WKB expansion for NLS

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:

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 $$j\in J$$ $$\begin{eqnarray} \partial_t\phi_j + \frac{1}{2} |\nabla \phi_j|^2=0. \end{eqnarray}$$

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 $$j$$.
• I think that $\Delta$ should be $\nabla$. – Jon Jan 8 '20 at 12:31