Estimations universelles pour les solutions d EDP elliptiques non linéaires

Transcription

1 Estimations universelles pour les solutions d EDP elliptiques non linéaires LAMFA, UMR CNRS 6140 Université de Picardie Jules Verne (Amiens)

2 Large solutions in probability, geometry, and PDE 1. Singular Yamabe Problem: can an open set of Euclidean space be conformally transformed into a complete manifold of prescribed curvature? 2. Extinction of super Brownian motion: for which open sets of Euclidean space can we guarantee that super Brownian motion never reaches the boundary? 3. : can we obtain (pointwise) a priori estimates for solutions of a nonlinear elliptic PDE, without knowing their boundary values?

3 A bit of conformal geometry Definition A map ψ : Ω R N R N is conformal if it preserves angles. Given a metric g on a manifold M, a metric g is conformal to g if for some ρ : M (0, + ), g ij = ρ g ij

4 Example: the TGV metric

5 How is curvature affected by a conformal transformation? Assume N = 2. If g = ρ g = e 2u g, then the Gaussian curvature equation holds: gu + K g = K g e 2u in M Assume N 3. If g = ρ g = u 4 N 2 g, then the scalar curvature or Yamabe equation holds: N 2 4(N 1). gu + c N R gu = c N R g u N+2 N 2 in M, where c N = In particular, if M = Ω is an open set of Euclidean space, u = c N R g u N+2 N 2 in Ω.

6 The singular Yamabe problem Recall that we want to find a complete metric g ij = ρδ ij on an open set Ω of Euclidean space. This implies (case N 3): ( u = c N R g u N+2 N 2 in Ω. u = + on Ω. A solution to the above equation is called a BBUS or large solution. Remark If Ω is smooth, there is no solution having positive scalar curvature.

7 The Poincaré metric Let Ω = B denote the unit ball of R N and try to solve the singular Yamabe problem for constant negative curvature R g. Then, if N = 2, the problem becomes ( u = 4e 2u in B. u = + on B, or, if N 3, ( u = N(N 2)u N+2 N 2 in B. u = + Let R be the hyperbolic radius, defined by ( e u R := ρ 1/2 = on B. if N = 2 u 2 N 2 if N 3 Then, there exists a unique BBUS explicitly given by (B, g ) is the Poincaré disk. See for more. R = 1 x 2.

8 Brownian motion Let (B t) t 0 be Brownian motion in R N, i.e. B t takes values in R N. Reducing to the case N = 1, t B t is almost surely continuous B t has independent increments i.e. for 0 s < t, (B t B s) (B r ) 0 r s B t has stationary increments (B t B s) L = B t s B 0 B t N (rt, σ 2 t). Normalization: r = 0, σ = 1.

9 Exit time and the Dirichlet problem Let Ω be a bounded open set of R N. Consider the exit time of Brownian motion. τ Ω = inf{t > 0 : B t Ω c } Problem if B t=0 = x Ω, is it true that τ Ω = 0 a.s.? Take g C( Ω) and for x Ω, set u(x) = E x (g(b τω )). Then, and τ Ω = 0 iff for all such g, u = 0, in Ω u(x) = g(x), on Ω. Theorem (Wiener criterion) Ω is Dirichlet regular iff Z 1 0 Cap(B(x, r) Ω c ) dr r N 2 r = +, for all x Ω.

10 Brownian snake Fix x R N and set W x = {finite path started at x} = {w C([0, ζ w], R N ) : w(0) = x}. ζ w is the lifetime of w, and ŵ = w(ζ w) its tip. Heuristicaly, the Brownian snake satisfies W t takes values in W x the lifetime ζ t = ζ Wt is b.m. (reflected at 0) when ζ t decreases, the path W t is decreased from its tip When ζ t increases, the path W t is extended by adding little pieces of N-dim. b.m. at its tip.

11 Brownian snake ζ s ζ s time W s m ζ (s,s ) W s R d s s x The Brownian snake (W s ) s 0 is the Markov process with values in W W x = t takes {finite values paths started in W x at x} such that: the The lifetime lifetime ζ t ζis b.m. s := ζ (reflected Ws is a linear atbrownian 0) motion reflected at 0. conditionnally Conditionallyon (ζ (ζ t) s ) t 0 s 0,, if(w 0 s ) s 0 s < iss time-inhomogeneous, Markov, and if s < s, Ws (t) =WW s (t) s(t) for= every W s (t), 0 t mfor ζ (s, 0s ) := r min min ] ζ r ζ p =: m (Ws (m ζ (s, s )+t) W s (m ζ (s, s p [s,s ] ))) 0 t ζs m ζ (s,s ) is distributed as a Brownian (W s (m motion + t) in WR s d (m)) independent 0 t ζs m of Wis s. b.m. in R N W s Jean-FrançoisLe Gall (Université Paris-Sud) Spatial branching processes Göteborg 9 / 28

12 Exit set Given an open set Ω R N, and a finite path w W x, set Exit points from a domain τ(w) = inf{t 0 : w(t) Ω c.} Classical theory of relations between Brownian motion and PDEs : A key role is played by the first exit point of Brownian motion from a domain D. Here one constructs a measure supported on the set of exit points of the paths W s from D (assuming that the initial point x D) The set of exit points of b.s. is given by x E Ω = {W t(τ(w t)) : t 0, τ(w t) < }} D For every finite path w W x, set τ(w) =inf{t 0 : w(t) / D} and E D = {W s(τ(w s)) : τ(w s) < } (exit points of the paths W s) There exists a (σ finite) measure (the excursion Jean-François measure Le Gall (Université of b.s.) Paris-Sud) s.t. Spatial u(x) branching = processes N x (E Ω ) solves Göteborg 16 / 28 ( u = 4u 2, in Ω, u = + on Ω whenever Ω is Dirichlet regular.

13 Ω Assume u solves a nonlinear PDE (no BC s) of the form u = f (u) in Ω. Fix a subdomain ω Ω. Want C = C(ω, Ω, f, N) s.t. u C in ω, More precisely, does there exist a function g = g N,f such that u(x) g(dist(x, Ω)) for all x Ω?

14 The Lane-Emden equation Fix p > 1, Ω an open set of R N, and consider u = u p 1 u in Ω. A solution u C 2 (Ω) can be seen as a critical point of the energy functional E ω(u) = 1 Z u 2 dx 1 Z u p+1 dx, where ω Ω. 2 ω p + 1 ω i.e. for ϕ C 1 c (Ω), Z DE(u)(ϕ) = ( u u p 1 u)ϕ dx Compute the second variation of energy: Z D 2 E(u)(ϕ) = ϕ 2 p u p 1 ϕ 2 dx. Ω Ω

15 Stability Definition A solution is said to be stable whenever D 2 E(u)(ϕ) 0 for all ϕ C 1 c (Ω). More generally, u has finite Morse index k N if k is the maximal dimension of any subspace X C 1 c (Ω) s.t. D 2 E(u)(ϕ) < 0 for all ϕ X \ {0}. Remark If Ω is bounded and u C 2 (Ω), k is the number of negative eigenvalues of the linearized operator p u p 1. So, every solution u C 2 (Ω) has finite Morse index on ω Ω.

16 Universal estimate for the Lane-Emden equation Theorem (Davila-D-Farina) Assume 1 N 10 and take any p > 1, p N+2. Let u N 2 C2 (Ω) be a solution of finite Morse index k. There exists a constant C = C(N, p, k) s.t. u(x) C dist(x, Ω) 2 p 1, for all x Ω.

17 Discussion: critical dimension, critical exponent if N 11, the universal estimate holds iff p < p c(n). For p p c(n), consider ( u = λ(1 + u) p in B u = 0 on B. u L (B) u L (B) u L (B) λ λ λ λ λ λ 2(N 2) Then ([Joseph-Lundgren]), 1 N 2 3 N 9 N 10

18 Discussion: Liouville, Yamabe, weak solutions the universal estimate implies the Liouville property. if p = N+2, the estimate fails. Take N 2 «u(x) N 2 2 = c N λ + x 2. λ Note: the stereographic projection π : S N R N { } is conformal and g ij = u N 2 4 δ ij is the pullback of the standard metric on S N. the universal estimate extends to weak solutions u H 1 (Ω) L p (Ω). It fails for weak solutions belonging to L p (Ω). Counter-example: u(x) = c N,p x p 1 2.

19 Discussion: p p c (N)? Theorem (Davila-D-Farina) Let u H 1 (Ω) L p (Ω), u > 0, have finite Morse index k and take N 11, p p c(n). Then, u C (Ω \ Σ), where where γ = 2p + 2 p p(p 1) 1. Remark p = p c(n) solves H dim (Σ) N 2 p + γ p 1, N 2 p + γ p 1 = 0.

20 Discussion: without Morse index? If p ( N+2, pc(n)), ([Joseph-Lundgren]), N 2 u L (B) u L (B) λ λ 2(N 2) λ λ λ λ If 1 < p < N+2 and u > 0, then all solutions have bounded Morse index N 2 ([Gidas-Spruck, Serrin-Zhou]). Fails again for sign-changing solutions ([Rabinowitz, Ambrosetti-Rabinowitz]). 2 3 N 9 N 10

21 Boundary blow-up solutions Assume f 0 and consider u = f (u) in Ω Theorem (Keller, Osserman) Let Ω open set of R N, N 1. Assume f C(Ω), f 0, f increasing. if f > 0, there exists g = g N,f s.t. u(x) g(dist(x, Ω)) for all x Ω, u iff (KO) Z + dt p F (t) <, where F = f. if f (0) = 0, there exists g = g N,f s.t. iff (KO) holds. 0 u(x) g(dist(x, Ω)) for all x Ω, u 0

22 Discussion: Liouville, regularity of f, BBUS Liouville property: lim d 0 g = + lim g = d + lim g = + d 0 g = 0 lim d + need not assume f continuous ([Olofsson]) existence of BBUS: take Ω bounded, f (0) = 0. If Ω is Dirichlet regular, we can easily solve ( uk = f (u k ) in Ω, u k = k on Ω. Then u k U, where U is a BBUS. Theorem (Dhersin-Legal) Take f (u) = u 2. BBUS 0 N 3 or N 4 and Z 1 0 C 2,2 (Ω c B(x, r)) dr r N 2 r = +, for all x Ω

23 Discussion: f not increasing, Liouville again The previous result can be extended to f (u) = u p, p > 1 ([Labutin]), with critical exponent p = N and capacity C N 2 2,p. If f not increasing, the universal estimate fails. Simply take f (u) = u 2 sin 2 (u). However, Theorem (Dumont-D-Goubet-Radulescu) If f 0, loc. Lip., f (0) = 0, Ω Dirichlet regular bounded BBUS 0 (KO). Theorem (D) Universal Harnack inequality? Liouville property (KO).

24 Conjectures on BBUS ([McKenna]): if f increasing, then the BBUS is unique ([Brezis]): if f loc. Lip., Ω = B, then every BBUS is radial ([Lazer-McKenna]) : characterize all f s for which universal blow-up occurs i.e. any BBUS satisfies u(x) = g(dist(x, Ω)) + o(1), where ( g = f (g) in (0, 1) g(0 + ) = + Theorem (Costin-D) Assume f loc. Lip., Ω = B. Let U 1, U 2 be two BBUS. Then, U 1 U 2 C If K s.t. t f (t) + Kt increasing, then Z + U 1 (U 1 U 2 ) C. dt F (t).

25 Partial solutions to conjectures Uniqueness Let u 1, u 2 denote two solutions. It suffices to prove that u 1 u 2. If not, let ω = {x B : u 1 (x) > u 2 (x)}. Then, ( (u2 u 1 ) = f (u 2 ) f (u 1 ) 0 in ω, u 2 u 1 = 0 on ω. By the maximum principle, u 2 u 1 0 in ω, a contradiction. Symmetry The gradient estimate implies r u >> θ u. Then, apply the moving plane method ([Porretta-Véron]). Universal blow-up Set g = F. Then, universal blow-up always holds when and always fails when lim inf > 0. Z + lim g(u) G(t) u + u g(t) dt = 0 3

26 From power series to Borel summation To solve an ODE (or a PDE or a difference equation), say y y = 0 one can look for a solution in the form of a power series ỹ = X a k t k. That is, provided the solution is analytic, we replace an analytic problem (solving the ODE) with an algebraic problem that can be solved algorithmically. Here: ka k a k 1 = 0. In so doing, we use fundamentally the isomorphism between convergent power series and analytic functions y = ỹ. Problem For most equations, formal asymptotic expansions thus obtained are (factorially) divergent. How can we make sense of such expressions?

27 Euler s example What is the meaning for large t of ỹ = + X k=0 k! ( t) k+1 Euler argues for the preservation of the usual properties of summation: formally, ỹ solves the ODE ỹ ỹ = 1/t. Thus, denoting y the sum of ỹ, we should have y y = 1 t all solutions of this ODE are given by y = e t Ei( t) + Ce t, where Ei(x) = R x s 1 e s ds. formally, ỹ 0 as t +. The only solution y having this property is y = e t Ei( t). Thus, e t Ei( t) is the sum of ỹ.

28 Borel summation: heuristics In practice, divergent series have coefficients a k = O(k! n ). A change of variable allows to choose n = 1. E.g. k! 2 t k (2k)!( t) 2k = j!/y j. Laplace transform LY (t) = R + 0 e pt Y (p) dp. Note that Thus formally, k=0 L 1 k! = pk t k+1 X+ L 1 k! ( t) = X+ ( p) k = 1 k p. Still formally, LL 1 = Id, so P + k=0 = LL 1 + X k=0 k=0 k! ( t) k+1 = k! ( t) = L 1 Z + k p = e pt p dp = et Ei( t)

29 Borel summation: rigorous definition The Borel transform B acts on formal power series: The Borel summation LB is given by This requires that ỹ = X y k t k+1 Bỹ = X y k k! pk y = LBỹ = Z + 0 e pt Bỹ(p) dp Bỹ has nonzero radius of convergence (true after our change of variables) Bỹ continues analytically in p R + : serious restriction. Take e.g. ỹ = P k!. Then, Bỹ = 1. t k+1 1 p Bỹ is exponentially bounded on R +.

30 Generalized Borel summation

31 Application to BBUS: examples Let d(x) = 1 r, p > 1 and k 0 = [2/(p 1)] + 1. Consider Then, the blow-up solution satisfies u = d 2 p 1 u = u p. k 0 X k=0 a k d k + o(1) Now, for get u = e 1 d u = u(ln u) 4, + X k=0 a k d k + exp. small

32 The general case: starting point Let v = du/dr and multiply the equation by v. We get «d v 2 + N 1 v 2 = d dr 2 r dr (F(u)). We define the resulting error term by g := v F(u), which satisfies the D.E. dg = N 1 v 2 Change independent variable u = u(r). dr r Thinking of g as a function of the variable u, we have Solving for v, we finally obtain 8 < dg du = N 1 v. r v = N 1 p 2(F(u) g), r dg/du = N 1 r : dr/du = 1/v = 1/ p 2(F(u) g).