which implies that { u n } is bounded in L 2 ( R N ) . By the logarithm Sobolev inequality (1.4), we can obtain that { u n } is bounded in H . One can check that for any ε > 0 , there exists C ε > 0 such that B ( ∣ u ∣ ) ≤ C ε ∣ u ∣ 2 + ε . Using u n ∈ N , we have

∫ R N A ( ∣ u n ∣ ) d x = ∫ R N B ( ∣ u n ∣ ) d x − ‖ u n ‖ s 2 ≤ C ε ∫ R N ∣ u n ∣ 2 + ε d x ≤ C .

It follows from Lemma 2.1 in [5] that { u n } is bounded in L A ( R N ) . Therefore, { u n } is bounded in W s ( R N ) .

We claim that there exists { y n } such that

lim inf n → ∞ ∫ B 1 ( y n ) u n 2 d x > 0 .

If not, then we have

lim n → ∞ sup y ∈ R N ∫ B 1 ( y n ) u n 2 d x = 0 ,

which implies that u n → 0 in L p ( R N ) for 2 < p < 2 s ∗ . Using u n ∈ N again, we have

‖ u n ‖ s 2 = ∫ R N u n 2 log u n 2 d x ≤ ∫ R N ( u n 2 log u n 2 ) + d x ≤ C p ∫ R N ∣ u n ∣ p d x ,

which implies that ‖ u n ‖ L p ≥ C > 0 . This is a contradiction.

Let v n ( x ) = u n ( x + y n ) . Noting that for any y n ∈ R N , { v n } is still a bounded minimizing sequence of c , we may assume that, up to a subsequence, there exists v ∈ H \ { 0 } such that

v n ⇀ v in H , v n → v in L loc p ( R N ) , 2 ≤ p < 2 s ∗ , v n → v a.e. on R N .

By Lemma 3.1, we can know that v ∈ W s ( R N ) .

Now, we intend to show that J ( v ) = 0 and I ( v ) = c . If J ( v ) < 0 , then there exists 0 < λ < 1 such that J ( λ v ) = 0 . Direct computation yields that

c ≤ I ( λ v ) = λ 2 2 ∫ R N v 2 d x < lim m → ∞ 1 2 ∫ R N v n 2 d x = c ,

which leads to a contradiction. If J ( v ) > 0 , then we claim that J ( v n − v ) < 0 for sufficiently large n .

Indeed, by Lemma 3.1, we have

lim n → ∞ J ( v n − v ) = lim n → ∞ J ( v n ) − J ( v ) = − J ( v ) < 0 .

Thus, by a similar argument above, we also have

c ≤ lim m → ∞ 1 2 ∫ R N ∣ v n − v ∣ 2 d x = c − 1 2 ∫ R N v 2 d x < c ,

which is a contradiction.

As a result, we have J ( v ) = 0 . Consequently, we have

c ≤ J ( v ) ≤ lim m → ∞ 1 2 ∫ R N v n 2 d x = c .

Thus, we complete the proof.□

Lemma 3.4

If potential V ( x ) satisfies ( V 3 ) , then m cannot be achieved.

Proof

If u ∈ N , then J ( ∣ u ∣ ) ≤ J ( u ) = 0 . Thus, there exists t ∈ ( 0 , 1 ] such that J ( t ∣ u ∣ ) = 0 and I ( t ∣ u ∣ ) ≤ I ( u ) . Hence, we have

c = inf { I ( u ) : u ∈ N and u ≥ 0 in R N } .

By density argument, one obtains

c = inf { I ( u ) : u ∈ N ∩ C 0 ∞ ( R N ) and u ≥ 0 in R N } .

Therefore, for any ε > 0 , we can choose a nonnegative function u ∈ N such that

I ( u ) < c + ε .

Without loss of generality, we may assume that supp ( u ) ⊂ B R ( 0 ) . Define

u ¯ ≔ u ( x + k e 1 ) − u ( x − k e 1 ) ,

where e 1 = ( 1 , 0 , … , 0 ) ∈ R N . By Lemma 2.1, there exists ( α , β ) ∈ R + × R + such that

α u ( x + k e 1 ) − β u ( x − k e 1 ) = α u ¯ + + β u ¯ − ∈ ℳ .

Combining this with u ∈ N , direct computation yields that

(3.8) α 2 log α 2 ∫ R N u 2 d x = − α β ∫ R N ∫ R N u ¯ + ( x ) u ¯ − ( y ) + u ¯ − ( x ) u ¯ + ( y ) ∣ x − y ∣ N + 2 s d x d y > 0

and

(3.9) β 2 log β 2 ∫ R N u 2 d x = − α β ∫ R N ∫ R N u ¯ + ( x ) u ¯ − ( y ) + u ¯ − ( x ) u ¯ + ( y ) ∣ x − y ∣ N + 2 s d x d y > 0 .

Hence, α = β > 1 . On the other hand, it follows from supp ( u ) ⊂ B R ( 0 ) that

∫ R N ∫ R N u ¯ + ( x ) u ¯ − ( y ) ∣ x − y ∣ N + 2 s d x d y = ∫ B R ( − 2 k e 1 ) ∫ B R ( 2 k e 1 ) u ¯ ( x ) u ¯ − ( y ) ∣ x − y ∣ N + 2 s d x d y = O 1 ∣ k − R ∣ N + 2 s .

Let k → ∞ , by (3.8) and (3.9), we have

log α 2 ∫ R N u 2 d x = o k ( 1 ) = log β 2 ∫ R N u 2 d x ,

where o k ( 1 ) → 0 as k → + ∞ . Therefore, α → 1 and β → 1 as k → ∞ . It follows from Lemma 2.3 that

m ≤ I ( α u ¯ + + β u ¯ − ) = 2 I ( u ) + o k ( 1 ) ≤ 2 c + ε + o k ( 1 ) .

Since ε is arbitrary, we deduce that

m ≤ 2 c ,

which is contradiction to Theorem 1.1. Thus, we complete the proof.□

3.3 ( V ) α potential

In this subsection, we prove c and m are achieved under conditions ( V 4 ) and ( V 5 ) . In order to overcome the difficulties caused by the loss of compactness, we first study the following problem in bounded domain, the solutions will be used as minimizing sequences.

(3.10) ( − Δ ) s u + V ( x ) u = u log u 2 , x ∈ B R , u = 0 , x ∈ R N \ B R ,

where B R = { x ∈ R N ∣ ∣ x ∣ < R } . We now define

c R = inf u ∈ N R I ( u ) , m R = inf u ∈ ℳ R I ( u ) ,

where

N R = N ∩ X 0 s ( B R ) , ℳ R = ℳ ∩ X 0 s ( B R )

and

X 0 s ( Ω ) = { u ∈ H s ( R N ) ∣ u = 0 a.e. on R N \ Ω } .

Obviously, I is well defined on X 0 s ( B R ) and I ∈ C 1 ( X 0 s ( B R ) , R ) . Therefore, one can easily verify the following result.

Lemma 3.5

c R , m R are achieved. Assume that u ∈ N R and I ( u ) = c R , or u ∈ ℳ R and I ( u ) = m R , then u is a solution of equation (3.10).

Lemma 3.6

Both c R and m R are decreasing with respect to R and converge to c , m , respectively, as R → ∞ .

Proof

Obviously, both c R and m R are decreasing with respect to R , and c R ≥ c , m R ≥ m . Now, we check that m R → m as R → ∞ . By the definition of m , for any ε > 0 , there exists u ∈ ℳ such that I ( u ) ≤ m + ε . Choose a cutoff function ξ R satisfying

ξ R ≡ 1 on B R 2 , supp ( ξ R ) ⊂ B R , 0 ≤ ξ R ≤ 1 , ∣ ∇ ξ R ∣ ≤ C R .

Define u R ≔ ξ R u , then we have

(3.11) ∫ R N ∣ u R + ∣ 2 log ∣ u R + ∣ 2 d x → ∫ R N ∣ u + ∣ 2 log ∣ u + ∣ 2 d x , as n → ∞

and

(3.12) ∫ R N ∣ u R − ∣ 2 log ∣ u R − ∣ 2 d x → ∫ R N ∣ u − ∣ 2 log ∣ u − ∣ 2 d x , as n → ∞ .

Now, we calculate that

(3.13) [ u R ] s 2 = ∫ ∣ x ∣ ≤ R 2 ∫ ∣ y ∣ ≤ R 2 ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + 2 ∫ ∣ x ∣ ≤ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ u R ( x ) − u R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x ∣ ≥ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ u R ( x ) − u R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y ≔ I 1 + I 2 + I 3 .

By a direct computation, we have

(3.14) I 2 ≤ 4 ∫ ∣ x ∣ ≤ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ ξ R ( x ) − ξ R ( y ) ∣ 2 ∣ u ( x ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x ∣ ≤ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ u ( x ) − u ( y ) ∣ 2 ∣ ξ R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y ≤ C ∫ ∣ x ∣ ≤ R 2 ∣ u ( x ) ∣ 2 ∫ ∣ y ∣ ≥ R 2 min ( 1 , ∣ x − y ∣ 2 / R 2 ) ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ y ∣ ≥ R 2 ∫ ∣ x ∣ ≤ R 2 ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y ≤ C O ( R − 2 s ) + ∫ ∣ y ∣ ≥ R 2 ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y = O ( R − 2 s ) + o R ( 1 ) ,

where o R ( 1 ) → 0 as R → ∞ . The last equality is because of

∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y < ∞ .

Similarly, we can deduce that

(3.15) I 3 = o R ( 1 ) , as R → + ∞ .

It follows from (3.13)–(3.15) that

(3.16) ∫ R N ∫ R N ∣ u R ( x ) − u R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y → ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y , as R → + ∞ .

Moreover, it also holds that

(3.17) ∫ R N ∫ R N u R + ( x ) u R − ( y ) + u R − ( x ) u R + ( y ) ∣ x − y ∣ N + 2 s d x d y → ∫ R N ∫ R N u + ( x ) u − ( y ) + u − ( x ) u + ( y ) ∣ x − y ∣ N + 2 s d x d y .

Therefore,

⟨ I ′ ( u R ) , u R + ⟩ → ⟨ I ′ ( u ) , u + ⟩ = 0 , ⟨ I ′ ( u R ) , u R − ⟩ → ⟨ I ′ ( u ) , u − ⟩ = 0 , as R → + ∞ .

By using Lemma 2.1, there exists α u R , β u R > 0 such that α u R u R + + β u R u R − ∈ ℳ R and α u R → 1 , β u R → 1 as R → + ∞ . Thus,

m R ≤ I ( α u R u R + + β u R u R − ) ≤ I ( α u R u + + β u R u − ) + C ∫ ∣ x ∣ ≤ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ u R ( x ) − u R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x ∣ ≥ R 2 ∫ ∣ y ∣ ≥ R 2 ∣ u R ( x ) − u R ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x ∣ ≥ R 2 V ( x ) u 2 + ∣ u 2 log u 2 ∣ d x ≤ I ( u + + u − ) + o R ( 1 ) ≤ m + ε + o R ( 1 ) ,

which implies lim R → + ∞ m R ≤ m . Considering m R ≥ m , we then conclude that lim R → + ∞ m R = m .

Similarly, we can prove lim R → + ∞ c R = c .□

Next, we need to study the following limiting functional:

I ∞ ( u ) = 1 2 ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + 1 2 ∫ R N ( V ∞ + 1 ) u 2 d x − 1 2 ∫ R N u 2 log u 2 d x , u ∈ W s ( R N ) .

Define

J ∞ ( u ) ≔ ∫ R N ∫ R N ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ∞ u 2 d x − ∫ R N u 2 log u 2 d x , u ∈ W s ( R N ) ,

and

N ∞ ≔ { u ∈ W s ( R N ) ⧹ { 0 } ∣ J ∞ ( u ) = 0 } , c ∞ ≔ inf u ∈ N ∞ I ∞ ( u ) .

It follows from Lemma 3.3 that c ∞ > 0 is achieved.

Lemma 3.7

Let u R be a solution of equation (3.10). Suppose that u R is bounded in W s ( R N ) with respect to R , and up to a subsequence R n → ∞ , that

∫ B R n ∣ u n ∣ p d x → λ ∈ ( 0 , ∞ ) ,

where u n ≔ u R n and 2 < p < 2 s ∗ . Then there exist β ∈ ( 0 , 1 ] and { x n } satisfying that for any ε > 0 , there exists r ε > 0 such that, for any r ′ ≥ r ≥ r ε > 0 ,

(3.18) lim inf n → ∞ ∫ B r ( x n ) ∣ u n ∣ p d x ≥ β λ − ε

and

(3.19) lim inf n → ∞ ∫ R N \ B r ′ ( x n ) ∣ u n ∣ p d x ≥ ( 1 − β ) λ − ε .

Moreover, if β < 1 , then lim inf n → ∞ I ( u n ) ≥ c + c ∞ .

Proof

Since { u n } is bounded in H s ( R N ) and ∫ B R n ∣ u n ∣ p d x is bounded away from zero, then the existence of such a number β ∈ ( 0 , 1 ] follows from a result of P.-L. Lions (see Lemma I.1. [20] or Lemma 4.3 [25]).

Now, we assume that β < 1 . Choose ε n → 0 and r n ′ ≥ r n → + ∞ such that, up to a subsequence, we may assume

(3.20) lim inf n → ∞ ∫ B r n ( x n ) ∣ u n ∣ p d x ≥ β λ − ε n , lim inf n → ∞ ∫ R N \ B r n ′ ( x n ) ∣ u n ∣ p d x ≥ ( 1 − β ) λ − ε n .

Choose a cutoff function ξ ( x ) satisfying

ξ ≡ 1 on B 3 ( 0 ) \ B 2 ( 0 ) , supp ( ξ ) ⊂ B 4 ( 0 ) \ B 1 ( 0 ) , 0 ≤ ξ ≤ 1 , ∣ ∇ ξ ∣ ≤ 2 .

Define ξ n ( x ) ≔ ξ ( ∣ x − x n ∣ / r n ) , then ξ n u n ∈ W s ( R N ) . Direct computation yields

(3.21) ∫ R N ∫ R N ∣ u n ( x ) − u n ( y ) ∣ 2 ξ n ( x ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ( x ) u n 2 ξ n d x − ∫ R N ξ n ( u n 2 log u n 2 ) − d x = ∫ R N ξ n ( u n 2 log u n 2 ) + d x + ∫ R N ∫ R N ( u n ( x ) − u n ( y ) ) ( ξ n ( x ) − ξ n ( y ) ) u n ( y ) ∣ x − y ∣ N + 2 s d x d y ≤ C ∫ R N ξ n ∣ u n ∣ p d x + [ u n ] s ∫ R N ∫ R N ∣ ξ n ( x ) − ξ n ( y ) ∣ 2 ∣ u n ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y 1 2 ≤ C ∫ R N ξ n ∣ u n ∣ p d x + C [ u n ] s ∫ R N ∣ u n ( y ) ∣ 2 ∫ ∣ x − y ∣ ≤ r n ∣ x − y ∣ 2 / r n 2 ∣ x − y ∣ N + 2 s d x d y + ∫ R N ∣ u n ( y ) ∣ 2 ∫ ∣ x − y ∣ ≥ r n 1 ∣ x − y ∣ N + 2 s d x d y 1 2 ≤ C ∫ R N ξ n ∣ u n ∣ p d x + O ( r n − s ) .

Thus,

(3.22) ∫ 2 r n ≤ ∣ x − x n ∣ ≤ 3 r n ∫ R N ∣ u n ( x ) − u n ( y ) ∣ 2 ∣ x − y ∣ N + 2 s d x d y + ∫ 2 r n ≤ ∣ x − x n ∣ ≤ 3 r n V ( x ) u n 2 ξ n + ∣ u n 2 log u n 2 ∣ d x = o n ( 1 ) .

Take another cutoff function η ( x ) satisfying

η ≡ 1 on B 2 ( 0 ) , supp ( η ) ⊂ B 3 ( 0 ) , 0 ≤ η ≤ 1 , ∣ ∇ η ∣ ≤ 2 .

Define

w n ( x ) = η ∣ x − x n ∣ r n u n ( x ) , v n ( x ) = 1 − η ∣ x − x n ∣ r n u n ( x ) .

Using (3.20), we can infer

(3.23) ∫ R N ∣ w n ∣ p d x ≥ ∫ B 2 r n ( x n ) ∣ u n ∣ p d x ≥ β λ − ε n

and

(3.24) ∫ R N ∣ v n ∣ p d x ≥ ∫ R N \ B 3 r n ( x n ) ∣ u n ∣ p d x ≥ ( 1 − β ) λ − ε n .

On the other hand, since supp ( v n ) ⊂ R N ⧹ B 2 r n ( x n ) , one can deduce that

(3.25) ∫ R N ∫ R N ( w n ( x ) − w n ( y ) ) ( v n ( x ) − v n ( y ) ) ∣ x − y ∣ N + 2 s d x d y = ∫ ∣ x − x n ∣ ≥ 2 r n ∫ ∣ y − x n ∣ ≥ 2 r n ( w n ( x ) − w n ( y ) ) ( v n ( x ) − v n ( y ) ) ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x − x n ∣ ≥ 2 r n ∫ ∣ y − x n ∣ < 2 r n ( w n ( x ) − w n ( y ) ) ( v n ( x ) − v n ( y ) ) ∣ x − y ∣ N + 2 s d x d y + ∫ ∣ x − x n ∣ < 2 r n ∫ ∣ y − x n ∣ ≥ 2 r n ( w n ( x ) − w n ( y ) ) ( v n ( x ) − v n ( y ) ) ∣ x − y ∣ N + 2 s d x d y = o n ( 1 ) .

It follows from (3.22)–(3.25) that

(3.26) I ( u n ) = I ( w n ) + I ( v n ) + o n ( 1 ) .

Moreover, by using (3.25) again, we derive that

J ( w n ) = ⟨ I ′ ( u n ) , w n ⟩ + o n ( 1 ) = o n ( 1 )

and

J ( v n ) = ⟨ I ′ ( u n ) , v n ⟩ + o n ( 1 ) = o n ( 1 ) .

Therefore, there exist two sequences { α n } , { β n } such that α n w n ∈ N , β n v n ∈ N , and α n → 1 , β n → 1 as n → ∞ .

If { x n } is bounded, then

(3.27) lim inf n → ∞ I ( β n v n ) = lim inf n → ∞ I ∞ ( β n v n ) ≥ c ∞ .

Furthermore, it follows from (3.26) and (3.27) that

lim inf n → ∞ I ( u n ) = lim inf n → ∞ I ( α n w n ) + lim inf n → ∞ I ( β n v n ) ≥ c + c ∞ .

If { x n } is unbounded, then

(3.28) lim inf n → ∞ I ( α n w n ) = lim inf n → ∞ I ∞ ( α n w n ) ≥ c ∞ ,

which upon combining with (3.26) also yields lim inf n → ∞ I ( u n ) ≥ c + c ∞ .□

Lemma 3.8

Assume ( V 4 ) holds, then c is achieved.

Proof

First, we claim that c < c ∞ . It follows from Lemma 3.3 that there exists w ∈ N ∞ such that I ∞ ( w ) = c ∞ and J ∞ ( w ) = 0 . Moreover, by a similar argument of the proof of Theorem 1.1, we conclude that w ( x ) > 0 for each x ∈ R N .

It follows from ( V 4 ) that

c ≤ max α ≥ 0 I ( α w ) = I ( α w w ) < I ∞ ( α w w ) ≤ max α ≥ 0 I ∞ ( α w ) = I ∞ ( w ) = c ∞ ,

where α w is the constant such that max α ≥ 0 I ( α w ) = I ( α w w ) .

Now, it follows from Lemmas 3.5 and 3.6 that there exists a sequence { u n } ⊂ N R n such that I ( u n ) = c n → c as n → ∞ . Since J ( u n ) = 0 , we can deduce that { u n } is uniformly bounded in W s ( R N ) . Thus, for any ε > 0 , by Lemma 3.7 there exist r > 0 and a sequence { x n } ⊂ R N such that

lim inf n → ∞ ∫ B r ( x n ) ∣ u n ∣ p d x ≥ λ − ε ,

where

λ ≔ lim n → ∞ ∫ R N ∣ u n ∣ p d x > 0 .

Now, we claim that { x n } is bounded in R N . If not, then we have

J ∞ ( u n ) = J ( u n ) + o n ( 1 ) ,

which implies the existence of α n > 0 such that α n → 1 and J ∞ ( α n u n ) = 0 . Thus, we have

c ∞ ≤ lim inf n → ∞ I ∞ ( α n u n ) = lim inf n → ∞ I ∞ ( u n ) = lim inf n → ∞ I ( u n ) = c ,

which leads to a contradiction. Therefore, { x n } is bounded and thus u n → u in L p ( R N ) for 2 < p < 2 s ∗ . Proceeding as the arguments in the proof of Lemma 3.2, we can prove that c is achieved.□

Lemma 3.9

Assume ( V 4 ) holds, and let u ( x ) be a positive ground state solution of equation (1.1), then

u ( x ) ≤ C 1 + ∣ x ∣ N + 2 s .

Proof

By Lemma 2.4 and Proposition 2.5, we know that u ∈ C 1 , 2 s + σ 0 − 1 ( R N ) ∩ L ∞ ( R N ) if 2 s + σ 0 > 1 , while u ∈ C 0 , 2 s + σ 0 ( R N ) ∩ L ∞ ( R N ) if 2 s + σ 0 ≤ 1 . Then, by a similar argument to the proof of Lemma 2.6 in [26], one can derive

lim ∣ x ∣ → ∞ u ( x ) = 0 .

Thus, there exists R > 0 such that

( − Δ ) s u + V 0 u ≤ ( − Δ ) s u + V ( x ) u = u log u 2 ≤ 0 , x ∈ R N \ B R .

It follows from Lemma C.1 in [27] that the fundamental solution G s , V 0 satisfies

( ( − Δ ) s + V 0 ) G s , V 0 = δ 0 , x ∈ R N .

Furthermore, G s , V 0 is radial, positive, strictly decreasing in ∣ x ∣ and satisfies

(3.29) lim ∣ x ∣ → ∞ ∣ x ∣ N + 2 s G s , V 0 = V 0 − 2 C N , s

with some positive constant C N , s . Hence, there exists a constant c 0 > 0 such that G s , V 0 ≥ c 0 > 0 in B R . Let C 0 = c 0 − 1 ‖ u ‖ L ∞ , then we have

w ( x ) ≔ C 0 G s , V 0 − u ≥ 0 , x ∈ B R

and

( − Δ ) s w + V 0 w = ( ( − Δ ) s + V 0 ) ( C 0 G s , V 0 − u ) ≥ 0 , x ∈ R N \ B R .

Note that the function w is continuous away from the origin and w ( x ) > 0 on B R holds. We claim that w ( x ) > 0 on B R c as well. Suppose on the contrary that w is strictly negative somewhere in B R c . Since w ( x ) → 0 as ∣ x ∣ → + ∞ and w ( x ) > 0 on B R , this implies that w attains a strict global minimum at some point x 0 ∈ B R c with w ( x 0 ) < 0 . By using the singular integral expression for ( − Δ ) s , it is easy to see that ( ( − Δ ) s w ) ( x 0 ) < 0 . On the other hand, we have ( − Δ ) s w + V 0 w ≥ 0 on B R c , which implies that ( ( − Δ ) s w ) ( x 0 ) > 0 . This is a contradiction, and we conclude that w ( x ) ≥ 0 on R N . Thus, we conclude

(3.30) u ≤ C 0 G s , V 0 .

Therefore, the result is directly from (3.29) and (3.30).□

From Lemma 3.8, there exists u ∈ N , w ∈ N ∞ such that I ( u ) = c and I ∞ ( w ) = c ∞ . Furthermore, with the same argument as that in proving Theorem 1.1, we may assume u > 0 and w > 0 . Similar to Lemma 3.9, one obtains

w ( x ) ≤ C 1 + ∣ x ∣ N + 2 s , for some constant C > 0 .

Denote w R ( x ) ≔ w ( x 1 + 2 R , x ′ ) with ( x 1 , x ′ ) ∈ R N . We have the following estimate.

Lemma 3.10

Suppose that V ( x ) satisfies ( V 4 )–( V 5 ), then m < c + c ∞ .

Proof

Direct computation yields

(3.31) I ( α u + β w R ) = α 2 2 [ u ] s 2 + ∫ R N ( V ( x ) + 1 ) u 2 d x + β 2 2 [ w R ] s 2 + ∫ R N ( V ( x ) + 1 ) w R 2 d x + α β ∫ R N ( V ( x ) + 1 ) u w R d x − 1 2 ∫ R N ( α u + β w R ) 2 log ( α u + β w R ) 2 d x + α β ∫ R N ∫ R N ( u ( x ) − u ( y ) ) ( w R ( x ) − w R ( y ) ) ∣ x − y ∣ N + 2 s d x d y .

To prove this lemma, it is sufficient to show that

(3.32) sup ( α , β ) ∈ R 2 I ( α u + β w R ) < c + c ∞ .

In fact, by Miranda theorem, there exist α ¯ > 0 and β ¯ < 0 such that α ¯ u + β ¯ w R ∈ ℳ . Thus, m < c + c ∞ .

Claim 1. There exists R 0 , r 0 > 0 such that for all R ≥ R 0 , α 2 + β 2 = r 2 > r 0 2 ,

sup ( α , β ) ∈ R 2 I ( α u + β w R ) ≤ 0 .

Indeed, let α ˜ = α / r , β ˜ = β / r and φ = α ˜ u + β ˜ w R . We first find R ′ > 0 such that for all R > R ′ , ‖ φ ‖ s and ∫ R N φ 2 log φ 2 d x are bounded from above and below by two positive constants. Thus, we have

I ( α u + β w R ) = I ( r φ ) = r 2 I ( φ ) − 1 2 r 2 log r 2 ∫ R N φ 2 d x ,

which implies the conclusion if r > 0 is large enough.

Claim 2. There exists ε > 0 small and R 1 > 0 large such that for R > R 1 , it holds

(3.33) ∫ R N ∫ R N ( u ( x ) − u ( y ) ) ( w R ( x ) − w R ( y ) ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N ( V ( x ) + 1 ) u w R d x = O ( R − 2 s + ε ( N + 2 s ) ) .

In fact, w R > 0 solves the equation

( − Δ ) s w R + V ∞ w R = w R log w R 2 , x ∈ R N .

Thus,

∫ R N ∫ R N ( u ( x ) − u ( y ) ) ( w R ( x ) − w R ( y ) ) ∣ x − y ∣ N + 2 s d x d y + ∫ R N V ∞ u w R d x = ∫ R N u w R log w R 2 d x .

For arbitrarily ε small, there exists C ε > 0 such that ∣ t log t 2 ∣ ≤ C ε t 1 − ε for all 0 < t < 1 . Thus, applying Lemma 3.9, we derive

∫ R N u w R log w R 2 d x ≤ ∫ R N \ B R ∣ u w R log w R 2 ∣ d x + ∫ B R ∣ u w R log w R 2 ∣ d x ≤ C ∫ R N \ B R ∣ u ∣ d x + ∫ R N \ B R ∣ w log w 2 ∣ d x ≤ C R − 2 s + ∫ R N \ B R ∣ x ∣ − ( 1 − ε ) ( N + 2 s ) d x ≤ C R − 2 s + ε ( N + 2 s ) .

Therefore, Claim 2 follows.

Claim 3. For α and β bounded, there exists R 2 > 0 such that, for R > R 2 , it holds

∫ R N ( α u + β w R ) 2 log ( α u + β w R ) 2 d x = ∫ R N ( α u ) 2 log ( α u ) 2 d x + ∫ R N ( β w R ) 2 log ( β w R ) 2 d x + O ( R − 2 s ) .

In fact, applying Lemma 3.9, one can obtain

∫ R N \ B R ( 0 ) ( α u ) 2 log ( α u ) 2 d x = O ∫ R N \ B R ( 0 ) ∣ u ∣ 2 − ε d x = O ( R − N − 4 s + ε ( N + 2 s ) ) , ∫ R N \ B R ( 2 R e 1 ) ( β w R ) 2 log ( β w R ) 2 d x = O ∫ R N \ B R ( 0 ) ∣ w ∣ 2 − ε d x = O ( R − N − 4 s + ε ( N + 2 s ) ) ,

and

∫ R N \ ( B R ( 0 ) ∪ B R ( 2 Re 1 ) ) ( α u + β w R ) 2 log ( α u + β w R ) 2 d x = O ( R − N − 4 s + ε ( N + 2 s ) ) .

Furthermore, direct computation yields

∫ B R ( 0 ) ( α u + β w R ) 2 log ( α u + β w R ) 2 d x − ∫ B R ( 0 ) ( α u ) 2 log ( α u ) 2 d x ≤ ∫ B R ( 0 ) ( α u ) 2 ( log ( α u + β w R ) 2 − log ( α u ) 2 ) d x + ∫ B R ( 0 ) ( 2 α β u w R + β 2 w R 2 ) log ( α u + β w R ) 2 d x = O ( R − 2 s ) .

Hence, Claim 3 follows.

Now, we prove (3.32). From Claim 1, we may suppose that both α and β are bounded. Combining Claim 2, Claim 3 with (3.31), we can deduce that for R > max { R 0 , R 1 , R 2 } ,

I ( α u + β w R ) = I ( α u ) + I ( β w R ) + O ( R − 2 s + ε ( N + 2 s ) ) .

It is easy to check that J ( w R ) → 0 as R → ∞ . Thus, there exists β R satisfying β R → 1 as R → + ∞ and J ( β R w R ) = 0 . Moreover, for R is large enough, by ( V 3 ) and ( V 4 ) , we deduce that

I ( α u + β w R ) = I ( α u ) + I ( β w R ) + O ( R − 2 s + ε ( N + 2 s ) ) ≤ I ( u ) + I ∞ ( β R w R ) + 1 2 β R 2 ∫ R N ( V ( x ) − V ∞ ) w R 2 d x + O ( R − 2 s + ε ( N + 2 s ) ) ≤ c + c ∞ − 1 2 ∫ B 1 ( 0 ) ( V ∞ − V ( x − 2 R e 1 ) w 2 d x + O ( R − 2 s + ε ( N + 2 s ) ) ≤ c + c ∞ − C ∫ B 1 ( 0 ) w 2 ∣ x − 2 R e 1 ∣ m d x + O ( R − 2 s + ε ( N + 2 s ) ) ≤ c + c ∞ − C R m + O ( R − 2 s + ε ( N + 2 s ) ) < c + c ∞ .

Therefore, we complete the proof.□

Proposition 3.11

Assume ( V 4 )–( V 5 ) hold, then m is achieved.

Proof

It follows from Lemmas 3.5 and 3.6 that there exists a sequence { u n } such that u n ∈ ℳ R n and I ( u n ) = m R n → m as n → ∞ . Since J ( u n ) = 0 , we deduce that { u n } is bounded in W s ( R N ) . Thus, by Lemma 3.7, we find a sequence { x n } ⊂ R N such that for any ε > 0 , there exists r > 0 such that

lim inf n → ∞ ∫ B r ( x n ) ∣ u n ∣ p d x ≥ λ − ε ,

where

λ ≔ lim n → ∞ ∫ R N ∣ u n ∣ p d x > 0 .

Now, we claim that { x n } is bounded. If not, then

⟨ I ′ ( u n ) , ( u n ) + ⟩ = ⟨ I ∞ ′ ( u n ) , ( u n ) + ⟩ + o n ( 1 )

and

⟨ I ′ ( u n ) , ( u n ) − ⟩ = ⟨ I ∞ ′ ( u n ) , ( u n ) − ⟩ + o n ( 1 ) .

Thus, there exist α n , β n > 0 with α n → 1 , β n → 1 such that

⟨ I ∞ ′ ( α n ( u n ) + + β n ( u n ) − ) , α n ( u n ) + ⟩ = 0

and

⟨ I ∞ ′ ( α n ( u n ) + + β n ( u n ) − ) , β n ( u n ) − ⟩ = 0 .

This implies α n ( u n ) + + β n ( u n ) − ∈ ℳ ∞ . Arguing as we prove Theorem 1.1, we infer 2 c ∞ ≤ m ∞ . Therefore,

c + c ∞ ≤ 2 c ∞ ≤ lim inf n → ∞ I ∞ ( α n ( u n ) + + β n ( u n ) − ) ≤ lim inf n → ∞ I ∞ ( u n ) = lim inf n → ∞ I ( u n ) = m ,

which leads to a contradiction. This implies that { x n } is bounded. Therefore, u n → u in L p ( R N ) for 2 < p < 2 s ∗ . Proceeding as the proof of Lemma 3.2, we can verify that m is achieved.□

Acknowledgements

The research was supported by the Natural Science Foundation of China (11831009, 12071170).

Conflict of interest: The authors declare that there is no conflict of interest.

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Received: 2021-08-16

Accepted: 2022-01-14

Published Online: 2022-03-07

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

Articles in the same Issue

https://doi.org/10.1515/ans-2022-0002

Keywords for this article

fractional logarithmic Schrödinger equations; positive ground state solutions; sign-changing solutions; variational methods

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