Evolution of statistics of four-wave systems

In this section we are going to calculate the four-wave analog of PBP equation. These calculations are similar in spirit to those presented in the orevious section, but with the four-wave equation of motion (15). The calculation will be slightly more involved due to the frequency renormalisation, but it will be significantly simplified by our knowledge that (for the same reason as for three-wave case) all $\mu$ dependent terms will not contribute to the final result. Indeed, any term with nonzero $\mu$ would have less summations and therefore will be of lower order. On the diagrammatic language that would mean that all diagrams with non-zero valence can be discarded, as it was explained in the previous section. Below, we will only keep the leading order terms which are $\sim N^2$ and wich correspond to the zero-valence diagrams. Thus, to calculate $Z$ we start with the $J_1 - J_5$ terms (47),(48),(49),(50),(51) in which we put $\mu=0$ and substutute into them the values of $a^{(1)}$ and $a^{(2)}$ from (37) and (38). For the terms proportional to $\epsilon$ we have

$$\langle a_j^{(1)}\bar a_j^{(0)}\rangle_\psi=$$

$$- i \sum_{\alpha\mu\nu} W^{j\alpha}_{\mu\nu} \langle \bar a_j \ba... ...\Delta^{j \alpha}_{\mu\nu} + i \Omega_j \langle \vert a_j\vert^2 \rangle_\psi T$$

$$=- 2 i \sum_{\alpha} W^{j\alpha}_{j\alpha} A_j^2 A_\alpha^2 \cdot T + i \Omega_j A_j^2 T.$$ (64)

where we have used the fact that $\Delta(0)=T$. We see that our choice of the frequency renormalisation (16) makes the contribution of $\langle a_j^{(1)}\bar a_k^{(0)}\rangle_\psi$ equal to zero, and this is the main reason for making this choice. Also, this choice of the frequency correction simplifies the ${\epsilon}^2$ order and, most importantly, eliminates from them the $T^2$ cumulative growth, e.g.

$$\langle (a_j^{(1)}\bar a_j^{(0)})^2 \rangle_\psi= -2 W^{jj }_{jj ... ... \alpha} A_j A_\alpha T )^2\to 0 {\rm\ in \; the \ }N\to\infty {\rm\ limit. \ }$$

Finally we get,

$$J_1 = 0,$$

$$J_2 = \sum_{j\alpha\mu\nu}\lambda_j(1 + \lambda_j A_j^2)\vert W_{\mu\nu... ...{\mu\nu}^{j\alpha}\vert^2 \delta_{\mu\nu}^{j\alpha} A_\mu^2 A_\alpha^2 A_\nu^2,$$

$$J_3 = 2\sum_{j\alpha\mu\nu} \lambda_j\vert W_{\mu\nu}^{j\alpha}\vert^2 ... ... A_j^2 A_\nu^2 (A_\mu^2 -2A_\alpha^2 ) E(0,\tilde {\omega}_{\mu\nu}^{j\alpha}),$$

$$J_4 = 0,$$

$$J_5 = \sum_{j\neq k}\lambda_j\lambda_k\left[-2\vert W_{k\nu}^{j\alpha}\... ...lta_{\mu\nu}^{jk}\vert^2\delta_{\mu\nu}^{jk}A_\mu^2 \right]A_j^2 A_k^2 A_\nu^2.$$ (65)

By putting everything together in (45) and (46) and explointing the symmeteries introduced by the summation variables, we finally obtain in $N \to \infty$ limit followed by $T \to \infty$

$$\dot Z = 4 \pi {\epsilon}^2 \int \vert W^{jl}_{nm}\vert^2\delta(\... ...lambda_j + \lambda_l - \lambda_m - \lambda_n) Z \right] \, dk_j dk_l dk_m dk_n.$$ (66)

By applying to this equation the inverse Laplace transform, we get a four-wave analog of the PBP equation for the $N$-mode PDF:

$$\dot {\cal P} = { \pi {\epsilon}^2 } \int \vert W^{jl}_{nm}\vert^... ...left[{\delta \over \delta s} \right]_4 {\cal P} \right) \, dk_j dk_l dk_m dk_n,$$ (67)

where

$$\left[{\delta \over \delta s} \right]_4 = {\delta \over \delta s_... ...delta \over \delta s_l} - {\delta \over \delta s_m} -{\delta \over \delta s_n}.$$ (68)

This equation can be easily written in the continuity equation form (65) with the flux given in this case by

$${\cal F}_j = -{4 s_j \pi {\epsilon}^2 } \int \vert W^{jl}_{nm}\ve... ... s_ls_m s_n \left[{\delta \over \delta s} \right]_4 {\cal P} \, dk_l dk_m dk_n,$$ (69)