A non-local critical problem involving the fractional Laplacian operator

Transcription

1 Eduardo Colorado p. 1/23 A non-local critical problem involving the fractional Laplacian operator Eduardo Colorado Universidad Carlos III de Madrid UGR, Granada Febrero 212

2 Eduardo Colorado p. 2/23

3 Eduardo Colorado p. 3/23 The main results of the talk are in collaboration with: B. Barrios, UAM-ICMAT. A. de Pablo, UC3M. U. Sánchez, UC3M. B. Barrios, E. C., A. de Pablo, U. Sánchez, On Some critical problems for the fractional Laplacian operator. Preprint 211, arxiv:

4 (P λ ) { The main problem ( ) α/2 u = f λ (u), u > in Ω, u = on Ω,

5 (P λ ) { The main problem ( ) α/2 u = f λ (u), u > in Ω, u = on Ω, where f λ (u) = λu q +u r, λ >, < q < r N+α N α = 2 α 1 and Ω R N, with N > α, < α < 2.

6 Eduardo Colorado p. 4/23 (P λ ) { The main problem ( ) α/2 u = f λ (u), u > in Ω, u = on Ω, where f λ (u) = λu q +u r, λ >, < q < r N+α N α = 2 α 1 and Ω R N, with N > α, < α < 2. Subcritical case 1 < r < N+α N α, and q < 1 [BCdPS] C. Brändle, E.C., A. de Pablo, U. Sánchez, A concave-convex elliptic problem involving the fractional Laplacian. To appear in Proc. Roy. Soc. Edinburgh. Critical case r = N+α N α [BCPS] B. Barrios, E.C., A. de Pablo, U. Sánchez, On Some critical problems for the fractional Laplacian operator. Preprint 211, arxiv:

7 Scheme of the talk Definition of the Fractional Laplacian (through the spectral decomposition)

8 Scheme of the talk Definition of the Fractional Laplacian (through the spectral decomposition) Extended problem to one more variable

9 Scheme of the talk Definition of the Fractional Laplacian (through the spectral decomposition) Extended problem to one more variable Trace and Sobolev Inequalities

10 Scheme of the talk Definition of the Fractional Laplacian (through the spectral decomposition) Extended problem to one more variable Trace and Sobolev Inequalities Main results

11 Eduardo Colorado p. 5/23 Scheme of the talk Definition of the Fractional Laplacian (through the spectral decomposition) Extended problem to one more variable Trace and Sobolev Inequalities Main results Some ideas of the proofs

12 Definition of the Fractional Laplacian Powers of Laplacian operator ( ): Let (λ n,ϕ n ) be the eigenvalues and eigenfunctions of ( ) in Ω with zero Dirichlet boundary data. Then (λ α/2 n,ϕ n ) are the eigenvalues and eigenfunctions of ( ) α/2, also with zero Dirichlet boundary conditions.

13 Definition of the Fractional Laplacian Powers of Laplacian operator ( ): Let (λ n,ϕ n ) be the eigenvalues and eigenfunctions of ( ) in Ω with zero Dirichlet boundary data. Then (λ α/2 n,ϕ n ) are the eigenvalues and eigenfunctions of ( ) α/2, also with zero Dirichlet boundary conditions. The fractional Laplacian ( ) α/2 is well defined in the space H α/2 (Ω) = { u = a n ϕ n L 2 (Ω) : u 2 H α/2 = a 2 nλ α/2 n (Ω) } <.

14 Eduardo Colorado p. 6/23 Definition of the Fractional Laplacian Powers of Laplacian operator ( ): Let (λ n,ϕ n ) be the eigenvalues and eigenfunctions of ( ) in Ω with zero Dirichlet boundary data. Then (λ α/2 n,ϕ n ) are the eigenvalues and eigenfunctions of ( ) α/2, also with zero Dirichlet boundary conditions. The fractional Laplacian ( ) α/2 is well defined in the space H α/2 (Ω) = { u = a n ϕ n L 2 (Ω) : u 2 H α/2 = a 2 nλ α/2 n (Ω) } <. As a consequence, ( ) α/2 u = λ α/2 n a n ϕ n. Note that then u H α/2 (Ω) = ( )α/4 u L 2 (Ω).

15 Definition of the Fractional Laplacian We now consider the general problem (P) ( ) α/2 u = f(x,u) in Ω, u = on Ω.

16 Eduardo Colorado p. 7/23 Definition of the Fractional Laplacian We now consider the general problem (P) ( ) α/2 u = f(x,u) in Ω, u = on Ω. We say that u H α/2 (Ω) is an energy solution of (P) if the identity holds for ϕ H α/2 (Ω). Ω ( ) α/4 u( ) α/4 ϕdx = Ω f(x,u)ϕdx

17 Definition of the Fractional Laplacian (P λ ) ( ) α/2 u = λu q +u N+α N α, u > in Ω, u = on Ω, where λ >, < q < N+α N α = 2 α 1 and Ω RN, with N > α, < α < 2.

18 Definition of the Fractional Laplacian (P λ ) ( ) α/2 u = λu q +u N+α N α, u > in Ω, u = on Ω, where λ >, < q < N+α N α = 2 α 1 and Ω RN, with N > α, < α < 2. By the definition of solution, if f λ (u) = λu q +u N+α N α Ω ( ) α/4 u( ) α/4 ϕdx = Ω f λ (u)ϕdx, ϕ H α/2 (Ω).

19 Eduardo Colorado p. 8/23 Definition of the Fractional Laplacian (P λ ) ( ) α/2 u = λu q +u N+α N α, u > in Ω, u = on Ω, where λ >, < q < N+α N α = 2 α 1 and Ω RN, with N > α, < α < 2. By the definition of solution, if f λ (u) = λu q +u N+α N α Ω ( ) α/4 u( ) α/4 ϕdx = Since u H α/2 (Ω) f(u) L 2N N+α(Ω) H α/2 (Ω). Then f λ (u)ϕ L 1 (Ω). Ω f λ (u)ϕdx, ϕ H α/2 (Ω). Associated energy functional I(u) = 1 2 Ω ( ) α/4 u 2 dx λ q +1 Ω u q+1 dx 1 2 α Ω u 2 αdx which is well defined in H α/2 (Ω). Clearly, the critical points of I correspond to solutions to (P λ ).

20 Extended problems to one more variable Consider the cylinder C Ω = Ω (, ) R N+1 +. Given u Hα/2 (Ω), we define its α-harmonic extension w = Eα(u) to the cylinder C Ω as the solution to the problem div(y 1 α w) = in C Ω, w = on L C Ω = Ω (, ), w = u on Ω {y = }.

21 Extended problems to one more variable Consider the cylinder C Ω = Ω (, ) R N+1 +. Given u Hα/2 (Ω), we define its α-harmonic extension w = Eα(u) to the cylinder C Ω as the solution to the problem div(y 1 α w) = in C Ω, w = on L C Ω = Ω (, ), w = u on Ω {y = }. The extension function belongs to the space X α (C Ω) defined as the completion of {z C (C Ω ) : z = on L C Ω } with the norm z X α (C Ω ) = (κ α C Ω y 1 α z 2 dxdy ) 1/2 where κ α is a normalization constant.

22 Extended problems to one more variable Consider the cylinder C Ω = Ω (, ) R N+1 +. Given u Hα/2 (Ω), we define its α-harmonic extension w = Eα(u) to the cylinder C Ω as the solution to the problem div(y 1 α w) = in C Ω, w = on L C Ω = Ω (, ), w = u on Ω {y = }. With that κ α, the extension operator is an isometry Eα(ψ) X α (C Ω ) = ψ α/2, ψ Hα/2 H (Ω) (Ω).

23 Eduardo Colorado p. 9/23 Extended problems to one more variable Consider the cylinder C Ω = Ω (, ) R N+1 +. Given u Hα/2 (Ω), we define its α-harmonic extension w = Eα(u) to the cylinder C Ω as the solution to the problem div(y 1 α w) = in C Ω, w = on L C Ω = Ω (, ), w = u on Ω {y = }. With that κ α, the extension operator is an isometry Eα(ψ) X α (C Ω ) = ψ α/2, ψ Hα/2 H (Ω) (Ω). Moreover, for any ϕ X α (C Ω), we have the following trace inequality ϕ X α (C Ω ) ϕ(,) H α/2 (Ω).

24 Extended problems to one more variable The relevance of the extension function w is that it is related to the fractional Laplacian of the original function u through the formula κ α lim y +y1 α w y (x,y) = ( )α/2 u(x),

25 Eduardo Colorado p. 1/23 Extended problems to one more variable The relevance of the extension function w is that it is related to the fractional Laplacian of the original function u through the formula κ α lim y +y1 α w y (x,y) = ( )α/2 u(x), See: [CS] L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. Comm. Partial Differential Equations, 27. See also: [BCdPS] C. Brändle, E.C., A. de Pablo, U. Sánchez, To appear in Proc. Roy. Soc. Edinburgh. [CT] X. Cabré, J. Tan, Adv. Math., 21. [CDDS] A. Capella, J. Dávila, L. Dupaigne, Y. Sire, To appear in Comm. Partial Differential Equations.

26 Extended problems to one more variable When Ω = R N, the above Dirichlet to Neumann procedure provides a formula to the fractional Laplacian in the whole space equivalent to the one by Fourier Transform, (( ) α/2 )g)ˆ(ξ) = ξ α ĝ(ξ). See [CS] L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. Comm. Partial Differential Equations, 27.

27 Eduardo Colorado p. 11/23 Extended problems to one more variable When Ω = R N, the above Dirichlet to Neumann procedure provides a formula to the fractional Laplacian in the whole space equivalent to the one by Fourier Transform, (( ) α/2 )g)ˆ(ξ) = ξ α ĝ(ξ). See [CS] L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. Comm. Partial Differential Equations, 27. In that case there are explicit expressions to the α-harmonic extension and the fractional Laplacian in terms of the Poisson and Riesz kernels, resp. w(x,y) = P α y u(x) = c N,α y α ( ) α/2 u(x) = d N,α P.V. R N u(s) ( x s 2 +y 2 ) N+α 2 u(x) u(s) ds. R N x s N+α ds, αc N,α κ α = d N,α

28 Denoting Extended problems to one more variable L α w := div(y 1 α w), w ν α := κ α lim y +y1 α w y we can reformulate (P λ ) with the new variable as (P λ ) L α w = in C Ω w = on L C Ω w ν α = λwq +w N+α N α in Ω {y = }.

29 Denoting Extended problems to one more variable L α w := div(y 1 α w), w ν α := κ α lim y +y1 α w y we can reformulate (P λ ) with the new variable as (P λ ) L α w = in C Ω w = on L C Ω w ν α = λwq +w N+α N α in Ω {y = }. w X α (C Ω) is an energy solution if κ α C Ω y 1 α w, ϕ dxdy = Ω ( λw q +w N+α N α ) ϕdx, ϕ X α (C Ω ).

30 Denoting Extended problems to one more variable L α w := div(y 1 α w), w ν α := κ α lim y +y1 α w y we can reformulate (P λ ) with the new variable as (P λ ) L α w = in C Ω w = on L C Ω w ν α = λwq +w N+α N α in Ω {y = }. w X α (C Ω) is an energy solution if κ α C Ω y 1 α w, ϕ dxdy = Ω ( ) λw q +w N+α N α ϕdx, ϕ X α (C Ω ). Energy functional: J(w) = κ α 2 C Ω y 1 α w 2 dxdy λ q +1 Ω w q+1 dx 1 2 α Ω w 2 α dx.

31 Eduardo Colorado p. 12/23 Denoting Extended problems to one more variable L α w := div(y 1 α w), w ν α := κ α lim y +y1 α w y we can reformulate (P λ ) with the new variable as (P λ ) L α w = in C Ω w = on L C Ω w ν α = λwq +w N+α N α in Ω {y = }. w X α (C Ω) is an energy solution if κ α C Ω y 1 α w, ϕ dxdy = Ω ( ) λw q +w N+α N α ϕdx, ϕ X α (C Ω ). Energy functional: J(w) = κ α 2 C Ω y 1 α w 2 dxdy λ q +1 Ω w q+1 dx 1 2 α Ω w 2 α dx. Note that critical points of J in X α (C Ω) correspond to critical points of I in H α/2 (Ω). Even more, minima of J also correspond to minima of I.

32 Sobolev and Trace inequalities Assume N > α, there exists a positive constant C = C(α,r,N,Ω) such that for 1 r 2 α = 2N N α, C Ω y 1 α z(x,y) 2 dxdy C ( Ω ) 2/r z(x,) r dx for any z X α (C Ω).

33 Eduardo Colorado p. 13/23 Sobolev and Trace inequalities Assume N > α, there exists a positive constant C = C(α,r,N,Ω) such that for 1 r 2 α = 2N N α, C Ω y 1 α z(x,y) 2 dxdy C ( Ω ) 2/r z(x,) r dx for any z X α (C Ω). Also, for any v H α/2 (Ω). Ω ( ) α/4 v 2 dx C ( Ω ) 2/r v r dx

34 Sobolev and Trace inequalities When Ω = R N, r = 2 α, there exists a constant S(α,N) > such that R N+1 + ( 2/2 y 1 α z(x,y) 2 dxdy S(α,N) z(x,) 2 α α dx), z X α (R N+1 R N + ). The constant is achieved when z(,) = u( ) takes the form: u(x) = u ε (x) = ε (N α)/2 ( x 2 +ε 2 ) (N α)/2.

35 Sobolev and Trace inequalities When Ω = R N, r = 2 α, there exists a constant S(α,N) > such that R N+1 + ( 2/2 y 1 α z(x,y) 2 dxdy S(α,N) z(x,) 2 α α dx), z X α (R N+1 R N + ). The constant is achieved when z(,) = u( ) takes the form: u(x) = u ε (x) = ε (N α)/2 ( x 2 +ε 2 ) (N α)/2. Also we have the corresponding Sobolev inequality for any v H α/2 (R N ). ( ( ) α/4 v 2 dx κ α S(α,N) v 2 α dx R N R N ) 2/2 α

36 Eduardo Colorado p. 14/23 Sobolev and Trace inequalities When Ω = R N, r = 2 α, there exists a constant S(α,N) > such that R N+1 + ( 2/2 y 1 α z(x,y) 2 dxdy S(α,N) z(x,) 2 α α dx), z X α (R N+1 R N + ). The constant is achieved when z(,) = u( ) takes the form: u(x) = u ε (x) = ε (N α)/2 ( x 2 +ε 2 ) (N α)/2. Also we have the corresponding Sobolev inequality for any v H α/2 (R N ). ( ( ) α/4 v 2 dx κ α S(α,N) v 2 α dx R N R N ) 2/2 α Note that these constants are achived on R N, but are not attained in any bounded domain.

37 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω,

38 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω, Theorem 1 Let < q < 1, 1 α < 2. There exists < Λ < such that the problem (P λ ) 1. has no solution for λ > Λ; 2. has at least two solutions for each < λ < Λ; (1 α < 2) 3. has a solution for λ = Λ.

39 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω, Theorem 1 Let < q < 1, 1 α < 2. There exists < Λ < such that the problem (P λ ) 1. has no solution for λ > Λ; 2. has at least two solutions for each < λ < Λ; (1 α < 2) 3. has a solution for λ = Λ. In the subcritical case [BCdPS] the same restriction on α appeared. The difficulty was to find a Liouville-type theorem for < α < 1. Here, due to the lack of regularity, it is not clear how to separate the solutions in the appropriate way, see [CP,D] for more details. [BCdPS] C. Brändle, E.C., A. de Pablo, U. Sánchez, To appear in Proc. Roy. Soc. Edinburgh. [CP] E. C., I. Peral J. Funct. Anal. 23. [D] J. Dávila, J. Funct. Anal. 21.

40 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω, Theorem 1 Let < q < 1, 1 α < 2. There exists < Λ < such that the problem (P λ ) 1. has no solution for λ > Λ; 2. has at least two solutions for each < λ < Λ; (1 α < 2) 3. has a solution for λ = Λ. Theorem 2 Let q = 1, < α < 2 and N 2α. Then the problem (P λ ) 1. has no solution for λ λ 1 ; 2. has a solution for each < λ < λ 1.

41 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω, Theorem 1 Let < q < 1, 1 α < 2. There exists < Λ < such that the problem (P λ ) 1. has no solution for λ > Λ; 2. has at least two solutions for each < λ < Λ; (1 α < 2) 3. has a solution for λ = Λ. Theorem 2 Let q = 1, < α < 2 and N 2α. Then the problem (P λ ) 1. has no solution for λ λ 1 ; 2. has a solution for each < λ < λ 1. We have left open the range α < N < 2α. See the special case α = 2 and N = 3 in [BN]. If α = 1 this range is empty, see [T]. [BN] H. Brezis, L. Nirenberg, Comm. Pure Appl. Math [T] J. Tan, Calc. Var. Partial Differential Equations 211.

42 Eduardo Colorado p. 15/23 Remember the problem (P λ ) Main Results ( ) α/2 u = λu q +u 2 1, u > in Ω, u = on Ω, Theorem 1 Let < q < 1, 1 α < 2. There exists < Λ < such that the problem (P λ ) 1. has no solution for λ > Λ; 2. has at least two solutions for each < λ < Λ; (1 α < 2) 3. has a solution for λ = Λ. Theorem 2 Let q = 1, < α < 2 and N 2α. Then the problem (P λ ) 1. has no solution for λ λ 1 ; 2. has a solution for each < λ < λ 1. Theorem 3 Let 1 < q < 2 α 1, < α < 2 and N > α(1+(1/q)). Then the problem (P λ) has a solution for any λ >.

43 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ).

44 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). The proof follows by the Moser iterative method ([GT]) with appropriate test functions. [GT] D. Gilbarg, N.S. Trudinger, "Elliptic partial differential equations of second order" Springer-Verlag, Berlin 21.

45 Eduardo Colorado p. 16/23 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). Proposition 2 Let u be a solution of (P λ ). (i) If α = 1 and q 1 then u C (Ω). (ii) If α = 1 and q < 1 then u C 1,q (Ω). (iii) If α < 1 then u C,α (Ω). (iv) If α > 1 then u C 1,α 1 (Ω).

46 Eduardo Colorado p. 16/23 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). Proposition 2 Let u be a solution of (P λ ). (i) If α = 1 and q 1 then u C (Ω). (ii) If α = 1 and q < 1 then u C 1,q (Ω). (iii) If α < 1 then u C,α (Ω). (iv) If α > 1 then u C 1,α 1 (Ω). Proof: (i) By Proposition 1 and [CT] we get that u C,γ (Ω), for some γ < 1. Since q 1 then f λ (u) C,γ (Ω). Again by [CT], it follows that u C 1,γ (Ω). Iterating the process we conclude that u C (Ω). [CT] X. Cabré, J. Tan, Adv. Math. 21.

47 Eduardo Colorado p. 16/23 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). Proposition 2 Let u be a solution of (P λ ). (i) If α = 1 and q 1 then u C (Ω). (ii) If α = 1 and q < 1 then u C 1,q (Ω). (iii) If α < 1 then u C,α (Ω). (iv) If α > 1 then u C 1,α 1 (Ω). Proof: (ii) As before we have u C,γ (Ω), for some γ < 1. Therefore f λ (u) C,qγ (Ω). It follows that u C 1,qγ (Ω), which gives f λ (u) C,q (Ω). Finally this implies u C 1,q (Ω).

48 Eduardo Colorado p. 16/23 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). Proposition 2 Let u be a solution of (P λ ). (i) If α = 1 and q 1 then u C (Ω). (ii) If α = 1 and q < 1 then u C 1,q (Ω). (iii) If α < 1 then u C,α (Ω). (iv) If α > 1 then u C 1,α 1 (Ω). Proof: (iii) By [CDDS] we obtain that u C,γ (Ω) for all γ (,α). This implies that f λ (u) C,r (Ω) for every r < min{qα, α}. Therefore, again by another result in [CDDS], we get that u C,α (Ω). [CDDS] A. Capella, J. Dávila, L. Dupaigne, Y. Sire, To appear in Comm. Partial Differential Equations, arxiv:

49 Eduardo Colorado p. 16/23 Auxiliary results (regularity) Proposition 1 Let u H α/2 (Ω) be a solution to the problem ( ) α/2 u = f(x,u) inω, u > inω, u = on Ω with f(x,s) C(1+ s p ) (x,s) Ω R, and some < p 2 α 1. Then u L (Ω) with u L (Ω) C( u H α/2 (Ω) ). Proposition 2 Let u be a solution of (P λ ). (i) If α = 1 and q 1 then u C (Ω). (ii) If α = 1 and q < 1 then u C 1,q (Ω). (iii) If α < 1 then u C,α (Ω). (iv) If α > 1 then u C 1,α 1 (Ω). Proof: (iv) Since α > 1, we can write problem (P λ ) as follows ( ) 1/2 u = s in Ω, ( ) (α 1)/2 s = f λ (u) in Ω, u = s = on Ω. Reasoning as before, we obtain the desired regularity in two steps, using [CT] and [CDDS].

50 Auxiliary results (concentration-compactness) Following the classical result by P. L. Lions in [L]. [L] P. L. Lions Rev. Mat. Iberoamericana Part II, 1985.

51 Eduardo Colorado p. 17/23 Auxiliary results (concentration-compactness) Following the classical result by P. L. Lions in [L]. [L] P. L. Lions Rev. Mat. Iberoamericana Part II, Proposition 3 Let {w n } n N be a weakly convergent sequence to w in X α (C Ω), such that the sequence { y 1 α w n 2} n N is tight. Let u n = Tr(w n ) and u = Tr(w). Assume that µ, ν are two non negative measures such that y 1 α w n 2 µ and u n 2 α ν, as n (.2) in the sense of measures. Then there exist an at most countable set I, points {x k } k I Ω and real positive numbers µ k, ν k such that 1. µ y 1 α w 2 + k Iµ k δ xk, 2. ν = u 2 α + k Iν k δ xk, 3. µ k S(α,N)ν 2 2 α k.

52 Main ideas/steps of the proofs Theorem 1 Let < q < 1, 1 α < 2. Then, there exists < Λ < such that Problem (P λ ) 1. has no positive solution for λ > Λ; 2. has at least two positive solutions for each < λ < Λ; 3. has a positive solution for λ = Λ.

53 Main ideas/steps of the proofs Theorem 1 Let < q < 1, 1 α < 2. Then, there exists < Λ < such that Problem (P λ ) 1. has no positive solution for λ > Λ; 2. has at least two positive solutions for each < λ < Λ; 3. has a positive solution for λ = Λ. Proof of Theorem 1 1. Denoting (λ 1,ϕ 1 ) the first eigenvalue and an associated positive eigenfunction to the classical Laplace operator, we have that Ω ( ) λu q +u N+α N α ϕ 1 dx = λ α/2 1 uϕ 1 dx. Ω Observe that there exist positive constants c,δ such that λt q +t N+α N α > cλ δ t, for any t >, hence by the previous integral identity, cλ δ < λ α/2 1 Λ <.

54 Main ideas/steps of the proofs Theorem 1 Let < q < 1, 1 α < 2. Then, there exists < Λ < such that Problem (P λ ) 1. has no positive solution for λ > Λ; 2. has at least two positive solutions for each < λ < Λ; 3. has a positive solution for λ = Λ. Proof of Theorem 1 To prove Λ >, for λ > sufficiently small, one can use the iteration method of sub-supersolutions, starting with the subsolution and obtaining a minimal one. See for example, the pioneering works [GP,ABC] for the p-laplacian, Laplacian resp. among others... Moreover, it is easy to see that we have an interval of minimal solutions increasing with respect to λ for any < λ < Λ. [GP] J. García-Azorero, I. Peral Trans. Amer. Math. Soc [ABC] A. Ambrosetti, H. Brezis, G. Cerami J. Func. Analysis 1994.

55 Main ideas/steps of the proofs Theorem 1 Let < q < 1, 1 α < 2. Then, there exists < Λ < such that Problem (P λ ) 1. has no positive solution for λ > Λ; 2. has at least two positive solutions for each < λ < Λ; 3. has a positive solution for λ = Λ. Proof of Theorem 1 3. The idea consist (like in [ABC]) on passing to the limit as λ ր Λ on the sequence of minimal solutions w n = w λn. Clearly J λn (w n ) <, hence > J λn (w n ) 1 ( 1 2 J λ n (w n ),w n = κ α α 2 1 ) ( 1 2 w n 2 X α α (C Ω) λ n q +1 1 ) 2 wn q+1 dx. α Ω By the Sobolev and Trace inequalities, the sequence is bounded, w n X α C. Then there exist a subsequence w n w Λ X α (C Ω). Moreover, by comparison w Λ w λ > for any < λ < Λ. A non trivial solution to (P Λ ).

56 Eduardo Colorado p. 18/23 Main ideas/steps of the proofs Theorem 1 Let < q < 1, 1 α < 2. Then, there exists < Λ < such that Problem (P λ ) 1. has no positive solution for λ > Λ; 2. has at least two positive solutions for each < λ < Λ; 3. has a positive solution for λ = Λ. Proof of Theorem 1 2. Ideas/Steps: (i) The minimal solution is a local minimum for the functional. (ii) So we can use the Mountain Pass Theorem, obtaining a minimax sequence. (iii) In order to find a second solution, we need to prove a Palais-Smale (PS) condition under a critical level. (iv) Arguing by contradiction, if the local minimum would be the unique critical point, then the functional satisfies a local (PS) c condition for c under a critical level. To do that we construct a path by localizing the minimizers of the Trace/Sobolev inequalities at the possible Dirac Deltas given by the concentration-compactness result. Here appear new difficulties: 1.- It is not known how the fractional Laplacian acts on products of functions. 2.- By that, we work on the extended functional, but the minimizers have not an explicit expression on R N Also we need to prove that there is neither vanishing, nor dichotomy, since we pass to the infinite cylinder in the y-variable.

57 Main ideas/steps of the proofs In order to prove that the minimal solution is a local minimum, first we show that is a local minimum in the C 1 -topology. To start with we prove a separation lemma in that topology.

58 Main ideas/steps of the proofs In order to prove that the minimal solution is a local minimum, first we show that is a local minimum in the C 1 -topology. To start with we prove a separation lemma in that topology. Lemma 1 Let < µ 1 < λ < µ 2 < Λ. Let z µ1, z λ and z µ2 be the corresponding minimal solutions to (P λ ), λ = µ 1, λ and µ 2 respectively. If X = {z C 1 (Ω) z µ 1 z z µ2 }, then there exists ε > such that where B 1 is the unit ball in C 1 (Ω). {z λ }+εb 1 X,

59 Eduardo Colorado p. 19/23 Main ideas/steps of the proofs In order to prove that the minimal solution is a local minimum, first we show that is a local minimum in the C 1 -topology. To start with we prove a separation lemma in that topology. Lemma 1 Let < µ 1 < λ < µ 2 < Λ. Let z µ1, z λ and z µ2 be the corresponding minimal solutions to (P λ ), λ = µ 1, λ and µ 2 respectively. If X = {z C 1 (Ω) z µ 1 z z µ2 }, then there exists ε > such that where B 1 is the unit ball in C 1 (Ω). {z λ }+εb 1 X, Proof. Since α 1, by Proposition 2, < γ < 1 such that any solution to (P λ ) is in C 1,γ for any < λ < Λ. Then u(x) C dist (x, Ω), x Ω. By comparison with a positive first eigenfunction of the Laplacian, we get u(x) c dist (x, Ω), x Ω.

60 Main ideas/steps of the proofs Lemma 2 For all λ (,Λ) there exists a solution for (P λ ) which is a local minimum of the functional I in the C 1 -topology.

61 Eduardo Colorado p. 2/23 Main ideas/steps of the proofs Lemma 2 For all λ (,Λ) there exists a solution for (P λ ) which is a local minimum of the functional I in the C 1 -topology. Proof. Given < µ 1 < λ < µ 2 < Λ, let z µ1 and z µ2 be the minimal solutions of (P µ1 ) and (P µ2 ) respectively, we set f (x,s) = f λ (z µ1 (x)) if s z µ1, f λ (s) if z µ1 s z µ2, f λ (z µ2 (x)) if z µ2 s, and F (x,z) = z f (x,s)ds I (z) = 1 2 z H α/2 F (x,u)dx. (Ω) Ω This functional achieves its global minimum. By comparison with our functional in X and Lemma 1 we get a minimum in C 1 (Ω).

62 Main ideas/steps of the proofs We check that the theorem in [BN2] is easy to prove in our setting. Proposition 3 Let z H α/2 (Ω) be a local minimum of I in C 1 (Ω), i.e., there exists r > such that I(z ) I(z +z) z C 1 (Ω) with z C 1 (Ω) r. (.3) Then z is a local minimum of I in H α/2 (Ω), that is, there exists ε > such that I(z ) I(z +z) z H α/2 (Ω) with z H α/2 (Ω) ε.

63 Main ideas/steps of the proofs We check that the theorem in [BN2] is easy to prove in our setting. Proposition 3 Let z H α/2 (Ω) be a local minimum of I in C 1 (Ω), i.e., there exists r > such that I(z ) I(z +z) z C 1 (Ω) with z C 1 (Ω) r. (.4) Then z is a local minimum of I in H α/2 (Ω), that is, there exists ε > such that I(z ) I(z +z) z H α/2 (Ω) with z H α/2 (Ω) ε. [BN2] H. Brezis, L. Nirenberg H 1 versus C 1 local minimizers CRAS 1993.

64 Main ideas/steps of the proofs We check that the theorem in [BN2] is easy to prove in our setting. Proposition 3 Let z H α/2 (Ω) be a local minimum of I in C 1 (Ω), i.e., there exists r > such that I(z ) I(z +z) z C 1 (Ω) with z C 1 (Ω) r. (.5) Then z is a local minimum of I in H α/2 (Ω), that is, there exists ε > such that I(z ) I(z +z) z H α/2 (Ω) with z H α/2 (Ω) ε. We make a translation in the nonlinearity of the functional in order to get that minimum at the origin. Lemma 3 (i) The translated functional has a local minimum at the origin in H α/2 (Ω). Moreover, (ii) the extended functional has a local minimum at the origin in X α (C Ω).

65 Eduardo Colorado p. 21/23 Main ideas/steps of the proofs We check that the theorem in [BN2] is easy to prove in our setting. Proposition 3 Let z H α/2 (Ω) be a local minimum of I in C 1 (Ω), i.e., there exists r > such that I(z ) I(z +z) z C 1 (Ω) with z C 1 (Ω) r. (.6) Then z is a local minimum of I in H α/2 (Ω), that is, there exists ε > such that I(z ) I(z +z) z H α/2 (Ω) with z H α/2 (Ω) ε. We make a translation in the nonlinearity of the functional in order to get that minimum at the origin. Lemma 3 (i) The translated functional has a local minimum at the origin in H α/2 (Ω). Moreover, (ii) the extended functional has a local minimum at the origin in X α (C Ω). The part (i) follows by simple computations as in the classical case, [ABC]. The second part (ii) follows by using the isometry between H α/2 (Ω) and X α (C Ω), and the fact that the α-harmonic extension minimize the norm in X α (C Ω). [ABC] A. Ambrosetti, H. Brezis, G. Cerami J. Funct. Analysis 1994.

66 Eduardo Colorado p. 22/23 Main ideas/steps of the proofs Lemma 4 If v = is the only critical point of J in X α (C Ω) then J satisfies a local (PS) c condition for any c < c.

67 Eduardo Colorado p. 22/23 Main ideas/steps of the proofs Lemma 4 If v = is the only critical point of J in X α (C Ω) then J satisfies a local (PS) c condition for any c < c. In order to prove it, first we show that the (PS) c sequence of the previous lemma is tight. Lemma 5 For any η > there exists ρ > such that {y>ρ } Ω y 1 α z n 2 dxdy < η, n N.

68 Eduardo Colorado p. 22/23 Main ideas/steps of the proofs Lemma 4 If v = is the only critical point of J in X α (C Ω) then J satisfies a local (PS) c condition for any c < c. Proof of Lemma 4. Let {w n } be a (PS) c : J(wn ) c < c, J (w n ). By Lemma 5 and Proposition 3 (concentration-compactness), there exists an index set I (at most countable) and a sequence of points {x k } Ω, k I and real positive numbers µ k, ν k such that (up to a subsequence) y 1 α w n 2 µ y 1 α w 2 + k Iµ k δ xk and w n (,) 2 α ν = w (,) 2 α + k Iν k δ xk in the sense of measures, and moreover, µ k S(α,N)νk, for every k I. 2 2 α

69 Eduardo Colorado p. 22/23 Main ideas/steps of the proofs Lemma 4 If v = is the only critical point of J in X α (C Ω) then J satisfies a local (PS) c condition for any c < c. Proof of Lemma 4. Let φ be a regular non-increasing cut-off function, φ = 1 in B 1, φ = in B2 c. Then φ ε(x,y) = φ(x/ε,y/ε), it is clear that φ ε C. We denote ε Γ 2ε = B + 2ε (x k ) {y = }. Clearly, = lim n ( w n 2 αφ ε dx+λ Γ 2ε κ α lim y 1 α w n, φ ε w n dxdy n C Ω Γ 2ε w n q+1 φ ε dx κ α B + 2ε (x k ) y 1 α w n 2 φ ε dxdy ). Passing to the limit we obtain lim ε [ φ ε dν +λ w q+1 φ ε dx κ α Γ 2ε Γ 2ε B + 2ε (x k ) φ ε dµ ] = ν k κ α µ k. 2 2 α Since µ k S(α,N)νk, we get that ν k = or ν k (κ α S(α,N)) N α, k I.

70 Eduardo Colorado p. 22/23 Main ideas/steps of the proofs Lemma 4 If v = is the only critical point of J in X α (C Ω) then J satisfies a local (PS) c condition for any c < c. Proof of Lemma 4. Suppose that ν k for some k I. Then c = lim n J(w n) 1 2 J (w n ),w n α 2N Ω w 2 α dx+ α ( 1 2N ν k +λ 2 1 ) w q+1 dx q +1 Ω α 2N (κ αs(α,n)) N/α = c, a contradiction. Therefore ν k = k I, so u n u strongly in L 2 α(ω) and we conclude easily.

71 Main ideas/steps of the proofs Now the idea is the following. One consider the minimizers to the Sobolev inequality ε u ε (x) = (N α)/2 and its α-harmonic extension w ε, not explicit. Consider an ( x 2 +ε 2 ) (N α)/2 appropriate cut-off function φ in C Ω centered at the origin (we can assume Ω), and denote ψ ε = φw ε φw ε. Define Γ ε = {γ C([,1],X α (C Ω )) : γ() =, γ(1) = t ε ψ ε } for some t ε > such that J(t ε ψ ε ) <. And consider the minimax value Then we are going to prove that for ε 1, c ε = inf γ Γ ε max J(γ(t)) : t 1. c ε sup t J(tψ ε ) < c = α 2N (κ αs(α,n)) N/α. By the Mountain Pass Theorem, there exists a (PS) sequence {w n } X α (C Ω) verifying J(w n ) c ε < c, J (w n ). So by Lemma 4 we finish.

72 Main ideas/steps of the proofs Lemma 6 With the above notation, taking ε 1, φw ε 2 X α (C Ω) = w ε 2 X α (C Ω) +O(εN α ), φu ε 2 L 2 (Ω) = cε α +O(ε N α ) if N > 2α, cε α log(1/ε)+o(ε α ) if N = 2α, and if r = N+α N α = 2 α 1, φu ε r L r (Ω) cεn α 2, α < N < 2α.

73 Main ideas/steps of the proofs Lemma 6 With the above notation, taking ε 1, φw ε 2 X α (C Ω) = w ε 2 X α (C Ω) +O(εN α ), φu ε 2 L 2 (Ω) = cε α +O(ε N α ) if N > 2α, cε α log(1/ε)+o(ε α ) if N = 2α, and if r = N+α N α = 2 α 1, φu ε r L r (Ω) cεn α 2, α < N < 2α. Taking into account that the family u ε and ( the ) Poisson kernel are self-similar u ε (x) = ε α N 2 u 1 (x/ε), Py α (x) = 1 y N P1 α x, this gives that the family w y ε is also self-similar, more precisely ( w ε (x,y) = ε α N x 2 w 1 ε, y ). ε We will denote w 1,α = w 1.

74 Main ideas/steps of the proofs Lemma 6 With the above notation, taking ε 1, φw ε 2 X α (C Ω) = w ε 2 X α (C Ω) +O(εN α ), φu ε 2 L 2 (Ω) = cε α +O(ε N α ) if N > 2α, cε α log(1/ε)+o(ε α ) if N = 2α, and if r = N+α N α = 2 α 1, φu ε r L r (Ω) cεn α 2, α < N < 2α. Lemma 7 w ε (x,y) = ε α N 2 w 1 ( x ε, y ε) ; w1,α = w 1. w 1,α (x,y) c y w 1,α(x,y), α >, (x,y) R N+1 + w 1,α (x,y) cw 1,α 1 (x,y), α > 1, (x,y) R N+1 +. w 1,α (x,y) Cε N α, 1 2ε (x,y) 1 ε.

75 Main ideas/steps of the proofs Lemma 6 With the above notation, taking ε 1, φw ε 2 X α (C Ω) = w ε 2 X α (C Ω) +O(εN α ), φu ε 2 L 2 (Ω) = cε α +O(ε N α ) if N > 2α, cε α log(1/ε)+o(ε α ) if N = 2α, and if r = N+α N α = 2 α 1, φu ε r L r (Ω) cεn α 2, α < N < 2α. Lemma 8 For ε 1, sup t> J(tψ ε ) < c.

76 Eduardo Colorado p. 23/23 Main ideas/steps of the proofs Lemma 6 With the above notation, taking ε 1, φw ε 2 X α (C Ω) = w ε 2 X α (C Ω) +O(εN α ), φu ε 2 L 2 (Ω) = cε α +O(ε N α ) if N > 2α, cε α log(1/ε)+o(ε α ) if N = 2α, and if r = N+α N α = 2 α 1, φu ε r L r (Ω) cεn α 2, α < N < 2α. Lemma 8 For ε 1, sup t> J(tψ ε ) < c. For example for N > 2α, after some computations, sup t> J(tψ ε ) c cε α +O(ε N α ) < c.