= Solution
Use positive real local <wavenumbers>
$$
k_1(x)=\frac{\omega_{pe0}}{cL}\sqrt{x^2-x_T^2}\quad(x<a),\qquad
k_2(x)=\kappa(x)=\frac{\omega_{pe0}}{cL}\sqrt{x_T^2-x^2}\quad(a<x<b),\qquad
k_3(x)=\frac{\omega_{pe0}}{cL}\sqrt{x^2-x_T^2}\quad(x>b),
$$
within $|x|\le L$. If the exterior is <vacuum>, continue $k_1,k_3$ as $\omega/c$ outside the column. Define the positive <phase integrals> and barrier integral
$$
\Phi(x)=\int_x^a k_1(s)\,ds,\qquad
I(x)=\int_a^x\kappa(s)\,ds,\qquad
\Psi(x)=\int_b^x k_3(s)\,ds.
$$
The three <WKB approximations>, with the common $e^{-i\omega t}$ suppressed, are
$$
\boxed{\begin{aligned}
E_{\rm I}&=\frac{A_{\rm in}e^{-i(\Phi-\pi/4)}+A_{\rm r}e^{i(\Phi-\pi/4)}}{\sqrt{k_1}},\\
E_{\rm II}&=\frac{D e^{-I}+G e^{I}}{\sqrt\kappa},\\
E_{\rm III}&=\frac{A_{\rm t}e^{i(\Psi-\pi/4)}+A_{\rm back}e^{-i(\Psi-\pi/4)}}{\sqrt{k_3}}.
\end{aligned}}
$$
The <flux-normalized WKB amplitudes> make squared travelling-wave coefficients proportional to directed <Poynting flux>. Since $\Phi'=-k_1$ and $\Psi'=k_3$, $A_{\rm in}$ and $A_{\rm t}$ travel to the right, while $A_{\rm r}$ and $A_{\rm back}$ travel to the left. The constants $\pi/4$ are a convenient <Airy turning-point connection formula> convention.
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