The Existence of Solution of a Critical Fractional Equation

In this paper, we study the existence of solution of a critical fractional equation ; we will use a variational approach to find the solution. Firstly, we will find a suitable functional to our problem; next, by using the classical concept and properties of the genus, we construct a mini-max class of critical points.

• Variational Approach
• Fractional Laplacian
• Minimax Principle
• Genus

1. Introduction

In this paper, we focus our attention on the following problem:

{ ( − Δ ) s u = λ V ( x ) | u | p − 1 + β K ( x ) | u | 2 s * − 1 in   Ω u = 0   in   R n \ Ω (1.1)

where Ω is a bounded domain in R n , λ > 0 , 0 < s < 1 and n > 2 s , 1 < p < 2 s * , K ( x ) ∈ C ( R n ) ∩ L ∞ ( R n ) , V ( x ) ≥ 0 and V ( x ) ∈ C ( R n ) ∩ L q ( R n ) with q = 2 s * 2 s * − p here ( − Δ ) s denotes the fractional Laplace operator defined, up to a normalization factor, by

( − Δ ) s u ( x ) = ∫ R n u ( x ) − u ( y ) | x − y | n + 2 s d y , x ∈ R n . (1.2)

The aim of this paper is to study the existence of solutions, we will see that if 1 < p < 2 , then by concentration-compactness principle, together with mini-max arguments, we can prove the existence of solutions for (1.1). We now summarize the main result of the paper.

Theorem 1.1. Let 1 < p < 2 , K ( x ) ∈ C ( R n ) ∩ L ∞ ( R n ) and 0 ≤ V ( x ) ∈ C ( R n ) ∩ L q ( R n ) with q = 2 s * 2 s * − p . Moreover, V ( x ) > 0 is bounded on Ω . Then

1) For any λ > 0 , there exists β ˜ > 0 , then for any 0 < β < β ˜ , (1.1) has a consequence of weak solutions { u n } .

2) For any β > 0 , there exist λ ˜ > 0 , then for any 0 < λ < λ ˜ , (1.1) has a consequence of weak solutions { u n } .

We denote by H s ( R n ) the usual fractional Sobolev space endowed with the so-called Gagliardo norm

‖ u ‖ 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 ) 1 2 , (1.3)

Then we defined

X 0 s ( Ω ) = { u ∈ H s ( R n ) : u = 0 a . e . in   R n \ Ω } (1.4)

endowed with the norm

‖ u ‖ X 0 s ( Ω ) = ( ∫ R n × R n | u ( x ) − u ( y ) | 2 | x − y | n + 2 s d x d y ) 1 2 , (1.5)

we refer to [1] for a general definition of X 0 s ( Ω ) and its properties.

Observe that by [ [2] , Proposition 3.6] we have the following identity

‖ u ‖ X 0 s ( Ω ) = ‖ ( − Δ ) s 2 u ‖ L 2 ( R n ) . (1.6)

In this work, the Sobolev constant is given by (can be seen in [ [3] , theorem 7.58])

S ( n , s ) : = inf u ∈ H s ( R n ) \ { 0 } Q n , s ( u ) > 0 , (1.7)

where

Q n , s ( u ) : = ∫ R n × R n | u ( x ) − u ( y ) 2 | | x − y | n + 2 s d x d y ( ∫ R n × R n | u ( x ) | 2 s * d x ) 2 2 s * , u ∈ H s ( R n ) (1.8)

2. Statements of the Result

We will use a variational approach to find a solution of (1.1). Firstly, we will associate a suitable functional to our problem, the Euler-Lagrange functional related to problem (1) is given by J : X 0 s ( Ω ) → R defined as follow

J ( u n ) = 1 2 ‖ u n ‖ X 0 s ( Ω ) 2 − λ p ∫ Ω V ( x ) | u n | p d x − β 2 s * ∫ Ω K ( x ) | u n | 2 s * d x . (2.1)

To proof that J satisfy the Palais Smale condition at level c, we need the following lemma.

Lemma 2.1 [4] Letting ϕ be a regular function that satisfies that for some c ˜ > 0

| ϕ ( x ) | ≤ c ˜ 1 + | x | n + s , x ∈ R n (2.2)

and

| ∇ ϕ ( x ) | ≤ c ˜ 1 + | x | n + s , x ∈ R n (2.3)

Let B : X 0 s 2 ( Ω ) × X 0 s 2 ( Ω ) → R be a bilinear form defined by

B ( f , g ) ( x ) : = 2 ∫ R ( f ( x ) − f ( y ) ) ( g ( x ) − g ( y ) ) | x − y | n + s d y . (2.4)

then, for every s ∈ ( 0 , 1 ) , there exist positive constant c 1 and c 2 , such that for x ∈ R n , one has

| ( − Δ ) s 2 ϕ ( x ) | ≤ c 1 + | x | n + s and | B ( ϕ , ϕ ) ( x ) | ≤ c 1 + | x | n + s . (2.5)

To establish the next auxiliary result we consider a radial, nonincreasing cut-off function

ϕ ∈ C 0 ∞ ( R n ) and ϕ ε ( x ) : = ϕ ( x ε ) (2.6)

Lemma 2.2. [4] Letting { u m } be a uniformly bounded in X 0 s ( Ω ) and ϕ ε ∈ C 0 ∞ ( R n ) the function defined in (2.6). Then,

lim ε → 0 lim m → 0 | ∫ R n u m ( x ) ( − Δ ) s 2 ϕ ε ( x ) ( − Δ ) s 2 u m ( x ) d x | = 0. (2.7)

Lemma 2.3. [4] With the same assumptions of Lemma 2.8 we have that

lim ε → 0 lim m → 0 | ∫ R n ( − Δ ) s 2 u m ( x ) d x B ( u m , ϕ ε ) ( x ) | = 0. (2.8)

where B is defined in (2.4).

Lemma 2.4. [5] (Minimax principle) Assume that E ∈ C ( X , ℝ ) , and A is a family of nonempty subset of X, denote

c = inf A ∈ A sup x ∈ A E ( x ) (2.9)

If the following conditions holds:

1) c is a finite real number;

2) there exists an ε ¯ > 0 , such that A is invariant with respect to the family of mappings;

T = { T ∈ ( X , X ) | T ( x ) = x ,   if   E ( x ) < c − ε ¯ } , (2.10)

that is, for any T ∈ T , there holds

A ∈ A ⇒ T ( A ) ∈ A

Then, E possesses a ( P S ) c sequence at level c define as (6.1.1); Furthermore, if E satisfies the ( P S ) c condition (or the ( P S ) c condition at level c), then c is a critical value of E.

3. Proof of Theorem 1.1

Firstly, recalling that J is said to satisfy the Palais Smale condition at level c if any sequence { u n } ∈ X 0 s ( Ω ) such that J ( u n ) → c and J ′ ( u ) → 0 has a convergent subsequence.

Lemma 3.1. The ( P S ) c sequence { u n } for J is bounded.

Proof. Note that { u n } ⊂ X 0 s ( Ω ) satisfies

J ( u n ) = 1 2 ‖ u n ‖ X 0 s ( Ω ) 2 − λ p ∫ Ω V ( x ) | u n | p d x − β 2 s * ∫ Ω K ( x ) | u n | 2 s * d x = c + o n ( 1 ) (3.1)

and

〈 J ′ ( u n ) , ϕ 〉 = ∫ Ω ( − Δ ) s u n d x − λ ∫ Ω V ( x ) | u n | p − 2 u ϕ d x − β ∫ Ω K ( x ) | u n | 2 s * − 2 u ϕ d x = o n ( 1 ) ‖ ϕ ‖ X 0 s ( Ω ) , ∀ ϕ ∈ X 0 s ( Ω ) (3.2)

where o n ( 1 ) → 0 as n → ∞ . Choose ϕ = u n ∈ X 0 s ( Ω ) as test function in (3.2), we get that

o n ( 1 ) ‖ u n ‖ X 0 s ( Ω ) = 〈 J ′ ( u n ) , u n 〉 = ‖ u n ‖ X 0 s ( Ω ) 2 − λ ∫ Ω V ( x ) | u n | p d x − β ∫ Ω K ( x ) | u n | 2 s * d x = c + o n ( 1 ) . (3.3)

therefore, by (3.1) and (3.2), we have

c + o n ( 1 ) − 1 2 s * o n ( 1 ) ‖ u n ‖ X 0 s ( Ω ) = 1 2 ‖ u n ‖ X 0 s ( Ω ) 2 − λ p ∫ Ω V ( x ) | u n | p d x − β 2 s * ∫ Ω K ( x ) | u n | 2 s * d x         − 1 2 s * ‖ u n ‖ X 0 s ( Ω ) 2 − λ 2 s * ∫ Ω V ( x ) | u n | p d x − β 2 s * ∫ Ω K ( x ) | u n | 2 s * d x ≥ s n ‖ u n ‖ X 0 s ( Ω ) 2 − ( λ p − 1 2 s * ) ‖ V ( x ) ‖ L q ( Ω ) ‖ u n ‖ L 2 s * p ≥ s n ‖ u n ‖ X 0 s ( Ω ) 2 − ( λ p − 1 2 s * ) S ( n , s ) − p 2 ‖ V ( x ) ‖ L q ( Ω ) ‖ u n ‖ X 0 s ( Ω ) p . (3.4)

which yields the boundeness of { u n } in X 0 s ( Ω ) ,since 1 < p < 2 .

If K ( x ) ∈ L ∞ ( ℝ n ) , then for 2 < p < 2 s * , similar to the proof of 1 < p < 2 , we get

c + o n ( 1 ) + o n ( 1 ) ‖ u n ‖ X 0 s ( Ω ) ≥ ( p − 2 2 p ) ‖ u n ‖ X 0 s ( Ω ) 2 − ( p − 2 s * ) β 2 s * S − 2 s * 2 ‖ u n ‖ X 0 s ( Ω ) 2 s *

Which also yields the boundedness of ( P S ) c sequence { u n } .

Lemma 3.2. Assume that c < 0 . Then

1) For any λ > 0 , there exists β 0 > 0 , such that for any 0 < β < β 0 , then J satisfies ( P S ) c .

2) For any β > 0 there exists λ 0 > 0 such that for any 0 < λ < λ 0 , then J satisfies ( P S ) c .

Proof. By Lemma3.1 { u n } is bounded in X 0 s ( Ω ) , up to a subsequence, we get that

u n → u x ∈ X 0 s ( Ω ) .

u n → u x ∈ L r ( Ω ) , 1 ≤ r < 2 s * . (3.5)

u n → u a.e. x ∈ Ω .

Following [6] it is easy to prove that X 0 s ( Ω ) could also be the X 0 s ( Ω ) -norm. Applying [ [7] , Theorem1.5], we have that the exist an index. Set I ⊆ N a sequence of point { x k } x ∈ I ⊂ Ω and two sequences of nonnegative real numbers { μ k } k ∈ I , { v k } k ∈ I , such that

| ( − Δ ) s 2 u n | 2 → μ | ( − Δ ) s 2 u | 2 + ∑ k ∈ I μ k δ x k . (3.6)

moreover

| u n | 2 s * → μ | u | 2 s * + ∑ k ∈ I v k δ x k . (3.7)

in the sense of measures, with

v k ≤ S ( s , n ) − 2 s * 2 μ k 2 s * 2 for every k ∈ I (3.8)

here δ x k denotes the Dirac Delta at x k , while S ( n , s ) is the constant given in (1.7), we consider ϕ ∈ C 0 ∞ ( R n ) a nonincreasing cut-off function satisfying

ϕ = 1 in   B 1 ( x k 0 ) and ϕ = 0       in   B 2 ( x k 0 ) c (3.9)

Set ϕ ε ( x ) = ϕ ( x ε ) ,   x ∈ R n taking the derivative of (1.6), for any u , ϕ ∈ X 0 s ( Ω ) . We obtain that

∫ R n × R n ( u ( x ) − u ( y ) ) ( ϕ ( x ) − ϕ ( y ) ) | x − y | n + 2 s d x d y = ∫ R n ϕ ( x ) ( − Δ ) s u ( x ) d x (3.10)

Then, taking ϕ ε u n as a test function in J ′ ( u n ) → 0

lim n → 0 ∫ ℝ n ϕ ε u n ( − Δ ) u n d x − ( λ ∫ B 2 ε ( x k 0 ) V ( x ) u n p ϕ ε d x + β ∫ B 2 ε ( x k 0 ) K ( x ) u n 2 s * ϕ ε d x ) = 0 (3.11)

by (3.10), we have

lim n → ∞ ∫ ℝ n u n ( x ) ( − Δ ) s 2 u n ( x ) ( − Δ ) s 2 ϕ ε ( x ) d x   − 2 ∫ ℝ n ( − Δ ) s 2 u n ( x ) ∫ ℝ n ( ϕ ε ( x ) − ϕ ε ( y ) ) ( u n ( x ) − u n ( y ) ) | x − y | n + s d x d y = lim n → ∞ λ ∫ B 2 ε ( x k 0 ) V ( x ) | u n | p ( x ) ϕ ε ( x ) d x + β ∫ B 2 ε ( x k 0 ) K ( x ) | u n | 2 s * ( x ) ϕ ε ( x ) d x − ∫ B 2 ε ( x k 0 ) ( ( − Δ ) s 2 u n ) 2 ϕ ε ( x ) d x . (3.12)

therefore, by (3.5) (3.6) and (3.7) we get

lim ε → 0 lim n → ∞ ∫ ℝ n u n ( x ) ( − Δ ) s 2 u n ( x ) ( − Δ ) s 2 ϕ ε ( x ) d x   − 2 ∫ ℝ n ( − Δ ) s 2 u n ( x ) ∫ ℝ n ( ϕ ε ( x ) − ϕ ε ( y ) ) ( u n ( x ) − u n ( y ) ) | x − y | n + s d x d y = lim ε → 0 λ ∫ B 2 ε ( x k 0 ) V ( x ) | u n | p ( x ) ϕ ε ( x ) d x + ∫ B 2 ε ( x k 0 ) ϕ ε ( x ) d v − β ∫ B 2 ε ( x k 0 ) K ( x ) ϕ ε ( x ) d μ . (3.13)

Since ϕ is regular function with compact support, it is easy to see that it satisfies the hypothesis of Lemma 2.1, by Lemma 2.2 and Lemma 2.3 applied to the sequence { u n } , it follows that the left hand side of (3.13) goes to zero. We obtain that

lim ε → 0 ( λ ∫ B 2 ε ( x k 0 ) V ( x ) | u n | p ( x ) ϕ ε ( x ) d x + ∫ B 2 ε ( x k 0 ) ϕ ε ( x ) d v − β ∫ B 2 ε ( x k 0 ) K ( x ) ϕ ε ( x ) ) d μ = β K ( x k 0 ) v k 0 − μ k 0 = 0. (3.14)

Clearly, if K ( x ) ≤ 0 , we get μ k 0 = v k 0 = 0 ; if K ( x k 0 ) > 0 , by (3.8), we get v k 0 = 0 or v k 0 ≥ [ S ( n , s ) β K ( x k 0 ) ] n 2 s .

suppose that v k 0 ≠ 0 , we know that

0 > c = lim n → ∞ [ J ( u n ) − 1 2 s * 〈 J ′ ( u n ) , u n 〉 ] (3.15)

according to the embedded theorem, we have

0 > c ≥ ( 1 2 − 1 2 s * ) ‖ u n ‖ X 0 s ( Ω ) 2 − ( λ p − λ 2 s * ) ∫ Ω V ( x ) | u n | p d x = s n ‖ u n ‖ X 0 s ( Ω ) 2 − ( λ p − λ 2 s * ) ∫ Ω V ( x ) | u n | p d x ≥ s n S − 1 ( n , s ) ‖ u n ‖ L 2 s * ( Ω ) 2 − ( λ p − λ 2 s * ) S − p 2 ( n , s ) ‖ V ( x ) ‖ L q ( Ω ) ‖ u n ‖ L 2 s * p . (3.16)

This yields that

‖ u ‖ L 2 s * ( Ω ) ≤ C λ 1 2 − p . (3.17)

Thus, if v k 0 ≥ [ S ( n , s ) β K ( x k 0 ) ] n 2 s , we get that

0 > c = lim n → ∞ [ J ( u n ) − 1 2 s * 〈 J ′ ( u n ) , u n 〉 ]                 ≥ ( 1 2 − 1 2 s * ) ‖ u ‖ X 0 s ( Ω ) 2 + s n μ k 0 − ( λ p − λ 2 s * ) ∫ Ω V ( x ) | u | p d x                 ≥ s n S − 1 ( n , s ) ‖ u ‖ L 2 s * ( Ω ) 2 + s n μ k 0 − ( λ p − λ 2 s * ) S − p 2 ( n , s ) ‖ V ( x ) ‖ L q ( Ω ) ‖ u ‖ L 2 s * p                 ≥ s n S − 1 ( n , s ) ‖ u ‖ L 2 s * ( Ω ) 2 + s n μ k 0 − ( λ p − λ 2 s * ) S − p 2 ( n , s ) ‖ V ( x ) ‖ L q ( Ω ) ‖ u ‖ L 2 s * p                 ≥ s n S ( n , s ) v k 0 − 2 s * 2 − ( λ p − λ 2 s * ) S − p 2 ( n , s ) ‖ V ( x ) ‖ L p ( Ω ) ‖ u ‖ L 2 s * p                 ≥ s n S n 2 s ( n , s ) [ β K ( x k 0 ) ] 2 s − n 2 s − C λ 2 2 − p . (3.18)

However, if β > 0 is given, we can choose λ 0 > 0 so small for every 0 < λ < λ 0 that last term on the right-hand side above is greater than 0 which is contradiction when 2 < p < 2 s *

0 > c = lim n → ∞ [ J ( u n ) − 1 p 〈 J ′ ( u n ) , u n 〉 ] = ( 1 2 − 1 p ) ‖ u ‖ X 0 s ( Ω ) 2 − ( β 2 s * − β p ) ∫ Ω K ( x ) | u | 2 s * d x ≥ ( 1 2 − 1 p ) ‖ u ‖ X 0 s ( Ω ) 2 − ( β 2 s * − β p ) ∫ Ω ∩ { K ( x ) < 0 } K ( x ) | u | 2 s * d x ≥ ( 1 2 − 1 p ) ‖ u ‖ X 0 s ( Ω ) 2 − ( β 2 s * − β p ) ‖ K ( x ) ‖ L ∞ ‖ u ‖ L 2 s * 2 s *

β is the same as λ greater than 0. We see that v k 0 ≥ [ S ( n , s ) β K ( x k 0 ) ] n 2 s cannot occur if λ 0 or β 0 are choose properly. Thus μ k = v k = 0 . As consequence, we obtain that ( u n ) + − u → 0 in L 2 s * ( Ω ) , that is lim n → ∞ ∫ R n | ( u n ) + | 2 s * d x = ∫ R n | u | 2 s * d x . This implies convergence of λ V ( x ) | u n | p − 1 + β K ( x ) | u n | 2 s * − 1 in L 2 s * ( Ω ) . Finally using the continuity of the inverse operator ( − Δ ) s . We obtain strong convergence of u n in X 0 s ( Ω ) . #

Next, by using the classical concept and properties of the genus, we construct a min-max class of the critical point.

For a Banach space X, We define the set

A = { A ⊂ X \ { 0 } : A is closed in X and symmetric with respect to the orign }

For A ∈ A , define

γ ( A ) : = inf { m ∈ N , ∃ ϕ ∈ C ( A , R m \ { 0 } ) , ϕ ( x ) = − ϕ ( − x ) } (3.19)

If there is no mapping ϕ as above for any m ∈ N , there γ ( A ) = + ∞ . we refer to [8] for the properties of the genus.

Proposition 3.3. [8] Let A , B ⊂ Α ,

1) If there exists an odd map f ∈ C ( A , B ) , then γ ( A ) ≤ γ ( B ) ;

2) If A ⊂ B , then γ ( A ) ≤ γ ( B ) ;

3) γ ( A ∪ B ) ≤ γ ( A ) + γ ( B ) ;

4) If S is a sphere centered at the origin in R m , then γ ( s ) = m ;

5) If A is compact, there exists a symmetric Neighborhood N of A, such that γ ( N ¯ ) = γ ( A ) .

According Holder inequality, we get that

J ( u ) = 1 2 ‖ u ‖ X 0 s 2 − λ p ∫ Ω V ( x ) | u | p d x − β 2 s * ∫ Ω K ( x ) | u | 2 s * d x ≥ 1 2 ‖ u ‖ X 0 s 2 − C 1 λ ‖ u ‖ X 0 s p − C 2 β ‖ u ‖ X 0 s 2 s * (3.20)

We define the function

Q ( t ) : = 1 2 t 2 − C 1 λ t p − C 2 β t 2 s * (3.21)

Then it is easy to see that given β > 0 , there exists λ 1 > 0 so small that for every 0 < λ < λ 1 , there exists 0 < T 0 < T 1 such that Q ( t ) < 0 for 0 ≤ t ≤ T 0 , Q ( t ) > 0 for T 0 < t < T 1 . and Q ( t ) < 0 t > T 1 . Analogously, for given λ > 0 , we can choose β 1 > 0 with the property that T 0 , T 1 as above for each 0 < β < β 1 . Clearly, Q ( T 0 ) = Q ( T 1 ) = 0 .

As in [9] , Let τ : ℝ + → [ 0 , 1 ] be a nonincreasing C ∞ function such that τ ( t ) = 1 if 0 ≤ τ ≤ T 0 and τ ( t ) = 0 . if τ ≥ T 0 . Set Ψ ( u ) = τ ( ‖ u ‖ X 0 s ( Ω ) ) , we make the following truncation of the function J:

J ˜ ( u ) = 1 2 ‖ u ‖ X 0 s 2 − λ p ∫ Ω V ( x ) | u | p d x − β 2 s * ψ ( u ) ∫ Ω K ( x ) | u | 2 s * d x (3.22)

then

J ˜ ( u ) ≥ Q ˜ ‖ u ‖ X 0 s ( Ω ) . (3.23)

where Q ˜ ( t ) : = 1 2 t 2 − C 1 λ t p − C 2 β t 2 s * ψ ( t ) .

It is clear that J ˜ ( u ) ∈ C 1 and is bounded from below.

Lemma 3.4. [10] 1) For any λ > 0 and 0 < β < β 1 or any β > 0 and 0 < λ < λ 1 , if J ˜ ( u ) < 0 , then ‖ u ‖ X 0 s ( Ω ) < T 0 and J ˜ ( u ) = J ( u ) .

2) For any λ > 0 , there exists such that if 0 < β < β ¯ and c < 0 then J ˜ satisfies ( P S ) c .

3) For any β > 0 ,there exists λ ˜ > 0 ( λ ˜ ≤ λ 1 ) such that if 0 < λ < λ ˜ and c < 0 then J ˜ satisfies ( P S ) c .

Lemma 3.5. Denote J ˜ α : = { u ∈ X 0 s ( Ω ) , J ˜ ( u ) ≤ α } . Then for any m ∈ N , there is ε m < 0 such that γ ( J ˜ ε m ) ≥ m .

Proof. Denote by X 0 s ( Ω ) the closure of C 0 ∞ ( Ω ) with the respect to norm ‖ u ‖ X 0 s ( Ω ) = ( ∫ Ω | u ( x ) − u ( y ) | 2 | x − y | n + 2 s d x d y ) 1 2 , V ( x ) > 0 in Ω . Extending functions in

X 0 s ( Ω ) by 0 outside Ω . Let X m be a m-dimensional subspace of X 0 s ( Ω ) . For any u ∈ X m , u ≠ 0 . We write u = r m w with w ∈ X m and ‖ w ‖ X 0 s ( Ω ) = 1 . From the assumptions of V ( x ) , it is easy to see for every w ∈ X m with ‖ w ‖ X 0 s ( Ω ) = 1 that there exists d m > 0 such that

∫ Ω V ( x ) | w | p d x ≥ d m (3.24)

For 0 < r m < T 0 . Since all the norms are equivalent, we get

J ˜ ( u ) = J ( u ) = 1 2 ‖ u ‖ X 0 s ( Ω ) 2 − λ p ∫ Ω V ( x ) | u | p d x − β 2 s * ∫ Ω K ( x ) | u | 2 s * d x       ≤ 1 2 ‖ u ‖ X 0 s ( Ω ) 2 − λ p ∫ Ω V ( x ) | u | p d x + β 2 s * | ∫ Ω K ( x ) | u | 2 s * d x |   ≤ 1 2 r m 2 − λ c d m + c β r m 2 s * : = ε m .

Therefore for given λ and β . we can choose r m ∈ ( 0 , T 0 ) sufficiently small so that J ˜ ( u ) ≤ ε m < 0 .#

Let S r m = { u ∈ X 0 s ( Ω ) : ‖ u ‖ X 0 s ( Ω ) = r m } . Then S r m ∩ X m ⊂ J ˜ ε m , Hence by proposition 3.3 (2) and (4) r ( J ˜ ε m ) ≥ r ( S r m ∩ X m ) ≥ m .

We denote Γ m = { A ∈ Α : γ ( A ) ≥ m } and let

C m : = inf A ∈ Γ m sup u ∈ A J ( u ) (3.25)

then

− ∞ < C m ≤ ε m < 0 , m ∈ N (3.26)

because J ˜ ε m ∈ Γ m and J ˜ is bounded from below.

Proposition 3.6. Let λ , β be as in Lemma 3.5 (2) and (3). Then all c m given by (3.25) are critical values of J ˜ and c m → 0 as m → 0 .

Proof. Denote K ε = { u ∈ X 0 s ( Ω ) : J ˜ ( u ) = c , J ˜ ′ ( u ) = 0 } . Then by Lemma 3.4 (2) and (3), if c < 0 , K c is compact. It is clear that C m ≤ C m + 1 . By (3.26) C m < 0 . Hence C m → C ¯ ≤ 0 . Moreover, since ( P S ) c satisfied, it follows from a standard argument (see [11] ) that all C m are critical values of J ˜ . Now, we claim that c ¯ = 0 . If c ¯ < 0 because K c ¯ is compact and K c ¯ ∈ A , it follows from Proposition 3.3 (5) that γ ( K c ¯ ) = m 0 < + ∞ and there exists δ > 0 such that γ ( K c ¯ ) = γ ( N δ ( K c ¯ ) ) = m 0 . By the deformation Lemma [9] , there exists ε > 0 ( c ¯ + ε < 0 ) and an odd homeomorphism ς ( ⋅ ) : X 0 s ( Ω ) → X 0 s ( Ω ) such that

ς ( J ˜ c ¯ + ε \ N δ ( K c ¯ ) ) ⊂ J ˜ c ¯ − ε (3.27)

Since c m is increasing anad converges to c ¯ . there exists m ∈ N such that

c m > c ¯ − ε . (3.28)

And exists a A ∈ Γ m + m 0 such that

sup u ∈ A J ˜ ( u ) < c ¯ + ε (3.29)

By Proposition 3.3 (3), we obtain

γ ( A \ N δ ( K c ¯ ) ¯ ) ≥ γ ( A ) − γ ( N δ ( K c ¯ ) ) ≥ m (3.30)

By Proposition 3.3 (1), we obtain

γ ( ς ( A \ N δ ( K c ¯ ) ) ¯ ) ≥ m (3.31)

therefore

ς ( A \ N δ ( K c ¯ ) ) ∈ Γ m

consequently, from (3.28), we get

sup u ∈ ς ( A \ N δ ( K c ¯ ) ) J ˜ ( u ) ≥ c m > c ¯ − ε (3.32)

on the other hand, by (3.27) and (3.29)

ς ( A \ N δ ( K c ¯ ) ) ⊂ ς ( J ˜ c ¯ + ε \ N δ ( K c ¯ ) ) ⊂ J ˜ c ¯ − ε (3.33)

which implies that

sup u ∈ ς ( A \ N δ ( K c ¯ ) ) J ˜ ( u ) ≤ c ¯ − ε (3.34)

this contradicts to (3.32).Hence c m → 0 . #

By (1) of Lemma 3.4 J ˜ ( u ) = J ( u ) if J ˜ ( u ) < 0 . This and Proposition 3.6 give Theorem1.1.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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