Solution
= Solution
A <wave turning point> satisfies $k^2(x)=0$, equivalently $\omega_{pe}^2(x)=\omega^2$. Therefore
$$
\boxed{x=\pm x_T,\qquad x_T=L\sqrt{1-\frac{\omega^2}{\omega_{pe0}^2}}.}
$$
For $0<\omega<\omega_{pe0}$ the two <wave turning points> lie inside the column and are simple. Defining $a=-x_T$ and $b=x_T$, the local <dispersion relation> becomes
$$
k^2(x)=\frac{\omega_{pe0}^2}{c^2L^2}(x^2-x_T^2).
$$