Large Solutions for Fractional Laplacian Operators

Transcription

1 Some results of the PhD thesis Nicola Abatangelo Advisors: Louis Dupaigne, Enrico Valdinoci Université Libre de Bruxelles October 16 th, 2015

2 The operator Fix s (0, 1) and consider a measurable function u : R N R.

3 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y

4 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, where {e t } t>0 denotes the heat semigroup on R N

5 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 F[( ) s u](ξ) = ξ 2s Fu(ξ), where F denotes the Fourier transform

6 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 F[( ) s u](ξ) = ξ 2s Fu(ξ), 4 ( ) s u(x) = lim y 0 y 1 2s yu(x, y) where { div ( y 1 2s U ) = 0 in R N (0, + ) U(x, 0) = u(x) in R N {0}

7 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 F[( ) s u](ξ) = ξ 2s Fu(ξ), 4 ( ) s u(x) = lim y 0 y 1 2s yu(x, y) where { div ( y 1 2s U ) = 0 in R N (0, + ) 5 ( ) s u(x) E x[u(x t)] u(x) = lim, t 0 t U(x, 0) = u(x) in R N {0} where {X t} t>0 is a (2s)-stable process (with jumps)

8 The operator Fix s (0, 1) and consider a measurable function u : R N R. 1 ( ) s u(x) = 4s Γ(N/2 + s) s π N/2 Γ(1 s) p.v. u(x) u(y) dy R N N+2s x y 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 F[( ) s u](ξ) = ξ 2s Fu(ξ), 4 ( ) s u(x) = lim y 0 y 1 2s yu(x, y) where { div ( y 1 2s U ) = 0 in R N (0, + ) 5 ( ) s u(x) E x[u(x t)] u(x) = lim, t 0 t U(x, 0) = u(x) in R N {0}

9 Boundary blow-up solutions Consider a bounded R N and f C 1 (R) positive at +. A boundary blow-up (or large) solution is some u C 2 () satisfying u = f(u) in, lim u(x) = +. (L) x

10 Boundary blow-up solutions Consider a bounded R N and f C 1 (R) positive at +. A boundary blow-up (or large) solution is some u C 2 () satisfying u = f(u) in, lim u(x) = +. (L) x The Keller-Osserman condition Problem (L) has a solution if and only if + dt < +, where F = f. F (t)

11 Boundary blow-up solutions Consider a bounded R N and f C 1 (R) positive at +. A boundary blow-up (or large) solution is some u C 2 () satisfying u = f(u) in, lim u(x) = +. (L) x The Keller-Osserman condition Problem (L) has a solution if and only if + dt < +, where F = f. F (t) Ex. When f(u) = u p, the Keller-Osserman condition reads as p < 1, i.e. p > 1.

12 Previous results Ref. Felmer-Quaas, Chen-Felmer-Quaas ( ) s u = u p in, u = 0 in C, lim u(x) = +. x There exists τ 0 (s) ( 1, 0) such that: s < p < 1 2s τ 0 (s) : existence of a solution u δ 2s/(p 1), moreover it is the unique solution with this boundary behaviour 2 max { 1, 1 + 2τ 0 (s) 2s τ 0 (s) } < p < 1 2s τ 0 (s) : 3 p > 1: nonexistence of a solution u δ τ for any existence of infinitely many solutions u δ τ 0(s) τ ( 1, 0) \ { 2s/(p 1), τ 0 (s)}

13 The Dirichlet problem Ref. { ( ) s u = f in u = g in C : Barles-Chasseigne-Imbert, Chen-Véron, RosOton-Serra, Karlsen-Petitta-Ulusoy,...

14 The Dirichlet problem Ref. { ( ) s u = f in u = g in C : Barles-Chasseigne-Imbert, Chen-Véron, RosOton-Serra, Karlsen-Petitta-Ulusoy,... representation formula u(x) = G (x, y) f(y) dy + P (x, y) g(y) dy, C

15 The Dirichlet problem Ref. { ( ) s u = f in u = g in C : Barles-Chasseigne-Imbert, Chen-Véron, RosOton-Serra, Karlsen-Petitta-Ulusoy,... representation formula u(x) = G (x, y) f(y) dy + P (x, y) g(y) dy, C integration by parts formula: for any φ C(R N ), s.t. φ 0 in C and 1 ( ) s φ C c () u( ) s φ = fφ g( ) s φ. C

16 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy CB Γ(N/2) sin(πs) π 1+N/2.

17 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 Γ(N/2) sin(πs) π 1+N/2. and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy = CB = c(n, s) c(n, s + σ) dy (1 x 2 ) s CB x y N ( y 2 1) s+σ =

18 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 Γ(N/2) sin(πs) π 1+N/2. and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy = CB c(n, s) c(n, s + σ) dy (1 x 2 ) s+σ = (1 x 2 ) σ CB x y N ( y 2 1) s+σ =

19 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 Γ(N/2) sin(πs) π 1+N/2. and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy = CB c(n, s) c(n, s + σ) dy (1 x 2 ) s+σ = (1 x 2 ) σ CB x y N ( y 2 1) s+σ =

20 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 Γ(N/2) sin(πs) π 1+N/2. and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy = CB c(n, s) c(n, s + σ) dy (1 x 2 ) s+σ = (1 x 2 ) σ CB x y N ( y 2 1) s+σ =

21 An example on the ball The Poisson kernel for the ball B = B 1 (0) is given by ( c(n, s) 1 x 2 ) s P B (x, y) = x y N y 2, c(n, s) = 1 Ref. Landkof For any σ (0, 1 s), consider the external datum g σ(y) = c(n, s + σ) ( y 2 1) σ, y > 1 Γ(N/2) sin(πs) π 1+N/2. and compute the induced s-harmonic function u σ: u σ(x) = PB s (x, y) gσ(y) dy = CB c(n, s) c(n, s + σ) dy (1 x 2 ) s+σ = (1 x 2 ) σ CB x y N ( y 2 1) s+σ = c(n, s) (1 x 2 ) σ

22 The function u σ(x) = c(n, s) (1 x 2) σ c(n, s + σ) ( x 2 1 ) σ x < 1, σ (0, 1 s) x > 1 is s-harmonic in B for any σ (0, 1 s) and c(n, s + σ) = sin(πs + πσ) σ 1 s 0.

23 The function u σ(x) = c(n, s) (1 x 2) σ c(n, s + σ) ( x 2 1 ) σ x < 1, σ (0, 1 s) x > 1 is s-harmonic in B for any σ (0, 1 s) and c(n, s + σ) = sin(πs + πσ) σ 1 s 0. Letting σ 1 s, one gets s-harmonic in B, u 1 s (x) = { c(n, s) ( 1 x 2 ) (1 s) x < 1 0 x > 1

24 The function u σ(x) = c(n, s) (1 x 2) σ c(n, s + σ) ( x 2 1 ) σ x < 1, σ (0, 1 s) x > 1 is s-harmonic in B for any σ (0, 1 s) and c(n, s + σ) = sin(πs + πσ) σ 1 s 0. Letting σ 1 s, one gets s-harmonic in B, i.e. u 1 s (x) = { c(n, s) ( 1 x 2 ) (1 s) x < 1 0 x > 1 { ( ) s u 1 s = 0 in x < 1, u 1 s = 0 in x > 1. Ref. Bogdan, Ryznar

25 A maximum principle Theorem ) Consider u C 2s+α loc () L (R 1 N, (1 + x ) N 2s dx satisfying { ( ) s u 0 in, u 0 in C. Suppose in addition that there exists O R N such that O for which u C(O). Then u 0 in.

26 A maximum principle Theorem ) Consider u C 2s+α loc () L (R 1 N, (1 + x ) N 2s dx satisfying { ( ) s u 0 in, u 0 in C. Suppose in addition that there exists O R N such that O for which u C(O). Then u 0 in. Proof. Call + := {u > 0}: it holds +. Also, u admits a maximum in +, attained at some x +. So, ( ) s u(x ) u(x ) u(y) = p.v. dy > 0 R N N+2s x y which shows + =.

27 A second boundary datum Consider h C( ) and build the following: Consider the family x = θ(x) δ(x) δ(x), θ(x), {ϕ ε} ε 1D mollifier, { h(θ(x)) ϕε(δ(x)) δ(x) s f ε(x) = δ(x) < ε, 0 otherwise. { u ε(x) := G (x, y) f ε(y) dy } ε of solutions to { ( ) s u ε = f ε in, u ε = 0 in C.

28 On the one hand u ε(x) = G (x, ) f ε where M (x, θ) = ε 0 lim y y θ M (x, ) h dσ =: u(x), G (x, y) δ(y) s.

29 On the one hand u ε(x) = G (x, ) f ε where M (x, θ) = ε 0 lim y y θ M (x, ) h dσ =: u(x), G (x, y) δ(y) s. On the other hand, for any φ C(R N ), φ 0 in C, 1 ( ) s φ Cc u ε( ) s ε 0 φ = f εφ h φ δ s dσ. (), Rmk. Recall that (Bogdan, Grubb, RosOton-Serra) φ δ s C α ( ), if C 1,1.

30 The representation u = M (x, ) h dσ provides an s-harmonic function and u ( ) s φ = h φ δ s dσ.

31 The representation u = M (x, ) h dσ provides an s-harmonic function and u ( ) s φ = h φ δ s dσ. Moreover and h 1 (x) := M (x, θ) dσ(θ) δ(x) s 1 1 x M (x, θ) h(θ) dσ(θ) h 1 (x) h(y). x y

32 The test function space { T () := φ C(R N ) : φ 0 in C, 1 ( ) s φ Cc } ().

33 The test function space { T () := φ C(R N ) : φ 0 in C, 1 ( ) s φ Cc } (). 1 For any x R N \ ( ) s φ(x) = P (y, x) ( ) s φ(y) dx = ( ) s φ(x) ( ) s φ L ()δ(x) s min { 1, δ(x) N s},

34 The test function space { T () := φ C(R N ) : φ 0 in C, 1 ( ) s φ Cc } (). 1 For any x R N \ ( ) s φ(x) = P (y, x) ( ) s φ(y) dx = ( ) s φ(x) ( ) s φ L ()δ(x) s min { 1, δ(x) N s}, 2 for any θ φ δ s (θ) = M (y, θ) ( ) s φ(y) dy C( ).

35 The test function space { T () := φ C(R N ) : φ 0 in C, 1 ( ) s φ Cc } (). 1 For any x R N \ ( ) s φ(x) = P (y, x) ( ) s φ(y) dx = ( ) s φ(x) ( ) s φ L ()δ(x) s min { 1, δ(x) N s}, 2 for any θ φ δ s (θ) = M (y, θ) ( ) s φ(y) dy Rmk. Both are derived from the representation φ(x) = G (x, y) ( ) s φ(y) dy. C( ).

36 L 1 weak solutions Definition Let λ M(), µ M(C), ν M( ) be three Radon measures. A function u L 1 () is said to be an L 1 weak solution to ( ) s u = λ in u = µ in C (1) u = ν on h 1 if and only if for any φ T () u( ) s φ = φ dλ ( ) s φ dµ + C φ δ s dν.

37 L 1 weak solutions Definition Let λ M(), µ M(C), ν M( ) be three Radon measures. A function u L 1 () is said to be an L 1 weak solution to ( ) s u = λ in u = µ in C (1) u = ν on h 1 if and only if for any φ T () u( ) s φ = Theorem Problem (1) is well-posed when φ dλ ( ) s φ dµ + C φ δ s dν. δ s L 1 (, dλ), δ s min{1, δ N s } L 1 (C, dµ), ν ( ) <.

38 The nonlinear problem ( ) s u = f(u) in, f C(R), f(0) = 0, f increasing u = g in C, g 0, g δ s min{1, δ N s } < C u = h on, 0 h C( ) h 1 u( ) s φ = f(u)φ g ( ) s φ + h φ C δ s, φ T (). 1 If h 0: existence and uniqueness. 2 If h 0 and f(δ s 1 )δ s L 1 (): existence and uniqueness, f(δ s 1 )δ s L 1 (): nonexistence.

39 Very large solutions Consider the sequence {u k } k N of solutions to ( ) s u k = f(u k ) in, u k = 0 in C, u k = k on. h 1 Question. For what nonlinearities the sequence converges to a solution u of ( ) s u = f(u) in u = 0 in C? u = + on h 1

40 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k

41 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k 2. the candidate profile at the boundary is because heuristically ψ(δ s ), where ψ = f(ψ), ψ(0) = + ( ) s (ψ δ s ) ψ(δs ) δ 2s ψ (δ s ) = f(ψ(δ s )) so we must have:

42 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k 2. the candidate profile at the boundary is because heuristically ψ(δ s ), where ψ = f(ψ), ψ(0) = + ( ) s (ψ δ s ) ψ(δs ) δ 2s ψ (δ s ) = f(ψ(δ s )) so we must have: 2a. ψ(δ s ) L 1 () + ( t f(t) ) 1/(2s) dt <

43 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k 2. the candidate profile at the boundary is because heuristically ψ(δ s ), where ψ = f(ψ), ψ(0) = + ( ) s (ψ δ s ) ψ(δs ) δ 2s ψ (δ s ) = f(ψ(δ s )) so we must have: 2a. ψ(δ s ) L 1 () + ( t f(t) ) 1/(2s) dt < 2b. δ 1 s ψ(δ s ) + as δ 0 t (1+s)/(1 s) f(t) 0 as t,

44 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k 2. the candidate profile at the boundary is because heuristically ψ(δ s ), where ψ = f(ψ), ψ(0) = + ( ) s (ψ δ s ) ψ(δs ) δ 2s ψ (δ s ) = f(ψ(δ s )) so we must have: 2a. ψ(δ s ) L 1 () + ( t f(t) ) 1/(2s) dt < 2b. δ 1 s ψ(δ s ) + as δ 0 t (1+s)/(1 s) f(t) 0 as t,... is implied by 1.

45 1. f(δ s 1 ) δ s < +, otherwise it is not possible to build {u k } k 2. the candidate profile at the boundary is because heuristically ψ(δ s ), where ψ = f(ψ), ψ(0) = + ( ) s (ψ δ s ) ψ(δs ) δ 2s ψ (δ s ) = f(ψ(δ s )) so we must have: 2a. ψ(δ s ) L 1 () + ( t f(t) ) 1/(2s) dt < 2b. δ 1 s ψ(δ s ) + as δ 0 t (1+s)/(1 s) f(t) 0 as t,... is implied by 1. Theorem There exists U L 1 () such that U ψ(δ s ) and ( ) s U f(u).

46 What is still missing ( ) s u = u p u = 0 u = + h 1 in in C on nonexistence sequence converges? existence nonexistence 1 1+s sequence exits L 1 1+2s 1+s 1-s p black: proved, red: conjectures.

47 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R.

48 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R. 1 ( ) s u(x) = p.v. [u(x) u(y)] J(x, y) dy + κ(x)u(x), where J(x, y) κ(x) δ(x) δ(y) x y N+2s 1 δ(x) δ(y) + x y 2, 1 δ(x) 2s, x x, y

49 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R. 1 ( ) s u(x) = p.v. [u(x) u(y)] J(x, y) dy + κ(x)u(x), 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, where {e t } t>0 denotes the heat semigroup on

50 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R. 1 ( ) s u(x) = p.v. [u(x) u(y)] J(x, y) dy + κ(x)u(x), 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 ( ) s u(x) = λ s jûj ϕ j (x), û j = u ϕ j, j=1 where {(λ j, ϕ j )} j are eigenvalues and eigenfunctions of,

51 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R. 1 ( ) s u(x) = p.v. [u(x) u(y)] J(x, y) dy + κ(x)u(x), 2 ( ) s u(x) = 3 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, j=1 λ s jûj ϕ j (x), û j = u ϕ j, 4 ( ) s u(x) = lim y 1 2s yu(x, y) where y 0 div ( y 1 2s U ) = 0 in (0, + ) U(x, y) = 0 on (0, + ) U(x, 0) = u(x) on {0}

52 The spectral fractional Laplacian Fix s (0, 1), a bounded R N open and consider a measurable function u : R. 1 ( ) s u(x) = p.v. [u(x) u(y)] J(x, y) dy + κ(x)u(x), 2 ( ) s u(x) = s + u(x) e t u(x) Γ(1 s) 0 t 1+s dt, 3 ( ) s u(x) = λ s jûj ϕ j (x), û j = u ϕ j, j=1 4 ( ) s u(x) = lim y 1 2s yu(x, y) where y 0 div ( y 1 2s U ) = 0 in (0, + ) U(x, y) = 0 on (0, + ) U(x, 0) = u(x) on {0} 5 ( ) s u(x) E x[u( u(x) = lim t)], t 0 t where { X t} t>0 is a subordinated killed Brownian motion

53 Very large solutions ( ) s u = u p u = + h 2 in on nonexistence sequence converges? existence nonexistence 1 1 sequence exits L δ 1+s 1 1-s p black: proved, red: conjectures.

54