H\({}^{\infty}\)-well-posedness of two-sided problem for Schrödinger equation (Q920278)

The operator to be discussed here is: \[ (E)\quad P=P(D_ t,D_ x,D_ y)=-i\partial_ t-\partial^ 2_ x+\sum^{d}_{k=1}\partial^ 2_{y_ k}\text{ (Schrödinger op.).} \] The initial boundary value problem is considered in each of the following domains: \(\Omega(0,\infty)=\{(x,y);x>0,y\in R^ d\}\) (right half space), \(\Omega(-\infty,L)=\{(x,y);x<L,y\in R^ d\}\) (left half space) \(and\) \(\Omega(0,L)=\{(x,y);0<x<L,y\in R^ d\}\) (slab domain), where y denotes \((y_ 1,...,y_ d).\) Problem \(P(0,\infty)\) in the right half space means the initial boundary value problem with boundary condition \(B_ 1\) given on \(x=0\), and for Problem \(P(-\infty,L)\) in the left half space a boundary condition \(B_ 2\) is given on \(x=L\). Problem \(P(0,L)\) is the two sided initial boundary value problem where the boundary condition \(B_ 1\) and \(B_ 2\) are given on \(x=0\) and \(x=L\) respectively. It is generally supposed that if \(P(0,\infty)\) and \(P(-\infty,L)\) are solvable, then \(P(0,L)\) is solvable. \textit{R. Hersh} shows that if an operator is ``well-behaved'' then the above supposition is true, while giving the operator \(\partial_ t-i\partial^ 2_ x\), which is not ``well-behaved'', as a counter example [Studies and Essays Presented to Yu-Why Chen on his 60th Birthday 209-222 (1970; Zbl 0216.122)]. The author gives necessary and sufficient conditions for \(H^{\infty}\)- well-posedness of each of the above stated problem which impose on the boundary conditions \(B_ 1,B_ 2\). It is very interesting that the example given by Hersh may not be considered as a counter-example in the sense of \(H^{\infty}\)-well-posedness. The method of proof used here is that of Fourier-Laplace transform. The conditions on \(B_ 1\) and \(B_ 2\) are derived from properties of the Lopatinski determinant with respect to the boundary conditions.