Lectures on paracontrolled distributions with applications to - PowerPoint PPT Presentation

Lectures on paracontrolled distributions with applications to singular SPDEs Massimiliano Gubinelli CEREMADE Université Paris Dauphine Università Milano Bicocca – February 2nd–6th 2015 ( 1 / 63 )

Homogenisation of a random potential ⊲ Consider the linear heat equation with a small random time-independent periodic and smooth (Gaussian) potential V ∂ t U ( t , x ) = ∆ U ( t , x ) + ε 2 − α V ( x ) U ( t , x ) , t ≥ 0, x ∈ T d ε where ε > 0 is a small parameter, α < 2 and T ε = T / ε , T = R / ( 2 π Z ) ∼ [ 0, 2 π ) . ⊲ Introduce macroscopic variables u ε ( t , x ) = U ( t / ε 2 , x / ε ) with parabolic rescaling, then t ≥ 0, x ∈ T d ∂ t u ε ( t , x ) = ∆ u ε ( t , x ) + V ε ( x ) u ε ( t , x ) , with V ε ( x ) = ε − α V ( x / ε ) , x ∈ T d . Problem: Study the limit ε → 0 for u ε . ( 2 / 63 )

The random potential The covariance of the macroscopic noise is E [ V ε ( x ) V ε ( y )] = ε − 2 α C ε (( x − y ) / ε ) , x , y ∈ T d where C ε : T d ε → R is a smooth, positive–definite function on T d ε . Assume � ε C ε ( x ) dx = 1. T d Take smooth test functions ϕ , ψ ∈ S ( T d ) and let V ε ( ϕ ) = � T d ϕ ( x ) V ε ( x ) dx then � E [ V ε ( ϕ ) V ε ( ψ )] = ε − 2 α T d × T d ϕ ( x ) ψ ( y ) C ε (( x − y ) / ε ) dxdy � ∼ ε d − 2 α T d ϕ ( x ) ψ ( x ) dx as ε → 0. Lemma If d > 2 α then V ε → 0 in law. If d = 2 α then V ε converges in law to the space white noise ξ on T d . White noise on T d A family { ξ ( ϕ ) } ϕ ∈ S ( T d ) of r.v. such that ξ ( ϕ ) ∼ N ( 0, � ϕ � 2 L 2 ( T d ) ) . ( 3 / 63 )

Fourier representation On the covariance C ε we assume the form √ 2 π ) d ∑ � d k e i � x − y , k � R ( k ) → ε → 0 ( 2 π ) d /2 e i � x , k � R ( k ) C ε ( x − y ) = ( ε / R d k ∈ ε Z d where Z d 0 = Z d \{ 0 } and R ∈ S ( R d ) . There exists a family of centered complex Gaussian random variables { g ( k ) } k ∈ Z d such that g ( k ) ∗ = g ( − k ) and E [ g ( k ) g ( k ′ )] = I k + k ′ = 0 and ε d /2 − α � e i � x , k � 2 π ) d /2 ∑ V ε ( x ) = √ R ( ε k ) g ( k ) ( k ∈ Z d Taking α = d /2 we have (as distributions) � ξ ( x ) = ( 2 π ) − d /2 R ( 0 ) ∑ e i � x , k � g ( k ) . k ∈ Z d Exercise: Show that there exists a version of ξ taking values in S ′ . ( 4 / 63 )

Sobolev regularity Consider Sobolev spaces H σ over T d with norm � f � 2 H σ ( T d ) = ∑ ( 1 + | k | ) − 2 ρ | F T d f ( k ) | 2 . k ∈ Z d ε d − 2 α ( 1 + | k | ) − 2 ρ R ( ε k ) ∼ ε 2 ρ − 2 α → 0 E � V ε � 2 2 π ) d ∑ H − ρ = √ ( k ∈ Z d if ρ > α and d > 2 α . It stays bounded if d = 2 α and ρ > α . Similarly for E � X ε � 2 H 2 − ρ . The white noise ξ belongs to H − ρ ( T d ) for all ρ < d /2. It is possible to show that it is not better: a.s. � ξ � H − ρ = + ∞ for ρ ≥ d /2. ( 5 / 63 )

Guesswork As ε → 0 we guess that u ε → u where � if d > 2 α 0 L u = u ξ if d = 2 α with L = ∂ t − ∆ the heat operator. This would hold provided the solution map Ψ : η �→ v which sends potentials η to solutions of the parabolic Anderson model (PAM) L v = v η is continuous in an appropriate topology in which ( V ε ) ε converges. ( 6 / 63 )

Littlewood–Paley decomposition ϕ : R d → C with polynomial growth defines a Fourier multiplier ϕ ( D ) : S ′ → S ′ , ϕ ( D ) f = F − 1 ( ϕ F f ) . ⊲ Dyadic partition of unity: χ , ρ ∈ C ∞ ( R d , R + ) such that 1. supp ρ ⊆ B = {| x | � c } and supp ρ ⊆ A = { a � | x | � b } 2. χ + ∑ j � 0 ρ ( 2 − j · ) ≡ 1 3. supp ( χ ) ∩ supp ( ρ ( 2 − j · )) ≡ 0 for j � 1 and supp ( ρ ( 2 − i · )) ∩ supp ( ρ ( 2 − j · )) ≡ 0 for all i , j � 0 with | i − j | � 1. Write ρ − 1 = χ and ρ j = ρ ( 2 − j · ) for j � 0. ⊲ Littlewood–Paley blocks: ∆ j f = ρ j ( D ) f = F − 1 ( ρ j F f ) = K i ∗ f = F − 1 � � ρ j F f , j � − 1. where K i = ( 2 π ) − d /2 F − 1 ρ i = 2 id K ( 2 i · ) with K ∈ L 1 ( R d ) Littlewood–Paley decomposition f = ∑ for all f ∈ S ′ . ∆ j f = lim j → ∞ S j f j � − 1 ( 7 / 63 )

Hölder-Besov spaces For α ∈ R , the Hölder-Besov space C α is given by C α = B α ∞ , ∞ ( T d , R ) , where � � ( 2 j α � ∆ j f � L p ) q � 1/ q � f ∈ S ′ : � f � B α B α ∑ p , q = p , q = < ∞ . j � − 1 B α p , q is a Banach space and while the norm �·� B α p , q depends on ( χ , ρ ) , the space B α p , q does not and any other dyadic partition of unity corresponds to an equivalent norm. Notation: �·� α = �·� B α ∞ , ∞ . � ∆ i f � L ∞ � 2 − i α � f � α By Parseval B α 2,2 = H α . Example ∆ i δ 0 ( x ) = ( K i ∗ δ 0 )( x ) = K i ( x ) = 2 id K ( 2 i x ) ⇒ � ∆ i δ 0 � L ∞ ( T d ) ≃ 2 id so δ 0 ∈ C − d . ( 8 / 63 )

Tools Bernstein inequalities Let B be a ball and k ∈ N 0 . For any λ � 1, 1 � p � q � ∞ , and f ∈ L p with supp ( F f ) ⊆ λ B we have � � 1 p − 1 µ ∈ N d : | µ | = k � ∂ µ f � L q � k , B λ k + d max q � f � L p . Besov embedding Let 1 � p 1 � p 2 � ∞ and 1 � q 1 � q 2 � ∞ , and let α ∈ R . Then B α p 1 , q 1 is continuously embedded into B α − d ( 1/ p 1 − 1/ p 2 ) . p 2 , q 2 ( 9 / 63 )

An L 2 computation ε d /2 − α � e i � x , k � ρ i ( k ) 2 π ) d /2 ∑ ∆ i V ε ( x ) = √ R ( ε k ) g ( k ) ( k ∈ Z d so √ 2 π ) d ε − 2 α ∑ k ∈ Z d ρ i ( k ) 2 e i � x , k � R ( ε k ) E [ | ∆ i V ε ( x ) | 2 ] ε d ( = (1) ε d − 2 α 2 id sup k ∈ ε 2 i A R ( k ) , � where A is the annulus in which ρ is supported. Now if ε 2 i � 1 we have E [ | ∆ i V ε ( x ) | 2 ] � 2 id ε d − 2 α = ε β − 2 α 2 i β . The assumption d − 2 α � 0 then implies E [ | ∆ i V ε ( x ) | 2 ] � 2 ( 2 α + κ ) i ε κ for any 0 � κ � d − 2 α . In the case ε 2 i > 1 we use that � B ( 0,1 ) c R ( k ) d k < + ∞ to estimate � ε d ∑ R ( ε k ) � R d R ( k ) d k < + ∞ , k ∈ Z d and then E [ | ∆ i V ε ( x ) | 2 ] � ε − 2 α � 2 2 α i ( ε 2 i ) κ for any small κ > 0. Assume d − 2 α � 0. For any 0 � κ � d − 2 α E [ | ∆ i V ε ( x ) | 2 ] � 2 ( 2 α + κ ) i ε κ . ( 10 / 63 )

From L 2 to almost sure behavior ⊲ Note that ∆ i V ε ( x ) is a Gaussian r.v. so for any p � � E [ � V ε � p p , p ] = ∑ 2 − ip ρ T d dx E [ | ∆ i V ε ( x ) | p ] = C p ∑ 2 − ip ρ T d dx ( E [ | ∆ i V ε ( x ) | 2 ]) p /2 B − ρ i i 2 − ip ρ 2 p ( α + κ /2 ) i ε p κ /2 � ε p κ /2 � ∑ i for all ρ > α + κ /2. ⊲ By Besov embedding � V ε � B − ρ ∞ , ∞ � � V ε � B − ρ + d / p so p , p E [ � V ε � p ∞ , ∞ ] � E [ � V ε � p p , p ] � ε p κ /2 B − ρ B − ρ for all ρ > α + κ /2 + d / p . Note that κ and p are arbitrary. Theorem If d > 2 α then V ε → 0 in C − α − . While if d = 2 α then V ε converges to the space white noise on T d in C − α − . ( 11 / 63 )

Regularity of the solution map We are let to the study of the properties of the equation L v = η v with η ∈ C − α − . This stability is easy to establish when α < 1 by standard estimates in Besov spaces. We need two ingredients: ( γ = 2 − α − ) 1. Schauder estimates in Besov spaces for the parabolic equation L f = g in the form � f � γ � � g � γ − 2 2. Continuity of the product map ( η , v ) �→ v η in the form � v η � γ − 2 � � v � γ � η � γ − 2 v ∈ C γ − → v η ∈ C γ − 2 − → Γ ( v ) = L − 1 ( v η ) ∈ C γ ( 12 / 63 )

Schauder estimates Let Jf such that L Jf = f and Jf ( 0 ) = 0 then � t 0 e ∆ ( t − s ) f s d s . Jf ( t ) = � f ( t ) − f ( s ) � Consider C X = C ([ 0, T ] ; X ) and norms � f � C σ T X = sup 0 � s < t � T . Let | t − s | σ T = C T C σ ∩ C σ /2 L ∞ with the norm � · � L σ L σ T = max {� · � C T C σ , � · � C σ /2 L ∞ } . T T If σ ∈ ( 0, 2 ) then � Jf � L σ T � ( 1 + T ) � f � C T C σ − 2 � t �→ P t u � L σ T � � u � σ . ( 13 / 63 )

Product and paraproduct estimates Deconstruction of a product: f ∈ C ρ , g ∈ C γ fg = ∑ ∆ i f ∆ j g = f ≺ g + f ◦ g + f ≻ g i , j ≥− 1 f ≺ g = g ≻ f = ∑ f ◦ g = ∑ ∆ i f ∆ j g ∆ i f ∆ j g i < j − 1 | i − j |≤ 1 Paraproduct (Bony, Meyer et al.) f ≺ g ∈ C min ( γ + ρ , γ ) f ◦ g ∈ C γ + ρ only if γ + ρ > 0 ( 14 / 63 )

Proof. Recall f ∈ C ρ , g ∈ C γ . i ≪ j ⇒ supp F ( ∆ i f ∆ j g ) ⊆ 2 j A i ∼ j ⇒ supp F ( ∆ i f ∆ j g ) ⊆ 2 j B So if ρ > 0 ∆ q ( f ≺ g ) = ∑ = O ( 2 − q γ ) ⇒ f ≺ g ∈ C γ , j : j ∼ q ∑ ∆ q ( ∆ i f ∆ j g ) i : i < j − 1 � �� � O ( 2 − i ρ − j γ ) while if ρ < 0 ∆ q ( f ≺ g ) = ∑ = O ( 2 − q ( γ + ρ ) ) ⇒ f ≺ g ∈ C γ + ρ . j : j ∼ q ∑ ∆ q ( ∆ i f ∆ j g ) i : i < j − 1 � �� � O ( 2 − i ρ − j γ ) Finally for the resonant term we have ∆ q ( f ◦ g ) = ∑ ∆ q ( ∆ i f ∆ j g ) = ∑ O ( 2 − j ( ρ + γ ) ) ⇒ f ◦ g ∈ C γ + ρ i ∼ j � q i � q but only if the sum converges. ( 15 / 63 )