A^{(0)} + a$ to this background. Assuming that the modes depend only on $t$ and $x_2$ with $a_0=0$, the equations of motion become $$\
\begin{split}
\Box a_1 + 2\alpha k h ( e^{i k x_2} \partial_0 a_+ + e^{-i k x_2} \partial_0 \bar a_+ ) &=0\\
\Box a_2 &=0\\
\Box a_+ + i k \partial_2 a_+ - 4\alpha k h e^{-i k x_2} \partial_0 a_1 &=0\;,
\end{split}$$ where $\Box = -\partial_0^2 + \partial_2^2$ and $a_+= a_3 + i a_4$. The coefficients of the equations have $x_2$ dependence, which can be removed by using two real variables $\tilde a_3$ and $\tilde a_4$ such that $a_+= e^{-i k x_2} (\tilde a_3 + i \tilde a_4)$ in place of $a$. Assuming the $t$ and $x_2$ dependence of the fields $a_1$, $\tilde a_3$ and $\tilde a_4$ to be of the form $e^{-i \omega t + i q x_2}$ and that $a_2=0$, a non-trivial solution exists if and only if $$(\omega^2 - q^2)\left[\omega^4 - 2\omega^2 (q^2 + k (4\alpha E- k)) + q^2(q^2 - k ^2)\right]=0\;,$$ From the second factor, we see that the product of two solutions for $\omega^2$ is $q^2(q^2 - k^2)$, which is negative for $0<q<k$. Thus one of the two roots 