ODE solutions for the fractional Laplacian equations arising in co
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1 ODE solutions for the fractional Laplacian equations arising in conformal geometry Granada, November 7th, 2014
2 Background
3 Classical Yamabe problem (M n, g) = Riemannian manifold; n 3, g = u 4 n 2 g conformal metric with constant scalar curvature?
4 Classical Yamabe problem (M n, g) = Riemannian manifold; n 3, g = u 4 n 2 g conformal metric with constant scalar curvature? Or u C strictly positive solution for the following equation? c n g u + R g u = cu n+2 n 2, (1)
5 Definition of fractional Laplacian in R n Let γ (0, 1) ( ) γ f (x) = P.V. R n (f (x+y) f (x)) y (n+2γ) ( ) γ f = F 1 ( ξ 2γ (Ff )); dy, Let U(x, y) be solution of ; { div(y a U) = 0, x R n, y R +, U(x, 0) = f (x), x R n. ( ) γ f = d γ lim y 0 + y a y U, where a = 1 2γ and d γ = 22γ 1 Γ(γ). γγ( γ)
6 Geometric setting Let X n+1 be a smooth manifold with smooth boundary M n. Defining function: a function ρ on X n+1 which satisfies: ρ > 0 in X, ρ = 0 on M, dρ 0 on M. g + is conformally compact if ḡ = ρ 2 g + is smooth up to the boundary. Conformal infinity: (M n, [ĝ]); where ĝ = ḡ M. Conformal change of metric on M: ĝ w = w 4 n 2γ ĝ. Conformally compact Einstein manifold: Ric g + = ng +. Normal form: g + = dρ2 +g ρ ρ 2 with g ρ M = ĝ.
7 Conformal fractional Laplacian (Mazzeo-Melrose and Graham-Zworski) Given f C (M), s C such that Re(s) > n 2, there exists a solution of the form u = F ρ n s + Hρ s ; F, H C (X ), F ρ=0 = f, (2) for the eigenvalue problem g +u s(n s)u = 0, in X. (3)
8 Conformal fractional Laplacian (Mazzeo-Melrose and Graham-Zworski) Given f C (M), s C such that Re(s) > n 2, there exists a solution of the form u = F ρ n s + Hρ s ; F, H C (X ), F ρ=0 = f, (2) for the eigenvalue problem g +u s(n s)u = 0, in X. (3) Definition Scattering operator: S(s)f = H M. Conformally covariant fractional powers of the Laplacian: Pĝγ = d γ S( n 2 + γ); dγ = 22γ Γ(γ) Γ( γ) ; s = n 2 + γ; γ (0, n 2 ),
9 Conformal fractional Laplacian Properties of Pĝγ : If γ = 1, Pĝ 1 = L ĝ = ĝ + n 2 4(n 1) R ĝ (Conformal Laplacian). In general, Pĝγ ( ĝ ) γ + l.o.t. Case of S n (Branson) γ = Γ ( A 1/2 + γ + 1 ) 2 Γ ( A 1/2 γ + 1 ), A 1/2 = S n + ( n P g S n ) 2
10 Conformal fractional Laplacian Properties of Pĝγ : If γ = 1, Pĝ 1 = L ĝ = ĝ + n 2 4(n 1) R ĝ (Conformal Laplacian). In general, Pĝγ ( ĝ ) γ + l.o.t. Case of S n (Branson) γ = Γ ( A 1/2 + γ + 1 ) 2 Γ ( A 1/2 γ + 1 ), A 1/2 = S n + ( n P g S n ) 2 Conformal property: Pĝw γ φ = w n+2γ n 2γ Pĝγ (wφ), φ C (M), ĝ w = w 4 n 2γ ĝ.
11 Definition Fractional order curvature as: Qĝγ := Pĝγ (1).
12 Definition Fractional order curvature as: Qĝγ := Pĝγ (1). Known: Q 1 = scalar curvature, Q 2 = Q curvature, Q 1 2 = mean curvature. For γ (0, 1 2 ), Q γ =non-local curvature.
13 Definition Fractional order curvature as: Qĝγ := Pĝγ (1). Known: Q 1 = scalar curvature, Q 2 = Q curvature, Q 1 2 = mean curvature. For γ (0, 1 2 ), Q γ =non-local curvature. Examples: Q γ (S n ) = cst > 0, Q γ (R n ) = 0, Q γ (S n k 1 H k+1 ) = cst, (González-Mazzeo-Sire).
14 Definition Fractional order curvature as: Qĝγ := Pĝγ (1). Known: Q 1 = scalar curvature, Q 2 = Q curvature, Q 1 2 = mean curvature. For γ (0, 1 2 ), Q γ =non-local curvature. Examples: Q γ (S n ) = cst > 0, Q γ (R n ) = 0, Q γ (S n k 1 H k+1 ) = cst, (González-Mazzeo-Sire). Conformal property Pĝγ (w) = w n+2γ n 2γ Qĝw γ,
15 The extension problem on c.c.e. manifolds Another way for calculating P g γ : Th. (Chang-González, 2011) Let ḡ = ρ 2 g + { div(ρ a U) + E(ρ)U = 0 in (X, ḡ), Pĝγ f = d γ lim ρ 0 ρ a ρ U, U = f on M.
16 The extension problem on c.c.e. manifolds Another way for calculating P g γ : Th. (Chang-González, 2011) Let ḡ = ρ 2 g + { div(ρ a U) + E(ρ)U = 0 in (X, ḡ), Pĝγ f = d γ lim ρ 0 ρ a ρ U, U = f on M. 1 Dirichlet-to-Neumann 2 Degenerate elliptic local NON-local 3 Qĝγ f = 1
17 The extension problem on c.c.e. manifolds Another way for calculating P g γ : Th. (Chang-González, 2011) Let ḡ = ρ 2 g + { div(ρ a U) + E(ρ)U = 0 in (X, ḡ), Pĝγ f = d γ lim ρ 0 ρ a ρ U, U = f on M. 1 Dirichlet-to-Neumann 2 Degenerate elliptic local NON-local 3 Qĝγ f = 1 { Example: a = 0 γ = 1 2 cn ḡ U + Rḡ U = 0, P 1/2 = lim ρ U. ρ 0
18 Fractional Yamabe problem γ (0, 1), (X n+1, ḡ), (M n, [ĝ]); n 3, (González-Qing) ĝ w = w 4 n 2γ ĝ conformal metric with constant Q gw ˆ Or the extension problem div(ρ a U) + E(ρ)U = 0 in (X, ḡ), γ? d γ lim ρ a ρ U = cw n+2γ n 2γ where U M = w. ρ 0
19 Fractional Yamabe problem γ (0, 1), (X n+1, ḡ), (M n, [ĝ]); n 3, (González-Qing) ĝ w = w 4 n 2γ ĝ conformal metric with constant Q gw ˆ Or the extension problem div(ρ a U) + E(ρ)U = 0 in (X, ḡ), γ? d γ lim ρ a ρ U = cw n+2γ n 2γ where U M = w. ρ 0 Example: γ = 1 2 Yamabe problem with boundary (Escobar, Brendle, Chen, Marques, Han-Li...) ḡ u = u 4 (n 2) ḡ, u > 0; Rḡu = 0 and Hĝu =ctt? Variational method!
20 Main work
21 Joint with M.d.M González. M.d.M González; M. del Pino; J. Wei.
22 Delaunay solutions Problem: Singular fractional Yamabe problem with an isolated singularity ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0
23 Delaunay solutions Problem: Singular fractional Yamabe problem with an isolated singularity ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0 Caffarelli-Jin-Sire-Xiong: They proved that all the non-negative solutions for this problem are radially symmetric of the form: w(r) = r n 2γ 2 v(r) (cylinder)
24 Delaunay solutions Problem: Singular fractional Yamabe problem with an isolated singularity ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0 Caffarelli-Jin-Sire-Xiong: They proved that all the non-negative solutions for this problem are radially symmetric of the form: w(r) = r n 2γ 2 v(r) (cylinder) We look for periodic solutions v(t) (r = e t ).
25 Delaunay solutions Problem: Singular fractional Yamabe problem with an isolated singularity ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0 Caffarelli-Jin-Sire-Xiong: They proved that all the non-negative solutions for this problem are radially symmetric of the form: w(r) = r n 2γ 2 v(r) (cylinder) We look for periodic solutions v(t) (r = e t ). Interpretation: The metric g w = w 4 n 2γ dx 2 has constant fractional curvature.
26 Fractional case We extend the problem to X n+1
27 Fractional case We extend the problem to X n+1 Look for t-periodic solutions V (t, ρ) for the equation 1 (1 ρ2 4 ( ( ) 2 ρ ρ a 1 + ρ2 with boundary nonlinearity 4 ) ) ) 2 (1 ρ2 4 ρ V ρ a tt V + E(ρ)V = 0 lim ρ a ρ V = cv n+2γ n 2γ on ρ = 0 ρ 0 Difficulty: ( ) γ in radial coordinates
28 Why this extension?
29 Geometric interpretation Manifolds (X 1 = H n+1, g + H = dy 2 + dx 2 y 2 ), ( ) (X 2 = S 1 (L) R n, g + = ρ 2 dρ 2 + (1 ρ2 4 )2 g S n 1 + (1 + ρ2 4 )2 dt 2 Conformal infinities (M 1 = R + S n 1, dx 2 = dr 2 + r 2 g S n 1), (M 2 = S 1 (L) S n 1, ĝ = dt 2 + g S n 1). Note that (change r = e t ) dx 2 = e 2t dt 2 + e 2t g S n 1 = e 2t ĝ.
30 Geometric interpretation Manifolds (X 1 = H n+1, g + H = dy 2 + dx 2 y 2 ), ( ) (X 2 = S 1 (L) R n, g + = ρ 2 dρ 2 + (1 ρ2 4 )2 g S n 1 + (1 + ρ2 4 )2 dt 2 Conformal infinities (M 1 = R + S n 1, dx 2 = dr 2 + r 2 g S n 1), (M 2 = S 1 (L) S n 1, ĝ = dt 2 + g S n 1). Note that (change r = e t ) dx 2 = e 2t dt 2 + e 2t g S n 1 = e 2t ĝ. X 1 is a covering of X 2 (H n+1 /Z) and M 1 is a covering of M 2.
31 ODE study: classical case We look for radial solutions for g0 w = cw n+2 n 2, in R n \ {0}; w > 0.
32 ODE study: classical case We look for radial solutions for g0 w = cw n+2 n 2, in R n \ {0}; w > 0. Only solutions depending on t (r = e t ; w(r) = r n 2 2 v(r)): tt v = n(n 2) 4 v n+2 n 2 + (n 2)2 4 v. Critical points: ( v 0 = ( n 2 n )(n 2)/4, 0 ), and (0, 0). Constant Hamiltonian: H := 1 2 ( tv) 2 n 2 2n An explicit solution: (v 2n n (v 2 ) ) C. v = (cosh t) (n 2) 2. Minimal period at v 0 : λ = ± (n 2)i. Representation of the phase portrait.
33 ODE-type study (PDE): fractional case ( ) 2 ( ) ( ) ) 2 1 ρ2 4 ρ (ρ a 1 + ρ2 1 ρ2 4 4 ρv ρ a tt V + E(ρ)V = 0 in ρ (0, 2), t R, lim ρ 0 d γ ρ a n+2γ ρv = c n,γ V n 2γ on {ρ = 0}.
34 ODE-type study (PDE): fractional case ( ) 2 ( ) ( ) ) 2 1 ρ2 4 ρ (ρ a 1 + ρ2 1 ρ2 4 4 ρv ρ a tt V + E(ρ)V = 0 in ρ (0, 2), t R, lim ρ 0 d γ ρ a n+2γ ρv = c n,γ V n 2γ on {ρ = 0}. Theorem (dlt.-gonzález) Two critical points: (0, 0), (v 0, 0). A Hamiltonian quantity is preserved along trajectories: H(t) := cn,γ 2n dγ w n 2γ n 2γ + 2n ( 0 ρ a (1 + ρ2 ρ2 )(1 4 4 )2 ( ρv ) 2 ) ρ a (1 ρ2 4 )( t V )2 +(1 + ρ2 ρ2 )(1 4 4 )2 E(ρ)V 2 dρ cst. (4) We found an explicit solution (sphere) v = C(cosh t) (n 2γ) 2 with C > 1. (5) Linearized operator around v 0 has a pure complex eigenvalue ( minimal period!)
35 ODE vs PDE Figure: The phase portrait is difficult to find
36 The symbol Theorem (dlt.-gonzález ) Let µ k = k(k + n 1); k = 0, 1, 2... be the eigenvalues of S n 1. Then the Fourier symbol of the conformal fractional Laplacian P γ on M = R S n 1 over each eigenspace is given by Γ( γ 2 + Θ k γ(ξ) = 2 2γ Γ( 1 2 γ ξ 2 i) 2 + ξ 2 i) ( n 2 1)2 4µ k ( n 2 1)2 4µ k 2 2,
37 The symbol Theorem (dlt.-gonzález ) Let µ k = k(k + n 1); k = 0, 1, 2... be the eigenvalues of S n 1. Then the Fourier symbol of the conformal fractional Laplacian P γ on M = R S n 1 over each eigenspace is given by Γ( γ 2 + Θ k γ(ξ) = 2 2γ Γ( 1 2 γ ξ 2 i) 2 + ξ 2 i) ( n 2 1)2 4µ k ( n 2 1)2 4µ k 2 2, Corollary The principal symbol for the conformal Laplacian P m ; m N is ξ 2m.
38 Key ideas for the theorems Using Scattering equation. Change of variable. Computing the projection over each eigenspace of S n 1. Fourier transform. Using properties of the hypergeometric function and the regularity of the solution at the origin. The expansion of the solution u. The definition of the conformal fractional Laplacian. û k =A(1 z 2 ) n 4 γ 2 z 1 n 2 + ( n 2 1)2 µ k 2 Hyperg(a, b, c, z 2 ) +B(1 z 2 ) n 4 γ 2 z 1 n 2 ( n 2 1)2 µ k 2 Hyperg(ã, b, c, z 2 ).
39 Explicit solutions The linearized problem at v 0 1 : { g +u s(n s)u = 0; s = n 2 + γ, in Hn+1 P γ u = cu
40 Explicit solutions The linearized problem at v 0 1 : { g +u s(n s)u = 0; s = n 2 + γ, in Hn+1 P γ u = cu ( n 1 z ) z (1 z 2 ) z u(z, t) + (1 z 2 ) 2 zz u(z, t) + (1 z 2 ) tt u(z, t) + ( n2 4 γ2 )u(z, t) = 0. (6)
41 Explicit solutions The linearized problem at v 0 1 : { g +u s(n s)u = 0; s = n 2 + γ, in Hn+1 P γ u = cu ( n 1 z ) z (1 z 2 ) z u(z, t) + (1 z 2 ) 2 zz u(z, t) + (1 z 2 ) tt u(z, t) + ( n2 4 γ2 )u(z, t) = 0. (6) Working out the eigenvalue λ???
42 Explicit solutions The linearized problem at v 0 1 : { g +u s(n s)u = 0; s = n 2 + γ, in Hn+1 P γ u = cu ( n 1 z ) z (1 z 2 ) z u(z, t) + (1 z 2 ) 2 zz u(z, t) + (1 z 2 ) tt u(z, t) + ( n2 4 γ2 )u(z, t) = 0. (6) Working out the eigenvalue λ??? Γ( n + γ + λ 4 2 i) 2 2 Γ( n γ + = λ i) n + 2γ n 2γ Γ ( ( 1 n + γ)) ( Γ 1 ( n γ)). (7) (Expansion of u, Taylor s expansion, changes of variable and properties of some especial functions).
43 The variational approach The problem: Radial solutions for ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0 The fractional Laplacian ( ) γ u = c n,γ P.V. R n u(x) u(y) x y n+2γ dy We consider radially symmetric solutions of the form In radial coordinates ( ) γ u = c n,γ P.V. 0 u(r) = r n 2γ 2 v(r) r S n 2γ 2 v(r) s n 2γ n 1 2 v(s) r 2 +s 2 2rsθ σ n+2γ 2 s n 1 dσds
44 The variational approach The problem: Radial solutions for ( ) γ w = cw n+2γ n 2γ in R n \{0}, w > 0 The fractional Laplacian ( ) γ u = c n,γ P.V. R n u(x) u(y) x y n+2γ dy We consider radially symmetric solutions of the form In radial coordinates ( ) γ u = c n,γ P.V. 0 u(r) = r n 2γ 2 v(r) r S n 2γ 2 v(r) s n 2γ n 1 2 v(s) r 2 +s 2 2rsθ σ n+2γ 2 s n 1 dσds After some changes of variable (s = r s; r = e t, s = e τ ): n+2γ (v(t) v(τ))k(t τ)dτ + Av = v n 2γ K(t τ) = 1 dσ K(ξ) ξ 1 2γ (ξ 0). S n 1 (cosh(t τ) θσ) n 2γ 2
45 The variational approach We look for periodic solutions v(t + L) = v(t) The equation becomes L 0 (v(t) v(τ))k L (t τ)dτ + Av = v n+2γ where K L (t τ) = S n 1 + j= Note: K L is still well defined. n 2γ (*) 1 dσ. (cosh(t τ jl) θ,σ ) n+2γ 2 This is the Euler-Lagrange equation for the functional c(l) = inf v H γ L,v 0 L L 0 0 (v(t) v(τ))2 K L (t τ)dtdτ + A L ( L 0 v 2n n 2γ dt) n 2γ n 0 v 2 dt
46 Variational Formulation. Thus, solving our equation is equivalent to the minimization problem. c(l) = where inf v H γ L,v 0 L L 0 0 (v(t) v(τ))2 K L (t τ)dtdτ + A L ( L 0 v 2n n 2γ dt) n 2γ n 0 v 2 dt H γ L L L L = {v : R R; v(t + L) = v(t), (v(t) v(τ)) 2 K L (t τ)dτdt + v(t) 2 dt < + } (8) 0 0 0
47 Variational Formulation. Thus, solving our equation is equivalent to the minimization problem. c(l) = where inf v H γ L,v 0 L L 0 0 (v(t) v(τ))2 K L (t τ)dtdτ + A L ( L 0 v 2n n 2γ dt) n 2γ n 0 v 2 dt H γ L L L L = {v : R R; v(t + L) = v(t), (v(t) v(τ)) 2 K L (t τ)dτdt + v(t) 2 dt < + } (8) Note that The infimum is a finite number.
48 Variational Formulation. Thus, solving our equation is equivalent to the minimization problem. c(l) = where inf v H γ L,v 0 L L 0 0 (v(t) v(τ))2 K L (t τ)dtdτ + A L ( L 0 v 2n n 2γ dt) n 2γ n 0 v 2 dt H γ L L L L = {v : R R; v(t + L) = v(t), (v(t) v(τ)) 2 K L (t τ)dτdt + v(t) 2 dt < + } (8) Note that The infimum is a finite number. The functional has a minimizer.
49 Variational Formulation. Thus, solving our equation is equivalent to the minimization problem. c(l) = where inf v H γ L,v 0 L L 0 0 (v(t) v(τ))2 K L (t τ)dtdτ + A L ( L 0 v 2n n 2γ dt) n 2γ n 0 v 2 dt H γ L L L L = {v : R R; v(t + L) = v(t), (v(t) v(τ)) 2 K L (t τ)dτdt + v(t) 2 dt < + } (8) Note that The infimum is a finite number. The functional has a minimizer. The minimizer must be strictly positive.
50 Periodic solutions Using the variational formulation we can assert that: Theorem (dlt.-gonzález-del Pino-Wei) There is a unique L 0 > 0 such that c(l 0 ) is attained by a nonconstant minimizer when L L 0 and when L < L 0 c(l) is attained by the constant only. Note that L 0 is the minimal period!
51 Sketch of the proof For any L > 0 c(l) achieved by v L > 0 which solves (*). If c(l 0 ) is attained by a nonconstant function v L0 c(l) < c (L) = L 2/n (constant solution), for all L > L 0 : L small enough c(l) is attained by the constant: blow-up argument: L small v L is uniformly bounded. v L Constant: L large c(l) < c (L). L 0 := sup{l c(l) = c (l) for l (0, L)}.
52 Non uniqueness Uniqueness for the Yamabe problem?
53 Non uniqueness Uniqueness for the Yamabe problem? Classical case γ = 1: Let, µ(m, [g 0 ]) = inf{j[u]; u smooth} where J[u] = M (cn u 2 g +Rg u2 ) dvol g ( 2n M u n 2 dvol g ) n 2 n.
54 Non uniqueness Uniqueness for the Yamabe problem? Classical case γ = 1: Let, µ(m, [g 0 ]) = inf{j[u]; u smooth} where J[u] = M (cn u 2 g +Rg u2 ) dvol g ( 2n M u n 2 dvol g ) n 2 n. 1 µ(m) 0! 2 µ(m) > 0 non!
55 Non uniqueness Uniqueness for the Yamabe problem? Classical case γ = 1: Let, µ(m, [g 0 ]) = inf{j[u]; u smooth} where J[u] = M (cn u 2 g +Rg u2 ) dvol g ( 2n M u n 2 dvol g ) n 2 n. 1 µ(m) 0! 2 µ(m) > 0 non! Fractional case: Let, µ γ (M, [g 0 ]) = inf{i γ [w]; w smooth} where I γ [U] = d γ X n+1 (ρ a U 2 +E(ρ)U 2 ) dvolḡ M n (U n 2γ 2n dvolĝ ) n 2γ n.
56 Non uniqueness Uniqueness for the Yamabe problem? Classical case γ = 1: Let, µ(m, [g 0 ]) = inf{j[u]; u smooth} where J[u] = M (cn u 2 g +Rg u2 ) dvol g ( 2n M u n 2 dvol g ) n 2 n. 1 µ(m) 0! 2 µ(m) > 0 non! Fractional case: Let, µ γ (M, [g 0 ]) = inf{i γ [w]; w smooth} where I γ [U] = d γ X n+1 (ρ a U 2 +E(ρ)U 2 ) dvolḡ M n (U n 2γ 2n dvolĝ ) n 2γ n. 1 µ γ (M) 0! 2 µ γ (M) > 0 non!
57 Non uniqueness for Yamabe in S 1 (L) R n Delaunay in R n \ {0}
58 Open Questions Equality λ = L 0 Relation with the Steklov eigenvalue Gluing problems
59 Thanks!! :)
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