How to obtain this Pohozaev identity for the Gross-Pitaevskii equation? The Gross-Pitaveskii equation (after plugging in the traveling wave ansatz and writing in moving frame coordinates) reads \begin{equation} ic\partial_1 v +\Delta v +v(1-\vert v \vert^2)=0.    \end{equation}
Assume $v$ is a solution on $\Omega_n^N=[-n\pi,n\pi]^N$ and $v$ is $2\pi n$-periodic in every component.
In [1,p.623] the authors give Pohozaev's formula for $v$ without any proof \begin{align} \frac{N-2}{2} \int_{\Omega_n^N} \vert \nabla v \vert^2+\frac{N}{4}\int_{\Omega_n^N} (1-\vert v \vert^2)^2 - c\frac{N-1}{v} \int_{\Omega_n^N} \langle Jv, \zeta_1 \rangle \\ = n\pi \int_{\partial \Omega_n^N} \left( \frac{\vert \nabla v \vert^2}{2} + \frac{(1-\vert v \vert^2)^2}{4} \right) - \int_{\partial \Omega_n^N} \partial_\nu v \left( \sum_{j=1}^N x_j \partial_j v \right). \end{align}
Here the 2-form \begin{equation} Jv \equiv \sum_{1 \leq 1 < j \leq N} (\partial_i v \times \partial_j v) dx_i \wedge dx_j \end{equation} denotes the Jacobian of $v$ and $\zeta_1$ is the 2-form defined by \begin{equation} \zeta_1(x) \equiv - \frac{2}{N-1} \sum_{i=2}^N x_i dx_1 \wedge dx_i. \end{equation} Finally, $\langle \cdot, \cdot\rangle$ stands for the scalar product of 2-forms.
My Question is:

How do you obtain this "Pohozaev's formula"?

From what I have read on the internet already, it seems like you have to multiply the Gross-Pitaevskii equation by $(x|\nabla v(x))$ or similar and integrate by parts. But this doesn't help me too much. How do the 2-forms appear? I'm completly lost. Any hint would be much appreciated!

[1] Béthuel, F., P. Gravejat und J. C. Saut: Travelling waves for the Gross- Pitaevskii equation. II. Comm. Math. Phys., 285(2):567–651, 2009.
 A: Yes, you multiply by $x_i\partial_i v$ and integrate (summation convention where repeated indices are always summed from $1$ to $N$ is in effect throughout). Let me do them term by term (for ease of notation I assume $v$ is real. Minor changes with complex conjugations will get your the complex case.)
$$ \begin{align}
\int_\Omega -\triangle v x_i \partial_i v & = \int_\Omega -\partial_j\partial_j v x_i \partial_i v \\
&= - \int_{\partial\Omega} \nu_j\partial_j v x_i \partial_i v + \int_\Omega \partial_j v (\delta_{ij} \partial_i v + x_i \partial_i \partial_j v) \\
&= - \int_{\partial\Omega}\partial_\nu v x_i \partial_i v + \int_\Omega |\nabla v|^2 + \frac12 \int_\Omega x_i \partial_i |\nabla v|^2 \\
&= - \int_{\partial\Omega}\partial_\nu v x_i \partial_i v + \int_\Omega |\nabla v|^2 - \frac{N}{2} \int_{\Omega} |\nabla v|^2 + \frac12 \int_{\partial\Omega} \nu_i x_i |\nabla v|^2 \end{align}$$
where $\nu$ is the unit normal vector to the boundary. Using that $\Omega$ is a box $[-n\pi,n\pi]^N$ you have that $\nu_i x_i = n\pi$ on the boundary. 
$$\begin{align}
\int_\Omega - v(1-v^2) x_i \partial_i v &= -\frac12 \int_{\Omega} (1-v^2) x_i \partial_i v^2 \\
&= \frac12 \int_{\Omega}(1-v^2) x_i \partial_i(1-v^2) \\
&= \frac14 \int_{\Omega} x_i \partial_i (1-v^2)^2 \\
&= - \frac{N}4 \int_{\Omega} (1-v^2)^2 + \frac14 \int_{\partial\Omega} \nu_i x_i (1-v^2)^2 \end{align}$$
We've already taken care of the majority of the terms. The remaining one (the one involving the two form) comes from the $\partial_1$ term. This last derivation follows from Remark 4.2 which starts on page 615 of the paper you are referring to. 
