V}_\alpha({\boldsymbol\rho})=A\exp\left[\frac{-(x\mp B)^2-y^2}{2W^2}\right] \exp\left[i(C \pm Dx+Ex^2)\right],$$ where $A$ is the amplitude, $B$ is the component separation and $W$ determines the width. The phase $C$ is is related to the chemical potential and time according to $C=-\mu_\alpha^\mathcal{V} t/\hbar$, while $D$ and $E$ represent the wavenumber and chirp, respectively. The upper of $\mp$ and $\pm$ are for component $\alpha=1$ while the lower are for $\alpha=2$.

We utilize the Euler-Lagrange equations, $$\frac{\partial \mathcal L}{\partial \mathcal{P}_j} - \frac{d}{dt}\left(\frac{\partial \mathcal L}{\partial \dot{\mathcal{P}_j}} \right) = 0 ,$$ with $\mathcal{P}_j$ = {$A$, $B$, $C$, $D$, $E$, $W$}. The Lagrangian takes the form $$\mathcal{L} = \int_{-\infty}^\infty ( L_1 + L_2 + L_{12} ) d^2{\boldsymbol\rho},$$ where $$\begin{aligned}
L_\alpha &= \frac{\hbar^2}{2m} \left( \left|\frac{\partial \psi_\alpha^{\mathcal{V}}}{\partial x}\right|^2 + \left|\frac{\partial \psi_\alpha^{\mathcal{V}}}{\partial y}\right|^2 \right) \\
&+ \frac{m\omega_\rho^2}{2}\left(x^2+y^2 \right)\left|\psi_\alpha^{\mathcal{V}}\right|^2 + \frac{\hbar^2g}{2m} \left|\psi_\alpha^{\mathcal{V}}\right|^4 \\
&+ \frac{i\hbar}{2} \left(\psi_\alpha^\mathcal{V}\frac{\partial{\psi_\alpha^\mathcal{V}}^*}{\partial t} - {\psi^\mathcal{V}_\alpha}^*\frac{\partial\psi_\alpha^\mathcal{V}}{\partial t}\right) ,\end{aligned}$$ and the intercomponent coupling is given by $$L_{12} = \frac{\hbar^2g_{12}}{m}\left|\psi_1^\mathcal{V}\right|^2\left|\psi_2^\mathcal{V}\right|^2.$$ Recall that we have taken the symmetric situation $g_{11}=g_{22}\equiv g$ (and also $\mu_1=\mu_2\equiv \mu$), where $g_{\alpha\beta}=N_\beta\sqrt{8\pi}a_{\alpha\beta}/a_z$ with $N_\beta$ being the number of atoms in component $\beta$, $a_{ij}$ is the scattering length between component $\alpha$ and $\beta$, and $a_z = \sqrt{\hbar/m\omega_z}$ is the confinement length in the tight direction.

Equations of motion {#equations-of-motion .unnumbered}
-------------------

For 