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Example text

304) r+ (ζ) = r+ (ζ) (ζ ∈ R), α (ζk ) = α(ζk ) (k = 1, · · · , d, k = j), ζ − ζ0 b(ζ) (ζ ∈ R), H ζ − ζ¯0 ζk − ζ0 Ck = Ck (k = 1, · · · , d, k = j). 305) Proof. (1) ζ0 ∈ IP σ(L). 299) are not 0. Property 3 implies lim σ = ∞, lim σ = 0. 307) ⎠. Under the action of the Darboux transformation, the Jost solutions are changed to 1 (−iζI − S)ψr (x, t, ζ), −iζ + iζ¯0 1 (−iζI − S)ψl (x, t, ζ). ψl (x, t, ζ) = −iζ + iζ¯0 ψr (x, t, ζ) = Hence R = If ζ ∈ C+ , 1 (−iζI − S)R. 309) ⎞ 0 ⎟ r− (ζ) ⎠. 311) 62 DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS r− (ζ) has an additional zero ζ0 than r− (ζ).

196) (thus B(−λ) = −C(λ) holds automatically) and 1 pt = (B − C)x − λ(B + C). 197), we can obtain the recursion relations among aj ’s. They include two parts. 199) + 4aj p = 4 p x p x (j = n + m − 1, · · · , n + 1) are obtained from the coeﬃcients of negative powers of λ. Moreover, the term without λ leads to the equation pt − 1 4 an,x p + 4an p x + x an+1,x = 0. 200) 42 DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS The ﬁrst few aj ’s (0 ≤ j ≤ n) are a0 = α0 (t), 1 a1 = α0 (t)p2 + α1 (t), 2 1 1 3 1 a2 = α0 (t) ppxx − p2x + p4 + α1 (t)p2 + α2 (t), 4 8 8 2 ···.

300) and the solution is transformed by p = p + 2i (ζ¯0 − ζ0 )¯ σ . 301) The change of the scattering data under Darboux transformation is given by the following theorem . 300) (µ ( = 0, ζ0 ∈ C+ ), the scattering data are changed as follows: (1) If ζ0 is not an eigenvalue, then, after the action of the Darboux transformation, the number of eigenvalues increase one. All the original eigenvalues are not changed, and ζ0 is a unique additional eigenvalue. 302) α (ζζ0 ) = 1/µ, hence ζ − ζ¯0 b(ζ) (ζ ∈ R), H ζ − ζ0 ζk − ζ¯0 Ck (k = 1, · · · , d), H Ck = ζk − ζ0 ζ0 − ζ¯0 . 