We consider the equation (NLS)
\begin{eqnarray}\label{gnls}
i \epsilon\partial\_t u^{\epsilon} + \frac{\epsilon^2}{2}\Delta\_{\eta}u^{\epsilon} = \epsilon |u^{\epsilon}|^{2}u^{\epsilon}, \quad x \in \mathbb R^d
\end{eqnarray}
$d\geq 1$ The initial data is supposed to be given by a
 superposition of highly oscillatory plane waves, i.e.
\begin{eqnarray}\label{icscm}
u\_0^{\epsilon}(x) = \sum\_{j\in J\_0} \alpha\_j(x) e^{ik\_j\cdot x/ \epsilon},
\end{eqnarray}
where for some index set $J\_0 \subset \mathbb N$ we are given initial wave vectors  

$\Phi\_0= \{ k\_j \in \mathbb R^d: j\in J\_0\}$ with corresponding smooth, rapidly decaying amplitudes $\alpha\_j \in \mathcal{S}(\mathbb R^d).$
We seek an approximation of the exact solution $u^{\epsilon}$ in the following form 
\begin{eqnarray}\label{as}
u^{\epsilon}(t,x)\sim\_{\epsilon \to 0} u^{\epsilon}\_{app} (t,x) = \sum\_{j\in J} a\_{j} (t,x) e^{i \phi\_j(t,x)/\epsilon} \ \quad (\*)
\end{eqnarray}


I'm trying to understand the following [paragraph, p. 8-9](https://arxiv.org/pdf/1006.4701.pdf):


Formally, we plugging the approximation (\*) into (NLS), and comparing equal powers of $\epsilon$, we find that the leading order term is of order $ \mathcal{O}(\epsilon^0).$ It can be made identically zero, if for all $j\in J$
\begin{eqnarray}
\partial\_t\phi\_j + \frac{1}{2} |\Delta \phi\_j|^2=0.
\end{eqnarray}



> 
> **Questions are**: (A) What is the leading order term of equation that is formed after plugging the approximation (\*) into NLS ? (B) Why it can be made zero if if for all $j\in J$
> \begin{eqnarray}
> \partial\_t\phi\_j + \frac{1}{2} |\nabla \phi\_j|^2=0.
> \end{eqnarray}
> 
> 
> 



WKB expansion for NLS
**A.** Note that the time derivative $\partial u^\epsilon/\partial t$ and the spatial derivative $\partial u^\epsilon/\partial x$ are both of order $1/\epsilon$, since $u^\epsilon\propto e^{i\phi(t,x)/\epsilon}$. Since in the NLS the first-order time derivative has a prefactor $\epsilon$ and the second-order spatial derivative has a refactor $1/\epsilon^2$, the powers of $\epsilon$ cancel and to leading order the NLS is of order $\epsilon^0$. 


**B.** This term of order $\epsilon^0$ is 
$$\sum\_j a\_j(t,x)e^{i\phi\_j(t,x)/\epsilon}\left[\frac{\partial}{\partial t}\phi\_j+\frac{1}{2}\left(\frac{\partial}{\partial x}\phi\_j\right)^2\right],$$
so to make it vanish the term between square brackets should vanish for all $j$.


