The Brézis-Nirenberg Result for the Fractional Elliptic Problem with Singular Potential

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1 arxiv: v1 [math.ap] 23 May 2017 The Brézis-Nirenberg Result for the Fractional Elliptic Problem with Singular Potential Lingyu Jin, Lang Li and Shaomei Fang Department of Mathematics, South China Agricultural University, Guangzhou , P. R. China Abstract In this paper, we are concerned with the following type of fractional problems: ( s u µ u x 2s λu = u 2 s 2 u+f(x,u, in, ( u = 0 in R N \ where s (0,1, 2 s = 2N/(N 2s is the critical Sobolev exponent, f(x,u is a lower order perturbation of critical Sobolev nonlinearity. We obtain the existence of the solution for (* through variational methods. In particular we derive a Brézis- Nirenberg type result when f(x,u = 0. Key words and phrases. Fractional Laplacian, positive solution, critical Sobolev nonlinearity. AMS Classification: 35J10 35J20 35J60 1 Introduction Problems involving critical Sobolev nonlinearity have been an interesting topic for a long time. In the celebrated paper [2], Brézis and Nirenberg investigated the problem { u λu = u 2 2 u, in, (1.1 u = 0 on The research was supported by the Natural Science Foundation of China ( , and the China Scholarship Council (

2 where R N is a bounded domain with smooth boundary, N 3, 2 = 2N. They proved N 2 the following existence and nonexistence results. Let λ 1 is the first eigenvalue of with homogeneous Dirichlet boundary conditions. Then, A if N 4, problem (1.1 has a positive solution for 0 < λ < λ 1 ; B if N = 3, there exists a constant λ (0,λ 1 such that problem (1.1 has a positive solution for λ (λ,λ 1 ; C If N = 3 and is a ball, then λ = 1 4 λ 1, and problem (1.1 has no solution for λ (0,λ. Also, Jannelli in [11] obtained the Brézis-Nirenberg existence and nonexistence results for a class of problem with the critical Sobolev nonlinerity and the Hardy term as follows u µ u x λu = 2 u in, 2 u 2 (1.2 u = 0 on. For more results on (1.2, please see [4, 5, 6, 7, 11, 12] and the references therein. Recently Servadei and Valdinoci [14] extended the results of [2](part A to the following nonlocal fractional equation { ( s u λu = u 2 s 2 u, in, (1.3 u = 0 in R N \ where s (0,1, ( s is the fractional operator, which may be defined ( s u(x u(y u = 2c N,s P.V. R x y N+2sdy, x RN (1.4 N with c N,s = 2 2s 1 π N 2 Γ(N+2s 2 Γ( s (see [3]. Motivated by the above papers, in this paper we consider the following nonlocal fractional equation with singular potential ( s u µ u x λu = 2 2s u 2 s u+f(x,u, in, (1.5 u = 0 in R N \ where 0 R N is a bounded domain with smooth boundary, N 3, 0 < s < 1, ( s is the fractional operator defined in (1.4, 2 = 2N/(N 2s is the critical Sobolev exponent, and f(x, u is a lower order perturbation of critical Sobolev nonlinearity. Problem (1.5 can be considered as doubly critical due to the critical power u 2 s 2 in the semilinear term 2

3 u and the spectral anomaly of the Hardy potential x 2s. The fractional framework introduces nontrivial difficulties which have interest in themselves. In this paper we investigate the Brézis-Nirenberg type result for (1.5. Firstly we prove the existence of solutions for (1.5 through variational methods under some natural assumptions on f(x, u. Secondly we derive a Brézis-Nirenberg type result for the case f(x,u = 0. For other existence results involving (1.5, you can refer to [8, 9, 16]. is where Notice that equation (1.5 has a variational structure. The variational functional of (1.5 J λ (u = 1 (c R2N u(x u(y 2 N,s dxdy λ 2 x y N+2s 1 u(x 2 2 s dx F(x,udx, s F(x,t = t 0 u(x 2 u(x 2 dx µ x dx 2s (1.6 f(x,τdτ. (1.7 We say u(x is a weak solution of (1.5 if u(x satisfy that (u(x u(y(φ(x φ(y u(xφ(x c N,s dxdy µ dx λ u(xφ(xdx R x y 2N N+2s x 2s = u(x 2 s 2 u(xφ(xdx+ f(x,uφdx, φ H s (R N with φ = 0, a.e. in R N \, u H s (R N with u = 0 a.e. in R N \. As a result of Hardy inequality [8], it is easy to see that (1.8 ( u(x 2 RN Γ N,s R x dx u(x u(y 2 c N 2s N,s dy, u C x y N+2s 0 (RN (1.9 where Γ N,s = 2 2sΓ2 ( N+2s 4 Γ 2 ( N 2s,c N,s = 2 2s 1 π N Γ( N+2s 2 2 Γ( s, ( for 0 < µ < Γ N,s, we can define the first eigenvalue λ 1,µ of the following problem ( s u µ u = λu in, x 2s (1.11 u = 0 in R N \ as in (2.7. In this paper, we consider problem (1.5 in the case 0 < λ < λ 1,µ. As in the classical case of Laplacian, the main difficulty for nonlocal elliptic problems with critical nonlinearity is the lack of compactness. This would cause the functional J λ not to satisfy the Palais-Smale condition. Nevertheless, we are able to use the Mountain-Pass Theorem 3

4 without the Palais-Smale condition (as given in [2], Theorem 2.2 to overcome the lack compactness of Sobolev embedding. Before introducing our main results, we give some notations and assumptions. Notations and assumptions: Denote c and C as arbitrary constants. Let B r (x denote a ball centered at x with radius r, B r denote a ball centered at 0 with radius r and B C r = RN \B r. Let o n (1 be an infinitely small quantity such that o n (1 0 as n. Let f(x,u in (1.5 be a Carathéodary function f : R R satisfying the following conditions: f(x,u is continuous with respect to u; (1.12 f(x,u lim = 0 uniformly in x ; u 0 u (1.13 f(x,u lim = 0 uniformly in x. u u 2 s 1 (1.14 (1.12-(1.14 ensure that f(x, u is a lower order perturbation of the critical nonlinearity. From Lemma 5 and Lemma 6 in [14], for any ε > 0 there exist constants M(ε > 0,δ(ε > 0 such that for any u R,x, f(x,u ε u 2 1 +M(ε; f(x,u δ(ε u 2 1 +ε u. (1.15 We denote by H s (R N the usual fractional Sobolev space endowed with the norm u H s (R N = u L 2 (R N +(c N,s R 2N u(x u(y 2 x y N+2s dxdy 1/2. (1.16 Let X 0 be the functional space defined as with the norm u Xµ = X 0 = {u H s (R N, u = 0 a.e. in R N \} (1.17 (c R2N u(x u(y 2 u(x 2 1/2 N,s dxdy µ x y N+2s x dx, 2s which is equivalent to the norm H s (R N for 0 < µ < Γ N,s (see in Lemma 2.1. Define u(x u(y c 2 N,s dxdy µ u(x 2 dx R S = inf 2N x y N+2s R N x 2s u H s (R N \{0} (, (1.18 u(x R N 2 s dx 2/2 s the Euler equation associated to (1.18 is as follows ( s u µ u x 2s = u 2 s 2 u in R N. (1.19 4

5 In particular it has been showed in Theorem 1.2 of [8] that for any solution u(x H s (R N of (1.19, there exist two positive constants c,c such that c ( N 2s x 1 ηµ (1+ x 2ηµ 2 u(x C (, in R N \{0} (1.20 N 2s x 1 ηµ (1+ x 2ηµ 2 where η µ = 1 2α µ N 2s, (1.21 and α µ (0, N 2s 2 is a suitable parameter whose explicit value will be determined as the unique solution to the following equation ϕ s,n (α µ = 2 2sΓ(αµ+2s Γ( N αµ 2 2 Γ( N αµ 2s Γ( αµ = µ ( and ϕ is strictly increasing. Obviously any achieve function U(x of S also satisfies (1.20. Define the constant u(x u(y c 2 N,s dxdy µ u(x 2 dx λ R S λ = inf 2N x y N+2s x 2s u(x 2 dx u X 0 \{0} ( sdx 2 u(x 2 2 s (1.23 Obviously, S λ S because λ > 0. Just as in [2] and [11], we will actually focus on the case when the strict inequality occurs. The first results of the present paper is the following one: Theorem 1.1. Let s (0,1,N > 2s, be an open bounded set of R N, λ 1,µ is the first eigenvalue of (1.11, 0 < µ < Γ N,s (Γ N,s is defined in (1.10 and 0 < λ < λ 1,µ. Let f be a Carathéodory function verifying (1.12-(1.14, assume that there exists u 0 H s (R N \{0} with u 0 0 a.e. in such that supj λ (tu 0 < s t 0 N S N 2s, (1.24 problem (1.5 admits a solution u H s (R N, which is not identically zero, such that u = 0 a.e. in R N \. If assumption (1.24 is satisfied, then it follows Theorem 1.2. Theorem 1.2. Let s (0,1,N > 2s, be an open bounded set of R N, λ 1,µ is the first eigenvalue of (1.11, and µ ϕ 1 ( N 4s (ϕ is defined in (1.22. For any 0 < λ < λ 2 1,µ and f(x,u 0, problem (1.5 admits a solution u H s (R N, which is not identically zero, such that u = 0 a.e. in R N \. 5

6 Recall the results in [11] as the follows. Let λ 1 denote the first eigenvalue of the operator 1 x 2 in with Dirichlet boundary conditions and µ = ( N a If µ µ 1, problem (1.2 has a positive solution for 0 < λ < λ 1 ; b if µ 1 < µ < µ, there exists a constant λ (0, λ 1 such that problem (1.2 has a positive solution for λ (λ, λ 1 ; c If µ 1 < µ < µ and is a ball, then problem (1.2 has no solutions for λ λ. As for the results A, B, C for problem (1.3, the space dimension N plays a fundamental role when one seeks solutions of (1.3. In particular in [11] Jannelli point out the N = 3 is a critical dimension for (1.3, while for (1.2 it is only a matter of how µ close to µ. Theorem 1.2 in our paper is an extension of the results of [11] (part a. Here ϕ 1 ( N 4s = µ 1 when 2 s = 0. This paper is organized as follows. In Section 2, we give some preliminary lemmas. In Section 3 we prove the main results Theorem 1.1 and Theorem 1.2 by variational methods. In fact, we first prove Theorem 1.1 by Mountain-Pass Theorem. Then we reduce Theorem 1.2 to Theorem 1.1 by choosing a test function to verify the assumption of Theorem Preliminary lemmas In this section we give some preliminary lemmas. Lemma 2.1. Assume 0 < µ < Γ N,s. Then there exist two positive constants C and c such that for all u X 0 R2N C u 2 H s (R N c u(x u(y 2 u(x 2 N,s dxdy µ x y N+2s x dx 2s c u 2 H s (R N. (2.1 that is u Xµ = (c R2N u(x u(y 2 u(x 2 1/2 N,s dxdy µ x y N+2s x dx 2s is a norm on X 0 equivalent to the usual one defined in (1.16. (2.2 Proof. On one hand, from (1.9, for all 0 < µ < Γ N,s, u 2 H s (R N c N,s R2N u(x u(y 2 dxdy µ x y N+2s u(x 2 dx. (2.3 x 2s 6

7 On the other hand, c N,s u(x u(y 2 u(x 2 dxdy µ R x y 2N N+2s x R2N u(x u(y 2 (1 µ Γ N,s c N,s 2s dx x y N+2s dxdy Cc N,s R2N u(x u(y 2 x y N+2s dxdy (2.4 and ( u(x 2 dx (2 s 2/2 s u 2 s dx 2 u(x u(y 2 2 s c dxdy (2.5 R x y 2N N+2s then from (2.4 and (2.5 u H s (R N c ( c N,s u(x u(y 2 dxdy µ R x y 2N N+2s Collecting with (2.3 and (2.6, the proof is complete. u(x 2 x 2s dx 1/2. (2.6 Lemma 2.2. Let s (0,1,N > 2s, be an open bounded set of R N and 0 < µ < Γ N,s. Then problem (1.11 admits an eigenvalue λ 1,µ which is positive and that can be characterized as c N,s λ 1,µ = min u X 0 \{0} R 2N u(x u(y 2 dxdy µ x y N+2s u(x 2 dx u(x 2 x 2s dx Proof. It is remain to show that λ 1,µ can be attained and λ 1,µ > 0. Denote I(u = 1 2 Obviously (2.7 is equivalent to M = {u X 0, u L 2 ( = 1}, ( R2N u(x u(y 2 u(x 2 cn,s dxdy µ x y N+2s x dx = 1 2s 2 u 2 X µ. λ 1,µ = min u M (c N,s R2N u(x u(y 2 dxdy µ x y N+2s Let us take a minizing sequence u n M such that > 0. (2.7 u(x 2 dx > 0. (2.8 x 2s I(u n inf I(u 0 as n. (2.9 u M From Lemma 2.1, u n H s (R N is bounded. Then there exists u 0 such that (up to a subsequence, still denoted by u n u n u 0 in H s (R N,u n u 0 in L 2 loc(r N, and u n u 0 a.e. in R N as n. (2.10 7

8 So u 0 L 2 ( = 1 and u 0 = 0 a.e. in R N \, that is u 0 M. Denote v n = u n u 0, we have v n 0 in H s (R N,v n 0 in L 2 loc(r N, and v n 0 a.e. in R N as n, (2.11 then by Brézis-Lieb Lemma in [1], it follows R2N u n (x u n (y 2 R2N v n (x v n (y 2 R2N u 0 (x u 0 (y 2 dxdy = dxdy+ dxdy+o x y N+2s x y N+2s x y N+2s n (1, (2.12 u n (x 2 v n (x 2 u 0 (x 2 dx = dx+ dx+o x 2s x 2s x 2s n (1, (2.13 which implies that I(u n = 1 2 ( R2N u n (x u n (y 2 u n (x 2 cn,s dxdy µ dx x y N+2s x 2s = 1 2 v n 2 X µ u 0 2 X µ +o n (1 inf u M I(u(1+ v n 2 L 2 ( where the last inequality follows from (2.7 and (2.8. Then ( inf I(u = lim u M n 2 v n Xµ u 0 2 X µ inf I(u. (2.15 u M Thus inf I(u = I(u 0, (2.16 u M that is, u 0 is a minimizer of I(u. Since u 0 L 2 ( = 1, it is easy to obtain that λ 1,µ = I(u 0 > 0. The proof is complete. 3 The proof of main results In this section, we study the critical points of J λ which are solutions of problem (1.5. In order to find these critical points, we will apply a variant of Mountain-Pass Theorem without the Palais-Smale condition (refer [2]. Due to the lack of compactness in the embedding X 0 L 2 s (, the functional Jλ does not verify the Palais-Smale condition globally, but only in the energy range determined by the best fractional critical Sobolev constant S given in (1.18. First of all, we prove the functional J λ has the Mountain-Pass geometric structure. Proposition 3.1. Let λ (0,λ 1,µ and f be a Carathédory function satisfying conditions (1.12-(1.14. Then there exists ρ > 0, β > 0, e X 0 such that 1 for any u X 0 with u Xµ = ρ, J λ (u β; 2 e 0 a.e. in R N, e Xµ > ρ and J λ (e < β. 8

9 Proof. For all u X 0, by (2.7 and (1.15 we get for ε > 0 J λ 1 (c R2N u(x u(y 2 N,s dxdy λ u(x 2 dx µ 2 x y N+2s 1 u(x 2 2 s dx ε u 2 dx δ(ε u(x 2 s dx s 1 2 (1 λ (c R2N u(x u(y 2 N,s dxdy µ λ 1,µ x y N+2s ε 2 s 2 2 s ( u 2 s dx 2 s /2 (δ(ε+ 1 u(x 2s 2 s dx ( 1 2 (1 λ λ 1,µ ε 2 u(x 2 x 2s dx s 2 2 s S u 2 X µ (δ(ε+ 1 2 ss 2 s /2 u 2 s /2 Choose ε > 0 small enough, there exist α > 0 and ν > 0 such that u(x 2 x 2s dx X µ. J λ (u α u 2 X µ ν u 2 s X µ. (3.1 Now let u X 0, u Xµ = ρ > 0, since 2 s > 2, choose ρ small enough, then there exists β > 0 such that inf J λ β > 0. (3.2 u X 0, u Xµ =ρ Fix any u 0 X 0 with u 0 L 2 s( > 0. Without loss of generality, we can assume u 0 0 a.e. in R N (otherwise, we can replace any u 0 X 0 with its positive part, which belong to X 0 too, thanks to Lemma 5.2 in [13]. Since for t > 0 large enough it follows that Take J λ (u 0 ct 2 u 0 2 X µ t2 s F(x,tu 0 t2 s u 0 2 s, ( s 22 s u 0 2 s L 2 s as t. (3.4 e = t 0 u 0 (3.5 and t 0 sufficiently large, we have J(t 0 u 0 < 0 < β. The proof is complete. As a consequence of Proposition 3.1, J λ (u satisfies the geometry structure of Mountain- Pass Theorem. Set c =: inf sup J λ (γ(t, (3.6 γ Γ t [0,1] where Γ = {γ C([0,1],X 0 : γ(0 = 0,γ(1 = e X 0 }. Proof of Theorem 1.1 for all γ Γ, the function t γ(t Xµ is continuous, then there exists t (0,t such that γ( t Xµ = ρ. It follows that sup J λ (γ(t J λ (γ( t inf J λ (u β, (3.7 t [0,1] u X 0, u Xµ =ρ 9

10 then c β. Since te Γ, we have that 0 < β c = inf sup γ Γ t [0,1] sup J λ (te t [0,1] J λ (γ(t supj λ (ζu 0 < s ζ 0 N S N 2s. (3.8 By Theorem 2.2 in [2] and Proposition 3.1, there exists a Palais-Smale sequence {u n } X 0 a.e., J λ (u n c, J λ (u n 0. (3.9 From (1.15 and the definition of J λ, it follows which implies u n 2 s L 2 s( c +1 J λ (u n 1 2 < J λ(u n,u n > = ( s u n 2 s + (1 L 2 s( 2 f(x,u nu n F(x,u n dx ( s u n 2 s ε u L 2 s( n 2 s dx c u n dx c u n 2 s L 2 s( 2ε u n 2 s L 2 s( C is bounded for ε > 0 small enough. So c +1 J λ (u n 1 2 (1 λ u n 2 X λ µ 1 u 1,µ 2 n 2 s F(x,u L 2 s( n dx s 1 2 (1 λ u n 2 X λ µ 1 u 1,µ 2 n 2 s ε u L 2 s( n 2 L 2 ( C u n 2 s L 2 s( s 1 λ+2ε (1 u n 2 X 2 λ µ c. 1,µ (3.10 (3.11 Thus u n Xµ is bounded for ε > 0 small enough. Next we claim that there exists a solution u 0 X 0 of (1.5. Since u n is bounded in X 0, by Lemma 2.1, up to a subsequence, still denoted by u n, as n u n u 0 in H s (R N, (3.12 u n u 0 in L p loc (, for all p [1,2 s, (3.13 u n u 0 a.e. in R N. (3.14 Thus u 0 = 0 in R N \, u 0 X 0 and for all φ C0 (, as n (u n (x u n (y(φ(x φ(y (u 0 (x u 0 (y(φ(x φ(y c N,s dxdy c R x y 2N N+2s N,s dxdy, R x y 2N N+2s 10 (3.15

11 u n (xφ(x dx x 2s u n (xφ(xdx u 0 (xφ(x x 2s dx, (3.16 u 0 (xφ(xdx. (3.17 Denote g(u = u 2 s 2 u. From (3.12-(3.14, (1.15 and f(x,u,g(u is continuous in u, applying Lemma A.2 in [15], f(x,u n f(x,u 0 L 1 ( c u n u L q ( 0, g(u n g(u 0 L 1 ( c u n u L q ( 0, (3.18 where q = 2 s 1. Then it implies for all φ C 0 (, as n ( u n 2 s 2 u n u 0 2 s 2 u 0 φdx c g(u n g(u 0 L 1 ( 0, (3.19 (f(x,u n f(x,u 0 φdx c f(x,u n f(x,u 0 L 1 ( 0. (3.20 Thus u 0 X 0 is a solution of (1.5. To complete the proof of Theorem 1.1, it suffices to show that u 0 0. Suppose, in the contrary, u 0 0. From (3.12-(3.14 and (1.15 lim sup n f(x,u n u n dx εlimsup n u n 2 s dx+climsup n u n dx cε (3.21 where ε > 0 is an arbitrary constant. Letting ε 0, it follows that lim sup n f(x,u n u n dx 0. (3.22 Similar as (3.22, it follows Since < J λ (u u n (x u n (y 2 n,u n > = c N,s dxdy µ R x y 2N N+2s u n (x 2 s dx f(x,u n u n dx, thanks to (3.22 and (3.23, we have u n (x u n (y 2 c N,s dxdy µ R x y 2N N+2s lim sup F(x,u n dx 0. (3.23 n 11 u n (x 2 dx λ u x 2s n (x 2 dx (3.24 u n (x 2 dx u x 2s n (x 2 s dx 0. (3.25

12 Denote which implies and c N,s u n (x u n (y 2 dxdy µ R x y 2N N+2s u n (x 2 s dx L, It is easy to see that L > 0. From (1.18 we have u n (x 2 x 2s dx L 0 < β c = lim J λ (u n = ( 1 n 2 1 L. ( s L SL 2 2 s (3.27 then c s N S N 2s (3.28 which contradicts (3.8. Hence u 0 0 in. The proof of Theorem 1.1 is complete. For the proof of Theorem 1.2, the key step is to check the condition (1.24. Firstly we choose the test function v ε (x of (1.24 as the following. Let u(x = U(x U(x L 2 s(r N,u ε (x = ε N 2s 2 u( x. (3.29 ε Thus u ε L 2 s(r N = 1, and u ε(x is the achieve function of S (defined in (1.18. Let us fix δ > 0 such that B 4δ, and let η C 0 (B 2δ be such that 0 η 1 in R N, η 1 in B δ. For every ε > 0, define v ε (x = u ε (xη(x,x R N. (3.30 Obviously v ε X 0. Secondly we make some estimates for v ε. The main strategy of the estimations for v ε (x is similar to [14]. Proposition For all x B c r, it follows where r > 0 and c r is a positive constant depending on r. 2 For all y B C δ,x RN with x y < δ/2, it follows for all x,y Bδ C, it follows v ε u ε c r ε (N 2sηµ 2, (3.31 v ε c r ε (N 2sηµ 2 (3.32 v ε (x v ε (y c x y ε (N 2sηµ 2 ; (3.33 v ε (x v ε (y cmin{ x y,1}ε (N 2sηµ 2. (

13 Proof. 1From the definition of v ε (x and u ε (x, for x B c r, we have v ε (x u ε (x cε (N 2s/2 ( N 2s x/ε 1 ηµ (1+ x/ε 2ηµ 2 = cε ηµ(n 2s/2 ( N 2s x 1 ηµ (ε 2ηµ + x 2ηµ 2 cε ηµ(n 2s/2 r (1+ηµ(N 2s/2 c r ε ηµ(n 2s/2 (3.35 where c r is a positive constant depending on r. Since u ε cε (N 2s/2( x ε 1 ηµ (1+ x ε 2ηµ N 2s 2 1 ( x ε ηµ + x ε ηµ 1 ε = cε (N 2s/2( x (1+ x N 2s ε 1 ηµ ε 2ηµ 2 ( x + x ε ηµ ε ηµ 1 x (1+ x ε 1 ηµ ε 2ηµ ε ( x + x ε c u ε (x ηµ ε ηµ 1 x (1+ x ε 1 ηµ ε 2ηµ ε c u ε(x ( 1 x + ε2ηµ x 2ηµ+1 c r u ε (x (3.36 and v ε (x = η(xu ε (x+ u ε (xη(x c( u ε (x + u ε (x, (3.37 collecting (3.35-(3.37, we obtain (3.31 and ( Similar to Claim 10 in [14], it follows (3.33-(3.34. In fact, for all x R N,y B C δ, x y < δ 2, v ε (x v ε (y = v ε (ξ x y (3.38 where ξ R N with ξ = tx+(1 ty for some t (0,1. Since taking r = δ, from (3.32 it follows ( For all x B C δ,y BC δ, if x y δ 2, ξ = y +t(x y y t x y > δ 2, (3.39 v ε (x v ε (y v ε (x + v ε (y cmin{1, x y }ε ηµ(n 2s/2 ; if x y δ, (3.34 follows from ( Proposition 3.3. Let s (0,1, and N > 2s, then the following estimates holds v ε (x v ε (y 2 v ε (x 2 1 c N,s dxdy µ dx S +O(ε (N 2sηµ ; (3.40 R x y 2N N+2s x 2s 13

14 2 v ε (x 2 s = L 2 s( 1+O(εηµN ; (3.41 cε ηµ(n 2s cε 2s, if η µ < 2s, N 2s 3 v ε (x 2 dx c logε ε 2s, if η µ = 2s N 2s, (3.42 cε ηµ(n 2s +cε 2s, if η µ > 2s Proof. 1 Define N 2s. D = {(x,y R N R N x B δ,y B C δ, x y δ 2 } (3.43 E = {(x,y R N R N x B δ,y Bδ C, x y < δ }. ( Then v ε (x v ε (y 2 c N,s dxdy R x y 2N N+2s u ε (x u ε (y 2 = c N,s dxdy +c B δ B δ x y N+2s N,s u ε (x v ε (y 2 +c N,s dxdy +c x y N+2s N,s E D B C δ BC δ u ε (x v ε (y 2 dxdy x y N+2s v ε (x v ε (y 2 x y N+2s dxdy. (3.45 From (3.33 and E B δ B 2δ, we have c N,s E u ε (x v ε (y 2 dxdy c x y N+2s N,s B δ B 2δ ε (N 2sηµ x y N+2s 2dxdy c δε (N 2sηµ, (3.46 where c δ > 0 is a constant depending on δ. Similarly, from (3.34 c N,s B C δ BC δ u ε (x v ε (y 2 dxdy c x y N+2s N,s Bδ C BC δ c δ ε (N 2sηµ. ε (N 2sηµ min{1, x y 2 } dxdy x y N+2s (3.47 where c δ > 0 is a constant depending on δ. Now, it remains to estimate the integral on D of (3.45, u ε (x v ε (y 2 c N,s dxdy D x y N+2s u ε (x u ε (y 2 c N,s dxdy +c x y N+2s N,s D +c N,s D 2 v ε (y u ε (y u ε (x u ε (y dxdy x y N+2s D u ε (y v ε (y 2 dxdy x y N+2s (

15 Now we estimate the last term in the right hand side of (3.48. Once more by (3.31, u ε (y v ε (y 2 c N,s dxdy D x y N+2s u ε (y 2 + v ε (y 2 c N,s dxdy x y N+2s (3.49 D c x y δ/2 where c δ > 0 is a constant depending on δ. And for (x,y D, ε (N 2sηµ x y N+2sdxdy c δε (N 2sηµ. u ε (xv ε (y u ε (xu ε (y cε (N 2sηµ/2 ε (N 2s/2 ( N 2s x/ε 1 ηµ (1+ x/ε 2ηµ 2 c (, N 2s x 1 ηµ (1+ x/ε 2ηµ 2 (3.50 let x = x,z = y x, it follows ε u ε (xu ε (y c N,s dxdy D x y N+2s 1 c ( dxdy N 2s D x 1 ηµ (1+ x/ε 2ηµ 2 x y N+2s cε N 1 ( d xdz x δ/ε, z > δ N 2s 2 ε x 1 ηµ (1+ x 2ηµ 2 z N+2s 1 cε N (N 2s(1 ηµ 2 d x ( cε N (N 2s(1 ηµ 2 ( cε N (N 2s(1 ηµ 2 ( cε N (N 2s(1 ηµ 2 = O(ε (N 2sηµ. x δ/ε x 1 r 1 r 1 ( N 2s x 1 ηµ (1+ x 2ηµ 2 1 ( d x+ N 2s x 1 ηµ (1+ x 2ηµ 2 r (N 1 (1 ηµn 2s 2 dr + N 2s (1+r 2ηµ 2 1 r δ/ε 1 x δ/ε 1 ( d x N 2s x 1 ηµ (1+ x 2ηµ 2 r (N 1 (1+ηµ(N 2s 2 dr r (N 1 (1 ηµn 2s 2 dr +ε N+(1+ηµ(N 2s 2 +c (

16 It follows from (3.48-(3.51 that u ε (x v ε (y 2 c N,s dxdy c x y N+2s N,s D D u ε (x u ε (y 2 x y N+2s dxdy +O(ε (N 2sηµ. (3.52 Thus from (3.46, (3.47 and (3.52, v ε (x v ε (y 2 R2N u ε (x u ε (y 2 c N,s dxdy c R x y 2N N+2s N,s dxdy +O(ε (N 2sηµ. (3.53 x y N+2s Since RN (1 η(x 2 u ε (x 2 µ dx c x 2s = c and µ RN (1 η(x 2 u ε (x 2 x 2s dx c it gives = c c c x δ x δ x 2δ x 2δ x 2δ/ε r 2δ/ε u ε (x 2 x 2s dx ε (N 2s ( N 2s x dx x/ε 1 ηµ (1+ x/ε 2ηµ 2s 1 ( N 2s x dx x 1 ηµ (1+ x 2ηµ 2s r N 1 2s (1+ηµ(N 2s dr = O(ε (N 2sηµ, u ε (x 2 x 2s dx ε (N 2s (3.54 ( N 2s x dx = O(ε x/ε 1 ηµ (1+ x/ε (N 2sηµ, 2ηµ 2s RN v ε (x 2 RN u ε (x 2 RN (1 η(x 2 u ε (x 2 µ dx = µ dx µ dx x 2s x 2s x 2s RN u ε (x 2 = µ dx+o(ε (N 2sηµ. x 2s (3.55 (3.56 From (3.53 and (3.56, we have v ε (x v ε (y 2 RN v ε (x 2 c N,s dxdy µ dx S +O(ε (N 2sηµ. (3.57 R x y 2N N+2s x 2s 2 Similar to (3.56, we have R N v ε (x 2 s dx = R N u ε (x 2 s dx+o(ε Nηµ (

17 3Finally, we want to prove (3.42 v ε (x 2 dx u ε (x 2 dx c x δ x δ = ε 2s c ε (N 2s ( N 2s dx x/ε 1 ηµ (1+ x/ε 2ηµ x δ/ε ( 1 N 2s dx x 1 ηµ (1+ x 2ηµ δ/ε cε 2s r N 1 ( N 2s dr 1 r 1 ηµ (1+r 2ηµ cε ηµ(n 2s cε 2s, if η µ < 2s, N 2s = c logε ε 2s, if η µ = 2s N 2s, cε ηµ(n 2s +cε 2s, if η µ > 2s N 2s. (3.59 Proof of Theorem 1.2 From Theorem 1.1, we only need to verify (1.24. Denote t ε be the attaining point of max J λ(tv ε. Similar as the proof of Lemma 8.1 in [10], let t ε be the t>0 attaining point of max J λ(tv ε, we claim t ε is uniformly bounded for ε > 0 small. In fact, t>0 we consider the function g(t = J λ (tv ε = t2 2 ( v ε(x 2 X µ λ v ε 2 dx t2 s v 2 ε 2 s dx s t2 2 (1 λ (3.60 v ε (x 2 X λ µ t2 s v ε 2 s dx. 1,µ Since lim g(t = and g(t > 0 when t closed to 0, so that maxg(t is attained for t + t>0 t ε > 0. Then g (t ε = t ε ( v ε (x 2 X µ λ From (3.61 and Lemma 3.3, for ε sufficiently small, we have S 2 (1 λ t 2 s 2 ε λ µ which implies t ε is bounded for ε > 0 small enough. 2 s v ε 2 dx t 2 s 1 ε v ε 2 s dx = 0. (3.61 = v ε(x 2 X µ λ v ε 2 dx v ε 2 sdx From the definition of η µ, (1.21 and (1.22, we have < 2S. (3.62 µ ϕ 1 ( N 4s η µ 2s 4 N 2s. (

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