stationary NLS equation with chemical potential $ \mu $ is obtained by setting $ \phi(x,t)=\phi(x)\mathrm{e}^{-\mathrm{i}\mu t} $: $$\begin{aligned}
        (-\mu-\partial_x^2+F(|\phi|^2))\phi=0. \label{eq:nls2}
    \end{aligned}$$ As will be seen, the value of $ \mu $ is fixed by the asymptotic form of $ \phi $ . The stationary Bogoliubov equation with eigenenergy $ \epsilon $ is obtained by setting $ u(x,t)=u(x)\mathrm{e}^{-\mathrm{i}(\epsilon+\mu)t} $ and $  v(x,t)=v(x)\mathrm{e}^{-\mathrm{i}(\epsilon-\mu)t} $: $$\begin{aligned}
        \epsilon\begin{pmatrix}u\\v\end{pmatrix}=\mathcal{L}_\mu\begin{pmatrix}u\\v\end{pmatrix} \label{eq:bogos}
    \end{aligned}$$ with $$\begin{aligned}
        \mathcal{L}_\mu:= \mathcal{L}+\begin{pmatrix}-\mu &0 \\ 0& \mu \end{pmatrix}. \label{eq:bogos2}
    \end{aligned}$$

Bogoliubov phonons in a uniform condensate {#subsec:uniform}
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Let us derive the dispersion relation (the energy-momentum relation) of Bogoliubov phonons when the condensate is flowing uniformly: $ \phi(x) = \sqrt{\rho_\infty}\mathrm{e}^{\mathrm{i}(px+\varphi)} $. In order for this $ \phi(x) $ to be the solution of Eq. (\[eq:nls2\]), the chemical potential must be $$\begin{aligned}
        \mu=p^2+F(\rho_\infty). \label{eq:cp}
    \end{aligned}$$ The four solutions of Bogoliubov equation (\[eq:bogos\]) are given by $$\begin{aligned}
        w_i(x,\varphi):= \begin{pmatrix} \bar{u}_i\, \mathrm{e}^{\mathrm{i}(px+\varphi)} \\ \bar{v}_i\, \mathrm{e}^{-\mathrm{i}(px+\varphi)} \end{pmatrix} \mathrm{e}^{\mathrm{i}k_i x}, \label{eq:bogouniform}
    \end{aligned}$$ where $ i=1,\,2,\,3, \text{ and }4 $, and the wavenumber $ k_i $s are the roots of the following quartic equation: $$\begin{aligned}
        (\epsilon-2kp)^2=k^2(k^2+2\rho_\infty F'(\rho_\infty)). \label{eq:uniformdisp}
    \end{aligned}$$ Equation (\[eq:uniformdisp\]) gives the dispersion relation, and from this dispersion one can see that a half of the Landau’s critical 