In this paper, we consider the following quasilinear equation of choquard type

( − d i v ( A ( x , u ) | ∇ u | p − 2 ∇ u ) + 1 p A t ( x , u ) | ∇ u | p + V ( ε x ) | u | p − 2 u = ( I α ∗ | u | q ) | u | q − 2 u ,   x ∈ ℝ N u ( x ) → 0,   | x | → ∞ (1)

where A ( x , t ) is given real functions on ℝ N × ℝ and A t ( x , t ) = ∂ ∂ t A ( x , t ) with N ≥ 3 , 1 < p < N , max { N − 2 p , 1 } < α < N , p ( N + α ) 2 N ≤ q < p α * = p ( N + α ) 2 ( N − p ) , and ε > 0 is a small parameter, I α : ℝ N → ℝ is the Riesz potential defined by

I α ( x ) = Γ ( N − α 2 ) Γ ( α 2 ) π N 2 2 α | x | N − α : = A α | x | N − α

and Γ is the Gamma function.

In the mathematical literature, very few results are known about Equation (1). Since the classical variational approach fails if A t ( x , t ) ≡ 0 . In reference [1] , the author considered the following quasilinear equation

− d i v ( A ( x , u ) | ∇ u | p − 2 ∇ u ) + 1 p A t ( x , u ) | ∇ u | p + | u | p − 2 u = g ( x , u ) ,   x ∈ ℝ N , (2)

In fact, the natural functional associated to (2) is

J ( u ) = 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N | u | p d x − ∫ ℝ N     G ( x , u ) d x ,

which is not defined in W 1, p ( ℝ N ) for a general coefficient A ( x , t ) in the principal part. Moreover, even if A ( x , t ) is smooth strictly positive bounded function, the corresponding energy functional is well defined in W 1, p ( ℝ N ) , if A t ( x , t ) ≡ 0 it is Gâteaux differentiable only along directions of W 1, p ( ℝ N ) ∩ L ∞ ( ℝ N ) [1] . More recently, if Ω is a bounded subset of ℝ N a different approach has been developed which exploits the interaction between two different norm on W 0 1, p ( Ω ) ∩ L ∞ ( Ω ) [2] [3] [4] . So, use the interaction between the norm ‖   ⋅   ‖ x and the standard one on W 1, p ( ℝ N ) , if G ( x , t ) has a subcritical growth, we state that J satisfies a weaken version of the Cerami’s variant of the Palais-Smale condition in X r . We note that, in general, J cannot verify the standard Palais-Smale condition, or its Cerami’s variant, as Palais-Smale sequences may converge in the W 1, p ( ℝ N ) -norm but be unbounded in L ∞ ( ℝ N ) [4] . For this reason, we know that there is a bounded radial solution of Equation (2) under certain assumptions.

For the Choquard equation

− Δ u + V ( x ) u = ( I α ∗ | u | p ) | u | p − 2 u ,   x ∈ ℝ 3 . (3)

where p ∈ ( 2 N − α N , 2 N − α N − 2 ) . When the potential V is a positive constant, Lieb [5] obtained the existence and uniqueness of positive radial ground states for (3), Lions [6] established the existence of infinitely many radial solutions, Ma and Zhao [7] studied the radial symmetry and uniqueness of positive ground states for (3) in higher dimension space via the method of moving planes. For more related results, we refer to [2] [8] [9] [10] [11] [12] and references therein.

In this article, we consider the existence of sign-changing solutions for the quasilinear Choquard Equation (1) by using the method of invariant sets of descending flow and the perturbation method. Byeon-Wang type penalization method [13] can be used to deal with multiple localized nodal solutions for semiclassical Schrödinger equations. Additional coercive term [14] can be used to make the perturbed functional has necessary compactness properties in changed space.

From now on, Let A : ℝ N × ℝ → ℝ be such that

(A₀) A ( x , t ) is a C 1 Carathéodory function, i.e.

A ( ⋅ , t ) : x ∈ ℝ N ↦ A ( x , t ) ∈ ℝ is measurable for all t ∈ ℝ ,

A ( x , ⋅ ) : t ∈ ℝ ↦ A ( x , t ) ∈ ℝ is C 1 for a.e. x ∈ ℝ N ;

(A₁) A ( x , t ) and A t ( x , t ) are essentially bounded if t is bounded, i.e.

sup | t | ≤ r | A ( ⋅ , t ) | ∈ L ∞ ( ℝ N ) ,     sup | t | ≤ r | A t ( ⋅ , t ) | ∈ L ∞ ( ℝ N )   for   any   r > 0 ;

(A₂) Exists a constant c 1 ,   c 2 > 0 such that

c 1 ≤ A ( x , t ) ≤ c 2   a .e .   x ∈ ℝ N ;

(A₃) Exist constants R ≥ 1 and α 1 > 0 such that

p A ( x , t ) + A t ( x , t ) t ≥ α 1 A ( x , t )   a .e .   in   ℝ N ,   if   | t | ≥ R ;

(A₄) Exist constants μ > p and α 2 > 0 such that

( μ − p ) A ( x , t ) − A t ( x , t ) t ≥ α 2 A ( x , t )   a .e .   in   ℝ N ,   for   all   t ∈ ℝ .

The potential function V satisfies the following assumptions:

(V₁) V ∈ C 1 ( ℝ N , ℝ ) and there exist constants b > a > 0 such that

a ≤ V ( x ) ≤ b ,   x ∈ ℝ N ;

(V₂) There exists a bounded domain M ⊂ ℝ N with smooth boundary ∂ M such that

〈 n ( x ) ← , ∇ V ( x ) 〉 > 0,   x ∈ ∂ M .

where n ( x ) ← is the outer normal of ∂ M at x. according to condition (V₂) that

A = { x ∈ M | ∇ V ( x ) = 0 } ≠ ∅ . (see [15] )

and A is a closed subset of M , without losing generality, we assume that 0 ∈ A .

For any set B ∈ ℝ N and any δ > 0 , we denote

B δ = { x ∈ ℝ N | d i s t ( x , B ) : = inf y ∈ B | x − y | < δ } ,

B δ = { x ∈ ℝ N | δ x ∈ B } .

The main results of this paper are as follows:

Theorem 1.1 Assume that A ( x , t ) satisfy conditions (A₀)-(A₄), the potential function V satisfies (V₁) and (V₂), for any positive integer k, there exists ε k > 0 such that if 0 < ε < ε k , the problem (1) has at least k pairs of sign-changing solutions ± u j , ε , j = 1 , 2 , ⋯ , k . In addition, for any δ > 0 there exist μ > 0 ,   c = c k > 0 and ε k ( δ ) > 0 such that if 0 < ε < ε k ( δ ) , then it holds that

| u j , ε | ≤ c exp { − μ     d i s t ( x , ( A δ ) ε ) } ,   j = 1 , 2 , ⋯ , k ,   x ∈ ℝ N .

We note that Equation (1) corresponding energy functional is

I ε ( u ) = 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N     V ( ε x ) | u | p d x − 1 2 q ∫ ℝ N ( I α ∗ | u | q ) | u | q d x (4)

We use the penalization method due to Byeon and Wang [13] . Let ζ ∈ C ∞ be a cut-off function, ζ ( t ) = 0 for t ≤ 0 ; ζ ( t ) = 1 for t ≥ 1 ; 0 ≤ ζ ′ ( t ) ≤ 2 and 0 ≤ ζ ( t ) ≤ 1 . Define

χ ε ( x ) = ( 0 , x ∈ M ε ε − p ζ ( d i s t ( x , M ε ) ) , x ∉ M ε

In order to overcome the lack of compactness condition, we set the workspace as X ε = W 1, p ( ℝ N ) ∩ L ε m ( ℝ N ) , where L ε m ( ℝ N ) is a weighted space defined as

L ε m ( ℝ N ) = { u ∈ L m ( ℝ N ) | ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u | m d x < + ∞ } ,

The corresponding norm is defined as

‖ u ‖ L ε m ( ℝ N ) = ( ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u | m d x ) 1 m .

which p < m < min { 2 , q } if 1 < p < 2 ; p < m < q if p ≥ 2 .

Define

‖ u ‖ X ε = ‖ u ‖ W 1 , p ( ℝ N ) + ‖ u ‖ L ε m ( ℝ N ) .

Meanwhile, We add the forced disturbance term such that I ε has the necessary compactness property on X ε , introduce some auxiliary functions. Let ξ ∈ C ∞ ( ℝ , [ 0,1 ] ) be a smooth, even function, such that ξ ( t ) = 1 if | t | ≤ 1 , ξ ( t ) = 0 if | t | ≥ 2 , and ξ is decreasing in [ 1,2 ] . For ε ∈ ( 0,1 ] , x ∈ ℝ N , t ∈ ℝ , we define

b ε ( x , t ) = ξ ( ε exp { d i s t ( ε x , M ) } t ) ,   m ε ( x , t ) = ∫ 0 t     b ε ( x , τ ) d τ ,

k ε ( x , t ) = ( t m ε ( x , t ) ) m − p | t | p − 2 t   ,   K ε ( x , t ) = ∫ 0 t     k ε ( x , τ ) d τ .

Now, we define the perturbation functional:

Γ ε ( u ) = 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N     E ( ε x ) | u | p d x + σ ∫ ℝ N     K ε ( x , u ) d x                       + 1 p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 2 q ∫ ℝ N ( I α ∗ | u | q ) | u | q d x   ,   u ∈ X ε (5)

where p < p β < q , E ( x ) = V ( x ) − σ , σ > 0 sufficiently small, E satisfies the assumptions (V₁) and (V₂).

Note that for any v ∈ X ε

〈 Γ ′ ε ( u ) , v 〉 = ∫ ℝ N     A ( x , u ) | ∇ u | p − 2 ∇ u ∇ v d x + 1 p ∫ ℝ N     A t ( x , u ) v | ∇ u | p d x       + ∫ ℝ N     E ( ε x ) | u | p − 2 u v d x + σ ∫ ℝ N     k ε ( x , u ) v d x       + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u | p − 2 u v d x       − ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u v d x (6)

We will use the abstract critical point theorem to prove the existence of the infinitely variable sign solution of Equation (1). However, when verifying the conditions of the theorem, we encounter essential difficulties, that is, we can’t guarantee that the positive cone and the negative cone are flow-invariant, so we follow the idea of literature [14] [16] [17] and use truncation method to truncate the nonlocal term. First, we define the following auxiliary functions:

b λ ( t ) = ξ ( λ t ) ,     m λ ( t ) = ∫ 0 t     b λ ( τ ) d τ ,

g λ ( t ) = m λ ( t ) t ,     h λ ( t ) = g λ ( t ) + b λ ( t ) .

Define the perturbation functional:

Γ ε , λ ( u ) = 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N     E ( ε x ) | u | p d x + σ ∫ ℝ N     K ε ( x , u ) d x                           + 1 p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 2 q g λ ( ψ 1 2 ( u ) ) ψ ( u ) ,   u ∈ X ε (7)

where ψ ( u ) = ∫ ℝ N ( I α ∗ | u | q ) | u | q d x .

By the definition of b λ ,   g λ , it is easy to prove that g λ ( t ) satisfies the following properties:

Lemma 2.1 [15] For t > 0 ,   0 < λ < 1 , it holds

(1) g λ ( t ) = 1 ,   g ′ λ ( t ) = 0 if 0 < t < 1 λ ;

(2) g ′ λ ( t ) t + g λ ( t ) = b λ ( t ) ;

(3) g ′ λ ( t ) t + g λ ( t ) ≤ c λ t ,   b λ ( t ) t ≤ g λ ( t ) t ≤ c λ , where c λ = ∫ 0 ∞     ξ ( τ ) d τ λ .

∀     v ∈ X ε , since g ′ λ ( t ) t + g λ ( t ) = b λ ( t ) , we have

〈 Γ ′ ε , λ ( u ) , v 〉 = ∫ ℝ N     A ( x , u ) | ∇ u | p − 2 ∇ u ∇ v d x + 1 p ∫ ℝ N     A t ( x , u ) v | ∇ u | p d x       + ∫ ℝ N     E ( ε x ) | u | p − 2 u v d x + σ ∫ ℝ N     k ε ( x , u ) v d x       + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u | p − 2 u v d x       − 1 2 h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u v d x . (8)

According to Hardy-Littlewood-Sobolev inequality and Sobolev inequality

ψ 1 2 ( u ) ≤ c ‖ u ‖ W 1, p ( ℝ N ) q .

Therefore, when ‖ u ‖ W 1, p ( ℝ N ) ≤ ( 1 c λ ) 1 q , there is Γ ε , λ ( u ) = Γ ε ( u ) and Γ ′ ε , λ ( u ) = Γ ′ ε ( u ) ; when | u ( x ) | ≤ ε − 1 exp { − d i s t ( ε x , M ) } and ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + = 0 , there is Γ ε ( u ) = I ε ( u ) and Γ ′ ε ( u ) = I ′ ε ( u ) . Hence, finding the solution of the Equation (4) translates into finding the critical point of Γ ε , λ ( u ) .

We describe the abstract critical point theorem in detail below [15] .

Let X be a Banach space and f be an even C 1 functional on X. Let P ,   Q are two open convex sets of X, Q = − P . Set

W = P ∪ Q ,   Σ = ∂ P ∩ ∂ Q .

Assume

(I₁) f satisfies the (PS) condition.

(I₂) c ∗ = inf x ∈ Σ f ( x ) > 0 . And assume there exists an odd continuous map F : X → X satisfying the following:

(F₁) given c 0 , b 0 > 0 , there exists b = b ( c 0 , b 0 ) > 0 such that if ‖ f ′ ( x ) ‖ ≥ b 0 , | f ( x ) | ≤ c 0 , then

〈 f ′ ( x ) , x − F x 〉 ≥ b ‖ x − F x ‖ X > 0.

(F₂) F ( ∂ P ) ⊂ P , F ( ∂ Q ) ⊂ Q .

Define

Γ j = { E | E ⊂ X :   E   compact ,   − E = E ,   γ ( E ∩ η − 1 ( Σ ) ) ≥ j   for   η ∈ Λ } ,

Λ = { η | η ∈ C ( X , X ) :   η   odd ,   η ( P ) ⊂ P ,   η ( Q ) ⊂ Q , η ( x ) = x   if   f ( x ) < 0 } .

where γ is the genus of symmetric sets, defined by

γ ( E ) = inf { n | there   exists   an   odd   map   η : E → ℝ n \ { 0 } } .

Assume

(Γ) Γ j , j = 1 , 2 , ⋯ is nonempty.

We define

c j = inf E ∈ Γ j sup x ∈ E \ W f ( x ) ,   j = 1 , 2 , ⋯ ,

K c = { x | f ′ ( x ) = 0 ,   f ( x ) = c } ,   K c ∗ = K c \ W .

Theorem 2.2 ( [18] , Theorem 2.5) Assume (I₁), (I₂), (F₁), (F₂), (Γ) hold, then

(1) c j ≥ c ∗ , K c j ∗ ≠ ∅ ,

(2) c j → ∞ , for j → ∞ .

(3) γ ( K c ∗ ) ≥ k , if c j = c j + 1 = ⋯ = c j + k − 1 = c .

In this section, we first introduce some important properties of auxiliary functions, then prove that Γ ε , λ satisfies the (PS) condition, and then use Theorem B to prove the existence of the critical point of Γ ε , λ .

Lemma 3.1 ( [19] , Lemma 2.2), ( [20] , Lemma 2.1) For x ∈ ℝ N

(1) 0 ≤ b ε ( x , t ) ≤ m ε ( x , t ) t ≤ 1 ;

(2) If | t | < ε − 1 exp { − d i s t ( ε x , M ) } , then m ε ( x , t ) = t ;

If ε − 1 exp { − d i s t ( ε x , M ) } ≤ | t | ≤ 2 ε − 1 exp { − d i s t ( ε x , M ) } , then ε − 1 exp { − d i s t ( ε x , M ) } ≤ | m ε ( x , t ) | ≤ c ε − 1 exp { − d i s t ( ε x , M ) } ;

If | t | > 2 ε − 1 exp { − d i s t ( ε x , M ) } , then | m ε ( x , t ) | = c ε − 1 exp { − d i s t ( ε x , M ) } ; where c = ∫ 0 ∞     ξ ( τ ) d τ ;

(3) c 1 ( 1 + ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | t | m − p ) | t | p − 2 t ≤ k ε ( x , t ) ≤ c 2 ( 1 + ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | t | m − p ) | t | p − 2 t ;

(4) 1 m t k ε ( x , t ) ≤ K ε ( x , t ) ≤ 1 p t k ε ( x , t ) ;

(5) ( k ε ( x , t 1 ) − k ε ( x , t 2 ) ) ( t 1 − t 2 ) ≥ c ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | t 1 − t 2 | m , p ≥ 2 ;

( k ε ( x , t 1 ) − k ε ( x , t 2 ) ) ( t 1 − t 2 ) ≥ c ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | t 1 − t 2 | 2 | t 1 | 2 − m + | t 2 | 2 − m , 1 < p < 2 ;

(6) | k ε ( x , t 1 ) − k ε ( x , t 2 ) | ≤ c ( | t 1 | p − 2 + | t 2 | p − 2 + ε m − p exp { ( m − p ) d i s t ( ε x , M ) } ( | t 1 | m − 2 + | t 2 | m − 2 ) ) | t 1 − t 2 | , p ≥ 2 ;

| k ε ( x , t 1 ) − k ε ( x , t 2 ) | ≤ c ( | t 1 − t 2 | p − 1 + ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | t 1 − t 2 | m − 1 ) , 1 < p < 2 .

Lemma 3.2 Let { u n } ⊂ X ε be a Palais-Smale sequence of the functional Γ ε , λ , then { u n } is bounded in X ε .

Proof: According to (7), (8) and assumptions (A₄), (A₂) and Lemma 3.1 (4) (3), we have

μ   Γ ε , λ ( u n ) − 〈 Γ ′ ε , λ ( u n ) , u n 〉 = μ [ 1 p ∫ ℝ N     A ( x , u n ) | ∇ u n | p d x + 1 p ∫ ℝ N     E ( ε x ) | u n | p d x + σ ∫ ℝ N K ε ( x , u n ) d x         + 1 p β ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 2 q g λ ( ψ 1 2 ( u n ) ) ψ ( u n ) ]       − ∫ ℝ N     A ( x , u n ) | ∇ u n | p d x − 1 p ∫ ℝ N     A t ( x , u n ) u n | ∇ u n | p d x − ∫ ℝ N     E ( ε x ) | u n | p d x       − ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u n | p d x

      − σ ∫ ℝ N     k ε ( x , u n ) u n d x + 1 2 h λ ( ψ 1 2 ( u n ) ) ψ ( u n ) = 1 p ∫ ℝ N ( ( μ − p ) A ( x , u n ) − A t ( x , u n ) u n ) | ∇ u n | p d x + μ − p p ∫ ℝ N     E ( ε x ) | u n | p d x       + σ ∫ ℝ N ( μ K ε ( x , u n ) − k ε ( x , u n ) u n ) d x + μ p β ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β       − ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u n | p d x       + 1 2 h λ ( ψ 1 2 ( u n ) ) ψ ( u n ) − μ 2 q g λ ( ψ 1 2 ( u n ) ) ψ ( u n )

≥ α 2 p ∫ ℝ N     A ( x , u n ) | ∇ u n | p d x + μ − p p ∫ ℝ N     E ( ε x ) | u n | p d x       + σ ( μ m − 1 ) ∫ ℝ N     k ε ( x , u n ) u n d x + c ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − c ≥ c 1 α 2 p ∫ ℝ N | ∇ u n | p d x + μ − p p ∫ ℝ N     E ( ε x ) | u n | p d x + c ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β       − c + c ∫ ℝ N     c 1 ( 1 + ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | u n | m − p ) | u n | p d x

≥ c ( ‖ u n ‖ W 1 , p ( ℝ N ) p + ‖ u n ‖ L ε m ( ℝ N ) m ) + c ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − c ≥ c ‖ u n ‖ X ε ( ℝ N ) p + c ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − c

Thus, The (PS) sequence { u n } of functional Γ ε , λ is bounded in X ε . ■

Lemma 3.3 The embedding X ε ↪ L r ( ℝ N )   ( 1 ≤ r < p ∗ ) is compact.

Proof: Let { u n } be the (PS) sequence of functional Γ ε , λ , u n ⊂ X ε satisfy Γ ε , λ ( u n ) → c , Γ ′ ε , λ ( u n ) → 0 in ( X ε ) ′ , by Lemma 3.2, there is a constant η ^ L > 0 independent of ε such that ‖ u n ‖ ε ≤ η ^ L and ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β ≤ η ^ L . Up to subsequences, suppose u n ⇀ u in X ε and u n → u in L l o c r ( ℝ N ) ( 1 ≤ r < p * ) . First prove u n → u in L 1 ( ℝ N ) , for any R > 0 that

∫ ℝ N \ B ( 0 , R ) | u | d x = ∫ ℝ N \ B ( 0 , R ) exp { m − p m d i s t ( ε x , M ) } ⋅ exp { − m − p m d i s t ( ε x , M ) } ⋅ | u | d x ≤ ( ∫ ℝ N \ B ( 0 , R ) exp { ( m − p ) d i s t ( ε x , M ) } | u | m d x ) 1 m       ⋅ ( ∫ ℝ N \ B ( 0 , R ) exp { − m − p m − 1 d i s t ( ε x , M ) } d x ) m − 1 m ≤ ‖ u ‖ L ε M ( ℝ N ) ( ∫ ℝ N \ B ( 0 , R ) exp { − m − p m − 1 d i s t ( ε x , M ) } d x ) m − 1 m = o R ( 1 ) ,

Hence

∫ ℝ N | u n − u | d x = ∫ B ( 0 , R ) | u n − u | d x + ∫ ℝ N \ B ( 0 , R ) | u n − u | d x = o n ( 1 ) + o R ( 1 ) → 0.

For 1 < r < p * ,

∫ ℝ N | u n − u | r d x = ∫ ℝ N | u n − u | r θ + ( 1 − θ ) r d x ≤ ( ∫ ℝ N | u n − u | r θ ⋅ 1 r θ d x ) r θ ⋅ ( ∫ ℝ N | u n − u | ( 1 − θ ) r ⋅ p * ( 1 − θ ) r d x ) ( 1 − θ ) r p * ≤ c ( ∫ ℝ N | u n − u | d x ) r θ → 0.

where 0 < θ < 1 , 1 r = θ + 1 − θ p * . ■

Lemma 3.4 For every ε > 0 , Γ ε , λ satisfies the Palais-Smale condition.

Proof: Let { u n } ⊂ X ε be a Palais-Smale sequence of the functional Γ ε , λ , so | Γ ε , λ ( u n ) | ≤ c and Γ ′ ε , λ ( u n ) → 0 ( n → ∞ ) , We will prove that { u n } has a convergent subsequence in X ε . According to Lemma 3.2, u n ⇀ u in X ε , then by assuming A₂, A₁, Hardy-Littlewood-Sobolev inequality, Hölder inequality and Lemma 3.3, there are

o ( 1 ) = 〈 Γ ′ ε , λ ( u n ) − Γ ′ ε , λ ( u ) , u n − u 〉 = ∫ ℝ N ( A ( x , u n ) | ∇ u n | p − 2 ∇ u n − A ( x , u ) | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x       + 1 p ∫ ℝ N ( A t ( x , u n ) | ∇ u n | p − A t ( x , u ) | ∇ u | p , u n − u ) d x       + ∫ ℝ N     E ( ε x ) ( | u n | p − 2 u n − | u | p − 2 u , u n − u ) d x       + σ ∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x

+ ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u n | p − 2 u n ( u n − u ) d x − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u | p − 2 u ( u n − u ) d x − 1 2 h λ ( ψ 1 2 ( u n ) ) ∫ ℝ N ( I α ∗ | u n | q ) ( | u n | q − 2 u n ( u n − u ) ) d x + 1 2 h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) ( | u | q − 2 u ( u n − u ) ) d x

≥ c 1 ∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x         + ∫ ℝ N     E ( ε x ) ( | u n | p − 2 u n − | u | p − 2 u , u n − u ) d x         + σ ∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x + o ( 1 ) .

Thus

∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x → 0 ;

∫ ℝ N ( | u n | p − 2 u n − | u | p − 2 u , u n − u ) d x → 0 ;

∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x → 0.

In the following proof process, the following basic inequalities will be used [21] .

∀     y , z ∈ ℝ N , when p ≥ 2 there is

( ( | y | p − 2 y − | z | p − 2 z , y − z ) ≥ c | y − z | p , | | y | p − 2 y − | z | p − 2 z | ≤ c ( | y | p − 2 + | z | p − 2 ) | y − z | ; (9)

when 1 < p < 2 there is

( ( | y | p − 2 y − | z | p − 2 z , y − z ) ≥ c | y − z | 2 ( | y | 2 − p + | z | 2 − p ) − 1 , | | y | p − 2 y − | z | p − 2 z | ≤ c | y − z | p − 1 . (10)

For p ≥ 2 , by (9) we have

∫ ℝ N | ∇ ( u n − u ) | p ≤ c ∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x → 0 ,

∫ ℝ N | u n − u | p ≤ c ∫ ℝ N ( | u n | p − 2 u n − | u | p − 2 u , u n − u ) d x → 0.

In addition, it follows from Lemma 3.1 (5) that

∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u n − u | m ≤ c ∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x → 0.

For 1 < p < 2 , by (10) we have

∫ ℝ N | ∇ ( u n − u ) | p ≤ c ∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) p 2 ( | ∇ u n | 2 − p + | ∇ u | 2 − p ) p 2 d x ≤ c ( ∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x ) p 2       ⋅ ( ∫ ℝ N ( | ∇ u n | 2 − p + | ∇ u | 2 − p ) p 2 − p d x ) 2 − p 2 ≤ c ( ∫ ℝ N ( | ∇ u n | p − 2 ∇ u n − | ∇ u | p − 2 ∇ u , ∇ u n − ∇ u ) d x ) p 2 → 0 ,

Similarly,

∫ ℝ N | u n − u | p ≤ c ( ∫ ℝ N ( | u n | p − 2 u n − | u | p − 2 u , u n − u ) d x ) p 2 → 0.

It follows from Lemma 3.1 (5) that

( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) ≥ c ε m − p exp { ( m − p ) d i s t ( ε x , M ) } | u n − u | 2 | u n | 2 − m + | u | 2 − m .

then

exp { ( m − p ) d i s t ( ε x , M ) } | u n − u | m ≤ exp { ( 2 − m ) ( m − p ) 2 d i s t ( ε x , M ) } ⋅ ( | u n | 2 − m + | u | 2 − m ) m 2       ⋅ [ ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) ] m 2 .

Thus

∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u n − u | m ≤ c ( ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } ( | u n | 2 − m + | u | 2 − m ) m 2 − m d x ) 2 − m 2         ⋅ ( ∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x ) m 2 ≤ c ( ∫ ℝ N ( k ε ( x , u n ) − k ε ( x , u ) ) ( u n − u ) d x ) m 2 → 0.

So, u n → u in X ε , Γ ε , λ satisfies the Palais-Smale condition. ■

Next, we will prove the existence of the critical point of the functional Γ ε , λ by using the descending invariant set method Theorem 2.2. Before verifying the condition of Theorem 2.2, we will give a few important Lemmas:

Lemma 3.5 Let max { N − 2 p , 1 } < α < N , p < q < p α * , if u n ⊂ W 1, p ( ℝ N ) , u n ⇀ u in W 1, p ( ℝ N ) , and u n ( x ) → u ( x ) a.e. x ∈ ℝ N , then there is a subcolumn still marked as { u n } , which satisfies

(1) ∫ ℝ N ( I α ∗ | u n | q ) | u n | q = ∫ ℝ N ( I α ∗ | u n − u | q ) | u n − u | q + ∫ ℝ N ( I α ∗ | u | q ) | u | q + o n ( 1 ) ,

(11)

(2) ∫ ℝ N [ ( I α ∗ | u n | q ) | u n | q − 2 u n − ( I α ∗ | u n − u | q ) | u n − u | q − 2 ( u n − u ) − ( I α ∗ | u | q ) | u | q − 2 u ] φ = o n ( 1 ) ‖ φ ‖ W 1 , p ( ℝ N ) .

(12)

where φ ∈ c o ∞ ( ℝ N ) , as n → ∞ , o n ( 1 ) → 0 .

To prove Lemma 3.5, we also need the following results:

Lemma 3.6 ( [22] , Theorem 4.2.7) Let Ω ⊆ ℝ N be a domain and { u n } is bounded in L q ( Ω ) for some q > 1 , If u n ( x ) → u ( x ) a.e. x ∈ Ω , then u n ⇀ u in L q ( Ω ) .

Lemma 3.7 ( [23] , Theorem 2.5) If { u n } is bounded in W 1, p ( ℝ N ) , u n ⇀ u in W 1, p ( ℝ N ) and u n ( x ) → u ( x ) a.e. x ∈ ℝ N , then for p < q < p α *

(1) ∫ ℝ N | | u n | q − | u n − u | q − | u | q | 2 N N + α d x → 0 ,

(2) ∫ ℝ N | | u n | q + | u n − u | q − | u | q | 2 N N + α d x → 0 ,

(3) ∫ ℝ N | | u n | q − 2 u n − | u n − u | q − 2 ( u n − u ) − | u | q − 2 u | 2 N p ( N + α ) p − 2 N + 2 p d x → 0.

Lemma 3.8 ( [24] , Theorem 2.6) Let 0 < α < N , s ∈ ( 1 , N α ) , and let

{ u n } ⊂ L 1 ( ℝ N ) ∩ L s ( ℝ N ) be bounded and such that, up to a subsequence, for any bounded domain Ω ⊂ ℝ N , u n → 0 in L s ( Ω ) as n → ∞ . Then, up to a subsequence if necessary, ( I α ∗ u n ) ( x ) → 0 a.e. x ∈ ℝ N as n → ∞ .

Proof of Lemma 3.5:

(1) We note that:

∫ ℝ N ( I α ∗ | u n | q ) | u n | q − ( I α ∗ | u n − u | q ) | u n − u | q − ( I α ∗ | u | q ) | u | q = ∫ ℝ N ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) − ( I α ∗ | u | q ) | u | q ,

By the Hardy-Littlewood-Sobolev inequality, for sufficiently small δ > 0 , there exists K 1 > 0 , which makes

| ∫ Ω 1 ( I α ∗ | u | q ) | u | q | ≤ δ 6 ;   Ω 1 : = { x ∈ ℝ N : | u ( x ) | ≥ K 1 } .

Using the Hardy-Littlewood-Sobolev inequality again, we have

| ∫ Ω 1 ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) | ≤ C ( N , α ) ( ∫ ℝ N | | u n | q + | u n − u | q | 2 N N + α ) N + α 2 N ⋅ ( ∫ Ω 1 | | u n | q − | u n − u | q | 2 N N + α ) N + α 2 N ≤ C ( N , α ) ( ∫ Ω 1 | | u n | q − | u n − u | q | 2 N N + α ) N + α 2 N ≤ C ( N , α ) ( ∫ Ω 1 | u | 2 N N + α ) N + α 2 N ,

For the above δ , when K 1 is large enough, there is

| ∫ Ω 1 ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) | ≤ δ 6 .

Similary, let Ω 2 : = { x ∈ ℝ N : | x | ≥ R } \ Ω 1 , let R > 0 be large enough so that

| ∫ Ω 2 ( I α ∗ | u | q ) | u | q | ≤ δ 6 ,

and

| ∫ Ω 2 ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) | ≤ δ 6 .

For K 2 > K 1 , let Ω 3 ( n ) : = { x ∈ ℝ N : | u ( x ) | ≥ K 2 } \ ( Ω 1 ∪ Ω 2 ) . If Ω 3 ( n ) ≠ ϕ , then ∀     x ∈ Ω 3 ( n ) , there is | u ( x ) | < K 1 and | x | < R . Observed, u n ( x ) → u ( x ) a.e. x ∈ Ω . By Severini-Egoroff theorem, u n converges to u in B R ( 0 ) , thus | Ω 3 ( n ) | → 0 . So, for a large enough n

| ∫ Ω 3 ( n ) ( I α ∗ | u | q ) | u | q | ≤ δ 6 ,

and

| ∫ Ω 3 ( n ) ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) | ≤ δ 6 .

Finally, we estimate that

∫ Ω 4 ( n ) ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) − ( I α ∗ | u | q ) | u | q

where Ω 4 ( n ) = ℝ N \ ( Ω 1 ∪ Ω 2 ∪ Ω 3 ( n ) ) , Ω 4 ( n ) ⊂ B R ( 0 ) .

It follows from Lebesgue dominated convergence theorem that

lim n → ∞ ∫ Ω 4 ( n ) | u n − u | 2 N q N + α = 0   and   lim n → ∞ ∫ Ω 4 ( n ) | | u n | q − | u | q | 2 N N + α = 0.

This means that by the Hardy-Littlewood-Sobolev inequality,

| ∫ Ω 4 ( n ) ( I α ∗ ( | u n | q + | u n − u | q ) ) | u n − u | q | ≤ C ( N , α ) ( ∫ Ω 4 ( n ) | u n − u | 2 N q N + α ) N + α 2 N → 0,

| ∫ Ω 4 ( n ) ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u | q ) | ≤ C ( N , α ) ( ∫ Ω 4 ( n ) | | u n | q − | u | q | 2 N N + α ) N + α 2 N → 0.

Let H n = | u n | q + | u n − u | q − | u | q , then

lim n → ∞ ∫ Ω 4 ( n ) ( I α ∗ ( | u n | q + | u n − u | q ) ) ( | u n | q − | u n − u | q ) − ( I α ∗ | u | q ) | u | q = lim n → ∞ ∫ Ω 4 ( n ) ( I α ∗ H n ) | u | q .

Because H n is bounded in L 2 N N + α ( ℝ N ) and H n → 0 a.e. x ∈ ℝ N , by Lemma 3.6, H n ⇀ 0 in L 2 N N + α ( ℝ N ) , thus I α ∗ H n ⇀ 0 in L 2 N N − α ( ℝ N ) , so

lim n → ∞ ∫ Ω 4 ( n ) ( I α ∗ H n ) | u | q → 0.

Thus, summing up, we obtain that

lim sup n → ∞ | ∫ ℝ N ( I α ∗ | u n | q ) | u n | q − ( I α ∗ | u n − u | q ) | u n − u | q − ( I α ∗ | u | q ) | u | q | ≤ δ .

By the arbitrariness of δ , conclusion (1) is established.

(2) According to Lemma 3.7, we have

lim n → ∞ ∫ ℝ N | | u n | q − | u n − u | q − | u | q | 2 N N + α d y = 0,

thus

| ∫ ℝ N ( I α ∗ ( | u n | q − | u n − u | q − | u | q ) ) V n φ | = o n ( 1 ) ‖ φ ‖ W 1, p ( ℝ N ) , (13)

where V n = | u n | q − 2 u n ,   | u n − u | q − 2 ( u n − u ) or | u | q − 2 u .

Similarly,

lim n → ∞ ∫ ℝ N | | u n | q − 2 u n − | u n − u | q − 2 ( u n − u ) − | u | q − 2 u | 2 N p ( N + α ) p − 2 N + 2 p d x = 0 ,

thus

| ∫ ℝ N ( I α ∗ W n ) ( | u n | q − 2 u n − | u n − u | q − 2 ( u n − u ) − | u | q − 2 u ) φ | = o n ( 1 ) ‖ φ ‖ W 1, p ( ℝ N ) , (14)

where W n = | u n | q , | u n − u | q or | u | q .

The direct calculation of (13) + (14) is

∫ ℝ N [ ( I α ∗ | u n | q ) | u n | q − 2 u n − ( I α ∗ | u n − u | q ) | u n − u | q − 2 ( u n − u ) − ( I α ∗ | u | q ) | u | q − 2 u ] φ = ∫ ℝ N ( I α ∗ | u n − u | q ) | u | q − 2 u φ + ∫ ℝ N ( I α ∗ | u | q ) | u n − u | q − 2 ( u n − u ) φ       + o n ( 1 ) ‖ φ ‖ W 1 , p ( ℝ N ) .

According to Rellich theorem, there is a subcolumn, which may as well still be recorded as { u n } , for any bounded region Ω ⊂ ℝ N , we have | u n − u | q → 0 in L 2 N N + α ( Ω ) . By Lemma 3.8, I α ∗ | u n − u | q → 0 a.e. x ∈ ℝ N , from Hardy-Littlewood-Sobolev inequality it follows that

sup n ‖ I α ∗ | u n − u | q ‖ L 2 N N − α ( ℝ N ) ≤ sup n ‖ | u n − u | ‖ L 2 N q N + α ( ℝ N ) < ∞ .

By Lemma 3.6, we have ‖ I α ∗ | u n − u | q ‖ N p N p − N + p ⇀ 0 in L 2 N p − 2 N + 2 p ( N − α ) p ( ℝ N ) , with 2 N p − 2 N + 2 p ( N − α ) p > 1 .

Because | u | N p ( q − 1 ) N p − N + p ∈ L 2 N p − 2 N + 2 p ( N + α ) p − 2 N + 2 p ( ℝ N ) , so

∫ ℝ N ( I α ∗ | u n − u | q ) | u | q − 2 u φ ≤ ( ∫ ℝ N ( I α ∗ | u n − u | q ) N p N p − N + p | u | N p ( q − 1 ) N p − N + p d x ) N p − N + p N p ⋅ ‖ φ ‖ W 1 , p ( ℝ N ) = o n ( 1 ) ‖ φ ‖ W 1 , p ( ℝ N ) .

In a similar way, we can verify that

∫ ℝ N ( I α ∗ | u | q ) | u n − u | q − 2 ( u n − u ) φ = o n ( 1 ) ‖ φ ‖ W 1, p ( ℝ N ) .

Therefore, conclusion (2) is established. ■

Let

J ε ( u ) = 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N   E ( ε x ) | u | p d x + σ ∫ ℝ N   K ε ( x , u ) d x .

The definition operator F : X ε → X ε , v = F u ∈ X ε is the only solution of the following equation.

1 2 〈 J ′ ε ( u ) + J ′ ε ( v ) − d J ε ( u − v ) , η 〉 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v η d x = 1 2 h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u η d x ,   ∀   η ∈ X ε (15)

Lemma 3.9 ∀   u ,   v ∈ X ε

(1) For p ≥ 2 ,

〈 J ′ ε ( u ) − J ′ ε ( v ) , φ 〉 ≤ c ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ W 1 , p ( ℝ N ) ‖ φ ‖ W 1 , p ( ℝ N )         + c ( ‖ u ‖ L ε m ( ℝ N ) m − 2 + ‖ v ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v ‖ L ε m ( ℝ N ) ‖ φ ‖ L ε m ( ℝ N ) ,

For 1 < p < 2 ,

〈 J ′ ε ( u ) − J ′ ε ( v ) , φ 〉 ≤ c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 ‖ φ ‖ W 1 , p ( ℝ N ) + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ‖ φ ‖ L ε m ( ℝ N ) ) .

(2) For p > 1 ,

〈 J ′ ε ( u ) , φ 〉 ≤ c ( ‖ u ‖ W 1, p ( ℝ N ) p − 1 ‖ φ ‖ W 1, p ( ℝ N ) + ‖ u ‖ L ε m ( ℝ N ) m − 1 ‖ φ ‖ L ε m ( ℝ N ) ) .

(3) For p ≥ 2 ,

〈 J ′ ε ( u ) − J ′ ε ( v ) , u − v 〉 ≥ c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m ) ,

For 1 < p < 2 ,

〈 J ′ ε ( u ) − J ′ ε ( v ) , u − v 〉 ≥ c ‖ u − v ‖ W 1 , p ( ℝ N ) 2 ( ‖ u ‖ W 1 , p ( ℝ N ) 2 − p + ‖ v ‖ W 1 , p ( ℝ N ) 2 − p ) − 1         + c ‖ u − v ‖ L ε m ( ℝ N ) 2 ( ‖ u ‖ L ε m ( ℝ N ) 2 − m + ‖ v ‖ L ε m ( ℝ N ) 2 − m ) − 1 .

Proof: (1) From V 1 ,   A 1 and Hölder inequality it follows that

〈 J ′ ε ( u ) − J ′ ε ( v ) , φ 〉 = ∫ ℝ N ( A ( x , u ) | ∇ u | p − 2 ∇ u − A ( x , v ) | ∇ v | p − 2 ∇ v ) ∇ φ d x       + 1 p ∫ ℝ N ( A t ( x , u ) | ∇ u | p − A t ( x , v ) | ∇ v | p ) φ d x       + ∫ ℝ N     E ( ε x ) ( | u | p − 2 u − | v | p − 2 v ) φ d x + σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) φ d x

≤ c ∫ ℝ N | | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v | ⋅ | ∇ φ | d x + c ∫ ℝ N | | ∇ u | p − | ∇ v | p | ⋅ | φ | d x       + c ∫ ℝ N | | u | p − 2 u − | v | p − 2 v | ⋅ | φ | d x + σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) φ d x ≤ c ( ∫ ℝ N | | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v | p p − 1 d x ) p − 1 p ‖ φ ‖ W 1 , p ( ℝ N )       + c ( ∫ ℝ N | | u | p − 2 u − | v | p − 2 v | p p − 1 d x ) p − 1 p ‖ φ ‖ W 1 , p ( ℝ N )       + σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) φ d x .

For p ≥ 2 , from (9) we have

( ∫ ℝ N | | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v | p p − 1 d x ) p − 1 p ≤ c ( ∫ ℝ N ( | ∇ u | p − 2 + | ∇ v | p − 2 ) p p − 1 ⋅ | ∇ u − ∇ v | p p − 1 d x ) p − 1 p ≤ c ( ( ∫ ℝ N | ∇ u | p d x ) p − 2 p + ( ∫ ℝ N | ∇ v | p d x ) p − 2 p ) ⋅ ( ∫ ℝ N | ∇ u − ∇ v | p d x ) 1 p ≤ c ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ W 1 , p ( ℝ N ) ,

similarly,

( ∫ ℝ N | | u | p − 2 u − | v | p − 2 v | p p − 1 d x ) p − 1 p ≤ c ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ W 1 , p ( ℝ N ) .

For 1 < p < 2 , from (10) we have

( ∫ ℝ N | | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v | p p − 1 d x ) p − 1 p ≤ c ( ∫ ℝ N | ∇ u − ∇ v | p d x ) p − 1 p ≤ c ‖ u − v ‖ W 1, p ( ℝ N ) p − 1 ,

( ∫ ℝ N | | u | p − 2 u − | v | p − 2 v | p p − 1 d x ) p − 1 p ≤ c ( ∫ ℝ N | u − v | p d x ) p − 1 p ≤ c ‖ u − v ‖ W 1, p ( ℝ N ) p − 1 .

In addition, by Lemma 3.1 (6) and Hölder inequality, for p ≥ 2 ,

σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) φ d x ≤ c ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } ( | u | m − 2 + | v | m − 2 ) | u − v | | φ | d x ≤ c ( ‖ u ‖ L ε m ( ℝ N ) m − 2 + ‖ v ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v ‖ L ε m ( ℝ N ) ‖ φ ‖ L ε m ( ℝ N ) .

For 1 < p < 2 ,

σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) φ d x ≤ c ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u − v | m − 1 | φ | d x ≤ c   ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ‖ φ ‖ L ε m ( ℝ N ) .

(2) Take v = 0 in (1) to get (2).

(3) From assume A₁, A₂, we have

〈 J ′ ε ( u ) − J ′ ε ( v ) , u − v 〉 = ∫ ℝ N ( A ( x , u ) | ∇ u | p − 2 ∇ u − A ( x , v ) | ∇ v | p − 2 ∇ v ) ( ∇ u − ∇ v ) d x         + 1 p ∫ ℝ N ( A t ( x , u ) | ∇ u | p − A t ( x , v ) | ∇ v | p ) ( u − v ) d x         + ∫ ℝ N     E ( ε x ) ( | u | p − 2 u − | v | p − 2 v ) ( u − v ) d x         + σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) ( u − v ) d x

≥ c 1 ∫ ℝ N ( | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v ) ( ∇ u − ∇ v ) d x         + c ∫ ℝ N ( | u | p − 2 u − | v | p − 2 v ) ( u − v ) d x         + σ ∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) ( u − v ) d x .

For p ≥ 2 , from (9) we have

∫ ℝ N ( | ∇ u | p − 2 ∇ u − | ∇ v | p − 2 ∇ v ) ( ∇ u − ∇ v ) d x ≥ c ∫ ℝ N | ∇ u − ∇ v | p d x ≥ c ‖ u − v ‖ W 1 , p ( ℝ N ) p ,

∫ ℝ N ( | u | p − 2 u − | v | p − 2 v ) ( u − v ) d x ≥ c ‖ u − v ‖ W 1 , p ( ℝ N ) p .

From Lemma 3.1 (5), we have

∫ ℝ N ( k ε ( x , u ) − k ε ( x , v ) ) ( u − v ) d x ≥ c ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u − v | m d x = c ‖ u − v ‖ L ε m ( ℝ N ) m .

Similarly, from (10) and Lemma 3.1 (5) can prove the case of 1 < p < 2 , so (3) holds. ■

Lemma 3.10 If ‖ u ‖ X ε is bounded, then ‖ F u ‖ X ε = ‖ v ‖ X ε is bounded.

Proof: By (15)

‖ F u ‖ W 1 , p ( ℝ N ) p + ‖ F u ‖ L ε m ( ℝ N ) m ≤ c 〈 J ′ ε ( v ) , v 〉 ≤ c h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u v d x ≤ c ‖ u ‖ W 1 , p ( ℝ N ) 2 q − 1 ‖ F u ‖ W 1 , p ( ℝ N ) ≤ c ‖ F u ‖ W 1 , p ( ℝ N )

thus ‖ F u ‖ X ε is bounded. ■

Lemma 3.11 F is odd, well defined, and continuous on X ε .

Proof: From the definition of operator F, it is easy to know that F is an odd operator. Definition

G ( v ) = 1 2 J ε ( v ) + 1 2 J ε ( v − u ) + 1 2 〈 J ′ ε ( u ) , v 〉       + 1 p ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p d x       − 1 2 h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u v d x ,   ∀   v ∈ X ε

Equation (15) has a unique solution v = F u , which can be obtained by solving the minimization problem inf { G ( v ) | v ∈ X ε } . Since

G ( v ) ≥ 1 2 J ε ( v ) + 1 2 〈 J ′ ε ( u ) , v 〉 − c λ ≥ c 1 ‖ v ‖ W 1, p ( ℝ N ) p − c 2 ‖ v ‖ W 1, p ( ℝ N ) − c λ

thus G is coercive.

Let { v n } ⊂ X ε be a minimizing sequence for the functional G, v n → v in X ε . By the lower semicontinuity

G ( v ) ≤ lim inf n → ∞ G ( v n ) = inf { G ( v ) | v ∈ X ε }

so v is a solution of (15). Assume v 1 ,   v 2 are solutions of (15), taking v 1 − v 2 as the test function, we have

1 2 〈 J ′ ε ( v 1 ) − J ′ ε ( v 2 ) , v 1 − v 2 〉 + 1 2 〈 J ′ ε ( v 1 − u ) − J ′ ε ( v 2 − u ) , v 1 − v 2 〉 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | v 1 | p − 2 v 1 − | v 2 | p − 2 v 2 ) ( v 1 − v 2 ) d x = 0

Hence

〈 J ′ ε ( v 1 ) − J ′ ε ( v 2 ) , v 1 − v 2 〉 = 0.

By Lemma 3.9 (3), for p ≥ 2

〈 J ′ ε ( v 1 ) − J ′ ε ( v 2 ) , v 1 − v 2 〉 ≥ c ( ‖ v 1 − v 2 ‖ W 1 , p ( ℝ N ) p + ‖ v 1 − v 2 ‖ L ε m ( ℝ N ) m ) ,

for 1 < p < 2

〈 J ′ ε ( v 1 ) − J ′ ε ( v 2 ) , v 1 − v 2 〉 ≥ c ‖ v 1 − v 2 ‖ W 1 , p ( ℝ N ) 2 ( ‖ v 1 ‖ W 1 , p ( ℝ N ) 2 − p + ‖ v 2 ‖ W 1 , p ( ℝ N ) 2 − p ) − 1         + c ‖ v 1 − v 2 ‖ L ε m ( ℝ N ) 2 ( ‖ v 1 ‖ L ε m ( ℝ N ) 2 − m + ‖ v 2 ‖ L ε m ( ℝ N ) 2 − m ) − 1 .

Then v 1 = v 2 , we prove that Equation (15) has a unique solution of v = F u .

The following proves that F is continuous, taking η = v n − v in (15), we have

1 2 〈 J ′ ε ( u − v ) − J ′ ε ( u n − v n ) , ( u − v ) − ( u n − v n ) 〉 + 1 2 〈 J ′ ε ( v n ) − J ′ ε ( v ) , v n − v 〉 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | v n | p − 2 v n − | v | p − 2 v ) ( v n − v ) d x = 1 2 〈 J ′ ε ( u − v ) − J ′ ε ( u n − v n ) , u − u n 〉 + 1 2 〈 J ′ ε ( u n ) − J ′ ε ( u ) , v − v n 〉         + [ ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ] ∫ ℝ N     χ ε ( x ) | v | p − 2 v ( v − v n ) d x         + 1 2 h λ ( ψ 1 2 ( u n ) ) ∫ ℝ N ( ( I α ∗ | u n | q ) | u n | q − 2 u n − ( I α ∗ | u | q ) | u | q − 2 u ) ( v n − v ) d x         + 1 2 ( h λ ( ψ 1 2 ( u n ) ) − h λ ( ψ 1 2 ( u ) ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u ( v n − v ) d x . (16)

Suppose u n → u in X ε , by Lemma 3.9 and Lemma 3.10, for p ≥ 2 , we have

〈 J ′ ε ( u − v ) − J ′ ε ( u n − v n ) , u − u n 〉 + 〈 J ′ ε ( u n ) − J ′ ε ( u ) , v − v n 〉 ≤ c ( ‖ u − v ‖ W 1, p ( ℝ N ) p − 2 + ‖ u n − v n ‖ W 1, p ( ℝ N ) p − 2 ) ‖ u − v − u n + v n ‖ W 1, p ( ℝ N ) ‖ u − u n ‖ W 1, p ( ℝ N )         + c ( ‖ u − v ‖ L ε m ( ℝ N ) m − 2 + ‖ u n − v n ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v − u n + v n ‖ L ε m ( ℝ N ) ‖ u − u n ‖ L ε m ( ℝ N )         + c ( ‖ u n ‖ W 1, p ( ℝ N ) p − 2 + ‖ u ‖ W 1, p ( ℝ N ) p − 2 ) ‖ u n − u ‖ W 1, p ( ℝ N ) ‖ v − v n ‖ W 1, p ( ℝ N )         + c ( ‖ u n ‖ L ε m ( ℝ N ) m − 2 + ‖ u ‖ L ε m ( ℝ N ) m − 2 ) ‖ u n − u ‖ L ε m ( ℝ N ) ‖ v − v n ‖ L ε m ( ℝ N ) ≤ c ‖ u n − u ‖ X ε = o n ( 1 ) , (17)

Similarly, for 1 < p < 2 , we have

〈 J ′ ε ( u − v ) − J ′ ε ( u n − v n ) , u − u n 〉 + 〈 J ′ ε ( u n ) − J ′ ε ( u ) , v − v n 〉 ≤ c ( ‖ u − v − u n + v n ‖ W 1 , p ( ℝ N ) p − 1 ‖ u − u n ‖ W 1 , p ( ℝ N ) + ‖ u − v − u n + v n ‖ L ε m ( ℝ N ) m − 1 ‖ u − u n ‖ L ε m ( ℝ N ) )         + c ( ‖ u n − u ‖ W 1 , p ( ℝ N ) p − 1 ‖ v − v n ‖ W 1 , p ( ℝ N ) + ‖ u n − u ‖ L ε m ( ℝ N ) m − 1 ‖ v − v n ‖ L ε m ( ℝ N ) ) = o n ( 1 ) . (18)

By Lemma 3.10 and Hardy-Littlewood-Sobolev inequality, we obtain that

[ ( ∫ ℝ N     χ ε ( x ) | u n | p d x − 1 ) + β − 1 − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ] ∫ ℝ N     χ ε ( x ) | v | p − 2 v ( v − v n ) d x   + 1 2 ( h λ ( ψ 1 2 ( u n ) ) − ( ψ 1 2 ( u ) ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u ( v n − v ) d x = o n ( 1 ) . (19)

By Lemma 3.5 and Lemma 3.10, we have

1 2 h λ ( ψ 1 2 ( u n ) ) ∫ ℝ N ( ( I α ∗ | u n | q ) | u n | q − 2 u n − ( I α ∗ | u | q ) | u | q − 2 u ) ( v n − v ) d x = 1 2 h λ ( ψ 1 2 ( u n ) ) ∫ ℝ N ( I α ∗ | u n − u | q ) | u n − u | q − 2 ( u n − u ) ( v n − v ) d x = o n ( 1 ) . (20)

Thus the right-hand side of (16) satisfies

R H S = o n ( 1 ) . (21)

Next, we estimate the left-hand side of (16), for p ≥ 2 , by (9)

L H S ≥ 〈 J ′ ε ( v n ) − J ′ ε ( v ) , v n − v 〉 ≥ c ( ‖ v n − v ‖ W 1, p ( ℝ N ) p + ‖ v n − v ‖ L ε m ( ℝ N ) m ) , (22)

for 1 < p < 2 , by (10)

L H S ≥ 〈 J ′ ε ( v n ) − J ′ ε ( v ) , v n − v 〉 ≥ c ‖ v n − v ‖ W 1, p ( ℝ N ) 2 ( ‖ v n ‖ W 1, p ( ℝ N ) 2 − p + ‖ v ‖ W 1, p ( ℝ N ) 2 − p ) − 1       + ‖ v n − v ‖ L ε m ( ℝ N ) 2 ( ‖ v n ‖ L ε m ( ℝ N ) 2 − m + ‖ v ‖ L ε m ( ℝ N ) 2 − m ) − 1 . (23)

From (21) to (23), for p > 1 , we obtain

‖ v n − v ‖ X ε → 0,     as   n → ∞ .

Therefore, F is continuous. ■

We verify the condition (F₁) in Theorem 2.2

Lemma 3.12 If u ∈ X ε ,   v = F u , then

(1) 〈 Γ ′ ε , λ ( u ) , u − v 〉 ≥ c ( ‖ u − v ‖ W 1, p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m )

(2) ‖ Γ ′ ε , λ ( u ) ‖ ≤ c ( 1 + | Γ ε , λ ( u ) | + ‖ u − v ‖ X ε ) α ‖ u − v ‖ X ε ,   p ≥ 2 ‖ Γ ′ ε , λ ( u ) ‖ ≤ c ‖ u − v ‖ X ε α − 1 ,   1 < p < 2

where α = max { p , m , p β }

Proof: (1) By (15), ∀   η ∈ X ε we have

〈 Γ ′ ε , λ ( u ) , η 〉 = 〈 J ′ ε ( u ) , η 〉 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u | p − 2 u η d x       − 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) + J ′ ε ( v − u ) , η 〉       − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v η d x = 1 2 〈 J ′ ε ( u − v ) , u − v 〉 + 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) , u − v 〉       + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | u | p − 2 u − | v | p − 2 v ) η d x . (24)

Hence

〈 Γ ′ ε , λ ( u ) , u − v 〉 = 1 2 〈 J ′ ε ( u − v ) , u − v 〉 + 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) , u − v 〉         + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | u | p − 2 u − | v | p − 2 v ) ( u − v ) d x ≥ c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m ) .

(2) For p ≥ 2 , by (7) (15) we have

μ   Γ ε , λ ( u ) − 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) + J ′ ε ( u − v ) , u 〉 = μ   Γ ε , λ ( u ) − 1 q 〈 J ′ ε ( u ) , u 〉 + 1 2 h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q d x       − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v u d x = 1 p ∫ ℝ N ( ( μ − p ) A ( x , u ) − A t ( x , u ) ) | ∇ u | p d x + μ − p p ∫ ℝ N     E ( ε x ) | u | p d x       + σ μ ∫ ℝ N     K ε ( x , u ) d x − σ ∫ ℝ N     k ε ( x , u ) u + μ p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β

      − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v u d x       + 1 2 h λ ( ψ 1 2 ( u ) ) ψ ( u ) d x − μ 2 q g λ ( ψ 1 2 ( u ) ) ψ ( u ) d x ≥ c ‖ u ‖ W 1 , p ( ℝ N ) p + σ ( μ − m ) m ∫ ℝ N     k ε ( x , u ) u d x + μ p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β       − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v u d x ,

while

σ ( μ − m ) m ∫ ℝ N     k ε ( x , u ) u d x ≥ c ∫ ℝ N exp { ( m − p ) d i s t ( ε x , M ) } | u | m d x ≥ c ‖ u ‖ L ε m ( ℝ N ) m .

By Höder inequality and Young inequality, we obtain that

μ p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | v | p − 2 v u d x = μ p β ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) | u | p d x         + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | u | p − 2 u − | v | p − 2 v ) u d x ≥ c ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − c ( ∫ ℝ N     χ ε ( x ) ( | u | p − 2 + | v | p − 2 ) | u − v | | u | d x ) β − c ≥ c ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − c ( ∫ ℝ N     χ ε ( x ) | u − v | p d x ) β − c ,

Hence

μ   Γ ε , λ ( u ) − 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) + J ′ ε ( u − v ) , u 〉 ≥ c ( ‖ u ‖ W 1 , p ( ℝ N ) p + ‖ u ‖ L ε m ( ℝ N ) m ) + c ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − c ‖ u − v ‖ W 1 , p ( ℝ N ) p β − c . (25)

In addition, by Lemma 3.9

μ   Γ ε , λ ( u ) − 1 2 〈 J ′ ε ( u ) − J ′ ε ( v ) + J ′ ε ( u − v ) , u 〉 ≤ c | Γ ε , λ ( u ) | + c ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ W 1 , p ( ℝ N ) ‖ u ‖ W 1 , p ( ℝ N )         + c ( ‖ u ‖ L ε m ( ℝ N ) m − 2 + ‖ v ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v ‖ L ε m ( ℝ N ) ‖ u ‖ L ε m ( ℝ N )         + c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 ‖ u ‖ W 1 , p ( ℝ N ) + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ‖ u ‖ L ε m ( ℝ N ) )

≤ c | Γ ε , λ ( u ) | + c ‖ u ‖ W 1 , p ( ℝ N ) p − 1 ‖ u − v ‖ W 1 , p ( ℝ N ) + c ‖ u ‖ L ε m ( ℝ N ) m − 1 ‖ u − v ‖ L ε m ( ℝ N )         + c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 ‖ u ‖ W 1 , p ( ℝ N ) + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ‖ u ‖ L ε m ( ℝ N ) ) ≤ c | Γ ε , λ ( u ) | + c ( ‖ u ‖ W 1 , p ( ℝ N ) p + ‖ u ‖ L ε m ( ℝ N ) m ) + c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m ) , (26)

that is

c ( ‖ u ‖ W 1 , p ( ℝ N ) p + ‖ u ‖ L ε m ( ℝ N ) m ) + c ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − c ‖ u − v ‖ W 1 , p ( ℝ N ) p β − c ≤ c | Γ ε , λ ( u ) | + c ( ‖ u ‖ W 1 , p ( ℝ N ) p + ‖ u ‖ L ε m ( ℝ N ) m ) + c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m ) .

Thus

‖ u ‖ W 1 , p ( ℝ N ) p + ‖ u ‖ L ε m ( ℝ N ) m + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β ≤ c ( 1 + | Γ ε , λ ( u ) | + ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m + ‖ u − v ‖ W 1 , p ( ℝ N ) p β ) . (27)

By (24) and Lemma 3.9(1), (27) and Young inequality, we have

‖ Γ ′ ε , λ ( u ) ‖ ≤ 1 2 ‖ J ′ ε ( u − v ) ‖ + 1 2 ‖ J ′ ε ( u ) − J ′ ε ( v ) ‖       + | ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | u | p − 2 u − | v | p − 2 v ) d x | ≤ c ( 1 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ) ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ W 1 , p ( ℝ N )       + c ( ‖ u ‖ L ε m ( ℝ N ) m − 2 + ‖ v ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v ‖ L ε m ( ℝ N ) + c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 )

≤ c ( 1 + ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ) ( ‖ u ‖ W 1 , p ( ℝ N ) p − 2 + ‖ v ‖ W 1 , p ( ℝ N ) p − 2 ) ‖ u − v ‖ X ε         + c ( ‖ u ‖ L ε m ( ℝ N ) m − 2 + ‖ v ‖ L ε m ( ℝ N ) m − 2 ) ‖ u − v ‖ X ε ≤ c ( 1 + | Γ ε , λ ( u ) | + ‖ u − v ‖ W 1 , p ( ℝ N ) p + ‖ u − v ‖ L ε m ( ℝ N ) m + ‖ u − v ‖ W 1 , p ( ℝ N ) p β ) 2 ‖ u − v ‖ X ε ≤ c ( 1 + | Γ ε , λ ( u ) | + ‖ u − v ‖ X ε ) α ‖ u − v ‖ X ε .

where α = max { p β , p , m } .

For 1 < p < 2 , by (24) and Lemma 3.9 (2), we have

‖ Γ ′ ε , λ ( u ) ‖ ≤ 1 2 ‖ J ′ ε ( u − v ) ‖ + 1 2 ‖ J ′ ε ( u ) − J ′ ε ( v ) ‖       + | ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ∫ ℝ N     χ ε ( x ) ( | u | p − 2 u − | v | p − 2 v ) d x | ≤ c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ) + c ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β − 1 ≤ c ( ‖ u − v ‖ W 1 , p ( ℝ N ) p − 1 + ‖ u − v ‖ L ε m ( ℝ N ) m − 1 ) ≤ c ‖ u − v ‖ X ε α − 1 .

■

From Lemma 3.12, we can get the following corollary:

Corollary 3.13 For all b 0 ,   c 0 > 0 , there exists b = b ( b 0 ,   c 0 ) > 0 such that if

| Γ ε , λ ( u ) | ≤ c 0   ,   ‖ Γ ′ ε , λ ( u ) ‖ ≥ b 0

then

〈 Γ ′ ε , λ ( u ) , u − F u 〉 ≥ b ‖ u − F u ‖ X ε > 0 ,   u − F u ≠ 0.

we denote

D ( f , g ) = ∫ ℝ N ∫ ℝ N f ( x ) g ( y ) | x − y | N − α d x d y

Lemma 3.14 ( [5] , Theorem 9.8) If N ≥ 3 ,   0 < α < N , and D ( f , f ) ,   D ( g , g ) < ∞ , then

| D ( f , g ) | 2 ≤ D ( f , f ) ⋅ D ( g , g )

only when g ≠ 0 ,   f = c g , the equal sign holds, where c is a constant.

For δ > 0 , define convex cones P and Q.

P : = { u | u ∈ X ε ( ℝ N ) , ‖ u + ‖ W 1 , p ( ℝ N ) < δ } Q : = { u | u ∈ X ε ( ℝ N ) , ‖ u − ‖ W 1 , p ( ℝ N ) < δ }

We verify the condition (F₂) in Theorem 2.2

Lemma 3.15 For 0 < λ < 1 , there exists δ λ > 0 , such that for 0 < δ < δ λ , we have

F ( ∂ P ) ⊂ P   ,   F ( ∂ Q ) ⊂ Q   .

Proof: We only prove F ( ∂ P ) ⊂ P ,   ∀   u ∈ ∂ P , let v = F u , taking η = v + in (15), by Hardy-Littlewood-Sobolev inequality and Lemma 3.14 have

‖ v + ‖ W 1 , p ( ℝ N ) p ≤ ‖ v + ‖ X ε p ≤ 〈 J ′ ( v ) , v + 〉 ≤ c h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u | q − 2 u v + d x ≤ c h λ ( ψ 1 2 ( u ) ) ∫ ℝ N ( I α ∗ | u | q ) | u + | q − 2 u + v + d x ≤ c h λ ( ψ 1 2 ( u ) ) ( ∫ ℝ N ( I α ∗ ( | u + | q − 2 u + v + ) ) | u + | q − 2 u + v + d x ) 1 2 ( ∫ ℝ N ( I α ∗ | u | q ) | u | q d x ) 1 2 = c h λ ( ψ 1 2 ( u ) ) ψ 1 2 ( u ) ( ∫ ℝ N ( I α ∗ ( | u + | q − 2 u + v + ) ) | u + | q − 2 u + v + d x ) 1 2 ≤ c λ ‖ u + ‖ L q r ( ℝ N ) q − 1 ‖ v + ‖ L q r ( ℝ N )

then ‖ v + ‖ L q r ( ℝ N ) p − 1 ≤ c λ ‖ u + ‖ L q r ( ℝ N ) q − 1 . Just taking δ λ = ( 1 c λ ) 1 q − p , then for 0 < δ < δ λ

‖ v + ‖ L q r ( ℝ N ) ≤ ( c λ ‖ u + ‖ L q r ( ℝ N ) q − 1 ) 1 p − 1 ≤ ( c λ δ q − 1 ) 1 p − 1 < δ

Hence F ( ∂ P ) ⊂ P

Similarly available: F ( ∂ Q ) ⊂ Q ■

We verify the condition (I₂) in Theorem 2.2

Lemma 3.16 There exist δ 0 > 0 and 0 < c * = c * ( δ ) , such that for any 0 < δ < δ 0 have

Γ ε , λ ( u ) ≥ c * ,   ∀   u ∈ ∂ P ∩ ∂ Q .

Proof: ∀   u ∈ ∂ P ∩ ∂ Q , we have

Γ ε , λ ( u ) ≥ 1 p ∫ ℝ N     A ( x , u ) | ∇ u | p d x + 1 p ∫ ℝ N     E ( ε x ) | u | p d x − 1 2 q g λ ( ψ 1 2 ( u ) ) ψ ( u ) ≥ min { α 0 , a } p ‖ u ‖ W 1 , p ( ℝ N ) p − c 0 ‖ u ‖ L q r ( ℝ N ) 2 q ≥ min { α 0 , a } p ‖ u ‖ W 1 , p ( ℝ N ) p − c 0 δ 2 q − p ‖ u ‖ L q r ( ℝ N ) p ≥ ( min { α 0 , a } c p − c 0 δ 2 q − p ) ‖ u ‖ W 1 , p ( ℝ N ) p

Taking δ 0 is small enough to make c 0 δ 0 2 q − p ≤ min { α 0 , a } 2 c p , then for 0 < δ < δ 0 , we have

Γ ε , λ ( u ) ≥ min { α 0 , a } 2 c p ‖ u ‖ W 1 , p ( ℝ N ) p ≥ min { α 0 , a } 2 c p δ 0 p : = c *

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Finally, we verify the condition Γ in Theorem 2.2

Let

J 0 ( u ) = 1 p ∫ B 1 ( 0 ) ( a | ∇ u | p + b | u | p ) d x + σ ∫ B 1 ( 0 ) exp { ( m − p ) | x | } | u | m d x     − c ˜ 2 q g λ ( ψ 1 2 ( u ˜ ) ) ψ ( u ˜ )

where c ˜ = ∫ 0 ∞     ξ ( τ ) d τ , B 1 ( 0 ) = { x ∈ ℝ N | | x | < 1 } , u ∈ X ε ( B 1 ( 0 ) ) , u ˜ ≡ u in B 1 ( 0 ) , u ˜ ≡ 0 in ( B 1 ( 0 ) ) c . Let { e n } n = 1 ∞ be a family of linearly independent functions in C 0 ∞ ( B 1 ( 0 ) ) , exist an increasing sequence R n so that

J 0 ( u ) < 0 ,   ∀   u ∈ H n ,   ‖ u ‖ ≥ R n

where H n : = s p a n { e 1 , ⋯ , e n } , choose the appropriate ε to make B 1 ( 0 ) ⊂ M ε .

Define

φ n ∈ c ( B n , C 0 ∞ ( B 1 ( 0 ) ) )

φ n ( t ) = R n ∑ i = 1 n     t i e i ,   t = ( t 1 , ⋯ , t n ) ∈ B n = { t | t ∈ ℝ N , | t | ≤ 1 }

Note

Γ j = { E | E ⊂ X ε ,   E   is   compact   set ,   − E = E ,   for   η ∈ Λ ,   γ ( E ∩ η − 1 ( Σ ) ) ≥ j } ,

Λ = { η | η ∈ C ( X ε , X ε ) ,   η   is   odd   function ,   η ( P ) ⊂ P ,   η ( Q ) ⊂ Q , η ( u ) = u   if   Γ ε , λ ( u ) ≤ 0 } .

Lemma 3.17 The set Γ j ,   j = 1 , 2 , ⋯ is nonempty.

Proof: The proof is similar to ( [25] , Lemma 4.2).

So far, we have verified all the conditions of theorem 2.2 and obtained the following existence theorem.

Theorem 3.18 Assuming that the conditions V₁ and V₂ hold, then there exist 0 < ε ˜ < 1 and 0 < λ ˜ < 1 , such that when 0 < ε < ε ˜ , 0 < λ < λ ˜ , the functional Γ ε , λ has infinitely many sign-changing critical points, the corresponding critical values are

c j ( ε , λ ) = inf E ∈ Γ j sup u ∈ E \ W Γ ε , λ ( u ) ,   j = 1 , 2 , ⋯ (28)

Moreover,

(1) Existence of m j ,   j = 1 , 2 , ⋯ independent of ε , λ such that

c j ( ε , λ ) ≤ m j ,   j = 1 , 2 , ⋯ (29)

(2) If c j ( ε , λ ) = ⋯ = c j + k − 1 ( ε , λ ) = c , then γ ( K c * ) ≥ k , with

K c * = K c \ W , K c = { x | d Γ ε , λ ( u ) = 0 , Γ ε , λ ( u ) = c } .

Proof: All the assumptions of Theorem 2.2 are satisfied, by Theorem 2.2 we know (2) is true. We only need to prove (29). It’s easy to verify that { c j } is incremental, and E j = φ j + 1 ( B j + 1 ) ∈ Γ j , for t ∈ B j + 1 , u ∈ φ j + 1 ( t ) , exists 0 < ε ˜ < 1 , 0 < λ ˜ < 1 , such that ( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + β = 0 , then if 0 < ε < ε ˜ , 0 < λ < λ ˜ , for u ∈ φ j + 1 ( B j + 1 ) has Γ ε , λ ≤ J 0 ( u ) , thus

c j ( ε , λ ) ≤ m j : = sup u ∈ E j J 0 ( u )

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In this section, we will prove the main result of this paper, that is, when ε , λ is small enough, the critical point of Γ ε , λ is also the critical point of I ε . In order to prove the perturbation functional Γ ε , λ and the original functional I ε have a common critical point, we will prove the following theorem.

Theorem 4.1 (1) Suppose Γ ε , λ ( u ) ≤ L ,   Γ ′ ε , λ ( u ) = 0 , then there is a constant H = H ( L ) such that ‖ u ‖ W 1, p ( ℝ N ) ≤ H ;

(2) Suppose Γ ε ( u ) ≤ L ,   Γ ′ ε ( u ) = 0 , then exists constant μ > 0 , c = c ( L ) , For any δ > 0 , there is ε = ε ( δ ) > 0 , and when 0 < ε < ε ( δ ) , we have

| u ( x ) | ≤ c exp { − μ     d i s t ( x , ( A δ ) ε ) } ,   ∀   x ∈ ℝ N .

Proof (1) It is easily obtained by Lemma 3.2.

(2) Refer to ( [19] , Lemma 5.1-Lemma 5.5), results (2). It is necessary to estimate the uniform boundedness of the solution of the truncation problem. With the help of the technique of section decomposition ( [26] , Theorem 2.1, Theorem 3.3), through Moser iteration [14] [15] [16] [17] , we can get, if Γ ε , λ ( u n ) ≤ L , Γ ′ ε , λ ( u n ) = 0 , then exists c , μ independent of n makes:

| u n ( x ) | ≤ c exp { − μ R } ,   x ∈ Ω R ( n ) .

we note R n ( x ) = min { | x − y n , k | | k ∈ Λ } , then

| u n ( x ) | ≤ c exp { − μ R n ( x ) } , ∀   x ∈ Ω R n ( n ) .

Because ε n y n , k → y k * ∈ A , there is ε ( δ ) > 0 for any δ > 0 , when ε n ≤ ε ( δ ) , we have ε n y n , k ∈ A δ , thus

| u n ( x ) | ≤ c exp { − μ R n } ≤ c exp { − μ     d i s t ( x , ( A δ ) ε ) } ,   ∀   x ∈ ℝ N . ■

Corollary 4.2 (1) Suppose Γ ε , λ ( u ) ≤ L ,   Γ ′ ε , λ ( u ) = 0 , then there is a constant λ ¯ = λ ¯ ( L ) such that Γ ε , λ ( u ) = Γ ε ( u ) and Γ ′ ε ( u ) = 0 if 0 < λ < λ ¯ ;

(2) Suppose Γ ε ( u ) ≤ L ,   Γ ′ ε ( u ) = 0 , then there is a constant ε ¯ = ε ¯ ( L ) such that Γ ε ( u ) = I ε ( u ) and I ′ ε ( u ) = 0 if 0 < ε < ε ¯ .

Proof (1) By Theorem 4.1(1), if 0 < λ < λ ¯ ( L ) = 1 c H q , then

‖ u ‖ W 1 , p ( ℝ N ) ≤ H = ( 1 c λ ¯ ( L ) ) 1 q ≤ ( 1 c λ ) 1 q

thus Γ ε , λ ( u ) = Γ ε ( u ) and Γ ′ ε ( u ) = 0 .

(2) By Theorem 4.1(2), there exists μ > 0 , c = c ( L ) such that for any δ > 0 , there is ε = ε ( δ ) > 0 , when 0 < ε < ε ( δ ) have

| u n ( x ) | ≤ c exp { − μ     d i s t ( x , ( A δ ) ε ) } ≤ c exp { − μ     d i s t ( x , M ε ) }

we note d = d i s t ( A δ , ∂ M ) , then for x ∉ M ε , we have

d i s t ( x , ( A δ ) ε ) ≥ d i s t ( x , M ε ) + d ε − 1

thus, as ε → 0

∫ ℝ N     χ ε ( x ) | u | p d x ≤ c ε − p ∫ ℝ N \ M ε | u | p d x ≤ c ε − p ∫ ℝ N \ M ε exp { − μ p     d i s t ( x , ( A δ ) ε ) } d x ≤ c ε − p exp { − μ d ε − 1 } ∫ ℝ N \ M ε exp { − μ ( p − 1 ) d i s t ( x , ( A δ ) ε ) } d x ≤ c ε − p exp { − μ d ε − 1 } ∫ ℝ N \ M exp { − μ ( p − 1 ) d i s t ( x , A δ ) } d x ≤ c ε − p exp { − μ d ε − 1 } → 0

In particular, there is a ε ¯ so that when 0 < ε ≤ ε ¯ , we have

( ∫ ℝ N     χ ε ( x ) | u | p d x − 1 ) + = 0

thus I ε ( u ) = Γ ε ( u ) and Γ ′ ε ( u ) = I ′ ε ( u ) = 0 . ■

Proof of Theorem 1.1 Given a positive integer k, by Theorem 3.18, there is 0 < ε ˜ < 1 and 0 < λ ˜ < 1 , such that when 0 < ε < ε ˜ , 0 < λ < λ ˜ , the functional Γ ε , λ has k pairs of sign-changing critical points ± u j , ε , j = 1 , 2 , ⋯ , k , and the corresponding critical values satisfy:

0 < c 1 ( ε , λ ) ≤ ⋯ ≤ c k ( ε , λ ) ≤ m k

By Corollary 4.2(2), there is ε k = ε k ( m k ) > 0 , when 0 < ε < ε ˜ k = min { ε k , ε ˜ } , Γ ε ( u ) ≤ m k and Γ ′ ε ( u ) = 0 , we have

Γ ε ( u ) = I ε ( u ) ,   I ′ ε ( u ) = 0.

By Corollary 4.2(1), there is λ k = λ k ( m k ) > 0 , when 0 < λ < λ ˜ k = min { λ k , λ ˜ } , Γ ε , λ ( u ) ≤ m k and Γ ′ ε , λ ( u ) = 0 , we have

Γ ε , λ ( u ) = Γ ε ( u ) ,   Γ ′ ε ( u ) = 0.

Hence, when 0 < ε < ε ˜ k ,   0 < λ < λ ˜ k , which u j , ε = u j ( ε , λ ) , j = 1 , 2 , ⋯ , k is also the critical point of functional I ε . Further, according to Theorem 4.1, there is a constant μ > 0 , c = c ( m k ) , so that for any δ > 0 , there is ε k ( δ ) > 0 , when 0 < ε < ε k ( δ ) , we have

| u j , ε | ≤ c exp { − μ     d i s t ( x , ( A δ ) ε ) } ,   ∀   x ∈ ℝ N .

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This work was supported partially by the National Natural Science Foundation of China (11961081).

The authors declare no conflicts of interest regarding the publication of this paper.

Wang, D., Wang, Y.Q. and Chen, S.X. (2023) Existence and Concentration of Sign-Changing Solutions of Quasilinear Choquard Equation. Journal of Applied Mathematics and Physics, 11, 1124-1151. https://doi.org/10.4236/jamp.2023.114074