Sharp bounds for the bottom of the spectrum of Laplacians (and their associated stability estimates)

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Sharp bounds for the bottom of the spectrum of Laplacians (and their associated stability estimates) Lorenzo Brasco I2M Aix-Marseille Université lorenzo.brasco@univ-amu.fr http://www.latp.univ-mrs.fr/ brasco/ Jyväskylä, 13 Toukokuu 2014

Foreword: a linear introduction Assumption R N open connected set with < Helmholtz equation u = λ u in u = 0 on has nontrivial solutions only for a discrete set of values 0 < λ 1 () < λ 2 () λ 3 ()... since the resolvent R : L 2 () L 2 () is positive, symmetric and compact Relevant facts λ i () are called Dirichlet eigenvalues of on corresponding solutions u i give orthonormal basis of L 2 () (once renormalized)

a first eigenfunction u 1 is positive and u 1 > 0 in any other eigenfunction has to be orthogonal to Vect({u 1 })...any other eigenfunction has to change sign Variational characterization (Courant-Fischer) ( ) ( ) λ k () = min max u 2 / u 2 dim(e) k u E Sharp lower bounds for the bottom of the spectrum λ 1 () is not smaller than λ 1 of a ball B with = B λ 2 () is not smaller than λ 2 of two disjoint balls Upper bounds? Take ε = [0, 1] [0, ε], then λ 1 ( ε ) as ε 0

Overview of the talk 1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

Critical points of Dirichlet integrals Define the eigenvalues as the critical points of u u p on the manifold S p () = { u W 1,p 0 () : Definition u S p () and λ R form an eigenpair if p u = λ u p 2 u Remark: a uniform L bound For (u, λ) eigenpair we have } u p = 1 in u L () C N,p λ N p 2 (Estimate is sharp for p 1)

A useful tool: Hidden Convexity Hidden Convexity Lemma Let p > 1, 1 q p and u 0, u 1 0 u p is convex along σ t = [ ] 1/q (1 t) u q 0 + t uq 1 Important consequence 1. u p is geodesically convex on S p () {u 0} 2. convexity is strict on functions having the Harnack property Remark This property is false for q > p This is equivalent to (usual) convexity of u u p u p 1

Yet another useful tool: Picone-type inequality A nonlinear variation on a theme by Picone (Allegretto-Huang) If p > 1 v 0 and u > 0 ( v (P) u p 2 p u, u p 1 ) v p This is Hidden Convexity in a different guise! Take σ t = [(1 t) u p + t v p ] 1/p, then σ t p u p t v p u p passing to the limit t 0 + we get (P)

The first eigenvalue If (u, λ) is an eigenpair, then necessarily u p = λ u p so that Properties λ 1 () = min u p dx u S p() A. first eigenfunctions have constant sign B. a unique minimizer u 1 (up to the choice of the sign) C. every other eigenfunction has to change sign Proof Idea for B. and C. strictly convex functions have at most 1 critical point (for C. this has been used in [B. - Franzina])

The second eigenvalue We introduce the set of loops then we define C 2 = {f : S 1 S p () : f odd & continuous} λ 2 () = inf f C 2 Is it really the second? max u p dx u Im(f ) 1. first of all, λ 2 () is an eigenvalue (general fact, see Drabek-Robinson) 2. moreover, we have λ 1 () < λ 2 () (see the topological argument at the blackboard)

The second eigenvalue (suite) 3. (u, λ) is an eigenpair and λ λ 1 ()......u has to change sign! Then u+ 0 and u 0 take the loop f (ϑ) = (cos ϑ) u + + (sin ϑ) u (cos ϑ) u + + (sin ϑ) u L p by using the equation u + p = λ u + p and u p = λ u p then λ λ 2 () max ϑ S 1 ( cos ϑ p cos ϑ p ) u + p + sin ϑ p u p = λ u + p + sin ϑ p u p

1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

1st guise: Mountain pass level Introduce the set of continuous paths Γ(u 1, u 1 ) = {γ : [0, 1] S p () : γ(0) = u 1 γ(1) = u 1 } then mountain pass characterization 1 λ 2 () = Sketch of the equivalence inf γ Γ(u 1, u 1 ) max u Im(γ) u p easy part : take a path γ Γ(u 1, u 1 ) and close it, i.e. consider {γ} { γ} which is admissible for λ 2 () some care is needed: take an optimal loop f : S 1 S p () and go down in the valley! without increasing the energy 1 Cuesta, De Figueiredo & Gossez, JDE 1999

Down in the valley (in a convex way) Lemma [B.-Franzina] Let u, v S p () such that u 0 and v 0 or v + has less energy than v There exists a continuous curve σ connecting v and u such that { } σ t max u p, v p Why is it useful? On every symmetric loop f, one can find v such that v + has less energy than v Remark The curve σ can be constructed by appealing again to Hidden Convexity

2nd guise: Optimal partition λ 2 () = min { } max{λ 1 ( + ), λ 1 ( )} : +, Hal() where Hal() = { ( +, ) : + + = } Sketch of the equivalence take ( +, ) Hal() and their 1st eigenfunctions u + and u, then construct the usual loop f (ϑ) = (cos ϑ) u + + (sin ϑ) u (cos ϑ) u + + (sin ϑ) u L p u = u + u eigenfunction corresponding to λ 2 (), then u ± p λ 2 () = = λ 1 ({u ± > 0}) u ± p

1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

The Faber-Krahn inequality Theorem [Faber-Krahn] Among sets of given volume, the ball is the only set minimizing λ 1 p/n λ 1 () B p/n λ 1 (B) 0 where B is a ball and with equality if and only if is a ball Remark The same result stays true for the first (p, q) eigenvalue { } λ 1,q () = u p : u q = 1 (with the same proof) min W 1,p 0 ()

Proof u S p () first eigenfunction and u its Schwarz symmetral Polya-Szegő inequality: ( u p Coarea = Jensen = 0 0 ( 0 Isoperimetry 0 u p {u=t} u {u=t} set µ(t) = {u > t} ) dσ dt u dσ u Perimeter ({u > t}) p ( µ (t)) p 1 dt ) p dt ( {u=t} u 1 dσ) p 1 Perimeter ({u > t}) p ( µ (t)) p 1 dt = u p and if λ 1 () = λ 1 ( ), level sets of u are balls

The Hong-Krahn-Szego inequality Among sets of given volume, the disjoint union of equal balls is the only set minimizing λ 2 Theorem [Hong-Krahn-Szego] p/n λ 2 () 2 p/n B p/n λ 1 (B) 0 where B is a ball and with equality if and only if is a disjoint union of equal balls Remark For B 1 B 2 with B 1 = B 2 and B 1 B 2 =, we have B 1 B 2 p/n λ 2 (B 1 B 2 ) = 2 p/n B i p/n λ 1 (B i )

Proof 1. by the optimal partition characterization we can find +, disjoint such that λ 2 () = max { λ 1 ( + ), λ 1 ( ) } 2. use Faber-Krahn inequality λ 2 () max{λ 1 (B + ), λ 1 (B )} with B + = + and B = 3. hence minimizer of λ 2 is a disjoint union of balls B 1 B 2 4. use the homogeneity of λ 2 to conclude that B 1 = B 2

1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

A quantitative Faber-Krahn inequality The following result is not sharp, but valid for every N 2 and every p > 1 (even p = 1, indeed) and effective Theorem There exists an explicit constant c N,p > 0 such that p/n λ 1 () B p/n λ 1 (B) c N,p A() 3 where A is the L 1 distance from optimizers, i.e. { } 1 1 B A() := inf L 1 : B ball, B = Remarks c N,p stays bounded and strictly positive as p 1 c N,p goes to 0 like p p as p goes to same result for the first (p, q) eigenvalue

Steps of the proof I General principle Back to the proof of the Pólya-Szegő inequality u p u p and add remainder terms in the estimates Crucial estimate we used so far... 0 Perimeter ({u > t}) p ( µ (t)) p 1...but we can be more precise B dt Isoperimetry 0 Perimeter ({u > t}) p ( µ (t)) p 1 dt Perimeter ({u > t}) Perimeter ({u > t}) + A({u > t}) 2 Fusco-Maggi-Pratelli [Ann. Math. 2008]

Steps of the proof II Then we end up with... Polya-Szegő with (ugly) remainder u p u p + What we get? B λ 1 () λ 1 (B) 0 0 A({u > t}) 2 ( µ dt (t)) p 1 A({u > t}) 2 ( µ dt (t)) p 1 Missing piece of information! How does asymmetry of propagate to super-level sets {u > t}? A()? A({u > t})

Steps of the proof III: an idea by Hansen & Nadirashvili for every subset U, we introduce its measure defect 0 δ(u) := \ U 1 Lemma δ(u) 1 4 A() = A(U) 1 2 A() how to use it?: in the remainder term brutally cut at level T such that δ({u > T }) = 1 4 A() 0 A({u > t}) 2 ( µ dt (t)) p 1 By monotonicity of t {u > t}, for every 0 < t < T we gain δ({u > t}) 1 4 A()...

... and thus we have A({u > t}) 1 2 A() for every 0 < t < T in the end? by cutting at level T, we get λ 1 () λ 1 (B) A() 2 T then by Jensen inequality we end up with 0 dt ( µ (t)) p 1 λ 1 () λ 1 (B) A() 3 p T p Small problem We do not have any control on T!!

Alternative for T Let τ > 0 be a small constant, we have an alternative 1) T τ A() or 2) T < τ A() Case 1) λ 1 () λ 1 (B) A() 3 p T p τ p A() 3 Case 2) General fact: we can compare λ 1 () and λ 1 ({u > T }) λ 1 () λ 1 ({u > T }) (u T ) p + Faber Krahn λ 1 (B) {u > T } p N (u T ) p +

by choice of T, we have {u > T } p N = ( 1 1 4 A()) p N 1 + p N A() in conclusion λ 1 () λ 1 (B) [ 1 + p ] [ ] N A() T ) (u p + are we happy? Not really...what if the term microscopic? (u T ) p + is Impossible! the level T is small! Remember T < τ A() thus we have (u T ) p + 1 p T 1 p τ A() We conclude by choosing τ 1 (depending only on p, N)

A quantitative Hong-Krahn-Szego inequality Theorem [B.-Franzina-Pratelli] There exists an explicit constant c N,p > 0 such that p/n λ 2 () 2 p/n B p/n λ 1 (B) c N,p A 2() 3 2 (N+1) where A 2 is the L 1 distance from optimizers, i.e. { 1 1 B1 B A 2 () := inf 2 L 1 : B 1 B 2 = 0 with B i = } 2 Remarks The exponent on the asymmetry depends on the dimension! the behaviour of c N,p with respect to p is the same as c N,p in Faber-Krahn

Steps of the proof Memento: we know that λ 2 () = max{λ 1 ( + ), λ 1 ( )} 1. first step use the quantitative Faber-Krahn so to obtain λ 2 () λ 2 (B 1 B 2 ) A( + ) 3 + 1 2 + +A( ) 3 + 1 2 which means In terms of the deficit, I can control how + and are far from being two balls having measure /2 2. second step a geometric estimate A 2 () (N+1)/2 A( + ) + 1 2 + + A( ) + 1 2

The geometric estimate is sharp A set such that A 2 ( ε ) (N+1)/2 ε (N+1)/2 A( ε +) + A( ε ) Figure : Set ε (bold line) and pair of optimal balls for A 2 (dashed line) Remark The same set shows that the sharp exponent on A 2 must depend on the dimension

1. The bottom of the spectrum of the p Laplacian 1.bis Intermezzo: the second eigenvalue in different guises 2. Sharp lower bounds 3. Stability estimates 4. Open problems & Miscellanea

Some open problems sharp quantitative Faber-Krahn inequality p/n λ 1 () B p/n λ 1 (B) c N,p A() 2 possibly with an explicit constant c N,p Result true for p = 2, with an unknown constant (B., De Philippis & Velichkov) sharp quantitative Hong-Krahn-Szego inequality p/n λ 1 () 2 p/n B p/n λ 1 (B) c N,p A()? nonlocal case: characterization of λ 2? Hong-Krahn-Szego? (some preliminary results with E. Parini)...quantitative Faber-Krahn?

References A pioneering paper on λ 1 P. Lindqvist, PAMS (1990) Quantitative Faber-Krahn T. Bhattacharya, EJDE (2001) N. Fusco, F. Maggi, A. Pratelli, Ann. SNS (2008) L. B., G. De Philippis, B. Velichkov, Submitted (2013) Quantitative Hong-Krahn-Szego L. B., A. Pratelli, GAFA (2012) L. B., G. Franzina, Manuscripta Math. (2013)