The Pohozaev identity for the fractional Laplacian

Transcript

1 The Pohozaev identity for the fractional Laplacian Xavier Ros-Oton Departament Matemàtica Aplicada I, Universitat Politècnica de Catalunya (joint work with Joaquim Serra) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

2 Outline of the talk The classical Pohozaev identity; applications The Dirichlet semilinear problem for the fractional Laplacian The Pohozaev identity for the fractional Laplacian Applications Sketch of the proof Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February 013 / 18

3 The classical Pohozaev identity bounded Lipschitz domain, u = f (u) in u = 0 on, (1) Theorem (Pohozaev) ( n) u f (u)dx + n F (u)dx = u (x ν)dσ Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

4 Applications of the classical Pohozaev identity ( n) u f (u)dx + n F (u)dx = u (x ν)dσ Nonexistence of solutions: critical exponent u = u n+ n Ground states in R n : monotonicity formulas, estimates Radial symmetry: proof of P.-L. Lions combining the Pohozaev identity with the isoperimetric inequality Stable solutions: uniqueness, H 1 interior regularity etc. Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

5 Proof of the classical Pohozaev identity First note that (x u) = u + x ( u). Then, integrating by parts twice and using that u 0 on, we obtain (x u) u = u u + u x ( u) + (x u)( u ν)dσ = ( n) u u (x u) u + u (x ν)dσ We have used that u ν = u on. Finally, since u = f (u), then (x u) u = x F (u) = n F (u), and the identity follows. Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

6 The Dirichlet semilinear problem with ( ) s bounded C 1,1 domain, δ(x) := dist (x, ), f C 1 ( ) s u = f (u) in u = 0 in R n \, ( ) s u = g Theorem (X.R., J. Serra) (i) u C s (R n ) (ii) u/δ s C α () (iii) [u] C β (B ρ/ ) Cρs β u 0 B ρ B ρ/ (iv) [ u/δ s] C β (B ρ/ ) Cρα β Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

7 The Pohozaev identity for the fractional Laplacian bounded C 1,1 domain, ( ) s u = f (u) in u = 0 in R n \, Theorem (X. R., J. Serra) Denote δ(x) := dist (x, ). Then u/δ s C α () and ( u ) (s n) uf (u)dx + n F (u)dx = Γ(1 + s) (x ν)dσ, δ s where Γ is the gamma function. Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

8 Corollary: nonexistence results bounded C 1,1 domain, ( ) s u = f (u) in u = 0 in R n \, Corollary Assume that is star-shaped and F (t) < n s n t f (t) for all t. Then the problem admits no nontrivial solution. For example, for f (u) = u p we obtain nonexistence for p n+s n s. For positive solutions, this was done by [Fall-Weth, 1] with moving planes. Existence for subcritical p by [Servadei-Valdinoci, 1]. Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

9 Pohozaev identity with ( ) s Proposition (X. R., J. Serra) Assume 1 bounded C 1,1 domain u C s (R n ), u 0 outside, u/δ s C α () 3 Interior C β estimates for u and u/δ s, β < 1 + s 4 ( ) s u is bounded in Then (x u)( ) s u = s n u( ) s u Γ(1 + s) ( u δ s ) (x ν) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

10 Main consequences Changing the origin in our identity, we deduce the following Theorem (X. R., J. Serra) Under the same hypotheses of the Proposition, ( ) s u v xi = u xi ( ) s v + Γ(1 + s) u v δ s δ s ν i It has a local boundary term! Note the contrast with the nonlocal flux in the formula for f (x, u) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

11 Sketch of the Proof (Star-shaped domains) 1 u λ (x) = u(λx) (x u)( ) s u = d dλ u λ ( ) s u star-shaped u λ vanishes outside for λ > 1 u λ ( ) s u = ( ) s uλ ( ) s u R n Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

12 R n ( ) s uλ ( ) s u = λ s where w = ( ) s u. Therefore, R n ( ( ) s u ) (λx)( ) s u(x) dx = λ R s w(λx)w(x) dx n = λ s n w(λ 1 y)w(λ 1 y) dy R n (x u)( ) s u = s n w + 1 d R dλ n R n w λ w 1/λ where w λ (x) = w(λx). Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

13 R n ( ) s uλ ( ) s u = λ s where w = ( ) s u. Therefore, (x u)( ) s u = s n where w λ (x) = w(λx). R n ( ( ) s u ) (λx)( ) s u(x) dx = λ R s w(λx)w(x) dx n = λ s n w(λ 1 y)w(λ 1 y) dy R n u( ) s u + 1 d dλ R n w λ w 1/λ Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

14 What about d dλ λ=1 + R n w λ w 1/λ? Important properties: I(ϕ) = d dλ ϕ(λx)ϕ(x/λ) dx R n 1 I(ϕ) 0 since ( ) 1 ( ) 1 ϕ(λx)ϕ(x/λ)dx ϕ (λx)dx ϕ (x/λ)dx = ϕ R n R n Rn R n ψ smooth I(ψ) = 0 3 If I(ψ) = 0 I(ϕ + ψ) = I(ϕ) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

15 What about d dλ λ=1 + R n w λ w 1/λ? Important properties: I(ϕ) = d dλ ϕ(λx)ϕ(x/λ) dx R n 1 I(ϕ) 0 since ( ) 1 ( ) 1 ϕ(λx)ϕ(x/λ)dx ϕ (λx)dx ϕ (x/λ)dx = ϕ R n R n Rn R n ψ smooth I(ψ) = 0 3 If I(ψ) = 0 I(ϕ + ψ) = I(ϕ) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

16 What about d dλ λ=1 + R n w λ w 1/λ? Important properties: I(ϕ) = d dλ ϕ(λx)ϕ(x/λ) dx R n 1 I(ϕ) 0 since ( ) 1 ( ) 1 ϕ(λx)ϕ(x/λ)dx ϕ (λx)dx ϕ (x/λ)dx = ϕ R n R n Rn R n ψ smooth I(ψ) = 0 3 If I(ψ) = 0 I(ϕ + ψ) = I(ϕ) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

17 What about d dλ λ=1 + R n w λ w 1/λ? Important properties: I(ϕ) = d dλ ϕ(λx)ϕ(x/λ) dx R n 1 I(ϕ) 0 since ( ) 1 ( ) 1 ϕ(λx)ϕ(x/λ)dx ϕ (λx)dx ϕ (x/λ)dx = ϕ R n R n Rn R n ψ smooth I(ψ) = 0 3 If I(ψ) = 0 I(ϕ + ψ) = I(ϕ) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

18 What about d dλ λ=1 + R n w λ w 1/λ? We want to compute: I(w) = d dλ w λ w 1/λ R n Reduce to a 1 D calculation Use star-shaped (t, z)-coordinates (/3, z) z (1/3, z) 0 (1/, z) z x = tz, z, t > 0 d dλ w λ w 1/λ = R n d dλ (z ν)dσ(z) 0 t n 1 w(λtz)w ( tz λ ) dt Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

19 What about d dλ λ=1 + R n w λ w 1/λ? We want to compute: I(w) = d dλ w λ w 1/λ R n Reduce to a 1 D calculation Use star-shaped (t, z)-coordinates (/3, z) z (1/3, z) 0 (1/, z) z x = tz, z, t > 0 d dλ w λ w 1/λ = R n (z ν)dσ(z) d dλ 0 t n 1 w(λtz)w ( tz λ ) dt Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

20 What do we know about w = ( ) s/ u? Proposition (X. R., J. Serra) Fix z. Then, where w(tz) = ( ) s/ { u(tz) = c 1 log t 1 + c χ (0,1) (t) } u (z) + h(t) δs d dλ 0 ( t t n 1 h(λt)h dt = 0 λ) c 1 = Γ(1 + s) sin ( ) πs, and c = π π tan ( πs ) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

21 Summarising... { w(tz) = c 1 log t 1 + c χ (0,1) (t) } u (z) + h(t) δs d dλ w λ w 1/λ = (z ν)dσ(z) d ( tz ) R n dλ t n 1 w(λtz)w dt 0 λ = (z ν)dσ(z) d ( u ) ( t ) dλ δ s (z) t n 1 φ s (λt)φ s dt 0 λ u ) = (z ν)dσ(z)( δ s (z) C(s) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

22 Summarising... w(tz) = φ s (t) u (z) + h(t) δs { where φ s (t) = c 1 log t 1 + c χ (0,1) (t) } d dλ w λ w 1/λ = (z ν)dσ(z) d ( tz ) R n dλ t n 1 w(λtz)w dt 0 λ = (z ν)dσ(z) d ( u ) ( t ) dλ δ s (z) t n 1 φ s (λt)φ s dt 0 λ u ) = (z ν)dσ(z)( δ s (z) C(s) Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

23 And if the domain is not star-shaped... Key observations: 1 Pohozaev identity is quadratic in u and it comes from a bilinear identity (x u)( )s u = s n u( )s u Γ(1+s) (x u)( )s v + (x v)( )s u = u( )s v + s n v( )s u Γ(1 + s) s n every C 1,1 domain is locally star-shaped ( u δ s ) (x ν) u δ s v δ s (x ν) 3 the bilinear identity holds easily when u and v have disjoint support Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

24 And if the domain is not star-shaped... Key observations: 1 Pohozaev identity is quadratic in u and it comes from a bilinear identity (x u)( )s u = s n u( )s u Γ(1+s) (x u)( )s v + (x v)( )s u = u( )s v + s n v( )s u Γ(1 + s) s n every C 1,1 domain is locally star-shaped ( u δ s ) (x ν) u δ s v δ s (x ν) 3 the bilinear identity holds easily when u and v have disjoint support Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

25 And if the domain is not star-shaped... Key observations: 1 Pohozaev identity is quadratic in u and it comes from a bilinear identity (x u)( )s u = s n u( )s u Γ(1+s) (x u)( )s v + (x v)( )s u = u( )s v + s n v( )s u Γ(1 + s) s n every C 1,1 domain is locally star-shaped ( u δ s ) (x ν) u δ s v δ s (x ν) 3 the bilinear identity holds easily when u and v have disjoint support Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

26 And if the domain is not star-shaped... Key observations: 1 Pohozaev identity is quadratic in u and it comes from a bilinear identity (x u)( )s u = s n u( )s u Γ(1+s) (x u)( )s v + (x v)( )s u = u( )s v + s n v( )s u Γ(1 + s) s n every C 1,1 domain is locally star-shaped ( u δ s ) (x ν) u δ s v δ s (x ν) 3 the bilinear identity holds easily when u and v have disjoint support Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18

27 The end Thank you! Xavier Ros-Oton (UPC) The Pohozaev identity for the fractional Laplacian BCAM, Bilbao, February / 18