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}


1 Answer 1


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 $j$.

  • $\begingroup$ I think that $\Delta$ should be $\nabla$. $\endgroup$
    – Jon
    Jan 8, 2020 at 12:31
  • $\begingroup$ indeed it should, thanks. $\endgroup$ Jan 8, 2020 at 12:33

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