Lopatinskii–Shapiro condition

Definition

Let $\Omega \subseteq \R^n$ be an open set with smooth boundary $\partial \Omega$, with $n \ge 2$, and let $m \ge 1$. Let $A$ be a linear differential operator of order $2m$ with smooth coefficients on $\overline{\Omega}$, with principal symbol $a(x, \xi)$. For $x \in \partial \Omega$, let $\nu(x)$ be the inner unit normal. Assume that $A$ is properly elliptic. That is, let $x \in \partial \Omega$ and let $\xi' \in \R^n$ be nonzero and tangent to $\partial \Omega$ at $x$. Then the polynomial $\tau \mapsto a(x, \xi' + \tau \nu(x))$ has exactly $m$ roots with positive imaginary part. The roots are counted with multiplicity. Let $B_1, \ldots, B_m$ be linear differential operators on $\partial \Omega$ with smooth coefficients, of orders $m_j < 2m$ and principal symbols $b_j(x, \xi)$. Write $D_t = -i \, d/dt$. The operators $B_1, \ldots, B_m$ satisfy the Lopatinskii–Shapiro condition with respect to $A$ if the following holds. Let $x \in \partial \Omega$ and let $\xi'$ be a nonzero tangent vector at $x$. The only bounded solution $v \colon [0, \infty) \to \C$ of $$a\big(x, \xi' + \nu(x) D_t\big) v(t) = 0 \quad (t > 0), \qquad b_j\big(x, \xi' + \nu(x) D_t\big) v(0) = 0 \quad (j = 1, \ldots, m)$$ is $v = 0$.