Time-scale separation analysis --

Consider weakly nonlinear dispersive waves in a periodic box. Here we consider quadratic nonlinearity and the linear dispersion relations $\omega_k$ which allow three-wave interactions. Example of such systems include surface capillary waves [4] and internal waves in the ocean [9]. In Fourier space, the general form for the Hamiltonian systems with quadratic nonlinearity looks as follows,

$${\cal H} = \sum_{n=1}^\infty \omega_n\vert c_n\vert^2 +\epsilon \sum_{l,m,n=1}^\infty \left( V^l_{mn} \bar c_{l} c_m c_n\delta^l_{m+n}+c.c.\right),$$

$$i\dot c_l = \frac{\partial {\cal H}}{\partial \bar c_l}, \ c_l=a_l e^{-i \omega_l t},$$

$$i \dot a_l = \epsilon \sum_{m,n=1}^\infty \left( V^l_{mn} a_{m} a_{n}e^{i\omega_{mn}^l t} \delta^l_{m+n} \right.$$

$$\left. \hspace{3cm} + 2 \bar{V}^{m}_{ln} \bar a_{n} a_{m} e^{-i\omega^m_{ln}t } \delta^m_{l+n}\right),$$ (2)

where $a_n=a(k_n)$ is the complex wave amplitude in the interaction representation, $k_n = 2 \pi n/L$, $L$ is the box side length, $n=(n_1,n_2)$ for 2D, or $n=(n_1,n_2, n_3)$ in 3D, (similar for $k_l$ and $k_m$), $\omega^l_{mn}\equiv\omega_{k_l}-\omega_{k_m}-\omega_{k_m}$ and $\omega_l=\omega_{k_l}$ is the wave linear dispersion relation. Here, $V^l_{mn} \sim 1$ is an interaction coefficient and $\epsilon$ is introduced as a formal small nonlinearity parameter.

In order to filter out fast oscillations at the wave period, let us seek for the solution at time $T$ such that $2 \pi / \omega \ll T \ll 1/\omega \epsilon^2$. The second condition ensures that $T$ is a lot less than the nonlinear evolution time. Now let us use a perturbation expansion in small $\epsilon$,

$$a_l(T)=a_l^{(0)}+\epsilon a_l^{(1)}+\epsilon^2 a_l^{(2)}.$$

Substituting this expansion in (2) we get in the zeroth order $a_l^{(0)}(T)=a_l(0)$, i.e. the zeroth order term is time independent. This corresponds to the fact that the interaction representation wave amplitudes are constant in the linear approximation. For simplicity, we will write $a^{(0)}_l(0)= a_l$, understanding that a quantity is taken at $T=0$ if its time argument is not mentioned explicitly. The first order is given by

$$a^{(1)}_l (T) = -i \sum_{m,n=1}^\infty \left( V^l_{mn} a_m a_n \Delta^l_{mn} \delta^l_{m+n}\right.$$

$$\left.\hskip 4cm + 2 \bar{V}^m_{ln}a_m\bar{a}_n \bar\Delta^m_{ln}\delta^m_{l+n} \right),$$ (3)

where $\Delta^l_{mn}=\int_0^T e^{i\omega^l_{mn}t}d t = ({e^{i\omega^l_{mn}T}-1})/{i \omega^l_{mn}}.$ Here we have taken into account that $a^{(0)}_l(T)= a_l$ and $a^{(1)}_k (0)=0$.

To calculate the second iterate, write

$$i\dot{ a}^{(2)}_l = \sum_{m,n=1}^\infty \Big[ V^l_{mn}\delta^l_{m+n} e^{i \omega^l_{mn} t} \left(a_m^{(0)} a_n^{(1)}+ a_m^{(1)} a_n^{(0)}\right)$$

$$\hspace{3cm}+ 2\bar{V}^m_{ln}\delta^m_{l+n} e^{ -i \omega^m_{ln} t} \left(a_m^{(1)} \bar{a}_n^{(0)}+ a_m^{(0)} \bar{a}_n^{(1)}\right) \Big].$$

We now have to substitute (3) into (4) and integrate over time to obtain

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$$a_l^{(2)} (T) = \sum_{m,n, \mu, \nu=1}^\infty \left[ 2 V^l_{mn} \left( -V^m_{\mu ... ...l \nu}_{n \mu},\omega^l_{mn}]\delta^\mu_{m + \nu}\right) \delta^l_{m+n} \right.$$

$$\left. + 2 \bar V^m_{ln} \left( -V^m_{\mu \nu}\bar a_n a_\mu a_\n... ...{n \nu l},-\omega^m_{l n}] \delta^\mu_{m + \nu} \right) \delta^m_{l+ n} \right.$$

$$\left. + 2 \bar V^m_{ln} \left( \bar V^n_{\mu \nu}a_m \bar a_\mu ... ...u l}_{\nu m}, -\omega^m_{ln}]\delta^\mu_{n + \nu}\right)\delta^m_{l+n} \right],$$

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where we used $a^{(2)}_k (0)=0$ and introduced $E(x,y)=\int_0^T \Delta(x-y)e^{i y t} d t .$