Partial identification of the potential from phase shifts
Consider the inverse scattering problem for the operator \[ \phi ^{^{\prime \prime }}(r,\lambda)-\frac{\lambda ^{2}-\frac{1 }{4}}{r^{2}}\phi (r,\lambda)+(1-q(r))\phi (r,\lambda)=0,\quad r\geq 0, \] \[ \phi (r,\lambda)=\gamma (\lambda) r^{\lambda + \frac{1}{2}}(1+o(1))\text{ as }r\rightarrow 0+, \] \[ \phi(r,\lambda)=\sin \left( r-\tfrac{\pi }{2}\left( \lambda - \tfrac{1}{2}\right) +\delta (\lambda)\right) +o(1) \text{ as }r\rightarrow +\infty \] with fixed energy \(k^{2}=1\). The quantities \(\delta \left( n+\frac{1}{2} \right) =\delta _{n}\) are called phase shifts and \(\gamma \left( n+\frac{1}{2 }\right) =\gamma _{n}\) for \(n\geq 0.\) The inverse scattering problem here consists of the recovery of the potential \(q\)\ from the phase shifts \(\delta _{n}.\) The scattering amplitude \(F(t)\) can be expressed by the phase shifts: \[ F(t)=\sum_{n=0}^{\infty }(2n+1) F_{n}P_{n}(t),\quad F_{n}=e^{i\delta _{n}}\sin \delta _{n}, \] where the functions \(P_{n}(t)\) are the Legendre polynomials. This paper shows that the tail of the potential can be determined by the asymptotical behavior of the scattering amplitude and of the phase shifts. The following results are obtained: Assume that the potential \(q\) satisfies \(rq(r),rq^{\ast }(r)\in L_{1}(0,\infty)\) and let \(0<a<\infty \). (a) If \(q=q^{\ast }\) a.e. on \((a,\infty)\) then \(F(t) -F^{\ast }(t)\) is an entire function and satisfies \[ |F(t)-F^{\ast }(t)| \leq c(1+|t|)\exp (a\sqrt{2|t|}). \] Conversely, if \(q\) and \(q^{\ast }\) have compact support, \(F(t) -F^{\ast }(t)\) is an entire function and \[ F(t) -F^{\ast }(t) =O(\exp (a_{1}\sqrt{ 2|t|})) \] for all \(a_{1}>a\), then \(q=q^{\ast }\) a.e. on \((a,\infty)\). (b) If \(q=q^{\ast }\) a.e. on \((a,\infty),\) then, for all sufficiently large \(n\), it holds \[ |\delta _{n}-\delta _{n}^{\ast }| \leq \frac{c}{n^{2}} \left( \frac{ae}{2n}\right) ^{2n}. \] Conversely, if \(q\) and \(q^{\ast }\) have compact support and \[ \delta_{n}-\delta _{n}^{\ast }=O\left( \left( \frac{a_{1}e}{2n}\right) ^{2n}\right)\text{ as } n\rightarrow \infty \] holds for all \(a_{1}>a\), then \(q=q^{\ast }\) a.e. on \((a,\infty)\).
scientific article; zbMATH DE number 5888558
scientific article