Taylor-Couette flow and is weaker the larger $k$ is. It is imposed so that the right-hand-side in in the end yields a small perturbation to the transport semigroup. It is not optimal and, indeed, as shown in [@Zhang2015inviscid] if instead of estimating by absolute values one exploits signs and cancellations, a weaker condition can be obtained.

    The condition on $\kappa$ in our applications follows by \[H1\], since we choose the sets $I_{j}$ according to (dyadic) level sets of $U'$ and can control the growth and decay of $U'$.

By expanding the solution $\omega$ to as a Fourier series in the $x$ variable, namely $$\begin{aligned}
\omega(t,x,y)=\sum_{k\neq 0} \omega_k(t,y)\e^{ikx}, \qquad  \psi(t,x,y)=\sum_{k\neq 0} \psi_k(t,y)\e^{ikx},\end{aligned}$$ we can perform a $k$-by-$k$ analysis of the linearized equations. Thus, studying is equivalent to analyzing the collection of one-dimensional problems $$\begin{aligned}
\label{eq:Eulerk}
\begin{cases}
\de_t \omega +ik U(y) \omega = ik B(y)  \psi, \quad &\mbox{in } I, \\
-\Delta_k \psi:=-(-k^2+\de_{yy})\psi = \omega, &\mbox{in } I,\\
ik\psi= 0, &\mbox{on } \de I,
\end{cases}\end{aligned}$$ and $$\begin{aligned}
\omega(0,y)=\omega^{in}(y),  \quad \mbox{in } I,\end{aligned}$$ where we do not keep track of the index $k$ to simplify notation. It is clear that the $k=0$ mode is conserved by the above equation, and that the analysis 