Analysis of the Newton - Sabatier scheme for inverting fixed - energy phase shifts

Suppose that the inverse scattering problem is understood as follows: given fixed-energy phase shifts, corresponding to an unknown potential q = q(r) from a certain class, for example, q ∈ L 1,1 , recover this potential. Then it is proved that the Newton-Sabatier (NS) procedure does not solve the above problem. It is not a valid inversion method, in the following sense: 1) it is not possible to carry this procedure through for the phase shifts corresponding to a generic potential q ∈ L 1,1 , where L 1,1 := {q : q = q, ∞ 0 r|q(r)|dr < ∞} and recover the original potential: the basic integral equation, introduced by R. Newton without derivation, in general, may be not solvable for some r > 0, and if it is solvable for all r > 0, then the resulting potential is not equal to the original generic q ∈ L 1,1. Here a generic q is any q which is not a restriction to (0, ∞) of an analytic function. 2) the ansatz (*) K(r, s) = ∞ l=0 c l ϕ l (r)u l (s), used by R. Newton, is incorrect: the transformation operator I − K, corresponding to a generic q ∈ L 1,1 , does not have K of the form (*), and 3) the set of potentials q ∈ L 1,1 , that can possibly be obtained by NS procedure, is not dense in the set of all L 1,1 potentials in the norm of L 1,1. Therefore one cannot justify NS procedure even for approximate solution of the inverse scattering problem with fixed-energy phase shifts as data. Thus, the NS procedure, if considered as a method for solving the above inverse scattering problem, is based on an incorrect ansatz, the basic integral equation of NS procedure is, in general, not solvable for some r > 0, and in this case this procedure breaks down, and NS procedure is not an inversion theory: it cannot recover generic potentials q ∈ L 1,1 from their fixed-energy phase shifts. Suppose now that one considers another problem: given fixed-energy phase shifts, corresponding to some potential, find a potential which generates the same phase shifts. Then NS procedure does not solve this problem either: the basic integral equation, in general, may be not solvable for some r > 0, and then NS procedure breaks down.