Effective four-wave Hamiltonian

The Hamiltonian $\mbox{$\cal H$}$ in the normal variables $a_k$ is too complicated to work with. Our purpose is to simplify the Hamiltonian to the form:

$$ \mbox{$\cal H$} = \int \omega_k b_k b_k^* dk + \frac{1}{4}\int T^{k_1 k_2}_{k_3 k... ...{k_1}^*b_{k_2}^*b_{k_3}b_{k_4} \delta_{k_1+k_2-k_3-k_4}dk_1dk_2dk_3dk_4+ \cr \!$$ (4.1)

To do that we have to perform a transformation (Zakharov, 1975 [9], Krasitskii, 1990 [10])

$$a_k = b_k + \int \Gamma^{k}_{k_1 k_2}b_{k_1}b_{k_2}\delta_{k-k_1-k_2} - 2\int \Gamma^{k_2}_{k k_1}b^*_{k_1}b_{k_2}\delta_{k+k_1-k_2} +\cr$$ (4.2)

$$\Gamma^{k}_{k_1 k_2} = -\frac{1}{2}\frac{V^k_{k_1 k_2}} {\omega_{... ... k_2} = -\frac{1}{2}\frac{U_{k k_1 k_2}} {\omega_{k}+\omega_{k_1}+\omega_{k_2}}$$

$$B^{k k_1}_{k_2 k_3} = \Gamma^{k_1}_{k_2 k_1-k_2}\Gamma^{k_3}_{k k_3-k}+ \Gamma^{k_1}_{k_3 k_1-k_3}\Gamma^{k_2}_{k k_2-k}-\cr$$

Here $\tilde B^{k k_1}_{k_2 k_3}$ is an arbitrary function satisfying the conditions:

$$\tilde B^{k k_1}_{k_2 k_3} = \tilde B^{k_1 k}_{k_2 k_3} = \tilde B^{k k_1}_{k_3 k_2} = -(\tilde B^{k_2 k_3}_{k k_1})^*$$

The transformation (4.2) is canonical up to the terms of the order of $\vert b_k\vert^4$. It excludes from the Hamiltonian all cubic terms. The form of the $T^{k k_1}_{k_2 k_3}$ depends on the choice of the function $\tilde B^{k k_1}_{k_2 k_3}$. Let first $\tilde B^{k k_1}_{k_2 k_3} = 0$. Then $T^{k k_1}_{k_2 k_3} = \hat T^{k k_1}_{k_2 k_3}$ and

$$\hat T^{k k_1}_{k_2 k_3} = W^{k_1 k}_{k_2 k_3}- \hspace{9cm}\cr -... ..._{k}+\omega_{k_1}} +\frac{1}{\omega_{k_2+k_3}+\omega_{k_2}+\omega_{k_3}}\right]$$ (4.3)

The expression (4.3) in spite of its complexity has some remarkable properties. Let us consider all nontrivial solutions of the equations (1.1). They consist of the manifold (1.5) and of seven other manifolds which are obtained from (1.5) by the permutations

$$k \leftrightarrow k_1, \hspace{.5cm} k_2 \leftrightarrow k_3, \hspace{.5cm} (k, k_1) \leftrightarrow (k_2, k_3)$$

Direct analytic calculation shows that

$$\hat T^{k k_1}_{k_2 k_3}\equiv 0$$

on all these manifolds. Recently Craig and Worfolk [14] confirmed this cancellation by the independent calculation. Another remarkable feature of $\hat T^{k k_1}_{k_2 k_3}$ is the simplicity of its diagonal part. Let us denote

$$T_{k k_1} = \hat T^{k k_1}_{k k_1}$$

A simple, but long calculation ([6], [11]) shows that

$$T_{k k_1} = \frac{1}{4\pi^2} k k_1 min\{\vert k\vert,\vert k_1\vert\}$$

Let us consider the function

$$\tilde T^{k k_1}_{k_2 k_3} = [ \frac{1}{2}(T_{k k_2} + T_{k k_3} + T_{k_1 k_2} + T_{k_1 k_3}) -\cr$$

$$\theta(x) = \cases{1,&if $x\ge0$;\cr 0,&if $x<0.$}$$

Obviously

$$\tilde T^{k k_1}_{k k_1} = \hat T^{k k_1}_{k k_1} = T_{k k_1}$$

Let us choose

$$\tilde B^{k k_1}_{k_2 k_3} = \frac{\tilde T^{k k_1}_{k_2 k_3} - \hat T^{k k_1}_{k_2 k_3}} {\omega_{k} + \omega_{k_1} - \omega_{k_2} - \omega_{k_3}}$$

One can check that this transformation makes replacement of $\hat T^{k k_1}_{k_2 k_3}$ by $\tilde T^{k k_1}_{k_2 k_3}$. So, one can assume in future that

$$T^{k k_1}_{k_2 k_3} = \tilde T^{k k_1}_{k_2 k_3}$$

In the case of periodic boundary conditions

$$b_k = \sum_{n=-\infty}^{\infty} {b_n \delta_{k-k_0n}}$$

Introducing the notation

$$\omega_{k_0n} = \omega_n \cr T^{k_0n k_0n_1}_{k_0n_2 k_0n_3}$$

we can find that now

$$\frac{\partial b_n}{\partial t} + i\frac{\delta \mbox{$\cal H$}}{\delta b_n^*}=0.$$

$$\mbox{$\cal H$} = \sum_{n=-\infty}^{\infty} \omega_n \vert b_n\vert^2 + \frac{1}{4}... ...nfty} T^{n n_1}_{n_2 n_3} b_{n}^*b_{n_1}^*b_{n_2}b_{n_3} \delta_{n+n_1,n_2+n_3}$$

In the canonical transformation (4.2) all the integrals are replaced now by the discrete sums. In particulary instead of $\tilde B^{k k_1}_{k_2 k_3}$ we have now

$$\tilde B^{n n_1}_{n_2 n_3} = \tilde B^{k_0n k_0n_1}_{k_0n_2 k_0n_3}$$

Let us choose

$$\tilde B^{n n_1}_{n_2 n_3} = \frac {T^{n n_1}_{n_2 n_3} - T_{n n_... ...n_3}\delta_{n_1,n_2}} {\omega_{n} + \omega_{n_1} - \omega_{n_2} - \omega_{n_3}}$$ (4.4)

This expression has no singularities on the diagonals $n_2=n, n_3=n_1$ and $n_2=n_1, n_3=n$. The transformation (4.2) with (4.4) brings the Hamiltonian to the Birghoff's form

$$\mbox{$\cal H$} = \sum_{n=-\infty}^{\infty} \omega_n \vert b_n\vert^2 + \frac{1}{4}\sum_{n,n_1=-\infty}^{\infty} T_{n n_1} \vert b_{n}\vert^2\vert b_{n_1}\vert^2$$ (4.5)

This is a Hamiltonian of an integrable system. A level of nonlinearity, allowing the representation (4.5) has to be studied separately.