Appendix A: derivation WKB equations in presence of a condensate

Let us split $\phi$ into its real and imaginary parts $a = \Re \phi$ and $b = \Im \phi$. Then the equation () splits into two coupled equations

$$\partial_{t}a + \triangle b + 2\vec{v}\cdot\nabla a +\frac{\nabla\varrho}{\varrho}\cdot\nabla b +\rho(2 a b + b(a^2+b^2)) = 0,$$ (37)

$$\partial_{t}b - \triangle a +2a\varrho+2\vec{v}\cdot\nabla b -\frac{\nabla\varrho}{\varrho}\cdot\nabla a + {\varrho}(3 a^2+b^2 +a(a^2+b^2)) = 0,$$ (38)

where we have used the fact that $\frac{\nabla\psi_{0}}{\psi_{0}} = \frac{\nabla\varrho}{2\varrho}+\frac{i}{2} \vec{v}$, which follows from ().

Gabor transforming our two coupled equations () and () and using Taylor series to represent large-scale quantities,

$$\varrho(\vec{x}_{0}) = \varrho(\vec{x}) + (\vec{x}_{0}-\vec{x})\cdot\nabla\varrho(\vec{x}) + O(\varepsilon^{2}),$$

we find

$$\partial_{t}\hat{a} + \triangle \hat{b} + \frac{\nabla \varrho}{\varrho}\cdot \nabla \hat{b} + \vec{v}\cdot\nabla\hat{a} +{\cal G} \left[\rho(2 a b + b(a^2+b^2))\right] = 0,$$ (39)

$$\partial_{t}\hat{b} - \triangle \hat{a} - {\bf }\frac{\nabla \varrho}{\varrho}\cdot \nabla \hat{a} + 2{\bf }\vec{v}\cdot\nabla\hat{b} + 2\varrho\hat{a} + 2i{\bf }\nabla\varrho\cdot\partial_{k}\hat{a}$$

$$+ {\cal G}\left[( \varrho(3 a^2+b^2 +a(a^2+b^2)))\right] = 0.$$ (40)

Where ${\cal G}[f(x)]$ is the Gabor transform of $f(x)$. We have kept only $O(\varepsilon)$ terms and neglected the $O(\varepsilon^{2})$ and higher order terms. For generality, we have kept the nonlinear term.