Anisotropic Random Wave Models Anne Estrade & Julie Fournier - PowerPoint PPT Presentation

Anisotropic Random Wave Models Anne Estrade & Julie Fournier MAP5 - Universit´ e Paris Descartes Random waves in Oxford - June 2018

What is the talk about? k is a random vector in R d ( d ≥ 2 , k � = 0 ) Associate ◮ covariance function: t ∈ R d �→ E cos( k · t ) ◮ Gaussian random field on R d , say G k , that is stationary centered with such a covariance Question: ◮ links between anisotropy properties of G k and those of k ?

Outline of the talk 1. Random wavevector and associated covariance function 2. Level sets of Gaussian waves 3. Crest lines in the planar case without isotropy hypothesis

1. Random wavevector k is a random vector in R d such that P ( k = 0) = 0 (wavevector) Notations • matrix kk T = ( k i k j ) 1 ≤ i , j ≤ d • k = R � k with R = � k � and � k ∈ S d − 1 • d µ ( λ ): probability distribution of k on R d Vocabulary • k is isotropic if � k is uniformly distributed in S d − 1 • k is separable if � k � and � k are independent random variables

Particular cases ◮ � k � = κ, a . s . with κ constant > 0 (wavenumber) note that k is separable in that case

Particular cases ◮ � k � = κ, a . s . with κ constant > 0 (wavenumber) note that k is separable in that case ◮ d = 2, k separable, � k = (cos Θ , sin Θ) with ◮ Θ ∼ U ([0 , 2 π ]) (isotropic case) ◮ or Θ ∼ U ([ − δ, δ ]) (elementary case) ◮ or Θ ∼ C α | cos θ | α d θ (toy model)

Particular cases ◮ � k � = κ, a . s . with κ constant > 0 (wavenumber) note that k is separable in that case ◮ d = 2, k separable, � k = (cos Θ , sin Θ) with ◮ Θ ∼ U ([0 , 2 π ]) (isotropic case) ◮ or Θ ∼ U ([ − δ, δ ]) (elementary case) ◮ or Θ ∼ C α | cos θ | α d θ (toy model) ◮ d = 3 and k ∈ A = { x 2 + y 2 = z 4 } a . s . (Airy surface) Rmk: In examples 1 and 3, k is such that Pol ( k ) = 0

Single random wave Let k be a random wavevector in R d Let η be a r.v. independent of k with η ∼ U ([0 , 2 π ]) and √ 2 cos( k · t + η ) , t ∈ R d X ( t ) = Hence ◮ X is centered, variance 1 ◮ X is second order stationary with E [ X ( s ) X ( t )] = E cos( k · ( t − s )) ◮ X is not second order isotropic (unless k is isotropic)

Gaussian random wave associated with a wavevector Let k be a random wavevector in R d Def: We call Gaussian random wave associated with k any Gaussian random field G on R d that is stationary and centered with covariance r ( t ) := E ( G ( t ) G (0)) = E cos( k · t ) , t ∈ R d Rmk: Var G (0) = 1 and � R d e i λ · t d µ ( s ) ( λ ) r ( t ) = with µ ( s ) = 1 2 ( µ + ˇ µ ) the spectral measure of G

Covariance function k is a random vector in R d and r ( t ) = E cos( k · t ) , t ∈ R d Fact: ◮ r is of class C m iff k admits finite moments of order m ◮ for any j = ( j 1 , . . . , j d ), ∂ j r (0) = 0 if | j | is odd and ∂ j r (0) = ( − 1) | j | / 2 E k j if | j | is even ◮ E ( G ′ (0) G ′ (0) T ) = − r ′′ (0) = E ( kk T ) ( d × d matrix)

Partial Differential Equation � α j λ j P multivariate even polynomial: P ( λ ) = j ∈ N d ; | j | even � ( − 1) | j | / 2 α j ∂ j : differential operator L P = j ∈ N d ; | j | even Let k be a wavevector in R d and G associated Gaussian wave G is an a.s. solution of L P ( G ) = 0 ⇔ P ( k ) = 0 a . s . ⇔ spectral measure of G supported by { λ ∈ R d : P ( λ ) = 0 }

Examples ◮ Berry random wave: � k � = κ a . s . with κ constant > 0 Gaussian wave G satisfies ∆ G + κ 2 G = 0 a . s .

Examples ◮ Berry random wave: � k � = κ a . s . with κ constant > 0 Gaussian wave G satisfies ∆ G + κ 2 G = 0 a . s . ◮ Sea waves: k in R 3 with ( k x ) 2 + ( k y ) 2 = ( k t ) 4 , a . s . Gaussian wave G on R 2 × R : height at point ( x , y ) at time t . It satisfies ∆ G + ∂ 4 ∂ t 4 G = 0 a . s .

Examples ◮ Berry random wave: � k � = κ a . s . with κ constant > 0 Gaussian wave G satisfies ∆ G + κ 2 G = 0 a . s . ◮ Sea waves: k in R 3 with ( k x ) 2 + ( k y ) 2 = ( k t ) 4 , a . s . Gaussian wave G on R 2 × R : height at point ( x , y ) at time t . It satisfies ∆ G + ∂ 4 ∂ t 4 G = 0 a . s . ◮ Acoustic/optical waves in heterogeneous media, ...

2. Level sets Let k random wavevector in R d , G associated Gaussian random field defined on R d , a ∈ R fixed level G − 1 ( a ) = { t ∈ R d : G ( t ) = a } , ◮ submanifold of R d , dimension d − 1 ◮ nodal set in the case a = 0 ◮ ∀ t ∈ G − 1 k ( a ), tangent space T t G − 1 k ( a ) is ⊥ G ′ ( t ) question: ”favorite” orientation of T t G − 1 ( a )?

Favorite orientation of level sets def: favorite direction of V ( V : rdom in R d ) is any direction in Argmax { E ( V . u ) 2 ; u ∈ S d − 1 } But E ( V . u ) 2 = u . E ( VV T ) u and E ( G ′ (0) G ′ (0) T ) = E ( kk T ) so, morally: ”The favorite orientation(s) of the level sets G − 1 ( a ) is(are) orthogonal to the favorite direction(s) of k ” ”It becomes highly probable that the direction of the contour is near the principal direction” [Longuet-Higgins’57]

( d = 2) Favorite direction of level lines - examples Let k separable, so E ( kk T ) = ( E � k � 2 ) E ( � k � k T ) and let � k = (cos Θ , sin Θ) ◮ isotropic case: Θ ∼ U ([0 , 2 π ]) E ( � k � k T ) = I 2 then, no favorite direction ◮ toy model: Θ ∼ C α | cos θ | α d θ with some α > 0 � α + 1 � 0 E ( � k � k T ) = 1 α +2 0 1 favorite direction of level lines is ⊥ 0 ◮ elementary model Θ ∼ U ([ − δ, δ ]) with some δ ∈ (0 , π/ 2) � 1 + sinc(2 δ ) � 0 E ( � k � k T ) = 0 1 − sinc(2 δ ) favorite direction of level lines is ⊥ 0

Expected measure of level sets Let Q compact ⊂ R d . Kac-Rice formula yields � E [ H d − 1 ( G − 1 ( a ) ∩ Q )] = E [ � G ′ k ( t ) � | G k ( t ) = a ] p G k ( t ) ( a ) dt Q = H d ( Q ) e − a 2 / 2 E � G ′ √ k (0) � 2 π � with E � G ′ R d ( E ( kk T ) x · x ) 1 / 2 Φ d ( x ) dx k (0) � = Separable case: k = � k � � ⊥ � k with � k �⊥ k , then � k (0) � = ( E � k � 2 ) 1 / 2 R d ( E [ � k � E � G ′ k T ] x · x ) 1 / 2 Φ d ( x ) dx

Expected measure of level sets - Berry isotropic RW ◮ Berry isotropic case: � k � = κ and � k ∼ U ( S d − 1 ) E [ H d − 1 ( G − 1 ( a ) ∩ Q )] = H d ( Q ) e − a 2 / 2 κ Γ(( d + 1) / 2) √ Γ( d / 2) 2 π ◮ Berry isotropic planar case, nodal line ( d = 2 , a = 0) 1 E [ length ( G − 1 ( a ) ∩ Q )] = H 2 ( Q ) √ κ 2 2

Planar case - Mean length of level curves E [ length ( G − 1 ( a ) ∩ Q )] = H 2 ( Q ) e − a 2 / 2 E � G ′ √ k (0) � 2 π with � R 2 ( E ( kk T ) x · x ) 1 / 2 Φ 2 ( x ) dx E � G ′ k (0) � = � (1 − γ − /γ + ) 1 / 2 � = (2 /π ) 1 / 2 ( γ + ) 1 / 2 E , where � π/ 2 (1 − x 2 sin 2 θ ) 1 / 2 d θ , elliptic integral ◮ E ( x ) = 0 ◮ 0 ≤ γ − ≤ γ + are the eigenvalues of E ( kk T )

Mean length of level curves - separable case separable case: k = � k � � ⊥ � k with � k �⊥ k then ◮ E ( kk T ) = ( E � k � 2 ) E ( � k � k T ) γ − = Trace ( E ( � k � ◮ γ ± = ( E � k � 2 ) � k T )) = 1 γ ± and � γ + + � hence E [ length ( G − 1 ( a ) ∩ Q )] = H 2 ( Q ) e − a 2 / 2 ( E � k � 2 ) 1 / 2 F ( c ( � √ k )) π 2 �� 2 c � 1 / 2 � where the map F : c ∈ [0 , 1] �→ (1 + c ) 1 / 2 E 1+ c is strictly decreasing

Mean length of level curves - separable case separable case: k = � k � � ⊥ � k with � k �⊥ k then ◮ E ( kk T ) = ( E � k � 2 ) E ( � k � k T ) γ − = Trace ( E ( � k � ◮ γ ± = ( E � k � 2 ) � k T )) = 1 γ ± and � γ + + � hence E [ length ( G − 1 ( a ) ∩ Q )] = H 2 ( Q ) e − a 2 / 2 ( E � k � 2 ) 1 / 2 F ( c ( � √ k )) π 2 �� 2 c � 1 / 2 � where the map F : c ∈ [0 , 1] �→ (1 + c ) 1 / 2 E 1+ c is strictly decreasing ◮ but what about c ( � k )?

Coherency index γ + − γ − Def: the coherency index of matrix M is: γ + + γ − where 0 ≤ γ − ≤ γ + are the eigenvalues of M the coherency index of E ( kk T ) . c ( k ) = Result: if k is separable, ◮ c ( k ) = c ( � k ) only depends on the directional distrib. of k ◮ and E [ length ( G − 1 ( a ) ∩ Q ))] is a ց function of c ( � k )

Coherency index as anisotropy parameter (examples) separable case: k = � k � (cos Θ , sin Θ) with � k �⊥ ⊥ Θ ◮ Toy model: Θ ∼ C α | cos θ | α d θ c ( � k ) = α ( ր function of α ) ◮ Elementary model: Θ ∼ U ([ − δ, δ ] ∪ [ π − δ, π + δ ]) c ( � k ) = sinc(2 δ ) ( ց function of δ ∈ [0 , π/ 2])

3. Crest lines k a 2-dim rdom wavevector, G associated Gaussian wave ϕ ∈ [0 , π ) fixed, u ϕ = (cos ϕ, sin ϕ ) Z ϕ := G ′ · u ϕ = { G ′ ( t ) · u ϕ ; t ∈ R 2 } Z − 1 ϕ (0) = nodal line of Z ϕ :=crest line in direction ϕ Claim: Z ϕ Gaussian wave associated with rdom wavevector K ϕ K ϕ ∼ ( λ · u ϕ ) 2 d µ ( λ ) m 20 ( ϕ ) with � � ( λ · u ϕ ) i ( λ · u ϕ + π/ 2 ) j d µ ( λ ) = ( λ 1 ) i ( λ 2 ) j d µ ϕ ( λ ) m ij ( ϕ ) =

Mean length of crest lines 1 E [ length ( Z − 1 E � Z ′ ϕ (0) ∩ Q )] = H 2 ( Q ) √ ϕ (0) � 2 π m 20 ( ϕ )