Some Results on Integrable Algorithms X ING -B IAO H U ICMSEC, AMSS, - - PowerPoint PPT Presentation
CIRM, Luminy Sept. 28Oct.2, 2009 Some Results on Integrable Algorithms X ING -B IAO H U ICMSEC, AMSS, Chinese Academy of Sciences P.O.Box 2719, China, 100080 hxb@lsec.cc.ac.cn This is joint work with Yi HE, Hon-Wah TAM and Satoshi
k+1 − ε(n+1) k−1 )(ε(n+1) k
k ) = 1
−1 = 0,
k+1 − ρ(n+1) k−1 )(ρ(n+1) k
k ) = k
1
k+1 − η(n+1) k−1
k
k
1
k+1 − ε(n+1) k−1 )(ε(n+1) k
k ) = 1
−1 = 0,
2k =
2k+1 =
f(n)f(n+1) ,
1 2Dn + e 3 2Dn − e 1 2Dn]fn · fn = 0
n
n+1 − f m n f m+1 n+1 − δ[f m n−1f m+1 n+2 − f m n f m+1 n+1 ] = 0
n = fm
n−1fm+1 n+2
fm,nfm+1
n+1 ,
n
n = ˆ
n um n−1 − um+1 n
n+1 )
δ 1−δ.
k
k
k,i Sn+i .
k
k T (n) k−1 + b(n) k T (n+1) k−1
k
k
k,i Sn+kp+iJ,
k
k T (n+p) k−1
k T (n+q) k−1
k } be the sequence transformations obtained by the recursion scheme
k,j} such that
k
k
k,i Sn+kp+iJ,
0,0 = 1,
k,0 = a(n) k α(n+p) k−1,0,
k,i = a(n) k α(n+p) k−1,i + b(n) k α(n+q) k−1,i−1, i = 1, 2, . . . , k − 1,
k,k = b(n) k α(n+q) k−1,k−1.
i=0 α(n) k,i is independent of n, say k i=0 α(n) k,i = αk, if and only
k + b(n) k
k + b(n) k
i=0 γi.
k (σ)
k (σ)
i=0 α(n) k,i σ(zn+kp+iJ). Furthermore, assume that σ0, σ1, . . . , σk be elements of A(E, F)
k (σ) satisfies
k (σ0) = w(n) k ,
k (σi) = 0, i = 1, 2, . . . , k,
k } are arbitrary nonzero real or complex numbers.
k (σ)} have the presentation
k (σ) =
k
k (σ)} be the reference functional as
k (σ) = k
k,i σ(zn+kp+iJ)
k (σ)} can be computed recursively by
k (σ) = a(n) k T (n+p) k−1 (σ) + b(n) k T (n+q) k−1 (σ)
0 (σ) = σ(zn) and
k
k T (n+q) k−1 (σk)/d(n) k
k
k T (n+p) k−1 (σk)/d(n) k
k
k−1 T (n+q) k−1 (σk) − w(n+q) k−1 T (n+p) k−1 (σk)
k (σ)/w(n) k
k
k
k
k
k
k
k
k
k
k+m(σ) can be computed by
k+m(σ) = w(n) k+m
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k (σ) =
k
k
k
0,i
k
k−1
k−1
k−1
k−1,k − g(n+p) k−1,k
k−1,k,
k,i = g(n+p) k−1,i − g(n+q) k−1,i − g(n+p) k−1,i
k−1,k − g(n+p) k−1,k
k−1,k,
k,i } are defined by
k,i =
k (u) =
k
k (S)
k (1)
k (u)} can be computed by
k (u) =
k−1 (u) − Ψ(n+q) k−1 (u)
k−1 (gk+1) − Ψ(n+q) k−1 (gk+1)
k+1 − V n+p k+1
k
k
k+M+1V n+p+q k−M
k
k
0 = V n 1 = · · · = V n M−1 = 1,
M = g1(n), V n M+i = gi+1(n)
2M−1 =
2M = ∆Jg1(n + p)
M−1 = (−1)M−1V n (M−1)(M+1)+M.
k = τ n k+1
k
k+M+1τ n+p+q k−M
k
k+1 − τ n+q k
k+1
m as
m = f n k(M+1)+l = An k,l(ϕ; M),
k,l(ϕ; M) are given by
k,l(ϕ; M) =
i,j(j + l − 1; M)|1≤i,j≤k (k > 0)
i,j are defined as
i,j(j + l − 1; M) =
k satisfy the bilinear equation
k+M+1f n+p+q k−M
k
k+1 − f n+q k
k+1
0 = f n 1 = · · · = f n M = 1,
M+i = ϕi(n),
n, i = 1, 2, . . . , k, we derive a general transformation correspond-
k
n+kp xk n+kp+J · · · xk n+kp+kJ
n+kp xk n+kp+J · · · xk n+kp+kJ
k−1,k = (−1)k−1xn+kp · xn+kp+J · . . . · xn+kp+(k−1)J.
k
k−1
k−1
k
k , of degree at most k, which
k (xn+kp+iJ) = Sn+kp+iJ,
k
k
k
k
k−1,k.
k } and {s(n) k } can be recursively computed by
1
k+1 = s(n+q) k
k+1
k+1
k+1 = r(n+q) k
k
k
k
k−1 −
k−1 − ¯
k−1
k
k
k
1
k
k
k−1 − q(n+p) k
k+1 = e(n+J) k
k
k
k
−1 = 0,
2k+1 = ¯
2k−1 +
2k
2k
2k+2 = ¯
2k
2k+1 − ¯
2k+1
k
2k .
k(t).
2
4
6
2
4
6
2
4
6
k
i=0 α(n) k,i Sn+kp+iJ
k
k T (n+p) k−1
k T (n+q) k−1 .