Brownian sheet and the reflectionless potential. Setsuo TANIGUCHI. Faculty of Math., Kyushu Univ. Fukuoka , JAPAN

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1 Brownian sheet and the reflectionless potential Setsuo TANIGUCHI Faculty of Math., Kyushu Univ. Fukuoka , JAPAN June 1, 24, at Keio Univ. Partially supported by Grant-in-Aid for Scientific Research (A)(1)

2 The KdV equation v t = (3/2)vv x + (1/4)v xxx Nonlin. differ. eq. for a shallow water wave. n-soliton solutions; at t =, behaves n1 a j sech 2 ( c j (x c j + δ j )/2). For n N, the space of scattering data; S n = { (η j, m j ) (, ) 2n η i η j (i j) }. u s :reflectionless potential For s S n, u s (x) = 2(d/dx) 2 log det(i + G s (x)), ( mi m j e (η i+η j )x) where G s (x) = η i + η j If s(t) = (η j, m j exp[ 2ηj 3 t]), then v(x, t) = u s(t) (x) solves the KdV equation. 1 i,j n. 2

3 Kotani s homeo. Given λ >, Ξ cl ([ λ, ]) = {u s s S n, n N, Spec( (d/dx) 2 + u s ) [ λ, )}, Ξ([ λ, ]) = Ξ cl ([ λ, ]) unif conv on cpts, Σ([ λ, λ ]) = {σ meas. on [ λ, λ ] s.t. [ λ, λ ] {λ ζ 2 } 1 σ(dζ) 1}. Θ : Ξ([ λ, ]) Σ([ λ, λ ])(homeo) Let u Ξ([ λ, ]). 1) For λ C \ R,!f± u (x, λ) so that L u f± u = λf ± u, f ± u () = 1, f ± u L2 (R ± ). 2) σ± u s.t. R (1 + ξ2 ) 1 σ± u (dξ) < & ±(f± u ) (, λ) = α u ± + βu ± z + ( 1 R ξ z ξ ) 1+ξ σ u 2 ± (dξ) 3) σ u = Θ(u) is defined by (,] f( ± ξ ) 1 ξ σ u ± (dξ) = R f(ζ)σ u(ζ). 3

4 Marchenko and Dyson formulae: For x >, where u(x) = 2 d dx Ku + (x, x), (M) = 2 d2 u,x log det(i + F dx2 + ), (D) F u + (x, y) = ( 2 / x y) ( R (1 cos ξx)(1 cos ξy) ξ 2 σ+ u (dξ) x y). K+ u (x, y) + F + u (x, y) + x F+ u (y, t)ku + (x, t)dt =, F u,x + g = x F+ u (, z)g(z)dz. 4

5 Let σ u = Θ(u), {X σu (x)} x R be the centered Gaussian process with cov. R σu (x, y) R (eζ(x+y) e ζ x y )/(2ζ)σ u (dζ). Kotani showed R σu = F+ u, and hence u(x) = 4 d2 ( ( dx 2log exp 1 x ) ) X σu (y) 2 dy dp. 2 Let G = {{X(y)} y cent Gau pr with R σ, σ M (R)}, where M (R) = {σ meas. suppσ R}. Θ : Ξ([ λ, ]) G, λ > Question; Expressing u in terms of σ u. 5

6 For σ Σ n = { n j=1 c 2 j δ p j p i p j, c j > }, assume m, 1 j(1) < < j(m) n s.t. p j p j+1, (H) p j(l) >, p j(l)+1 = p j(l), and #{ p 1,..., p n } = n m. Let < r 1 < < r n m be the roots of nj=1 c 2 j /(p2 j r) + 1 =. Define < η 1 < < η n, < m 1,..., m n by {η 1,...,η n } = {p j(1),...,p j(m),r 1/2 1,...,r 1/2 c 2 j(l)+1 η k + η i 2η i c 2 η j(l) k i k η i p k + η i, m i = p k η i 2η i k i k j(l),j(l)+1 η k + η i η k η i if i = j(l), n p k + η i. p k η i k=1 n m }. ψ n : Σ n σ (η 1,..., η n, m 1,..., m n ) S n. 6

7 Theorem 1 (Ikeda-T.) For σ Σ n, set ξ σ (y) = e yd σ y X σ (y) = c, ξ σ (y), [ I σ (x) = exp 1 2 e zd σ db(z), x ] X σ (y) 2 dy dp, where B is an n-dim Bm on (, F, P ), D σ = diag(p j ), c = t (c 1,..., c n ). Then Θ 1 = ψ n on Σ n in the sense that u ψn (σ) (x) = 4(d/dx)2 log I σ (x), Question; Can one express the uniform convergence on compacts of generalized ref less potentials in terms of Gaussian processes? 7

8 Let {W (p, x)} (p,x) [, ) 2 be the Brownian sheet on (, F, P ), i.e., the centered Gaussian system with cov W (p, x)w (q, y)dp = (p q)(x y) The Wiener integral is defined so that L 2 ([, ) 2 ) h iso hdw L2 (P ) [, ) 2 s.t. dw = W (b, d) [, ) 2 χ [a,b) [c,d) W (a, d) W (b, c) + W (a, c), For any q < < q n, W q...q n (y) = ( {q j q j 1 } 1/2{ W (q j, y) W (q j 1, y) }) 1 j n is an n-dim Bm, and for f L 2 ([, ); R n ), y f(z), dw q...q n (z) = n j=1 [, ) 2 f j (z) qj q j 1 χ [qj 1,q j ) [,y)(q, z)w (dq, dz), 8

9 Let T > and take {h( ; t) t [, T ]} L 2 ([, ) 2 ) s.t. [, ) 2 h(v; t) h(v; s) 2 dv K sup s<t T t s <. Set X(y) = [, ) 2 h( ; y)dw. Then X(t) X(s) 2m dp (2m)! 2 m m! Km t s m Thus {X(y)} y [,T ] admits a conti. ver. Theorem 2. C m,t > s.t. for any {h( ; t) t [, T ]} as above, X(t) X(s) 2m sup s<t T t s m (3/2) dp C m,tk m, sup X(y) 2m dp C m,t K m. y [,T ] ) sup s<t T φ(t) φ(s) α / t s β 2 2 3α+2( β ) α T T φ(t) φ(s) α β 2 dtds. t s β 9

10 a, g : [, ) R; p.w.conti., cpt supp., h a,g (q, z; y) = e (y z)(q a) g(q)χ [,y) (z) X a,g (y) = h a,g( ; y)dw. [, ) 2 ˆp = {p n } n=1, ĉ = {c n} n=1 R. Suppose p n = p n(ˆp) if n n(ˆp), j=2 p j p j 1 <, j=1 c 2 j <, where n(ˆp) = min{k N p k+1 {p 1,..., p k }}. For a, b R with a b < p 1, h a,b,ˆp,ĉ (q, z; y) = e (y z)pj c j j=1 χ qj q [qj 1,q j 1 j ) [,y)(q, z), Y a,b,ˆp,ĉ (y) = h a,b,ˆp,ĉ( ; y)dw. [, ) 2 where q = b + a, q k = q + k j=1 p j p j 1 (p = b). 1

11 p = {p j } n j=1, c = {c j} n j=1 R with p i p j. For a, b R with a b < p 1, define q < q 1 < < q n as above. ξ a,b,p (y) = y e (y z)d(p) dw q...q n (z) X a,b,p,c (y) = c, ξ a,b,p (y), where D(p) = diag(p j ). Extend p, c to ˆp, ĉ, and observe X a,b,p,c (y) = Y a,b,ˆp,ĉ (y). Redefine the Wiener f nal rel to ref less pot; [ Ψ a,b,p,c (x) = exp 1 x ] X a,b,p,c (y) 2 dy dp. 2 Ψ a,b,p,c is C, and Ψ a,b,p,c (x) = 1 { 2 nj=1 c 2 4 j + + 2X a,b,p,c (x)x a,b,p,d(p)c (x) X a,b,p,c (x) 4} exp [ 2 1 x X a,b,p,c (y) 2 dy ] dp. 11

12 Covariances are given by X a,g (x)x a,g (y)dp = a g(q + a) 2eq(x+y) e q x y dq, 2q Y a,b,ˆp,ĉ (x)y a,b,ˆp,ĉ (y)dp = j=1 c 2 j 2p j { e (x+y)p j e x y p j }, X a,b,p,c (x)x a,b,p,c (y)dp = n j=1 c 2 j 2p j { e (x+y)p j e x y p j }. R σ (x, y) = {e ζ(x+y) e ζ x y }/(2ζ)σ(dζ); σ(dζ) =χ [ a, ) g(ζ + a) 2 dζ, n c 2 j δ p j (dζ), c 2 j δ p j (dζ). j=1 j=1 12

13 Theorem 3. Let a, b a, g : [, ) R; suppg [a + b, T ] ( T > ). Take m(1) < m(2) < N, p n = {p n j }m(n) j=1, cn = {c n j }m(n) j=1, s.t. (a) b < p n 1 < < pn m(n) = T a, (b) c n j = g(pn j + a){pn j pn j 1 }1/2, (c) max j {p n j pn j 1 } (pn = b). pw conti., Then, (i) X a,b,p n L2,cn(y) X a,g (y), (ii) X a,b,p n,c w n X a,g on C([, T ]; R), ( T ) (iii) Ψ (j) a,b,p n,c n Ψ (j) a,g uniformly on cpts, j =, 1, 2. { Ψ a,g (x) = 1 2 g(y) 2 dy 4 } + 2X a,g (x)x a, g (x) X a,g (x) 4 [ exp 1 2 x where g(y) = (y a)g(y). ] X a,g (y) 2 dy dp. 13

14 Theorem 4. Let a, T >, and σ a finite meas., suppσ ( a, T a). Take φ, C (R) with φ = 1, and set φ n (x) = nφ(nx), g n (x) = {φ n σ(x a)} 1/2. w Then (i) X a,gn Xσ on C([, T ]; R), where X σ is the cent Gau pr with cov R σ ; R σ (x, y) = {e ζ(x+y) e ζ x y }/(2ζ)σ(dζ) (ii) Ψ (j) a,g n Ψ (j) σ uniformly on cpts, j =, 1, 2. Cor. p n = {p n j }m(n) j=1, cn = {c n j }m(n) j=1 s.t. (i) X a,b,p n,c w n X σ on C([, T ]; R), (ii) Ψ (j) a,b,p n,c n Ψ (j) σ uniformly on cpts, j =, 1, 2. In particular, the ref less potential 4(d/dx) 2 log Φ a,b,p n,cn(x) tends to the generalized one 4(d/dx) 2 log Ψ σ (x) uniformly on cpts. 14

15 Key ingredients; 1) Exact expression of the second derivatives. 2) Estimation of Hölderian norms; a) (Thm.3) For Y a,b,ˆp,ĉ, Kˆp,ĉ,T = 2e 2T M(ˆp) {1 + T 2 M(ˆp) 2 }S(ĉ), M(ˆp) = sup p j, S(ĉ) = j c2 j. sup n Kˆp n,ĉ n,t <, sup n Kˆp n,d(ˆp n )ĉ n,t <. b) (Thm.4) For X a,g, K a,g,t = 2(T + a + 1) 2 (T 2 + 1) e 2T (T +a) g 2 dq. sup n K a,gn,t <, sup n K a, gn,t <. 15

16 {X(x)} x ; a conti Gau pr with cov R(x, y) {N(x)} x ; a 1-dim Bm indep of X G(x) = x X(y)dy + N(x), x. G x = σ(g(y) : y x), ˆX(x) = E[X(x) G x ], γ X (x) (X(x) ˆX(x)) 2 dp. K(x, y), L(x, y); the Volterra kernels K(x, y) + x K(x, z)r(z, y)dz = R(x, y), L(x, y) K(x, y) x y K(x, z)l(z, y)dz =. Then L(x, y) = χ [,x] (y) X(x)(X(y) ˆX(y))dP. In particular, γ X (x) = L(x, x) = K(x, x). Kleptsyna and Le Breton showed; x 2 exp[ 1 X(y) 2 dy]dp = exp[ 1 x 2 γ X (y)dy] Hence 4 d2 dx 2 log( exp[ 1 x 2 X(y) 2 dy ] dp ) = 2 d ( K(x, x)). dx Substitute X = X σu. The Marchenko formula recovered via the Gaussian process. 16

17 Let (η j, m j ) = ψ n (σ p,c ). U O(n) s.t. D(p) 2 + c c = UR 2 U 1 (R = diag(η)). Set φ(y, t) = U{cosh(yR + tr 3 ) sinh(yr + tr 3 )R 1 U 1 D(p)U}U 1, β t = (( y φ)φ 1 )(, t) D(p), q(x, t) = 1 x 2 X a,b,p,c (y) 2 dy 2 1 β tξ a,b,p (x), ξ a,b,p (x), v(x, t) = 4 2 x log( eq(x,t) dp ). Theorem 5 (I&T) v(x, t) solves the KdV. Theorem 6 (T) It holds that e ζq(x,t) dp = {e xtrd(p) det[c(x; B(ζ)) (ζβ t + D(p))s(x; B(ζ))]} 1/2, where c(x; A) = cosh(xa 1/2 ), s(x; A) = A 1/2 sinh(xa 1/2 ). Naïve question (open); If n, then? 17

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