stochastic analysis and the kdv equation setsuo taniguchi
play

STOCHASTIC ANALYSIS AND THE KdV EQUATION Setsuo TANIGUCHI Faculty - PowerPoint PPT Presentation

STOCHASTIC ANALYSIS AND THE KdV EQUATION Setsuo TANIGUCHI Faculty of Mathematics, Kyushu Univ. Joint work with N. Ikeda + http://www.math.kyushu-u.ac.jp/~taniguch/ 0 PDE and Stochastic Analysis u ( x, t ); ut = L V u, u ( , 0)


  1. STOCHASTIC ANALYSIS AND THE KdV EQUATION Setsuo TANIGUCHI Faculty of Mathematics, Kyushu Univ. Joint work with N. Ikeda + α http://www.math.kyushu-u.ac.jp/~taniguch/ 0

  2. PDE and Stochastic Analysis ✓ ✏ u ( x, t ); ut = L V u, u ( · , 0) = f Heat eq: ✒ ✑ � � � i,j aij∂x i ∂x j + � L V = 1 i bi∂x i + V 2 � f ( X ( t, x )) e Φ( x ; V ) � � � u ( x, t ) = E X ( t, x ) : L 0 -diff.pr. 1942 K.Itˆ o; stoch. integral, Itˆ o’s formula � d � 2 − x 2 1944 R.Cameron-W.Martin; 1 2 2 dx 1947 M.Kac; the Feynman-Kac formula ⋄ Refrectionless potential, generalized refrectionless potential, n -solitons of the KdV eq. vt = 3 2 vvx + 1 4 vxxx . 1

  3. v ( x, t ) = 2sech2( x + t − 2) v(x,y) 2 4 1.8 1.6 1.4 1.2 1 0.8 2 0.6 0.4 0.2 0 -10 0 -5 t-axis -2 0 5 -4 x-axis 10 2

  4. v ( x, t ) = 12 3+4 cosh(2 x +2 t )+cosh(4 x +16 t ) 3 cosh( x +7 t )+cosh(3 x +9 t ) v(x,y) 5 8 7 4 6 5 4 3 3 2 1 0 2 -15 -10 1 t-axis -5 0 0 5 -1 10 x-axis -2 15 3

  5. • u s ( s ∈ S ): reflectionless potential ✛ ✘ u s( x ) = − 2 d 2 dx 2 log det( I + G s( x )) , where ✚ ✙ � � � � S = { ηj, mj } 1 ≤ j ≤ n � n ∈ N , ηj, mj > 0 , ηi � = ηj �� mimj e − ( η i + η j ) x � G s( x ) = . ηi + ηj 1 ≤ i,j ≤ n o op. − d 2 • Schr¨ dx 2 + u s → Scattering data s ∈ S Ξ0 = { u s | s ∈ S } bijective S , ← → • Ξ ∋ u ⇔ ∃ µ > 0 , un ∈ Ξ0 s.t. Spec ( − d 2 dx 2 + un ) ⊂ [ − µ, ∞ ) , n = 1 , 2 , . . . un → u (unif on cpts) (Marchenko) 4

  6. • W = { w : [0 , ∞ ) → R | conti , w (0)=0 } ; 1-dim W.sp • X ( x ) : W → R : X ( x, w ) = w ( x ) , w ∈ W • Σ = { σ | finite meas on R with cpt supp } • P σ : the prob meas on W under which { X ( x ) } is the cent. Gaussian pr with cov fn � � eζ ( x + y ) − eζ | x − y | X ( x ) X ( y ) dP = σ ( dζ ) . 2 ζ R W • G = { P σ | σ ∈ Σ } bijective ← → Σ � � d X ( x )2 dP σ = e 2 ζxσ ( dζ ) ∵ ) dx W R 5

  7. � x � � � ψ ( P σ )( x )=4 d 2 − 1 X ( y )2 dy dP σ dx 2 log exp , x ≥ 0 2 0 W The Plan of talk ✬ ✩ ✬ ✩ G : Cen. Gauss Ξ : gen. rl. pot ψ P σ, σ ∈ Σ u ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ � � � � � � � � � � � � � � ❅ ❅ (3) � � � ❅ ❅ ❅ � ❅ � � ❅ ❅ � � � ❅ ❅ ❅ � ❅ unif conv. (2) on cpts ✬ ✩ ✬ ✩ Ξ0 : rl pot G 0 : P σ , � � ❅ � ❅ ❅ us, s ∈ S � ❅ � � ❅ ❅ σ = � ❅ � ❅ ❅ � � ❅ � ❅ ❅ � � j c 2 (1) jδp j ✫ ✪ ✫ ✪ ✫ ✪ ✫ ✪ Realization of P σ , Spelling out s ∈ S , Solitons 6

  8. reflectionless potential and n -soliton � � � σ = � n � j =1 c 2 Σ0= jδp j � n ∈ N , pj ∈ R , pj � = pi, cj > 0 • σ ∈ Σ0 . { b ( x ) } x ≥ 0 ; an n-dim B.m.on (Ω , F , P ) � x ξσ ( x ) = exD σ e − yD σ db ( y ) ( Dσ = diag [ pj ]) 0 Xσ ( x ) = � c , ξσ ( x ) � = � n i =1 ciξi σ ( x ) ( c=( ci ) ) P σ = P ◦ X − 1 σ ( Xσ : Ω → W ) • σ ∈ Σ0 ; ∃ m < n, 1 ≤ j (1) < · · · < j ( m ) ≤ n s.t. | pj | ≤ | pj +1 | , pj ( ℓ ) > 0 , pj ( ℓ )+1 = − pj ( ℓ ) # {| p 1 | , . . . , | pn |} = n − m . 7

  9. r r r r p 2 ψ : Σ0 ∋ σ �→ { ηj, mj } ∈ S ✉ ✉ ✉ ✉ n p 2 1 − 1 { η 1 < · · · < ηn } = { pj (1) ,. . ., pj ( m ) , √ r 1 ,. . ., √ rn − m } ( 0 < r 1 < · · · < rn − m : � n j =1 c 2 j/ ( p 2 j − r ) = − 1 )  c 2   � �  ηk + ηi pk + ηi j ( ℓ )+1   2 ηi ,   c 2  ηk − ηi pk − ηi   j ( ℓ ) k � = i k � = j ( ℓ ) ,j ( ℓ )+1  mi = if i = j ( ℓ ) ,    n  � �  ηk + ηi pk + ηi    − 2 ηi , otherwise.   ηk − ηi pk − ηi  k � = i k =1 8

  10. Thm 1. Let P σ ∈ G 0 = { P σ | σ ∈ Σ0 } . Then � x � � � − 1 X ( y )2 dy dP σ 4 log exp 2 0 W � � = − 2 log det I + Gψ ( σ )( x ) n � � � + 2 log det I + Gψ ( σ )(0) − 2 x ( pi + ηi ) . i =1 Moreover, ψ : G 0 → Ξ0 and ψ ( P σ ) = uψ ( σ ) . Finally, ψ : G 0 → Ξ0 is bijective. � • ψ ( P σ )( x ) = 4 d 2 exp( · · · ) dP σ dx 2 log W • ψ ( P σ ) = uψ ( σ ) on [0 , ∞ ) ; “ ψ ( G 0) ⊂ Ξ0 ” The real analyticity does the rest of job 9

  11. Cor. (i) If µ ( A ) = σ ( − A ) , then uψ ( σ )( x ) = ψ ( P µ )( − x ) for x ∈ ( −∞ , 0] . (ii) For y ≤ 0 , let b ( y ) = b ( − y ) , and � 0 ξσ ( y ) = − eyD σ e − zD σ db ( z ) y Xσ ( y ) = � c , ξσ ( y ) � . Then u = ψ ( P σ ) is represented as � � 0 ∨ x � � u ( x ) = 4 d 2 − 1 Xσ ( y )2 dy dx 2 log exp dP 2 Ω 0 ∧ x for every x ∈ R . 10

  12. • The τ -function of the KdV hierarchy is τ ( x,� t ) = det( I + A ( x,� t )) t = ( tj ) ∈ RN with # { tj � = 0 } < ∞ , where x ∈ R , � �� mimj � e −{ ζ i ( x,� t )+ ζ j ( x,� t ) } A ( x,� t ) = , ηi + ηj 1 ≤ i,j ≤ n ∞ � tαη 2 α +1 { ηj, mj } ∈ S , ζi = ζi ( x,� . t ) = xηi + i α =1 t = ( t, 0 , . . . ) , then v ( x, t ) = 2 ∂ 2 • If � x log τ ( x,� t ) solves the KdV eq; vt = 3 2 vvx + 1 4 vxxx . 11

  13. • For σ ∈ Σ0 , let ψ ( σ ) = { ηj, mj } ∈ Ξ0 . Define � ( ∂xφ ) φ − 1 � (0 ,� t ) , ζ = diag [ ζj ] , β� t = − � � cosh( ζ ) − sinh( ζ ) R − 1 U − 1 DσU U − 1 , φ ( x,� t ) = U σ + c ⊗ c = UR 2 U − 1 ( R = diag [ ηj ] ), U ∈ O ( n ) ; D 2 � � x � − 1 Xσ ( y )2 dy Iσ ( x,� t ) = exp 2 Ω 0 � +1 2 � ( β� t − Dσ ) ξσ ( x ) , ξσ ( x ) � dP. � � = − 1 Iσ ( x,� 2 log τ ( x,� Thm 2 (i) log t ) t ) � n +1 t ) − x 2 log τ (0 ,� i =1( pi + ηi ) 2 � � t )= − 4 ∂ 2 (ii) If � t = ( t, 0 , . . . ) , then vσ ( x,� x log Iσ is an n -soliton of the KdV eq. (Super pos) 12

  14. Change of variables formulae on W ; Prop. Let φ ( y ) ∈ R n × n be a sol of φ ′′ − Eσφ = 0 , where Eσ = D 2 σ + c ⊗ c . Let x > 0 and assume (A.1) det φ ( y ) � = 0 , (A.2) β ( y ) = − ( φ ′ φ − 1)( y ) is symm (0 ≤ y ≤ x ) . Then � � x � − 1 Xσ ( y )2 dy exp 2 Ω 0 � +1 2 � ( β (0) − Dσ ) ξσ ( x ) , ξσ ( x ) � dP � � 1 / 2 � � − 1 / 2 . ex tr D σ det φ ( x ) = det φ (0) 13

  15. γ : [0 , x ] → R n × n • OU pr � H.Osc; � x � � − 1 Xσ ( y )2 dy exp 2 Ω 0 � +1 2 � ( γ ( x ) − Dσ ) ξσ ( x ) , ξσ ( x ) � dP � � x � = e − tr D σ / 2 − 1 exp � Eσb ( y ) , b ( y ) � dy 2 Ω 0 � +1 2 � γ ( x ) b ( x ) , b ( x ) � dP • (C-M) Itˆ o ⊕ Girsanov; � x � � − 1 � ( γ 2 + γ ′ ) b ( y ) , b ( y ) � dy exp 2 Ω 0 � x � � 1 � +1 0 tr γ 2 � γ ( x ) b ( x ) , b ( x ) � dP = exp 2 γ 2 + γ ′ = Eσ , • γ ( x ) = β (0) − Dσ Cole-Hopf; γ ( y )= − ( φ ′ φ − 1)( x − y ) ⇒ φ ′′ − Eσφ = 0 14

  16. Pf of Thm 1: φ ′′− Eσφ =0 , φ (0)= I, φ ′ (0)= − Dσ ; φ ( y ) = cosh( yE 1 / 2 ) − E − 1 / 2 sinh( yE 1 / 2 ) Dσ σ σ σ (Case1) | pi | < | pi +1 | , i = 1 , . . . , n − 1 . � � φ ( y )= − 1 2 UV R − 1 B eyRB − 1 XC I + Gψ ( σ )( y ) � σ − rjI ) − 1 c |− 1 � , | ( D 2 V = diag � � n � , R = diag [ ηj ] , a ( i ) = sgn β =1( pβ − ηi ) � � � 1 / 2 , α � = i ( η 2 α − η 2 i ) � n b ( i ) = a ( i ) − 2 ηi β =1 ( p 2 β − η 2 i ) � � − 1 . B = diag [ b ( j )] , Xij = pj + ηi j = pj − ε � m (Case2) pε i =1 δj,ℓ ( i )+1 , ε → 0 . 15

  17. Pf of Thm 2: φ ( y ) = φ ( y,� t ) ; (A.1),(A.2) are fulfilled (Case1) | pi | < | pi +1 | , i = 1 , . . . , n − 1 . t ) } eζ ( y,� t ) = − 1 2 UR − 1 V B { I + A ( y,� t ) B − 1 XC φ ( y,� j = pj − ε � m (Case2) pε i =1 δj,ℓ ( i )+1 , ε → 0 . 16

  18. Bijectivity & Cor • Let u = u s ∈ Ξ0 (s ∈ S ) , and e +( x ; ζ ) be the right Jost sol of L = − ( d/dx )2 + u s ; L e +( ∗ ; ζ ) = ζ 2 e +( ∗ ; ζ ) , e +( x ; ζ ) ∼ eiζx ( x → ∞ ) Then ∃ λj ∈ C ∞ ( R ; R ) , 1 ≤ j ≤ n, s.t. ζ −√− 1 λ j ( x ) √− 1 ζx � e +( x ; ζ )= e ζ + √− 1 η j . j Define κ : Ξ0 → Σ0 by κ (s) = � j ( − λ ′ j (0)) δλ j (0) Then ψ ( κ (s)) = s , κ ( ψ ( σ )) = σ. • Let � u ( x ) = u s( − x ) . Then � s ( κ ( � s) = µ ) u = u � 17

  19. generalized reflectionless potentials • u ∈ Ξ ⇔ ∃ µ > 0 , un ∈ Ξ0 s.t. � � − d 2 Spec ⊂ [ − µ, ∞ ) , n = 1 , 2 , . . . dx 2 + un un → u (unif on cpts) � � x � � − 1 X ( y )2 dy dP σ • Φ σ ( x ) = exp 2 0 W ψ ( P σ ) = 4 d 2 dx 2 log Φ σ G ⊃ G 0 ∋ P σ �→ ψ ( P σ ) ∈ Ξ0 ⊂ Ξ : bijective Question: “ P σ n → P σ ” � “ un → u ”, ψ ( G ) � Ξ 18

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend