\documentclass[10pt,pdf,hyperref={unicode},aspectratio=43]{beamer}

%\documentclass[aspectratio=43]{beamer}
% \documentclass[aspectratio=1610]{beamer}
% \documentclass[aspectratio=169]{beamer}

%\usepackage{lmodern}
\usepackage{amsmath}
\usepackage{ulem}
\usepackage{xcolor}
\usepackage{cancel}
% подключаем кириллицу 
\usepackage[T2A]{fontenc}
\usepackage[koi8-r]{inputenc}
\usepackage{multimedia}
%\usepackage{movie15}
\usepackage{hyperref}

% отключить клавиши навигации
\setbeamertemplate{navigation symbols}{}

% тема оформления
%\usetheme{Pittsburgh}
%\usetheme{Warsaw}
\usetheme{Copenhagen}
%\usetheme{Rochester}

% цветовая схема
\usecolortheme{seahorse}

%\title{Integration of the NLSE with Periodic Boundary Conditions}
\title{Integration of the Nonlinear Schr\"odinger Equation with Periodic
Boundary Conditions}
%\subtitle{Use beamer everywhere you are}
\author{\underline{A.I. Dyachenko} }
\institute{Landau Institute for Theoretical Physics\\
Skolkovo Institute of Science and Technology\\
Higher School of Economics }
%}
\date{\ } 
% \logo{\includegraphics[height=5mm]{images/logo.png}\vspace{-7pt}}

\begin{document}

% титульный слайд
\begin{frame}
\titlepage
Grant RScF 19-72-30028-П
\end{frame} 

\begin{frame}
\frametitle{Summary. \hspace{30pt} Consider 1D NLSE for the field $q(x,t)$:} 
%\framesubtitle{subtitle}
\begin{raggedright}

%Consider 1D NLSE for the field $q(x,t)$:
%
\begin{equation}\nonumber
i \dot q +\frac{1}{2}q_{xx} + |q|^2q = 0
\end{equation}
with periodic boundary condition in space. For Fourier harmonics:
\begin{equation}\nonumber
i\dot q_n  - \frac{1}{2}k^2_n q_n + \sum_{n_1,n_2,n_3}
q^*_{n_1}q_{n_2}q_{n_3} 
\delta_{{n} + {n_1}, {n_2} + {n_3}} =0.
%
\end{equation}
\begin{equation}\nonumber
q(x,t) = \sum_n q_n(t) e^{ik_n t},\hspace{10pt}
q_n(t) = \frac{1}{L}\int^L_0 q(x,t) e^{-ik_n t}dt,\hspace{5pt} \mbox{here }
k_n =\frac{2\pi n}{L}.
\end{equation}
In search of some canonical transformation
$$
q_n \rightarrow b_n
$$
\vspace{-10pt}
\begin{block}{Linear Schrodinger Equation \hspace{40pt} New Hamiltonian }
\begin{equation}\nonumber
i\dot b_n  -\frac{1}{2}k^2_nb_n + 
\Omega_n b_n =0\hspace{30pt} H = \frac{1}{2}\sum_{n}k^2_n|b_n|^2 - \frac{1}{2}
\sum_{n,n_1}D_{{n}{n_1}} |b_{n}|^2|b_{n_1}|^2
\end{equation}
\end{block}{$\Omega_n$ {\bf does} depend on $n$ and initial conditions \hspace{7pt} and {\bf  does not} depend on $t$}
\end{raggedright}
\end{frame}
%
%
%
\begin{frame}
\frametitle{Resonant and Nonresonant Interactions} 
%\framesubtitle{subtitle}
\begin{raggedright}
\vspace{-15pt}
The nonlinear part of the Hamiltonian is divided into two parts, \center{Resonant and Nonresonant.}
\begin{equation}\nonumber
 H = \frac{1}{2}\sum_{n}k^2_n|q_n|^2 - \frac{1}{2}
\sum_{n,n_1,n_2,n_3}q^*_{n}q^*_{n_1}q_{n_2}q_{n_3} \delta_{{n} + {n_1}, {n_2} + {n_3}} 
\end{equation}
Resonant $R_4$ is:
$$
\hspace{-5pt} R_4 \hspace{0pt}\Leftarrow 
\begin{cases}
& k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3} =0 \\
& {n} + {n_1} - {n_2} - {n_3} =0.
\end{cases}
$$
%
$$
\sum_{n,n_1,n_2,n_3}
q^*_{n}q^*_{n_1}q_{n_2}q_{n_3}\delta_{{n} + {n_1}, {n_2} + {n_3}} = \sum_{n_i \in R^4}q^*_{n}q^*_{n_1}q_{n_2}q_{n_3}
+ \cancel{\sum_{n_i \notin R^4}q^*_{n}q^*_{n_1}q_{n_2}q_{n_3}}
$$
Of course, the canonical transformation generates additional, multiple wave interactions terms in the Hamiltonian.
\end{raggedright}
\end{frame}
%
%
\begin{frame}
\frametitle{Auxiliary Hamiltonian} 
%\framesubtitle{subtitle}
\begin{raggedright}
For Hamiltonian system with variable $b_n(t)$ the transformation
$$
b_n(0) \rightarrow b_n(t)
$$
is canonical. We will construct this transformation using an auxiliary Hamiltonian $\tilde H$(similar to Zakharov, Lvov and Falkovich,  1992)
\begin{eqnarray}\nonumber %\label{Haux}
&&\tilde H = \frac{i}{2} \!\!\!\! \sum_{n,n_1,n_2,n_3}
B^{{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}}+\cr
&&+\frac{i}{3} \!\!\!\!\!\!\!\!\!\!\sum_{n,n_1,n_2,n_3,n_4,n_5}\!\!\!\!\!\!\!\!\!\!
C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5}
\delta_{{n} + {n_1}+{n_2}, {n_3}+{n_4} + {n_5}}+\dots
\end{eqnarray}
%
Coefficients in the Hamiltonian have the following properties:
\begin{enumerate}
\item Permutations of upper and lower indexes do not change their values
\item $$B^{{n}{n_1}}_{{n_2}{n_3}}=
-B_{{n}{n_1}}^{{n_2}{n_3}}\hspace
{25pt}
C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}=
-C_{{n}{n_1}{n_2}}^{{n_3}{n_4}{n_5}}$$
\item $$B^{{n}{n_1}}_{{n}{n_1}}=
0\hspace
{25pt}
C^{{n}{n_1}{n_2}}_{{n}{n_1}{n_2}}=
0$$
\end{enumerate}

\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{Canonical Transformation} 
%\framesubtitle{subtitle}
\begin{raggedright}

This Hamiltonian helps to calculate coefficients of the series of canonical transformation.
One can express old variables $q_n$ in terms of $b_n(0)$:
\begin{equation}\nonumber
b_n(t) = b_n(0) + t\dot b_n(0)+\frac{t^2}{2}\ddot b_n(0)+\dots  
\hspace{20pt} 
%\mbox{Here} 
%\hspace{15pt}
\dot b_n(0) = -i\frac{\delta\tilde H}{\delta b^*_n}
\hspace{15pt}\ddot b_n(0) = -i\frac{\partial}{\partial t}\frac{\delta\tilde H}{\delta b^*_n}
\end{equation}
%
$$q_n(0)  \hspace{10pt} \Rightarrow \hspace{10pt}  b_n(0) \hspace{90pt} t=1$$
%
%
The general form of the canonical transformation in term of power series:
\begin{eqnarray}\nonumber
&&q_n = b_n + \sum_{n_1,n_2,n_3}
B^{{n}{n_1}}_{{n_2}{n_3}}
b^*_{n_1}b_{n_2}b_{n_3} 
\delta_{{n} + {n_1}, {n_2} + {n_3}}+\cr
&&+\frac{1}{2}\!\!\!\!\!\!\!\!\!\!\sum_{n,n_1,n_2,n_3,n_4,n_5}\!\!\!\!\!
{^2\!B}^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5} 
\delta_{{n} + {n_1} + {n_2}, {n_3}+{n_4} + {n_5}}+\cr
&&+\sum_{n,n_1,n_2,n_3,n_4,n_5}\!\!\!
C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5} 
\delta_{{n} + {n_1} + {n_2}, {n_3}+{n_4} + {n_5}}+\dots
\end{eqnarray}
We seek for the best choice for $B^{{n}{n_1}}_{{n_2}{n_3}}$, $C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}$, etc.
\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{Resonant Manifolds} 
%\framesubtitle{subtitle}
\begin{raggedright}



The Hamiltonian takes (after symmetrization) the following form:
\begin{eqnarray}\nonumber
H &=& \frac{1}{2}\sum_{n}k^2_n|b_n|^2 - 
 \frac{1}{2}\sum_{n,n_1,n_2,n_3}
 T^{{n}{n_1}}_{{n_2}{n_3}}
b^*_{n}b^*_{n_1}b_{n_2}b_{n_3}
%
\underline{
\delta_{{n} + {n_1}, {n_2} + {n_3}}
}- \cr
&-&\frac{1}{6}\sum_{n,n_1,n_2,n_3,n_4,n_5}
 T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5}
%
\underline{
\delta_{{n} + {n_1} +{n_2}, {n_3} + {n_4} + {n_5}}
}+\dots
\end{eqnarray}
Here
\begin{equation}\nonumber
\hspace{-60pt} T^{{n}{n_1}}_{{n_2}{n_3}} = \hspace{30pt}
1-\frac{1}{2} (k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3})
B^{{n}{n_1}}_{{n_2}{n_3}}
\end{equation}
\begin{equation}\nonumber
T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} =
R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} - (k^2_{n} + k^2_{n_1} + k^2_{n_2} - k^2_{n_3} - k^2_{n_4} - k^2_{n_5})C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} 
\end{equation}
%
%
\begin{block}{}
$$
\hspace{-50pt} R_4 \hspace{0pt}\Leftarrow 
\begin{cases}
& k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3} =0 \\
& {n} + {n_1} - {n_2} - {n_3} =0.
\end{cases}
$$
$$
R_6 \hspace{0pt}\Leftarrow 
\begin{cases}
& k^2_{n} + k^2_{n_1} + k^2_{n_2} - k^2_{n_3} - k^2_{n_4} - k^2_{n_5} =0 \\
& {n} + {n_1} + {n_2} - {n_3} - {n_4} - {n_5} =0.
\end{cases}
$$

\end{block}
%
%




\end{raggedright}
\end{frame}

\begin{frame}
%
\frametitle{$T^{{n}{n_1}}_{{n_2}{n_3}} =
1-\frac{1}{2} (k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3})
B^{{n}{n_1}}_{{n_2}{n_3}}$}%{How to choose $B^{{n}{n_1}}_{{n_2}{n_3}}$} 
%\framesubtitle{subtitle}
\begin{raggedright}
%
$$
\sum_{n,n_1,n_2,n_3}
 T^{{n}{n_1}}_{{n_2}{n_3}}
b^*_{n}b^*_{n_1}b_{n_2}b_{n_3} = \sum_{n_i \in R^4}
 T^{{n}{n_1}}_{{n_2}{n_3}}
b^*_{n}b^*_{n_1}b_{n_2}b_{n_3} + \cancel{\sum_{n_i \notin R^4}
 T^{{n}{n_1}}_{{n_2}{n_3}}
b^*_{n}b^*_{n_1}b_{n_2}b_{n_3}}
$$
%


%\vspace{-15pt}
%\begin{equation}\nonumber
%T^{{n}{n_1}}_{{n_2}{n_3}} =
%1-\frac{1}{2} (k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3})
%B^{{n}{n_1}}_{{n_2}{n_3}}
%\end{equation}
The best choice for $B^{{n}{n_1}}_{{n_2}{n_3}}$ is:
\begin{eqnarray}\nonumber
&&B^{{n}{n_1}}_{{n_2}{n_3}} =
\frac{2 \cdot {\bf 1}}
%2\frac{1+\delta_{{n}-{n_1}}\delta_{{n_2}-{n_3}}-\delta_{{n}-{n_2}}\delta_{{n_1}-{n_3}} -\delta_{{n}-{n_3}}\delta_{{n_1}-{n_2}}}
{k^2_{n}+k^2_{n_1} - k^2_{n_2} - k^2_{n_3}}  \hspace{20pt} \mbox{if } 
n_i \notin R^4,\cr
&&B^{{n}{n_1}}_{{n_2}{n_3}} = 0 \hspace{103pt} \mbox{if } 
n_i \in R^4
%2\frac{1+\delta_{{n}-{n_1}}\delta_{{n_2}-{n_3}}-\delta_{{n}-{n_2}}\delta_{{n_1}-{n_3}}
%-\delta_{{n}-{n_3}}\delta_{{n_1}-{n_2}}}{k^2_{n}+k^2_{n_1} - k^2_{n_2} - k^2_{n_3}}
\end{eqnarray}
%
$R^4$ has only trivial solutions:

\begin{block}{}
\begin{eqnarray}\nonumber
{n} = {n_2},\hspace{15pt}  {n_1} = {n_3}\hspace{60pt}
{n} = {n_3},\hspace{15pt}  {n_1} = {n_2}
\end{eqnarray}
\end{block}

%otherwise $B^{{n}{n_1}}_{{n_2}{n_3}} = 0$.
%
$$
T^{{n}{n_1}}_{{n_2}{n_3}} \hspace{15pt}\Rightarrow D_{{n}{n_1}} \hspace{15pt}= 
\begin{cases}
2 & \mbox{if } n\neq n_1 \leq 0,\\
1 & \mbox{if } n=n_1.
\end{cases}
$$
Note: $B^{{n}{n_1}}_{{n_2}{n_3}}$ satisfies 
symmetry conditions  and has no singularities.


%

\end{raggedright}
\end{frame}

\begin{frame}
%
\frametitle{$T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} =
R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} - (k^2_{n} + k^2_{n_1} + k^2_{n_2} - k^2_{n_3} - k^2_{n_4} - k^2_{n_5})C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}$}
%%
\begin{raggedright}
%
\begin{eqnarray}\nonumber
\sum_{n_i}
 T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5} &=& \sum_{n_i \in R^6}
 T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5} +\cr
&+&\cancel{\sum_{n_i \notin R^6}
 T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5}
}
\end{eqnarray}
%


%\vspace{-15pt}
%\begin{equation}\nonumber
%T^{{n}{n_1}}_{{n_2}{n_3}} =
%1-\frac{1}{2} (k^2_{n} + k^2_{n_1} - k^2_{n_2} - k^2_{n_3})
%B^{{n}{n_1}}_{{n_2}{n_3}}
%\end{equation}
The best choice for $C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}$ is:
\begin{eqnarray}\nonumber
&&C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} =
\frac{R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}}
%2\frac{1+\delta_{{n}-{n_1}}\delta_{{n_2}-{n_3}}-\delta_{{n}-{n_2}}\delta_{{n_1}-{n_3}} -\delta_{{n}-{n_3}}\delta_{{n_1}-{n_2}}}
{k^2_{n} + k^2_{n_1} + k^2_{n_2} - k^2_{n_3} - k^2_{n_4} - k^2_{n_5}}  \hspace{20pt} \mbox{if } 
n_i \notin R^6,\cr
&&C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} = 0 \hspace{153pt} \mbox{if } 
n_i \in R^6
%2\frac{1+\delta_{{n}-{n_1}}\delta_{{n_2}-{n_3}}-\delta_{{n}-{n_2}}\delta_{{n_1}-{n_3}}
%-\delta_{{n}-{n_3}}\delta_{{n_1}-{n_2}}}{k^2_{n}+k^2_{n_1} - k^2_{n_2} - k^2_{n_3}}
\end{eqnarray}
%
$R^6$ {\bf has} nontrivial solutions.
%\begin{eqnarray}\nonumber
%&& {n} = {n_2},\hspace{15pt}  {n_1} = {n_3}\cr
%&& {n} = {n_3},\hspace{15pt}  {n_1} = {n_2}
%\end{eqnarray}
%otherwise $B^{{n}{n_1}}_{{n_2}{n_3}} = 0$.
%
$$
T^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} \hspace{15pt} = 
\begin{cases}
R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}} & \mbox{if } n_i \in R^6,\\
0 & \mbox{if } n_i \notin R^6.
\end{cases}
$$
%
Note: $C^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}$ satisfies 
symmetry conditions  and has no singularities.

%

\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{New Hamiltonian} 
%\framesubtitle{subtitle}
\begin{raggedright}
%
\begin{block}{Six waves interactions}
$$R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}\equiv 0 \hspace{20pt} \mbox{if } n_i \in R^6$$
\end{block}
\vspace{15pt}
(Dyachenko, Kachulin, Zakharov, JETP Letters, 2013))
\vspace{-3pt}
\begin{equation}\nonumber
H = \frac{1}{2}\sum_{n}k^2_n|b_n|^2 - \frac{1}{2}
\sum_{n,n_1}D_{{n}{n_1}}
|b_{n}|^2|b_{n_1}|^2 - \cancel{
\frac{1}{6}\sum_{n_i \in R^6}R^{{n}{n_1}{n_2}}_{{n_3}{n_4}{n_5}}
b^*_{n}b^*_{n_1}b^*_{n_2}b_{n_3}b_{n_4}b_{n_5} +\dots
}
\end{equation}


Here
\vspace{-10pt}
\begin{equation}\nonumber
\sum_{n_1}D_{{n}{n_1}}|b_{n_1}|^2 = \Omega_n\hspace{15pt}
\end{equation}
{\bf does not} depend on $t$ and is defined by initial conditions only.
\end{raggedright}
\end{frame}
%
\begin{frame}
\frametitle{New Hamiltonian} 
%\framesubtitle{subtitle}
\begin{raggedright}
\begin{equation}\nonumber
H = \frac{1}{2}\sum_{n}k^2_n|b_n|^2 - \frac{1}{2}
\sum_{n,n_1}D_{{n}{n_1}}
|b_{n}|^2|b_{n_1}|^2
\end{equation}

%
and the equation for $b$:
\begin{equation}\nonumber
i\dot b_n  -\frac{1}{2}k^2_nb_n + 
\Big[\sum_{n_1}D_{{n}{n_1}}|b_{n_1}|^2\Big]b_n  =0
\end{equation}
\vspace{3pt}
The solution of this equation is simple (action-angle):
\begin{equation}\nonumber
b_n(t) = b_n(0)e^{i(\Omega_n -\frac{1}{2}k_n^2)t}
\end{equation}


%Zakharov and Shulman, "What is Integrability", Springer-Verlag, 1992
\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{Applicability} 
%\framesubtitle{subtitle}
\begin{raggedright}
\vspace{0pt}
Modulational instability
\begin{equation}\nonumber
q = q_0 + \delta q
\end{equation}
 For Fourier harmonics:
\begin{equation}\nonumber
\delta q_n \sim e^{\gamma_n t}
%
\end{equation}
\begin{block}{growth rate}
\vspace{-0pt}
\begin{equation}\nonumber
\gamma_n = k_n\sqrt{2|q_0|^2-k_n^2}
\end{equation}
\end{block}

%
\begin{block}{}
\center{$q_n^2 < {\Delta k}^2$}
\end{block}
\end{raggedright}
\end{frame}


\begin{frame}
\frametitle{Weakly Nonlinear Case \hspace{110pt} $\frac{b^2}{k^2}<<1$} 
%\framesubtitle{subtitle}
\begin{raggedright}
\vspace{-10pt}
\begin{equation}\nonumber
i \dot q +\frac{1}{2}q_{xx} + |q|^2q = 0
\end{equation}
 For Fourier harmonics:
\begin{equation}\nonumber
i\dot q_n  - \frac{1}{2}k^2_n q_n + \sum_{n_1,n_2,n_3}
q^*_{n_1}q_{n_2}q_{n_3} 
\delta_{{n} + {n_1}, {n_2} + {n_3}} =0.
%
\end{equation}
\begin{block}{Canonical Transformation}
\vspace{-10pt}
\begin{equation}\nonumber
q_n = b_n + \sum_{k^2_{n} + k^2_{n_1} \neq  k^2_{n_2} + k^2_{n_3}} 
%
\frac{b^*_{n_1}b_{n_2}b_{n_3}}{(k_{n_1} - k_{n_2})(k_{n_1} - k_{n_3})}\delta_{{n} + {n_1}, {n_2} + {n_3}}
\hspace{20pt} \frac{b^2}{k^2} <<1
\end{equation}
\end{block}
\begin{equation}\nonumber
H = \frac{1}{2}\sum_{n}k^2_n|b_n|^2 - \frac{1}{2}
\sum_{n,n_1}D_{{n}{n_1}}
|b_{n}|^2|b_{n_1}|^2 
\end{equation}
\begin{equation}\nonumber
i\dot b_n  -\frac{1}{2}k^2_n b_n + 
\Omega_n b_n =0\hspace{45pt} b_n(t) = b_n(0)e^{i(\Omega_n -\frac{1}{2}k_n^2)t}
\end{equation}
%
\begin{block}{}
$\Omega_n = \Big[\sum_{n_1}D_{{n}{n_1}}|b_{n_1}|^2\Big] =  \mbox{constant}$ - is almost Number of Waves
\end{block}
\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{Numerics (forward)\hspace{80pt} $L_d/L_{nl} = 0.05$} 
%\framesubtitle{subtitle}
\begin{raggedright}
\begin{figure}[H]
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/re_im_q_ini-eps-converted-to.pdf}\\ \vspace{-0.5cm}\small(a)}
\end{minipage}
\hfill
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/abs_q_ini.eps}\\ \vspace{-0.5cm}\small(b)}
\end{minipage}
\caption{ (a) -- Re and Im of the signal, (b) -- $|q|$}
\label{Re_Im_Abs_q_0_pic}
\end{figure}
%
%
\begin{figure}[H]
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/re_im_q_L_split_step.eps}\\ \vspace{-0.5cm}\small(a)}
\end{minipage}
\hfill
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/abs_q_L_split_step.eps}\\ \vspace{-0.5cm}\small(b)}
\end{minipage}
%\caption{(a) -- Re and Im of the signal, (b) -- $|q|$}
\label{Re_Im_Abs_q_L_pic}
\end{figure}
\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{Numerics (backward)} 
%\framesubtitle{subtitle}
\begin{raggedright}
\begin{figure}[H]
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/re_im_q_ini.eps}\\ \vspace{-0.5cm}\small(a)}
\end{minipage}
\hfill
\begin{minipage}[H]{0.4\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/abs_q_ini.eps}\\ \vspace{-0.5cm}\small(b)}
\end{minipage}
\caption{ (a) -- Re and Im of the signal, (b) -- $|q|$}
\label{Re_Im_Abs_q_0_pic}
\end{figure}
%
%
\begin{figure}[H]
\begin{minipage}[H]{0.40\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/re_im_q_ini_re_im_q_transf.eps}\\ \vspace{-0.5cm}\small(a)}
\end{minipage}
\hfill
\begin{minipage}[H]{0.40\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/abs_q_ini_abs_q_transf.eps}\\ \vspace{-0.5cm}\small(b)}
\end{minipage}
%\caption{Рисунок (а) -- вещественная и мнимые части. Рисунок (б) -- модуль}
\label{Re_Im_Abs_q_recover_pic}
\end{figure}
\end{raggedright}
\end{frame}

\begin{frame}
\frametitle{$\Omega_n$} 
%\framesubtitle{subtitle}
\begin{raggedright}
\begin{figure}[H]

\begin{minipage}[H]{0.9\linewidth}
\center{\includegraphics[width=1.0\textwidth]{./PICS/4report_2/pics/lambda.eps}\\ \vspace{-0.5cm}\small(b)}
\end{minipage}
\caption{ $\Omega_n$}
\label{lambda_pic}
\end{figure}
%
%

\end{raggedright}
\end{frame}




\end{document}

\begin{equation}\nonumber
T^{{n}{n_1}}_{{n_2}{n_3}} =
\delta_{{n}-{n_2}}\delta_{{n_1}-{n_3}}
+\delta_{{n}-{n_3}}\delta_{{n_1}-{n_2}} -\delta_{{n}-{n_1}}\delta_{{n_2}-{n_3}}
\end{equation}
