1989
DOI: 10.1002/cpa.3160420702
|View full text |Cite
|
Sign up to set email alerts
|

Direct and inverse scattering transforms with arbitrary spectral singularities

Abstract: IntroductionThe direct and inverse scattering problems for the n X n system (generalized the AKNS system) d -m + z [ J , m ] = q ( x ) m dx were solved by Beals and Coifman [3] for J diagonal with distinct eigenvalues and for potentials q, off-diagonal and generic. Partial solutions to the problems may be found in [13] and [7]. Certain bounded solutions m are piecewise meromorphic on the Riemann sphere relative to a contour 2, consisting of finitely many straight lines passing through the origin determined by… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

0
147
0

Year Published

1991
1991
2022
2022

Publication Types

Select...
8

Relationship

2
6

Authors

Journals

citations
Cited by 144 publications
(147 citation statements)
references
References 11 publications
0
147
0
Order By: Relevance
“…For instance, in the case of involution (2.4) the Riemann-Hilbert problem has zero index because the zeros appear in complex conjugate pairs: k j = k * j . It is known [19][20][21][22][23][24] (see also Ref. [14]) that in the generic case the spectral data include simple (distinct) zeros k 1 , .…”
Section: The Riemann-hilbert Problem Approach: Summarymentioning
confidence: 99%
“…For instance, in the case of involution (2.4) the Riemann-Hilbert problem has zero index because the zeros appear in complex conjugate pairs: k j = k * j . It is known [19][20][21][22][23][24] (see also Ref. [14]) that in the generic case the spectral data include simple (distinct) zeros k 1 , .…”
Section: The Riemann-hilbert Problem Approach: Summarymentioning
confidence: 99%
“…(2) a contour D which is the finite union of piecewise-smooth, simple curves (as closed sets) is said to be orientable if its complement, C \ D, can always be divided into two, possibly disconnected, disjoint open sets + + + and − − − , either of which has finitely many components, such that D admits an orientation so that it can either be viewed as a positively oriented boundary D + for + + + or as a negatively oriented boundary D − for − − − [71], that is, the (possibly disconnected) components of C \ D can be coloured by + or by − in such a way that the + regions do not share boundary with the − regions, except, possibly, at finitely many points [72]; (3) for each segment of an oriented contour D, according to the given orientation, the "+" side is to the left and the "−" side is to the right as one traverses the contour in the direction of orientation; so, for a matrix-valued function A(z), A ± (z) denote the nontangential limits A ± (z) := lim z →z…”
Section: Notational Conventionsmentioning
confidence: 99%
“…1 0 are, respectively, the raising and lowering matrices, 0 = 0 0 0 0 , R ± := {x ∈ R; ±x > 0}, C ± := {z ∈ C; ±Im(z) > 0}, and sgn(x) := 0 if x = 0 and x|x| −1 if x 0; (2) for a scalar ω and a 2×2 matrix Υ, ω ad(σ 3 ) Υ := ω σ 3 Υω −σ 3 ; (3) a contour D which is the finite union of piecewise-smooth, simple curves (as closed sets) is said to be orientable if its complement C \ D can always be divided into two, possibly disconnected, disjoint open sets ℧ + and ℧ − , either of which has finitely many components, such that D admits an orientation so that it can either be viewed as a positively oriented boundary D + for ℧ + or as a negatively oriented boundary D − for ℧ − [85], that is, the (possibly disconnected) components of C \ D can be coloured by + or − in such a way that the + regions do not share boundary with the − regions, except, possibly, at finitely many points [86]; (4) for each segment of an oriented contour D, according to the given orientation, the "+" side is to the left and the "-" side is to the right as one traverses the contour in the direction of orientation, that is, for a matrix…”
Section: Hyperelliptic Riemann Surfaces the Riemann-hilbert Problemsmentioning
confidence: 99%
“…♯ can be coloured by the two colours ±, C λ o ± are complementary projections [2,85,102,103], that is, (C − are complementary, the contour Γ ♯ can always be oriented in such a way that the ± regions lie on the ± sides of the contour, respectively.) The solution of the above (normalised at λ o ) RHP is given by the following integral representation.…”
Section: Following Lemma 321 Of [57] Setmentioning
confidence: 99%
See 1 more Smart Citation