of the (scalar)  (\[BVP2\]). Then we have the following.

\[ellipticity\] The  (\[BVP2\]) is an elliptic  with trivial kernel, that is if $g_\ell =0$ for $\ell=0,\dots,s-1$, then $S(g_0,\dots,g_{s-1})=0$.

In order to prove that the  (\[BVP2\]) is elliptic, we use the standard definition, see (10.1.1) in [@HO2]. Given any point $p\in b\Omega$ we need to consider a $C^\infty$ change of coordinates that takes $p$ into the origin, flattens the boundary, and such that the transformed vector fields at the origin coincide with the new basis vector fields. We write the new coordinates as $(x_0,x)\in [0,+\infty)\times{\Bbb R}^{2n-1}$. Then, the normal vector field is $\po$, and ${\partial}_1,\dots,{\partial}_{2n-1}$ are the tangential vector fileds. After taking the Fourier transform in the tangential directions, writing $\xi\in {\Bbb R}^{2n-1}$ for the variable dual to $x$, we need to show that the ordinary differential equation $$\label{ODE}
\begin{cases}
(-\po^2 +|\xi|^2)^s v & =0 \quad \text{on } [0,+\infty) \\
B_{s,\ell}\, v (0) & =0 \quad \ell=0,1,\dots,s-1
\end{cases}$$ admits the trivial solution as the only bounded solution on $[0,+\infty)$. Here $B_{s,\ell}$ denote the top order terms of the boundary operators in (\[BVP2\]) in our special chart, after freezing the coefficients and taking the Fourier transform.

We begin by describing the differential operators 