Strichartz estimates for the wave equation in convex domains Oana - PowerPoint PPT Presentation

Strichartz estimates for the wave equation in convex domains Oana Ivanovici Gilles Lebeau Fabrice Planchon Laboratoire Jean-Alexandre Dieudonn´ e Universit´ e de Nice Sophia-Antipolis Monastir June 2013

The wave equation (Ω , g ) = Riemannian manifold of dimension d ≥ 2, ∆ g = Laplace-Beltrami operator on (Ω , g ) . � ∂ 2 t u − ∆ g u = 0 , ( − T , T ) × Ω , ( W ) u | t = 0 = u 0 , ∂ t u | t = 0 = u 1 . If ∂ Ω � = ∅ , Dirichlet boundary condition: u | ( − T , T ) × ∂ Ω = 0 . (D) (in fact Neumann works as well..)

◮ Ω = R d : Strichartz estimates for ( W ) : � � ( S ) � u � L q ( R , L r ( R d )) ≤ C � u 0 � H γ ( R d ) + � u 1 � H γ − 1 ( R d ) , ( q , r ) = d - admissible , i.e. q , r ≥ 2, ( q , r , d ) � = ( 2 , ∞ , 3 ) , 1/q 1/q d=2 d=3 1/2 1/4 1/2 1/r 1/r 1/2 Here γ = γ ( d , q , r ) := d ( 1 2 − 1 r ) − 1 q (scaling condition)

1/q d=3 d=3 1/2 Energy estimate : � ∂ t , x u � L ∞ x � �∇ x u 0 � L 2 x + � u 1 � L 2 t L 2 x Strichartz endpoint : � u � L � � u 0 � ˙ x + � u 1 � L 2 2 + H 1 ∞− L x x t 1/r 1/2 ◮ On average (in time !), a solution to the wave equation has better integrability (almost L ∞ rather than L 6 ) than a given H 1 function ◮ Or, to have u ∈ L ∞ − (almost everywhere in time), we just need u ∈ H 1 rather than u ∈ H 3 / 2 ◮ Outside the admissibility set: Knapp counterexample (a correctly sized wave packet holds its shape on the correct time scale)

Dispersive estimates in R d ◮ Strichartz follows from energy conservation, interpolation and dispersion: ◮ A fundamental property of the half-wave operator � χ ( − λ − 2 ∆ R d ) e it √ − ∆ R d � L 1 ( R d ) → L ∞ ( R d ) � λ d + 1 2 | t | − d − 1 2 . ?? non flat background and/or boundaries ??

Boundary points: different scenarios transversal rays diffractive ray gliding ray highly ! multiply reflected rays

Dispersion in convex domains (Ω , g ) strictly convex , a ∈ Ω , δ a = the Dirac function: � χ ( λ − 1 D t ) e it √ ( d + 1 ) 2 | t | − ( d − 1 ) 1 − ∆ g ( δ a ) � L ∞ (Ω) � λ 2 ( λ | t | ) 4 ◮ Loss of 1 4 power of λ t compared to the Euclidian case ◮ The result is optimal because of the presence of swallowtail singularities in the wave front set. ◮ Currently: the Friedlander model case in full details (ILP) Theorem The above dispersive estimate holds in the Friedlander model case (actually, better bounds on the Green function that we shall use for Strichartz).

Strichartz (derived from previous dispersion) compared to preceding results: ILP to the left, Blair-Smith-Sogge to the right: ◮ d = 3 1/q 1/q d=3 1/2 1/2 3/8 3/8 1/3 1/3 1/2 1/r 1/2 1/r ◮ d = 4 1/q 1/q d=4 1/2 1/2 1/10 1/6 1/2 1/r 1/10 1/6 1/2 1/r

We do better than that: optimal loss for Strichartz = loss of 1/6(1/4−1/r) derivatives = loss is unavoidble (O.I. ’08,’09) = sharp Stricharz ( I−L−P ’12) = GAP to be filled 1/q 1/q d=3 d=2 1/2 5/12 1/3 1/4 1/4 5/24 ? 1/6 1/8 ? 1/4 1/2 1/r 1/2 1/4 ◮ open: (partially) fill the GAP (conjecture: it should be filled entirely !)

Theorem The above Strichartz estimates hold in the Friedlander model case. There is no loss (compared to the flat case) for ( q , r , d ) = ( 24 / 5 , ∞ , 2 ) , ( q , r , d ) = ( 12 / 5 , ∞ , 3 ) . In fact, we have optimal results for any d ≥ 2. Counterexamples are optimal at these endpoints. ◮ Requires a finer analysis to average out singularities ◮ d = 2 is already known to hold (for any smooth boundary) by previous work of Blair-Smith-Sogge

Model for convex boundaries Disk : r ≤ 1 Model domain : x ≥ 0, y ∈ R ∆ disk = ∂ 2 r + 1 r 2 ∂ 2 ∆ g = ∂ 2 x + ( 1 + x ) ∂ 2 y θ Model domain Disk Same to first order under x = 1 − r , y = θ .

Parametrix for waves starting at t = 0 from δ ( x = a , y = 0 ) , a ∈ Ω small A detailed description of the ”sphere” of center a and radius t , i.e. the set of points that can be reached following all the optical rays starting from a of length t . two main cases ◮ a � λ − 4 / 7 : use gallery modes (e.g. spectral decomposition); ◮ a > λ − 4 / 7 : write the flow as a superposition of waves essentially supported between two consecutive reflections; the only case which involves swallowtails In the last case, the parametrix is a sum of degenerate oscillatory integrals ◮ Non-tangential directions | ξ | ≫ λ − 1 / 4 : at worst, cusp-like singularities ◮ Main contribution comes from | ξ | � λ − 1 / 4 , where the swallowtail appears between every two consecutive reflexions

The Green function Explicit representation for the half-wave initial value problem with a Dirac at ( a , b ) as initial condition at time s : e ± i ( t − s ) √ � � λ k ( η ) e i ( y − b ) η e k ( x , η ) e k ( a , η ) d η G (( x , y , t ) , ( a , b , s )) = R k ≥ 1 where λ k ( η ) = η 2 + η 4 / 3 ω k , with ω k ≃ ( 3 k π 2 ) 2 / 3 ( 1 + O ( 1 k )) a zero of the Airy function and η 1 / 3 2 3 x − ω k ) , e k ( x , η ) = f k k 1 / 6 Ai ( η ( f k normalization constants) ( e k ) k ≥ 1 is an L 2 orthonormal basis (gallery modes) which provides a spectral decomposition of our Laplacian − ∂ 2 x + ( 1 + x ) η 2

Useful reductions ◮ reduce to d = 2 by rotational symmetry � ◮ spectral localization at λ k ( η ) ∼ λ : � insert ψ ( λ k ( η ) /λ ) = ψ ( D t /λ ) ◮ localization to tangent directions η ∼ λ : insert ψ ( η/λ ) this induces that k ≤ λ/ 100, as for the gallery modes (microlocally) ( η/λ ) 4 / 3 λ − 2 / 3 ω k ∼ ( ξ/λ ) 2 + x ( η/λ ) 2 and therefore k small compared to λ is equivalent to ξ small compared to η (at t = 0, x = a ) ◮ G is symmetric in x and a : we may restrict the computation of any L ∞ norm over ( x , a ) to the set { ( 0 ≤ ) x ≤ a }

Initial data very close to ∂ Ω (the case a small compared to λ − 4 / 7 ) u a ,λ ( t , x , y ) = e − it √ − ∆ g ψ 2 ( � − ∆ g /λ ) ψ 1 ( D y /λ ) δ x = a , y = 0 e i λ ( y η − t η √ λ � � 1 + ω k ( ηλ ) − 2 / 3 ) ψ ( η ) e k ( a , ηλ ) e k ( x , ηλ ) d η . = 2 π 1 ≤ k ≤ λ/ 100 Proposition ILP 2013, a � λ − 4 / 7 1 x , y ≤ C λ 2 min ( 1 , � 1 x ≤ a u a ,λ ( t , x , y ) � L ∞ ( λ | t | ) 1 / 3 ) . ◮ the geometry is irrelevant when a is very small (too many singularities in the WF). ◮ Finer analysis on the sum of gallery modes (inspired by exponential sum method). ◮ Why 4 / 7? for technical reasons not so clear yet..

Small time estimates � e i λ y η ˜ � u a ,λ ( t ) = λ 2 λ − 1 / 3 ψ ( η ) f ( k , t , x , a , ηλ ) d η + O ( λ −∞ ) , k ≃ a 3 / 2 λ k 1 / 3 χ ( ω 3 / 2 1 ηλ a 3 / 2 ) Ai (( ηλ ) 2 / 3 x − ω k ) Ai (( ηλ ) 2 / 3 a − ω k ) , k f ( k , . ) = where χ ∈ C ∞ is supported near 1. Lemma � f ( k ) | � L 1 / 3 . | k ≃ L ◮ For values L ≤ 1 t (where L = a 3 / 2 λ ) ⇒ OK ◮ It remains to study t > 1 a 3 / 2 λ .

ω 3 / 2 N ˆ Poisson formula � k f ( k ) = � f ( − 2 π N ) , ζ = k ηλ a 3 / 2 3 a 3 / 2 N ζ − t √ � ˆ e i λη ( 4 1 + a ζ 2 / 3 ) χ ( ζ ) f ( − 2 π N ) = a λ 2 / 3 × Ai ( a ( ηλ ) 2 / 3 ( 1 − ζ 2 / 3 )) Ai ( a ( ηλ ) 2 / 3 ( x a − ζ 2 / 3 )) d ζ. ◮ Stationary phase with ζ 1 / 3 t ≃ 4 N √ a ≃ 1 yields: c � a � t 3 e i λη ( y − t − 2 � 64 N 2 ) ψ ( η ) u a ,λ ( t ) = λ 2 λ − 1 / 6 3 t | 4 N √ a − t |≤ ct )) Ai ( a ( ηλ ) 2 / 3 ( x × Ai ( a ( ηλ ) 2 / 3 ( 1 − ζ 2 / 3 a − ζ 2 / 3 )) d η c c ◮ For a 2 λ < t we bound each Airy factor by a constant 1 ◮ For t > a 3 / 2 λ the supports in y are disjoint! 1 ◮ The condition t > a 3 / 2 λ ≥ a 2 λ gives a � λ − 4 / 7 .

Initial data localized at distance a > λ − 4 / 7 v N ( t , x , y , λ ) = λ 2 � e i λη y u N ( t , x , λη ) χ 0 ( η ) d η , ( 2 π ) 2 u N ( t , x , ηλ ) =( − i ) N λη � 3 e i λη (( t − t ′ ) ζ + ψ a ( t ′ )+ s ( x + 1 − ζ 2 )+ s 3 / 3 − 4 N / 3 ( ζ 2 − 1 ) 2 ) 2 π × χ ( ζ ) σ N ( t ′ , λη ) dt ′ dsd ζ , � v ( t , x , y , λ ) = v N ( t , x , y , λ ) . 0 ≤ N ≤ C 0 / √ a Proposition There exists C 0 , σ 0 such that the following holds true: 1. v is a solution to ✷ v = 0 for x > − 1; 2. its trace on the boundary, v ( t ∈ [ 0 , 1 ] , x = 0 ) is O ( λ −∞ ) ; 3. at time t = 0, we have v ( 0 , x , y , λ ) − ( λ � 2 π ) 2 e i λ ( η y +( x − a ) ξ ) χ 0 ( η ) χ 1 ( ξ/η ) d η d ξ = O ( λ −∞ ) .