2 The Zakharov equation

For a detailed derivation of the Zakharov equation, see references Zakharov (Reference Zakharov1968), Krasitskii (Reference Krasitskii1990), Zakharov, Lvov & Falkovich (Reference Zakharov, Lvov and Falkovich1992). A brief outline starting with the Hamiltonian (1.6) is given as follows:

1. (i) We introduce complex normal variables $a_{k}$ :

   (2.1a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D702}_{k}=\sqrt{{\displaystyle \frac{\unicode[STIX]{x1D714}_{k}}{2g}}}(a_{k}+a_{-k}^{\ast }),\quad \unicode[STIX]{x1D713}_{k}=-\text{i}\sqrt{{\displaystyle \frac{g}{2\unicode[STIX]{x1D714}_{k}}}}(a_{k}-a_{-k}^{\ast }).\end{eqnarray}$$
   Here, $\unicode[STIX]{x1D714}_{k}=\sqrt{g|k|}$ is the dispersion law for the gravity waves, $g$ is the gravitational acceleration constant and the Fourier transformations $\unicode[STIX]{x1D713}(x)\rightarrow \unicode[STIX]{x1D713}_{k}$ and $\unicode[STIX]{x1D702}(x)\rightarrow \unicode[STIX]{x1D702}_{k}$ are defined as follows:
   (2.2a,b ) $$\begin{eqnarray}f_{k}={\displaystyle \frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}}\int f(x)\text{e}^{-\text{i}kx}\,\text{d}x,\quad f(x)={\displaystyle \frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}}\int f_{k}\text{e}^{+\text{i}kx}\,\text{d}k.\end{eqnarray}$$
   In the new variables $a_{k}$ , the Hamiltonian takes the following form:
   (2.3) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}H_{2} & = & \displaystyle \int \unicode[STIX]{x1D714}_{k}a_{k}a_{k}^{\ast }\,\text{d}k,\\ H_{3} & = & \displaystyle \int V_{k_{1}k_{2}}^{k}\{a_{k}^{\ast }a_{k_{1}}a_{k_{2}}+a_{k}a_{k_{1}}^{\ast }a_{k_{2}}^{\ast }\}\unicode[STIX]{x1D6FF}_{k-k_{1}-k_{2}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2}\\ & & +\,{\displaystyle \frac{1}{3}}\displaystyle \int U_{kk_{1}k_{2}}\{a_{k}a_{k_{1}}a_{k_{2}}+a_{k}^{\ast }a_{k_{1}}^{\ast }a_{k_{2}}^{\ast }\}\unicode[STIX]{x1D6FF}_{k+k_{1}+k_{2}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2},\\ H_{4} & = & \displaystyle {\displaystyle \frac{1}{2}}\int W_{k_{1}k_{2}}^{k_{3}k_{4}}a_{k_{1}}^{\ast }a_{k_{2}}^{\ast }a_{k_{3}}a_{k_{4}}\unicode[STIX]{x1D6FF}_{k_{1}+k_{2}-k_{3}-k_{4}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}\,\text{d}k_{4}\\ & & \displaystyle +\,{\displaystyle \frac{1}{3}}\int G_{k_{1}k_{2}k_{3}}^{k_{4}}(a_{k_{1}}^{\ast }a_{k_{2}}^{\ast }a_{k_{3}}^{\ast }a_{k_{4}}+\text{c.c.})\unicode[STIX]{x1D6FF}_{k_{1}+k_{2}+k_{3}-k_{4}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}\,\text{d}k_{4}\\ & & \displaystyle +\,{\displaystyle \frac{1}{12}}\int R_{k_{1}k_{2}k_{3}k_{4}}(a_{k_{1}}^{\ast }a_{k_{2}}^{\ast }a_{k_{3}}^{\ast }a_{k_{4}}^{\ast }+\text{c.c.})\unicode[STIX]{x1D6FF}_{k_{1}+k_{2}+k_{3}+k_{4}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}\,\text{d}k_{4}.\end{array}\right\}\end{eqnarray}$$
   For our purposes, the exact expressions for the coefficients $V$ , $U$ , $W$ , $G$ and $R$ of the Hamiltonian are unimportant. Nevertheless, a careful reader can find them in  Zakharov (Reference Zakharov1998, Reference Zakharov1999), Dyachenko et al. (Reference Dyachenko, Kachulin, Zakharov, Pelinovsky and Kharif2015b ).

   The equations of motion (1.1) now yield the following:

   (2.4) $$\begin{eqnarray}{\displaystyle \frac{\unicode[STIX]{x2202}a_{k}}{\unicode[STIX]{x2202}t}}+\text{i}{\displaystyle \frac{\unicode[STIX]{x1D6FF}H}{\unicode[STIX]{x1D6FF}a_{k}^{\ast }}}=0.\end{eqnarray}$$

2. (ii) In the variables $a_{k}$ , the Hamiltonian contains non-resonant three-wave interactions, and hence, the variables are still suboptimal. We introduce yet another set of variables $b_{k}$ by another canonical transformation $a_{k}\rightarrow b_{k}$ to cancel all the non-resonant cubic and quartic terms in the new Hamiltonian. An efficient way to construct this transformation was offered in Zakharov et al. (Reference Zakharov, Lvov and Falkovich1992) and can be written as follows:

   (2.5) $$\begin{eqnarray}\displaystyle a_{k} & = & \displaystyle \displaystyle b_{k}+\int [2\tilde{V}_{kk_{2}}^{k_{1}}b_{k_{1}}b_{k_{2}}^{\ast }\unicode[STIX]{x1D6FF}_{k_{1}-k-k_{2}}-\tilde{V}_{k_{1}k_{2}}^{k}b_{k_{1}}b_{k_{2}}\unicode[STIX]{x1D6FF}_{k-k_{1}-k_{2}}-\tilde{U} _{kk_{1}k_{2}}b_{k_{1}}^{\ast }b_{k_{2}}^{\ast }\unicode[STIX]{x1D6FF}_{k+k_{1}+k_{2}}]\,\text{d}k_{1}\,\text{d}k_{2}\nonumber\\ \displaystyle & & \displaystyle +\int [A_{k_{1}k_{2}k_{3}}^{k}b_{k_{1}}b_{k_{2}}b_{k_{3}}+A_{k_{2}k_{3}}^{kk_{1}}b_{k_{1}}^{\ast }b_{k_{2}}b_{k_{3}}+A_{k_{3}}^{kk_{1}k_{2}}b_{k_{1}}^{\ast }b_{k_{2}}^{\ast }b_{k_{3}}+A^{kk_{1}k_{2}k_{3}}b_{k_{1}}^{\ast }b_{k_{2}}^{\ast }b_{k_{3}}^{\ast }]\nonumber\\ \displaystyle & & \displaystyle \times \,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3},\end{eqnarray}$$
   where the exact expressions for the coefficients of the transformation (2.5) can be found in Dyachenko et al. (Reference Dyachenko, Kachulin, Zakharov, Pelinovsky and Kharif2015b ). The resulting Hamiltonian after the transformation yields
   (2.6) $$\begin{eqnarray}\displaystyle H=\int \unicode[STIX]{x1D714}_{k}b_{k}b_{k}^{\ast }\,\text{d}k+{\displaystyle \frac{1}{2}}\int T_{kk_{1}}^{k_{2}k_{3}}b_{k}^{\ast }b_{k_{1}}^{\ast }b_{k_{2}}b_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}+\tilde{H}, & & \displaystyle\end{eqnarray}$$
   where $\tilde{H}$ is an infinite series in $b_{k},b_{k}^{\ast }$ starting from the fifth-order terms. The explicit (and cumbersome) expression for $T_{kk_{1}}^{k_{2}k_{3}}$ can be found in (Zakharov Reference Zakharov1968, Reference Zakharov1998, Reference Zakharov1999). The motion equation
   (2.7) $$\begin{eqnarray}{\displaystyle \frac{\unicode[STIX]{x2202}b_{k}}{\unicode[STIX]{x2202}t}}+\text{i}{\displaystyle \frac{\unicode[STIX]{x1D6FF}H}{\unicode[STIX]{x1D6FF}b_{k}^{\ast }}}=0,\end{eqnarray}$$
   (neglecting $\tilde{H}$ ) is the traditional Zakharov equation.

3 Canonical transformation for the Zakharov equation

A possibility for further simplification of (2.7) is based on the following remarkable fact, established in Dyachenko & Zakharov (Reference Dyachenko and Zakharov1994). Let us consider the resonance conditions for the four-wave interactions:

(3.1) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}k+k_{1}=\displaystyle k_{2}+k_{3},\\ \unicode[STIX]{x1D714}_{k}+\unicode[STIX]{x1D714}_{k_{1}}=\displaystyle \unicode[STIX]{x1D714}_{k_{2}}+\unicode[STIX]{x1D714}_{k_{3}}.\end{array}\right\} & & \displaystyle\end{eqnarray}$$

In the 1-D case, all solutions of this system of (3.1) can be divided into two parts – so-called ‘trivial’ and ‘non-trivial’ parts. The ‘non-trivial’ solution can be solved as follows:

(3.2) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}k=a(1+\unicode[STIX]{x1D701})^{2},\\ k_{1}=a(1+\unicode[STIX]{x1D701})^{2}\unicode[STIX]{x1D701}^{2},\\ k_{2}=-a\unicode[STIX]{x1D701}^{2},\\ k_{3}=a(1+\unicode[STIX]{x1D701}+\unicode[STIX]{x1D701}^{2})^{2}\quad \text{here }0<\unicode[STIX]{x1D701}<1.\end{array}\right\}\end{eqnarray}$$

Notice the product $kk_{1}k_{2}k_{3}<0$ . Now,

(3.3) $$\begin{eqnarray}T_{kk_{1}}^{k_{2}k_{3}}=F(a,\unicode[STIX]{x1D701})=a^{3}f(\unicode[STIX]{x1D701}).\end{eqnarray}$$

Direct calculation shows that for the ‘non-trivial’ resonance (3.2),

(3.4) $$\begin{eqnarray}f(\unicode[STIX]{x1D701})\equiv 0.\end{eqnarray}$$

This fact means that ‘non-trivial’ four-wave resonances are absent!

Moreover, $T_{kk_{1}}^{k_{2}k_{3}}\equiv 0$ if the product $kk_{1}k_{2}k_{3}\leqslant 0$ . In addition, it has a very simple form:

(3.5) $$\begin{eqnarray}\displaystyle T_{k_{2}k_{3}}^{kk_{1}} & = & \displaystyle \displaystyle \unicode[STIX]{x1D703}(kk_{1}k_{2}k_{3}){\displaystyle \frac{(kk_{1}k_{2}k_{3})^{1/2}}{4\unicode[STIX]{x03C0}}}\left[\left({\displaystyle \frac{\unicode[STIX]{x1D714}\unicode[STIX]{x1D714}_{1}}{\unicode[STIX]{x1D714}_{2}\unicode[STIX]{x1D714}_{3}}}\right)^{1/2}+\left({\displaystyle \frac{\unicode[STIX]{x1D714}_{2}\unicode[STIX]{x1D714}_{3}}{\unicode[STIX]{x1D714}\unicode[STIX]{x1D714}_{1}}}\right)^{1/2}\right]\text{min}(k,k_{1},k_{2},k_{3})\\ \displaystyle & & \displaystyle \qquad \text{min}(k,k_{1},k_{2},k_{3})\text{ is minimum of }(k,k_{1},k_{2},k_{3}).\end{eqnarray}$$

Here, $\unicode[STIX]{x1D703}(k)$ is the step function

(3.6) $$\begin{eqnarray}\unicode[STIX]{x1D703}(x)=\left\{\begin{array}{@{}ll@{}}0\quad & \text{if }x\leqslant 0,\\ 1\quad & \text{if }x>0.\end{array}\right.\end{eqnarray}$$

Obviously, for positive $k_{i}$ with resonant condition

(3.7) $$\begin{eqnarray}\displaystyle & \displaystyle k+k_{1}=k_{2}+k_{3}, & \displaystyle\end{eqnarray}$$

(3.8) $$\begin{eqnarray}\displaystyle & \displaystyle \text{min}(k,k_{1},k_{2},k_{3})={\textstyle \frac{1}{4}}(k+k_{1}+k_{2}+k_{3}-|k-k_{2}|-|k-k_{3}|-|k_{1}-k_{2}|-|k_{1}-k_{3}|).\qquad & \displaystyle\end{eqnarray}$$

This means that a system initially consisting of unidirectional waves retains this property for all times. Indeed, a wave with negative $k$ can appear only from the following equation with all positive $k$ on the right-hand side:

(3.9) $$\begin{eqnarray}\displaystyle \text{i}{\displaystyle \frac{\unicode[STIX]{x2202}b_{k}}{\unicode[STIX]{x2202}t}}=\unicode[STIX]{x1D714}_{k}b_{k}+\int T_{kk_{1}}^{k_{2}k_{3}}b_{k_{1}}^{\ast }b_{k_{2}}b_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}. & & \displaystyle\end{eqnarray}$$

However, $T_{kk_{1}}^{k_{2}k_{3}}$ for such a selection of $k$ is identically zero.

Now, one can see that the Zakharov equation is a compact equation but it can be further ‘improved’. In other words, the coefficient (3.5) can be simplified even further.

The system (3.1) also has the following ‘trivial’ solution:

(3.10a-d ) $$\begin{eqnarray}k_{2}=k_{1},\quad k_{3}=k,\quad \text{or}\quad k_{2}=k,\quad k_{3}=k_{1}.\end{eqnarray}$$

We introduce $T_{kk_{1}}$ (diagonal part) as the value of the four-wave coefficient on the trivial manifold (3.10). This was calculated in Zakharov (Reference Zakharov1968) and is equal to

(3.11) $$\begin{eqnarray}T_{kk_{1}}=T_{kk_{1}}^{kk_{1}}=\frac{1}{4\unicode[STIX]{x03C0}}|k||k_{1}|(|k+k_{1}|-|k-k_{1}|)=\frac{1}{2\unicode[STIX]{x03C0}}|k||k_{1}|\text{min}(|k|,|k_{1}|).\end{eqnarray}$$

Let us introduce $\tilde{T}_{kk_{1}}^{k_{2}k_{3}}$ as follows:

(3.12) $$\begin{eqnarray}\tilde{T}_{kk_{1}}^{k_{2}k_{3}}=\unicode[STIX]{x1D703}(kk_{1}k_{2}k_{3})[{\textstyle \frac{1}{2}}(T_{kk_{2}}+T_{kk_{3}}+T_{k_{1}k_{2}}+T_{k_{1}k_{3}})-{\textstyle \frac{1}{4}}(T_{kk}+T_{k_{1}k_{1}}+T_{k_{2}k_{2}}+T_{k_{3}k_{3}})].\end{eqnarray}$$

A canonical transformation of the second step replaces Zakharov’s $T_{kk_{1}}^{k_{2}k_{3}}$ from (2.6) with the simpler $\tilde{T}_{kk_{1}}^{k_{2}k_{3}}$ while keeping their diagonal part the same.

The simple method to construct the canonical transformation is based on the fact that a Hamiltonian system (with variable $\tilde{c}_{k}(t)$ ) is invariant under translation in time and that the transformation $\tilde{c}_{k}(0)\rightarrow \tilde{c}_{k}(\unicode[STIX]{x1D70F})$ is canonical. Let us construct this transformation (as a power series) using an auxiliary Hamiltonian $\tilde{{\mathcal{H}}}$ (starting from the quartic term) of the form:

(3.13) $$\begin{eqnarray}\tilde{{\mathcal{H}}}={\displaystyle \frac{1}{2}}\int \tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}\tilde{c}_{k}^{\ast }\tilde{c}_{k_{1}}^{\ast }\tilde{c}_{k_{2}}d_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}+\cdots \,,\end{eqnarray}$$

where the symmetry relations

(3.14) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}=\tilde{\boldsymbol{B}}_{\boldsymbol{ k}_{\mathbf{1}}k}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}=\tilde{\boldsymbol{B}}_{\boldsymbol{ k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{3}}\boldsymbol{k}_{\mathbf{2}}}=(\tilde{\boldsymbol{B}}_{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}^{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}})^{\ast } & & \displaystyle\end{eqnarray}$$

are necessary to obtain a real-valued $\tilde{{\mathcal{H}}}$ . Using a Taylor series, we can express the old canonical $b_{k}(\unicode[STIX]{x1D70F})=\tilde{c}_{k}(\unicode[STIX]{x1D70F})$ in terms of $\tilde{c}_{k}(0)$ :

(3.15) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\tilde{c}_{k}(\unicode[STIX]{x1D70F})=\tilde{c}_{k}(0)+\unicode[STIX]{x1D70F}{\displaystyle \frac{\unicode[STIX]{x2202}\tilde{c}_{k}(\unicode[STIX]{x1D70F})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D70F}}}\bigg|_{\unicode[STIX]{x1D70F}=0}+\cdots \,,\\ {\displaystyle \frac{\unicode[STIX]{x2202}\tilde{c}_{k}(\unicode[STIX]{x1D70F})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D70F}}}\bigg|_{\unicode[STIX]{x1D70F}=0}=-\text{i}{\displaystyle \frac{\unicode[STIX]{x1D6FF}\tilde{{\mathcal{H}}}(\tilde{c}_{k}(\unicode[STIX]{x1D70F}),\tilde{c}_{k}^{\ast }(\unicode[STIX]{x1D70F}))}{\unicode[STIX]{x1D6FF}\tilde{c}_{k}^{\ast }(\unicode[STIX]{x1D70F})}}\bigg|_{\unicode[STIX]{x1D70F}=0},\end{array}\right\}\end{eqnarray}$$

and

(3.16) $$\begin{eqnarray}b_{k}=\tilde{c}_{k}-\text{i}\int \tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}\tilde{c}_{k_{1}}^{\ast }\tilde{c}_{k_{2}}\tilde{c}_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}+\cdots\end{eqnarray}$$

is a canonical transformation. Now, we plug this transformation into the Hamiltonian (2.6) of the Zakharov equation and obtain the new Hamiltonian:

(3.17) $$\begin{eqnarray}\displaystyle H & = & \displaystyle \displaystyle \int \unicode[STIX]{x1D714}_{k}\tilde{c}_{k}\tilde{c}_{k}^{\ast }\,\text{d}k+{\displaystyle \frac{1}{2}}\int \left[T_{kk_{1}}^{k_{2}k_{3}}-\text{i}(\unicode[STIX]{x1D714}_{k}+\unicode[STIX]{x1D714}_{k_{1}}-\unicode[STIX]{x1D714}_{k_{2}}-\unicode[STIX]{x1D714}_{k_{3}})\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}\right]\nonumber\\ \displaystyle & & \displaystyle \times \,\tilde{c}_{k}^{\ast }\tilde{c}_{k_{1}}^{\ast }\tilde{c}_{k_{2}}d_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}+\cdots \,.\end{eqnarray}$$

The coefficient $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ of the auxiliary Hamiltonian is the same as the coefficient of the canonical transformation. It controls the four-wave coefficient $T_{kk_{1}}^{k_{2}k_{3}}$ in the Hamiltonian of the Zakharov equation (3.17). To replace Zakharov’s $T_{kk_{1}}^{k_{2}k_{3}}$ by the simpler $\tilde{T}_{kk_{1}}^{k_{2}k_{3}}$ , the coefficient $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ has to be equal to

(3.18) $$\begin{eqnarray}\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}=\text{i}{\displaystyle \frac{\tilde{T}_{kk_{1}}^{k_{2}k_{3}}-T_{kk_{1}}^{k_{2}k_{3}}}{\unicode[STIX]{x1D714}_{k}+\unicode[STIX]{x1D714}_{k_{1}}-\unicode[STIX]{x1D714}_{k_{2}}-\unicode[STIX]{x1D714}_{k_{3}}}}.\end{eqnarray}$$

One can check that $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ has no singularities at $k+k_{1}=k_{2}+k_{3}$ . Indeed, in the region where the product $kk_{1}k_{2}k_{3}\leqslant 0$ , the singularities are cancelled by virtue of the identity (3.4). In the region where the product $kk_{1}k_{2}k_{3}>0$ , the singularities are cancelled due to the special choice of $\tilde{T}_{kk_{1}}^{k_{2}k_{3}}$ . The exact expression for $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ was published in Dyachenko, Lvov & Zakharov (Reference Dyachenko, Lvov and Zakharov1995). This leads to the derivation of the ‘compact water wave equation’ (not yet the super compact).

Due to the absence of non-trivial resonances, waves moving in the same direction do not generate waves moving in the opposite direction, and hence, we can assume without loss of generality that for all wavenumbers $k_{i}>0$ , this leads to the following simplification:

(3.19) $$\begin{eqnarray}\displaystyle \tilde{T}_{kk_{1}}^{k_{2}k_{3}} & = & \displaystyle \displaystyle \left[-{\displaystyle \frac{1}{8\unicode[STIX]{x03C0}}}(kk_{2}|k-k_{2}|+kk_{3}|k-k_{3}|+k_{1}k_{2}|k_{1}-k_{2}|+k_{1}k_{3}|k_{1}-k_{3}|)\right.\nonumber\\ \displaystyle & & \displaystyle +\left.{\displaystyle \frac{1}{8\unicode[STIX]{x03C0}}}(kk_{1}(k+k_{1})+k_{2}k_{3}(k_{2}+k_{3}))\right]\unicode[STIX]{x1D703}(k)\unicode[STIX]{x1D703}(k_{1})\unicode[STIX]{x1D703}(k_{2})\unicode[STIX]{x1D703}(k_{3}).\end{eqnarray}$$

Returning from the Fourier space, we can write the following compact expression for the Hamiltonian in $x$ -space:

(3.20) $$\begin{eqnarray}\displaystyle H=\int \tilde{c}^{\ast }\hat{\unicode[STIX]{x1D714}}\tilde{c}\,\text{d}x+{\displaystyle \frac{1}{2}}\int \left|{\displaystyle \frac{\unicode[STIX]{x2202}\tilde{c}}{\unicode[STIX]{x2202}x}}\right|^{2}\left[{\displaystyle \frac{\text{i}}{2}}\left(\tilde{c}{\displaystyle \frac{\unicode[STIX]{x2202}\tilde{c}^{\ast }}{\unicode[STIX]{x2202}x}}-\tilde{c}^{\ast }{\displaystyle \frac{\unicode[STIX]{x2202}\tilde{c}}{\unicode[STIX]{x2202}x}}\right)-\hat{k}|\tilde{c}|^{2}\right]\,\text{d}x, & & \displaystyle\end{eqnarray}$$

where $\hat{\unicode[STIX]{x1D714}}$ denotes multiplication by $\sqrt{g|k|}$ in Fourier space. The compact equation, with Hamiltonian (3.20), was used for the numerical simulations in Dyachenko & Zakharov (Reference Dyachenko and Zakharov2012), Fedele & Dutykh (Reference Fedele and Dutykh2012a ,Reference Fedele and Dutykh b ).

4 Super compact equation

Note that the choice of (3.12) is not unique for introducing the new Hamiltonian. The conditions imposed on $\tilde{T}_{kk_{1}}^{k_{2}k_{3}}$ are rather loose:

1. (i) the symmetry conditions require that

   (4.1) $$\begin{eqnarray}\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}=\tilde{\boldsymbol{T}}_{\boldsymbol{k}_{\mathbf{1}}\boldsymbol{k}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}=\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{3}}\boldsymbol{ k}_{\mathbf{2}}}=\tilde{\boldsymbol{T}}_{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}^{\boldsymbol{ k}\boldsymbol{k}_{\mathbf{1}}};\end{eqnarray}$$

2. (ii) the diagonal part must be strictly defined as

   (4.2) $$\begin{eqnarray}\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{ k}_{\mathbf{2}}\boldsymbol{ k}_{\mathbf{3}}}=T_{kk_{1}}={\displaystyle \frac{1}{4\unicode[STIX]{x03C0}}}|k||k_{1}|(|k+k_{1}|-|k-k_{1}|)={\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}}|k||k_{1}|\text{min}(|k|,|k_{1}|).\end{eqnarray}$$

The symmetry conditions suggest that $\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ may be invariant under permutations of all $k_{i}$ . Let us choose $\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ as follows:

(4.3) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}} & = & \displaystyle \displaystyle {\displaystyle \frac{(kk_{1}k_{2}k_{3})^{1/2}}{2\unicode[STIX]{x03C0}}}\text{min}(k,k_{1},k_{2},k_{3})\unicode[STIX]{x1D703}_{k}\unicode[STIX]{x1D703}_{k_{1}}\unicode[STIX]{x1D703}_{k_{2}}\unicode[STIX]{x1D703}_{k_{3}},\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D703}_{k}\text{is the step function, }\unicode[STIX]{x1D703}_{k}=\unicode[STIX]{x1D703}(k)\end{eqnarray}$$

and min $(k,k_{1},k_{2},k_{3})$ is defined in (3.5) and (3.8). We substitute the new $\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ (4.3) into the coefficient $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ (3.18) and apply the canonical transformation (3.16). After that transformation, the functions $\tilde{c}_{k}$ satisfy the following equation:

(4.4) $$\begin{eqnarray}\displaystyle \text{i}\dot{\tilde{c_{k}}} & = & \displaystyle \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D6FF}H}{\unicode[STIX]{x1D6FF}\tilde{c}_{k}^{\ast }}}=\unicode[STIX]{x1D714}_{k}\tilde{c}_{k}+{\displaystyle \frac{k^{1/2}\unicode[STIX]{x1D703}_{k}}{2\unicode[STIX]{x03C0}}}\nonumber\\ \displaystyle & & \displaystyle \times \int \text{min}(k,k_{1},k_{2},k_{3})(k_{1}^{1/2}\unicode[STIX]{x1D703}_{k_{1}}\tilde{c}_{k_{1}}^{\ast })(k_{2}^{1/2}\unicode[STIX]{x1D703}_{k_{2}}\tilde{c}_{k_{2}})(k_{3}^{1/2}\unicode[STIX]{x1D703}_{k_{3}}\tilde{c}_{k_{3}})\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}.\qquad\end{eqnarray}$$

It is convenient to introduce a new Hamiltonian variable:

(4.5) $$\begin{eqnarray}c_{k}=k^{1/2}\unicode[STIX]{x1D703}_{k}\tilde{c}_{k}.\end{eqnarray}$$

$c_{k}$ is the Fourier transform of a function analytic in the upper complex half-plane. Note, the nonlinear term in (4.4) preserves the analyticity property. Multiplying (4.4) by $\text{i}k^{1/2}$ and using the definition of $c_{k}$ (4.5) results in

(4.6) $$\begin{eqnarray}{\dot{c}}_{k}+\text{i}k\unicode[STIX]{x1D703}_{k}\left[{\displaystyle \frac{\unicode[STIX]{x1D714}_{k}}{k}}c_{k}+{\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}}\int \text{min}(k,k_{1},k_{2},k_{3}),c_{k_{1}}^{\ast }c_{k_{2}}c_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}\right]=0\end{eqnarray}$$

which is exactly the super compact equation written in $k\text{-}\text{space}$ .

The expression in square brackets of (4.6) is the variational derivative of the following Hamiltonian:

(4.7) $$\begin{eqnarray}\displaystyle H=\int {\displaystyle \frac{\unicode[STIX]{x1D714}_{k}}{k}}|c_{k}|^{2}\,\text{d}k+{\displaystyle \frac{1}{4\unicode[STIX]{x03C0}}}\int \text{min}(k,k_{1},k_{2},k_{3})c_{k}^{\ast }c_{k_{1}}^{\ast }c_{k_{2}}c_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}. & & \displaystyle\end{eqnarray}$$

Using the following relations between $k$ -space and $x$ -space,

(4.8) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}kc_{k}^{\ast }\Leftrightarrow \text{i}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}}c^{\ast }(x),\quad kc_{k}\Leftrightarrow -\text{i}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}}c(x),\\ |k-k_{2}|c_{k}^{\ast }c_{k_{2}}\Leftrightarrow \hat{k}(|c(x)|^{2}),\quad (k+k_{1})c_{k}c_{k_{1}}\Leftrightarrow -\text{i}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}}(c(x)^{2}),\end{array}\right\}\end{eqnarray}$$

and definition of $\text{min}(k,k_{1},k_{2},k_{3})$ (3.8), the new Hamiltonian, whose fourth order is defined by the new coefficient $\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ (4.3) can be written in $x$ -space:

(4.9) $$\begin{eqnarray}H=\int c^{\ast }\hat{V}c\,\text{d}x+{\displaystyle \frac{1}{2}}\int \left[{\displaystyle \frac{\text{i}}{4}}\left(c^{2}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}}{c^{\ast }}^{2}-{c^{\ast }}^{2}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}}c^{2}\right)-|c|^{2}\hat{k}(|c|^{2})\right]\,\text{d}x.\end{eqnarray}$$

Here, the operator $\hat{V}$ is in $k$ -space so that $V_{k}=\unicode[STIX]{x1D714}_{k}/k$ . If one also introduces a bracket similar to the Gardner–Zakharov–Faddeev one (see Zakharov & Faddev Reference Zakharov and Faddev1971), then

(4.10) $$\begin{eqnarray}\unicode[STIX]{x2202}_{x}^{+}\Leftrightarrow \text{i}k\unicode[STIX]{x1D703}_{k}.\end{eqnarray}$$

Then, the equation of motion is the following:

(4.11) $$\begin{eqnarray}{\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}t}}+\unicode[STIX]{x2202}_{x}^{+}{\displaystyle \frac{\unicode[STIX]{x1D6FF}H}{\unicode[STIX]{x1D6FF}c^{\ast }}}=0.\end{eqnarray}$$

We introduce the advection velocity

(4.12) $$\begin{eqnarray}{\mathcal{U}}=\hat{k}|c|^{2},\end{eqnarray}$$

and taking a variational derivative, one can write (4.11) in the following form:

(4.13) $$\begin{eqnarray}{\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}t}}+\text{i}\hat{\unicode[STIX]{x1D714}}c-\text{i}\unicode[STIX]{x2202}_{x}^{+}\left(|c|^{2}{\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}x}}\right)=\unicode[STIX]{x2202}_{x}^{+}({\mathcal{U}}c).\end{eqnarray}$$

Note that $|c|^{2}$ has dimensions of potential. One can recognize two terms in the equation:

1. (i) nonlinear wave term: $\text{i}\hat{\unicode[STIX]{x1D714}}c-\text{i}\unicode[STIX]{x2202}_{x}^{+}(|c|^{2}(\unicode[STIX]{x2202}c/\unicode[STIX]{x2202}x))$ ;

2. (ii) advection term: $\unicode[STIX]{x2202}_{x}^{+}({\mathcal{U}}c)$ .

Along with the usual quantities, such as energy and both momenta, equation (4.13) conserves the action, or the number of waves:

(4.14) $$\begin{eqnarray}N=\int _{0}^{\infty }{\displaystyle \frac{|c_{k}|^{2}}{k}}\,\text{d}k.\end{eqnarray}$$

Equation (4.13) has an exact self-similar solution:

(4.15) $$\begin{eqnarray}c(x,t)=g(t_{0}-t)^{3/2}C\left({\displaystyle \frac{x}{g(t_{0}-t)^{2}}}\right).\end{eqnarray}$$

It is easy to check that $C(\unicode[STIX]{x1D709})$ satisfies the following condition:

(4.16) $$\begin{eqnarray}{\displaystyle \frac{3}{2}}C-2\unicode[STIX]{x1D709}{\displaystyle \frac{\unicode[STIX]{x2202}C}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}}+\text{i}\hat{K}^{1/2}C-\text{i}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}}\left(|C|^{2}{\displaystyle \frac{\unicode[STIX]{x2202}C}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}}\right)={\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}}((\hat{K}|C|^{2})C),\end{eqnarray}$$

where $C(\unicode[STIX]{x1D709})$ is a dimensionless function that is analytic in the upper half-plane and $\hat{K}$ is a dimensionless operator.

In $k$ -space, this solution (according to (4.6)) has the following form:

(4.17) $$\begin{eqnarray}c(k,t)=g^{2}(t_{0}-t)^{7/2}F(gk(t_{0}-t)^{2}).\end{eqnarray}$$

It is easy to check that the dimensionless function $F(\unicode[STIX]{x1D709})$ satisfies the following equation:

(4.18) $$\begin{eqnarray}{\displaystyle \frac{7}{2}}F+2\unicode[STIX]{x1D709}{\displaystyle \frac{\unicode[STIX]{x2202}F}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}}}=\text{i}\unicode[STIX]{x1D709}^{1/2}F+{\displaystyle \frac{\text{i}\unicode[STIX]{x1D709}}{2\unicode[STIX]{x03C0}}}\int \text{min}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D709}_{1},\unicode[STIX]{x1D709}_{2},\unicode[STIX]{x1D709}_{3})F^{\ast }(\unicode[STIX]{x1D709}_{1})F(\unicode[STIX]{x1D709}_{2})F(\unicode[STIX]{x1D709}_{3})\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D709}+\unicode[STIX]{x1D709}_{1}-\unicode[STIX]{x1D709}_{2}-\unicode[STIX]{x1D709}_{3}}\,\text{d}\unicode[STIX]{x1D709}_{1}\,\text{d}\unicode[STIX]{x1D709}_{2}\,\text{d}\unicode[STIX]{x1D709}_{3}.\end{eqnarray}$$

This equation may have a solution with singularities, but this has not been studied yet.

5 Back to $\unicode[STIX]{x1D702}$ and $\unicode[STIX]{x1D713}$

The physical variables, $\unicode[STIX]{x1D702}_{k}$ and $\unicode[STIX]{x1D713}_{k}$ , are hidden in the normal complex variable $c_{k}$ and must be recovered. This is necessary when comparing the theory with experiment.

According to canonical transformation (2.5), $\unicode[STIX]{x1D702}_{k}$ and $\unicode[STIX]{x1D713}_{k}$ are power series of $b_{k}$ . On the other hand, using (3.16), the definition of $\tilde{\boldsymbol{B}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ (3.18) with $\tilde{\boldsymbol{T}}_{\boldsymbol{k}\boldsymbol{k}_{\mathbf{1}}}^{\boldsymbol{k}_{\mathbf{2}}\boldsymbol{k}_{\mathbf{3}}}$ (4.3) and relation (4.5), can be easily written as power series of $c_{k}$ . The accuracy of the super compact equation provides power series up the third order:

(5.1a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D702}_{k}=\unicode[STIX]{x1D702}_{k}^{(1)}+\unicode[STIX]{x1D702}_{k}^{(2)}+\unicode[STIX]{x1D702}_{k}^{(3)},\quad \unicode[STIX]{x1D713}_{k}=\unicode[STIX]{x1D713}_{k}^{(1)}+\unicode[STIX]{x1D713}_{k}^{(2)}+\unicode[STIX]{x1D713}_{k}^{(3)}.\end{eqnarray}$$

The details on recovering the physical quantities $\unicode[STIX]{x1D702}(x,t)$ and $\unicode[STIX]{x1D713}(x,t)$ are given in Dyachenko et al. (Reference Dyachenko, Kachulin, Zakharov, Pelinovsky and Kharif2015b ). Here, we present only the linear and second-order terms. All of these terms can be written in $k\text{-}\text{space}$ in a compact form. This is an important property that allows one to recover physical values without multidimensional integrals.

(5.2a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D702}^{(1)}(x)={\displaystyle \frac{1}{\sqrt{2}g^{1/4}}}(\hat{k}^{-1/4}c(x)+\hat{k}^{-1/4}c(x)^{\ast }),\quad \unicode[STIX]{x1D713}^{(1)}(x)=-\text{i}{\displaystyle \frac{g^{1/4}}{\sqrt{2}}}(\hat{k}^{-3/4}c(x)-\hat{k}^{-3/4}c(x)^{\ast }).\end{eqnarray}$$

The operators $\hat{k}^{\unicode[STIX]{x1D6FC}}$ act in the Fourier space as multiplication by $|k|^{\unicode[STIX]{x1D6FC}}$ .

(5.3) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D702}^{(2)}(x) & = & \displaystyle {\displaystyle \frac{\hat{k}}{4\sqrt{g}}}[\hat{k}^{-1/4}c(x)-\hat{k}^{-1/4}c(x)^{\ast }]^{2},\\ \unicode[STIX]{x1D713}^{(2)}(x) & = & \displaystyle {\displaystyle \frac{\text{i}}{2}}[\hat{k}^{-1/4}c(x)^{\ast }\hat{k}^{1/4}c(x)^{\ast }-\hat{k}^{-1/4}c(x)\hat{k}^{1/4}c(x)]+\\ & & +\,{\displaystyle \frac{1}{2}}{\hat{H}}[\hat{k}^{-1/4}c(x)\hat{k}^{1/4}c(x)^{\ast }+\hat{k}^{-1/4}c(x)^{\ast }\hat{k}^{1/4}c(x)].\end{array}\right\} & & \displaystyle\end{eqnarray}$$

Here, ${\hat{H}}$ is the Hilbert transformation with eigenvalue $\text{i}\,\text{sign}(k)$ .

This accuracy (second-order power series) is sufficient to compare numerical data with the data in a flume.

6 Numerical simulation

6.1 Breather

The super compact equation (4.13) has a localized breather-type solution:

(6.1a,b ) $$\begin{eqnarray}\displaystyle c(x,t)=C(x-{\mathcal{V}}t)\text{e}^{\text{i}(k_{0}x-\unicode[STIX]{x1D714}_{0}t)}\quad \text{or}\quad c_{k}(t)=\text{e}^{\text{i}(\unicode[STIX]{x1D6FA}+{\mathcal{V}}k)t}\unicode[STIX]{x1D719}_{k} & & \displaystyle\end{eqnarray}$$

where $\unicode[STIX]{x1D719}_{k}$ satisfies the following equation:

(6.2) $$\begin{eqnarray}(\unicode[STIX]{x1D6FA}+{\mathcal{V}}k-\unicode[STIX]{x1D714}_{k})\unicode[STIX]{x1D719}_{k}={\displaystyle \frac{1}{2}}\int T_{kk_{1}}^{k_{2}k_{3}}\unicode[STIX]{x1D719}_{k_{1}}^{\ast }\unicode[STIX]{x1D719}_{k_{2}}\unicode[STIX]{x1D719}_{k_{3}}\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}.\end{eqnarray}$$

Here, ${\mathcal{V}}$ is the group velocity and $k_{0}$ and $\unicode[STIX]{x1D714}_{0}$ are the wavenumber and frequency of the carrier wave, respectively. $\unicode[STIX]{x1D6FA}$ is close to $\unicode[STIX]{x1D714}_{0}/2$ . This solution can be found numerically by the Petviashvili method (see Petviashvili Reference Petviashvili1976). A uniform grid is introduced in the periodic domain $x\in [0,L]$ . Therefore, the wavenumbers $k$ become discrete, with a step size of $\unicode[STIX]{x0394}k=2\unicode[STIX]{x03C0}/L$ , and all integrals over $k$ transform to sums over $k$ .

(6.3) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D719}_{k}^{n+1} & = & \displaystyle {\displaystyle \frac{NL_{k}^{n}}{M_{k}}}\left[{\displaystyle \frac{\displaystyle \mathop{\sum }_{k^{\prime }}(\unicode[STIX]{x1D719}_{k^{\prime }}^{n}NL_{k^{\prime }}^{n})}{\displaystyle \mathop{\sum }_{k^{\prime }}(\unicode[STIX]{x1D719}_{k^{\prime }}^{n}M_{k^{\prime }}\unicode[STIX]{x1D719}_{k^{\prime }}^{n})}}\right]^{-3/2},\quad M_{k}=\unicode[STIX]{x1D6FA}+{\mathcal{V}}k-\unicode[STIX]{x1D714}_{k},\\ NL^{n} & = & \displaystyle -{\displaystyle \frac{\unicode[STIX]{x2202}^{+}}{\unicode[STIX]{x2202}x}}\left(|\unicode[STIX]{x1D719}^{n}|^{2}{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{n}}{\unicode[STIX]{x2202}x}}\right)+\text{i}{\displaystyle \frac{\unicode[STIX]{x2202}^{+}}{\unicode[STIX]{x2202}x}}(\hat{k}(|\unicode[STIX]{x1D719}^{n}|^{2})\unicode[STIX]{x1D719}^{n}).\end{array}\right\}\end{eqnarray}$$

Here, $n$ is the number of iterations. $C(x-{\mathcal{V}}t)$ has an obvious association with a soliton for the nonlinear Schrödinger equation.

A free surface profile of the breather solution of this equation in the periodic domain $L=10$ km with $k_{0}=(2\unicode[STIX]{x03C0}/L)100$ is shown in figure 2. The gravity acceleration $g=9.81~\text{m}~\text{s}^{-2}$ . A breather is a very stable structure. A collision of two breathers moving with different velocities (or with $k_{0}=(2\unicode[STIX]{x03C0}/L)100$ and $k_{0}=(2\unicode[STIX]{x03C0}/L)200$ ) is shown in figure 3. An animation of this collision can be viewed in supplementary movie 1, available at https://doi.org/10.1017/jfm.2017.529.

Figure 2. Narrow breather with three crests. Free surface profile.

Figure 3. Snapshots of breathers collision.

6.2 Modulational instability

A freak wave appearing from a slowly modulated Stokes wave of small amplitude ( $\unicode[STIX]{x1D702}\simeq \unicode[STIX]{x1D702}_{0}\cos (k_{0}x-\unicode[STIX]{x1D714}_{k_{0}}t)$ ), with $k_{0}=(2\unicode[STIX]{x03C0}/L)100$ and $\unicode[STIX]{x1D702}_{0}\simeq 1.35~\text{m}$ , is shown in figure 4.

Figure 4. Amplitude of freak wave in the periodic domain $L=10$  km.

One can see the beginning of the wave breaking in figure 5: the wave is going to break to the right (the right slope of the wave is steeper than the left slope).

Figure 5. Three snapshots showing the beginning of the wave breaking (zoomed near $x=8.34$  km).

The animation of a typical freak wave arising can be found in supplementary movies 2 and 3.

The analytical study of the small-scale instabilities by the ‘frozen coefficient’ method allows one to conclude that the Cauchy problem for the super compact equation is a well-posed problem.

7 Spatial compact equation

The simplicity of the super compact equation enables an easy derivation of the spatial version of the equation. The details of this derivation can be found in Dyachenko & Zakharov (Reference Dyachenko and Zakharov2016). The idea of the derivation is based on the fact that the Fourier image (after transforming equation (4.13) in both space and time) $c_{k\unicode[STIX]{x1D714}}$ is supported on the shadowed area in the vicinity of the dispersion curve, as shown in figure 6. Note that for unidirectional waves, both $k$ and $\unicode[STIX]{x1D714}$ are positive. This equation (after multiplying by $(\unicode[STIX]{x1D714}+\sqrt{gk})$ ) looks like the following:

(7.1) $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D714}^{2}-gk)c_{k\unicode[STIX]{x1D714}} & = & \displaystyle \displaystyle {\displaystyle \frac{(\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{k})k\unicode[STIX]{x1D703}_{k}}{(2\unicode[STIX]{x03C0})^{2}}}\int _{k_{i},\unicode[STIX]{x1D714}_{i}>0}T_{k_{2}k_{3}}^{kk_{1}}c_{k_{1}\unicode[STIX]{x1D714}_{1}}^{\ast }c_{k_{2}\unicode[STIX]{x1D714}_{2}}c_{k_{3}\unicode[STIX]{x1D714}_{3}}\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D6FF}_{k+k_{1}-k_{2}-k_{3}}\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{1}-\unicode[STIX]{x1D714}_{2}-\unicode[STIX]{x1D714}_{3}}\,\text{d}k_{1}\,\text{d}k_{2}\,\text{d}k_{3}\text{d}\unicode[STIX]{x1D714}_{1}\,\text{d}\unicode[STIX]{x1D714}_{2}\,\text{d}\unicode[STIX]{x1D714}_{3}.\end{eqnarray}$$

For all $c_{k_{i}\unicode[STIX]{x1D714}_{i}}$ , the following relations for their arguments are valid:

(7.2) $$\begin{eqnarray}\unicode[STIX]{x1D714}_{i}=\sqrt{gk_{i}}+\tilde{\unicode[STIX]{x1D714}}_{nl}.\end{eqnarray}$$

Here, $\tilde{\unicode[STIX]{x1D714}}_{nl}$ is a nonlinear frequency shift, which can be estimated from (7.1) as

(7.3) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D714}}_{nl}\sim |c|^{2}.\end{eqnarray}$$

Figure 6. Domain (grey) in $k$ – $\unicode[STIX]{x1D714}$ space where all waves evolve.

Then, one can replace $T_{k_{2}k_{3}}^{kk_{1}}$ in $T_{\unicode[STIX]{x1D714}_{2}^{2}\unicode[STIX]{x1D714}_{3}^{2}}^{\unicode[STIX]{x1D714}^{2}\unicode[STIX]{x1D714}_{1}^{2}}$ and drop all the terms with $\tilde{\unicode[STIX]{x1D714}}_{nl}$ . After performing the backward Fourier transformation in $k$ -space, the following equation is derived:

(7.4) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}{\displaystyle \frac{\unicode[STIX]{x2202}c_{\unicode[STIX]{x1D714}}}{\unicode[STIX]{x2202}x}}-\text{i}{\displaystyle \frac{\unicode[STIX]{x1D714}^{2}}{g}}c_{\unicode[STIX]{x1D714}} & = & \displaystyle -{\displaystyle \frac{2\unicode[STIX]{x1D714}^{3}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}}}{g^{3}}}{\displaystyle \frac{\text{i}}{2\unicode[STIX]{x03C0}}}\int T_{\unicode[STIX]{x1D714}_{2}^{2}\unicode[STIX]{x1D714}_{3}^{2}}^{\unicode[STIX]{x1D714}^{2}\unicode[STIX]{x1D714}_{1}^{2}}c_{\unicode[STIX]{x1D714}_{1}}^{\ast }c_{\unicode[STIX]{x1D714}_{2}}c_{\unicode[STIX]{x1D714}_{3}}\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{1}-\unicode[STIX]{x1D714}_{2}-\unicode[STIX]{x1D714}_{3}}\,\text{d}\unicode[STIX]{x1D714}_{1}\,\text{d}\unicode[STIX]{x1D714}_{2}\,\text{d}\unicode[STIX]{x1D714}_{3},\\ T_{\unicode[STIX]{x1D714}_{2}^{2}\unicode[STIX]{x1D714}_{3}^{2}}^{\unicode[STIX]{x1D714}^{2}\unicode[STIX]{x1D714}_{1}^{2}} & = & \displaystyle {\displaystyle \frac{1}{4\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}_{1}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}_{2}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}_{3}}\text{min}(\unicode[STIX]{x1D714}^{2},\unicode[STIX]{x1D714}_{1}^{2},\unicode[STIX]{x1D714}_{2}^{2},\unicode[STIX]{x1D714}_{3}^{2}).\end{array}\right\}\end{eqnarray}$$

This is the Hamiltonian spatial equation for water waves with the Hamiltonian

(7.5) $$\begin{eqnarray}\displaystyle H={\displaystyle \frac{1}{g}}\int {\displaystyle \frac{1}{\unicode[STIX]{x1D714}}}|c_{\unicode[STIX]{x1D714}}|^{2}\,\text{d}\unicode[STIX]{x1D714}-{\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}}{\displaystyle \frac{1}{g^{3}}}\int T_{\unicode[STIX]{x1D714}_{2}^{2}\unicode[STIX]{x1D714}_{3}^{2}}^{\unicode[STIX]{x1D714}^{2}\unicode[STIX]{x1D714}_{1}^{2}}c_{\unicode[STIX]{x1D714}}^{\ast }c_{\unicode[STIX]{x1D714}_{1}}^{\ast }c_{\unicode[STIX]{x1D714}_{2}}c_{\unicode[STIX]{x1D714}_{3}}\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{1}-\unicode[STIX]{x1D714}_{2}-\unicode[STIX]{x1D714}_{3}}\,\text{d}\unicode[STIX]{x1D714}\,\text{d}\unicode[STIX]{x1D714}_{1}\,\text{d}\unicode[STIX]{x1D714}_{2}\,\text{d}\unicode[STIX]{x1D714}_{3}. & & \displaystyle\end{eqnarray}$$

The equation of motion $(\unicode[STIX]{x2202}/\unicode[STIX]{x2202}x)c_{\unicode[STIX]{x1D714}}=\text{i}\unicode[STIX]{x1D714}^{3}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}}(\unicode[STIX]{x1D6FF}H/\unicode[STIX]{x1D6FF}c_{\unicode[STIX]{x1D714}}^{\ast })$ is

(7.6) $$\begin{eqnarray}\displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}x}}+{\displaystyle \frac{\text{i}}{g}}{\displaystyle \frac{\unicode[STIX]{x2202}^{2}c}{\unicode[STIX]{x2202}t^{2}}} & = & \displaystyle \displaystyle {\displaystyle \frac{\hat{P}^{-}}{2g^{3}}}{\displaystyle \frac{\unicode[STIX]{x2202}^{3}}{\unicode[STIX]{x2202}t^{3}}}\left[{\displaystyle \frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}t^{2}}}(|c|^{2}c)+2|c|^{2}{\displaystyle \frac{\unicode[STIX]{x2202}^{2}c}{\unicode[STIX]{x2202}t^{2}}}+c^{2}{\displaystyle \frac{\unicode[STIX]{x2202}^{2}c^{\ast }}{\unicode[STIX]{x2202}t^{2}}}\right]\nonumber\\ \displaystyle & & \displaystyle \displaystyle +\,{\displaystyle \frac{\text{i}\hat{P}^{-}}{g^{3}}}{\displaystyle \frac{\unicode[STIX]{x2202}^{3}}{\unicode[STIX]{x2202}t^{3}}}\left[{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}}(c\hat{\unicode[STIX]{x1D714}}|c|^{2})+{\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}t}}\hat{\unicode[STIX]{x1D714}}|c|^{2}+c\hat{\unicode[STIX]{x1D714}}\left(c^{\ast }{\displaystyle \frac{\unicode[STIX]{x2202}c}{\unicode[STIX]{x2202}t}}-c{\displaystyle \frac{\unicode[STIX]{x2202}c^{\ast }}{\unicode[STIX]{x2202}t}}\right)\right].\end{eqnarray}$$

The operator $\hat{P}^{-}$ is the projection operator:

(7.7) $$\begin{eqnarray}\hat{P}^{-}={\textstyle \frac{1}{2}}(1-\text{i}{\hat{H}}),\text{ here},{\hat{H}}\text{ is the Hilbert transformation},\end{eqnarray}$$

and it is equal to $\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}}$ in Fourier space.

An analytical study of the small-scale instabilities by the ‘frozen coefficient’ method also allows one to conclude that the Cauchy problem for the compact spatial equation is a well-posed problem (although it includes a fifth derivative).

8 Some numerics for spatial equation

8.1 Breather

A breather is the localized solution of a spatial equation of the following type:

(8.1) $$\begin{eqnarray}\displaystyle c(x,t)=C\left(t-{\displaystyle \frac{x}{{\mathcal{V}}}}\right)\text{e}^{\text{i}(k_{0}x-\unicode[STIX]{x1D714}_{0}t)}. & & \displaystyle\end{eqnarray}$$

Fourier transforming over time, one can obtain:

(8.2) $$\begin{eqnarray}c_{\unicode[STIX]{x1D714}}(x)={\displaystyle \frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}}\int C\left(t-{\displaystyle \frac{x}{{\mathcal{V}}}}\right)\text{e}^{\text{i}k_{0}x-\text{i}(\unicode[STIX]{x1D714}_{0}-\unicode[STIX]{x1D714})t}\,\text{d}t={\displaystyle \frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}}\int C(\unicode[STIX]{x1D709})\text{e}^{-\text{i}(\unicode[STIX]{x1D714}_{0}-\unicode[STIX]{x1D714})\unicode[STIX]{x1D709}}\text{e}^{\text{i}k_{0}x-\text{i}(\unicode[STIX]{x1D714}_{0}-\unicode[STIX]{x1D714})x/{\mathcal{V}}}\,\text{d}\unicode[STIX]{x1D709},\end{eqnarray}$$

or

(8.3) $$\begin{eqnarray}c_{\unicode[STIX]{x1D714}}(x)=\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}}\text{e}^{\text{i}({\mathcal{K}}+\unicode[STIX]{x1D714}/{\mathcal{V}})x}.\end{eqnarray}$$

Here, ${\mathcal{K}}=k_{0}-(\unicode[STIX]{x1D714}_{0}/{\mathcal{V}})$ is close to $-\unicode[STIX]{x1D714}_{0}^{2}/g$ and $\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}}$ satisfies the following equation:

(8.4) $$\begin{eqnarray}\displaystyle \left({\mathcal{K}}+{\displaystyle \frac{\unicode[STIX]{x1D714}}{{\mathcal{V}}}}-{\displaystyle \frac{\unicode[STIX]{x1D714}^{2}}{g}}\right)\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}}=-{\displaystyle \frac{2\unicode[STIX]{x1D714}^{3}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D714}}}{g^{3}}}{\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}}\int T_{\unicode[STIX]{x1D714}^{2}\unicode[STIX]{x1D714}_{1}^{2}}^{\unicode[STIX]{x1D714}_{2}^{2}\unicode[STIX]{x1D714}_{3}^{2}}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}_{1}}^{\ast }\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}_{2}}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}_{3}}\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{1}-\unicode[STIX]{x1D714}_{2}-\unicode[STIX]{x1D714}_{3}}\,\text{d}\unicode[STIX]{x1D714}_{1}\,\text{d}\unicode[STIX]{x1D714}_{2}\,\text{d}\unicode[STIX]{x1D714}_{3}. & & \displaystyle\end{eqnarray}$$

This can be found by use of the iterative Petviashvili method ( $n$ is the number of iterations). A uniform grid is introduced in the periodic domain $t\in [0,T]$ . Therefore, the frequencies $\unicode[STIX]{x1D714}$ become discrete, with a step size of $\unicode[STIX]{x0394}\unicode[STIX]{x1D714}=2\unicode[STIX]{x03C0}/T$ , and all integrals over $\unicode[STIX]{x1D714}$ transform to sums over $\unicode[STIX]{x1D714}$ .

(8.5) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}}^{n+1} & = & \displaystyle {\displaystyle \frac{NL_{\unicode[STIX]{x1D714}}^{n}}{M_{\unicode[STIX]{x1D714}}}}\left[{\displaystyle \frac{\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D714}^{\prime }}(\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}^{\prime }}^{n}NL_{\unicode[STIX]{x1D714}^{\prime }}^{n})}{\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D714}^{\prime }}(\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}^{\prime }}^{n}M_{\unicode[STIX]{x1D714}^{\prime }}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D714}^{\prime }}^{n})}}\right]^{-3/2},\quad M_{\unicode[STIX]{x1D714}}={\mathcal{K}}+{\displaystyle \frac{\unicode[STIX]{x1D714}}{{\mathcal{V}}}}-{\displaystyle \frac{\unicode[STIX]{x1D714}^{2}}{g}},\\ NL^{n} & = & \displaystyle {\displaystyle \frac{-\text{i}\hat{P}^{-}}{2g^{3}}}{\displaystyle \frac{\unicode[STIX]{x2202}^{3}}{\unicode[STIX]{x2202}t^{3}}}\left[{\displaystyle \frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}t^{2}}}(|\unicode[STIX]{x1D719}^{n}|^{2}\unicode[STIX]{x1D719}^{n})+2|\unicode[STIX]{x1D719}^{n}|^{2}{\displaystyle \frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D719}^{n}}{\unicode[STIX]{x2202}t^{2}}}+{\displaystyle \frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D719}^{n\ast }}{\unicode[STIX]{x2202}t^{2}}}\unicode[STIX]{x1D719}^{n2}\right]\\ & & +{\displaystyle \frac{\hat{P}^{-}}{g^{3}}}{\displaystyle \frac{\unicode[STIX]{x2202}^{3}}{\unicode[STIX]{x2202}t^{3}}}\left[{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}}(\unicode[STIX]{x1D719}^{n}\hat{\unicode[STIX]{x1D714}}|\unicode[STIX]{x1D719}^{n}|^{2})+{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{n}}{\unicode[STIX]{x2202}t}}\hat{\unicode[STIX]{x1D714}}|\unicode[STIX]{x1D719}^{n}|^{2}+\unicode[STIX]{x1D719}^{n}\hat{\unicode[STIX]{x1D714}}\left({\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{n}}{\unicode[STIX]{x2202}t}}\unicode[STIX]{x1D719}^{n\ast }-\unicode[STIX]{x1D719}^{n}{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{n\ast }}{\unicode[STIX]{x2202}t}}\right)\right].\end{array}\right\}\end{eqnarray}$$

A free surface profile of the breather solution of this equation, in the periodic domain $T=320~\text{s}$ , with $\unicode[STIX]{x1D714}_{0}=0.78~(\text{s}^{-1})$ and ${\mathcal{K}}=-6.428\times 10^{-2}$ , is shown in figure 7. The Fourier harmonics $(|c_{\unicode[STIX]{x1D714}}|)$ of the breather solution in a logarithmic scale are shown in figure 8.

Figure 7. Breather solution. Free surface profile.

Figure 8. The Fourier harmonics $(|c_{\unicode[STIX]{x1D714}}|)$ of the breather solution.

An animation of breather generation in a ‘digital’ flume can be found in supplementary movie 4.

8.2 Modulational instability

In the spatial equation, modulational instability of a monochromatic wave also occurs. The monochromatic wave

(8.6a,b ) $$\begin{eqnarray}c(x,t)=c_{0}\text{e}^{\text{i}k_{0}x-\text{i}\unicode[STIX]{x1D714}_{0}t},\quad \text{or }c(x,\unicode[STIX]{x1D714})=\sqrt{2\unicode[STIX]{x03C0}}c_{0}\text{e}^{\text{i}k_{0}x}\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{0})\end{eqnarray}$$

is the simplest solution of (7.6) and (7.4). Substituting (8.6) into (7.4) yields the following relation:

(8.7) $$\begin{eqnarray}k_{0}={\displaystyle \frac{\unicode[STIX]{x1D714}_{0}^{2}}{g}}-{\displaystyle \frac{2\unicode[STIX]{x1D714}_{0}^{5}}{g^{3}}}|c_{0}|^{2},\end{eqnarray}$$

( $-2\unicode[STIX]{x1D714}_{0}^{5}/g^{3}|c_{0}|^{2}$ could be called a ‘nonlinear wavelength shift’). The perturbed solution has the following form:

(8.8) $$\begin{eqnarray}c(x,\unicode[STIX]{x1D714})=\sqrt{2\unicode[STIX]{x03C0}}\left(c_{0}\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{0})+c_{+}(x)\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{+})\text{e}^{\text{i}k_{+}x}+c_{-}(x)\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{-})\text{e}^{\text{i}k_{-}x}\right)\text{e}^{\text{i}k_{0}x}.\end{eqnarray}$$

Here, $\unicode[STIX]{x1D714}_{\pm }=\unicode[STIX]{x1D714}_{0}\pm \unicode[STIX]{x1D714}$ and $k_{+}=-k_{-}$ with the following condition:

(8.9) $$\begin{eqnarray}\displaystyle |c_{+}|,|c_{-}|\ll |c_{0}|. & & \displaystyle\end{eqnarray}$$

Substituting (8.8) into (7.4), one can obtain the sum of two independent equations:

(8.10) $$\begin{eqnarray}\displaystyle & & \displaystyle \bigg[{\displaystyle \frac{\unicode[STIX]{x2202}c_{+}}{\unicode[STIX]{x2202}x}}+\text{i}(k_{0}+k_{+})c_{+}-{\displaystyle \frac{\text{i}}{g}}\unicode[STIX]{x1D714}_{+}^{2}c_{+}+{\displaystyle \frac{4\text{i}\unicode[STIX]{x1D714}_{+}^{3}}{g^{3}}}T_{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{0}^{2}}|c_{0}|^{2}c_{+}+{\displaystyle \frac{2\text{i}\unicode[STIX]{x1D714}_{+}^{3}}{g^{3}}}T_{\unicode[STIX]{x1D714}_{0}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{-}^{2}}c_{0}^{2}c_{-}^{\ast }\bigg]\text{e}^{\text{i}k_{+}x}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\bigg[{\displaystyle \frac{\unicode[STIX]{x2202}c_{-}}{\unicode[STIX]{x2202}x}}+\text{i}(k_{0}+k_{-})c_{-}-{\displaystyle \frac{\text{i}}{g}}\unicode[STIX]{x1D714}_{-}^{2}c_{-}+{\displaystyle \frac{4\text{i}\unicode[STIX]{x1D714}_{-}^{3}}{g^{3}}}T_{\unicode[STIX]{x1D714}_{-}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{-}^{2}\unicode[STIX]{x1D714}_{0}^{2}}|c_{0}|^{2}c_{-}+{\displaystyle \frac{2\text{i}\unicode[STIX]{x1D714}_{-}^{3}}{g^{3}}}T_{\unicode[STIX]{x1D714}_{0}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{-}^{2}\unicode[STIX]{x1D714}_{+}^{2}}c_{0}^{2}c_{+}^{\ast }\bigg]\text{e}^{\text{i}k_{-}x}=0.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Expressions for $T_{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{0}^{2}}$ , $T_{\unicode[STIX]{x1D714}_{0}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{+}^{2}\unicode[STIX]{x1D714}_{-}^{2}}$ and $T_{\unicode[STIX]{x1D714}_{-}^{2}\unicode[STIX]{x1D714}_{0}^{2}}^{\unicode[STIX]{x1D714}_{-}^{2}\unicode[STIX]{x1D714}_{0}^{2}}$ can be easily obtained from (4.3). Suppose that $c_{\pm }$ grow according to

(8.11) $$\begin{eqnarray}\displaystyle c_{\pm }\Rightarrow c_{\pm }\text{e}^{\unicode[STIX]{x1D6FE}_{\unicode[STIX]{x1D714}}x}. & & \displaystyle\end{eqnarray}$$

Then, we can obtain the formula for $\unicode[STIX]{x1D6FE}_{\unicode[STIX]{x1D714}}$ given by a tenth-degree polynomial. By introducing steepness for the monochromatic wave $\unicode[STIX]{x1D702}(x)=\unicode[STIX]{x1D702}_{0}\cos k_{0}x$ as $\unicode[STIX]{x1D707}=\unicode[STIX]{x1D702}_{0}k_{0}$ , one can easily find (see § 5) that in terms of $c(x)=c_{0}\text{e}^{\text{i}k_{0}x}$ ,

(8.12) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}={\displaystyle \frac{\sqrt{2}|c_{0}|\unicode[STIX]{x1D714}_{0}^{3/2}}{g}}. & & \displaystyle\end{eqnarray}$$

The growth rate squared $\unicode[STIX]{x1D6FE}^{2}(\unicode[STIX]{x1D714})$ for $\unicode[STIX]{x1D714}_{0}=0.78~(\text{s}^{-1})$ , and the steepness of the carrier wave $\unicode[STIX]{x1D707}=0.1$ is shown in figure 9. Perturbations whose frequencies $\unicode[STIX]{x1D714}$ are such that $\unicode[STIX]{x1D6FE}^{2}(\unicode[STIX]{x1D714})>0$ are unstable, and they grow as $c_{\pm }\sim \text{e}^{\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D714})x}$ . Perturbations whose frequencies $\unicode[STIX]{x1D714}$ are such that $\unicode[STIX]{x1D6FE}^{2}(\unicode[STIX]{x1D714})<0$ are stable; therefore, they only change phase $c_{\pm }\sim \text{e}^{\text{i}\sqrt{-\unicode[STIX]{x1D6FE}^{2}(\unicode[STIX]{x1D714})}x}$ .

Figure 9. The growth rate squared $\unicode[STIX]{x1D6FE}^{2}(\unicode[STIX]{x1D714})$ of the perturbation $c_{\pm }$ of the monochromatic wave solution of the spatial equation (7.4). Here, $\unicode[STIX]{x1D714}_{0}=0.78~(\text{s}^{-1})$ and the steepness of the carrier wave $\unicode[STIX]{x1D707}\approx 0.1$ .