Journal of Applied Mathematics and Physics
Vol.07 No.01(2019), Article ID:90303,11 pages
10.4236/jamp.2019.71020

The Existence of Solution of a Critical Fractional Equation

Hui Chen

College of Science, University of Shanghai for Science and Technology, Shanghai, China

Copyright © 2019 by author(s) and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: January 4, 2019; Accepted: January 28, 2019; Published: January 31, 2019

ABSTRACT

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.

Keywords:

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 , ) , 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 ( 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 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 n u n ( x ) ( Δ ) s 2 u n ( x ) ( Δ ) s 2 ϕ ε ( x ) d x 2 n ( Δ ) s 2 u n ( x ) 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 n u n ( x ) ( Δ ) s 2 u n ( x ) ( Δ ) s 2 ϕ ε ( x ) d x 2 n ( Δ ) s 2 u n ( x ) 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 : + [ 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.

Cite this paper

Chen, H. (2019) The Existence of Solution of a Critical Fractional Equation. Journal of Applied Mathematics and Physics, 7, 243-253. https://doi.org/10.4236/jamp.2019.71020

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