The reducibility of the boundary conditions in the one-parameter family of elliptic linear boundary value problems. I (Q917774)

[For part II see the following review.] Let \(P_ 1\) and \(P_ 2\) be elliptic operators on \(R^ n\) with constant coefficients of order \(2\mu\) and \(2\nu\) with \(\mu >\nu\), respectively. Consider the boundary value problems: \[ (\epsilon \cdot P_ 1(D)+P_ 2(D))u=0\text{ in } R^ n_+,\quad 0<\epsilon <1;\quad b_{j_ k}(D)u|_{x_ 1\downarrow 0}=\phi_ k,\quad k=1,...,\mu,\quad j_ 1<...<j_{\mu}. \] Here \(R^ n_+=\{x\in R^ n\); \(x_ 1>0\}\) and \(\phi_ k\), \(k=1,...,\mu\), belong to \({\mathcal S}(R^{n-1})\). Under suitable condition, the solutions \(u_{\epsilon}\) converge to the solution \(u_ 0\) of \(P_ 2(D)\) \(u=0\) in \({\mathcal D}'(R^ n_+)\). Previous works studied only the case when \(u_{\epsilon}\to u_ 0\) in \(H^{j_{\nu}+1}(R^ n_+)\). In this case, \(u_ 0\) satisfies the boundary conditions for \(k=1,...,\nu\) by the continuity of the trace operator. By giving an affirmative example and a detailed analysis of a framework, called reducibility, it appears that the reason why \(u_ 0\) happens to satisfy the first \(\nu\) boundary condition and that \(u_ 0\) satisfies a different set of \(\nu\) boundary conditions in a different topology which does not ensure the continuity of the trace operator in general.