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\documentclass{article}  \usepackage[affil-it]{authblk}  \usepackage{graphicx}  \usepackage[space]{grffile}  \usepackage{latexsym}  \usepackage{textcomp}  \usepackage{longtable}  \usepackage{multirow,booktabs}  \usepackage{amsfonts,amsmath,amssymb}  \usepackage{natbib}  \usepackage{url}  \usepackage{hyperref}  \hypersetup{colorlinks=false,pdfborder={0 0 0}}  % You can conditionalize code for latexml or normal latex using this.  \newif\iflatexml\latexmlfalse  \usepackage[utf8]{inputenc}  \usepackage[T2A]{fontenc}  \usepackage[russian,english]{babel}  \begin{document}  \title{Вывод уравнения вихря}  \author{vladimir onoprienko}  \affil{Affiliation not available}  \date{\today}  \bibliographystyle{plain}  \maketitle  \selectlanguage{russian}Уравнения движения и уравнение неразрывности:  \selectlanguage{english}  \begin{align*}  \overset{(1)}{ \frac{\partial u}{\partial t}}  + \overset{(2)}{u\frac{\partial u}{\partial x}}  + \overset{(3)}{v\frac{\partial u}{\partial y}}  - \overset{(4)}{K\left(\frac{{\partial }^2u}{\partial x^2}+\frac{{\partial }^2u}{\partial y^2}\right)}  &= 0 \\  %newline  \frac{\partial v}{\partial t }+ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}-K\left(\frac{{\partial }^2v}{\partial x^2}+\frac{{\partial }^2v}{\partial y^2}\right)&=0 \\  %newline  \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} &= 0  \end{align*}  \selectlanguage{russian}На основании уравнения неразрывности можно ввести потенциал \selectlanguage{english}$\psi$, \selectlanguage{russian}так что \selectlanguage{english}$$u=-\frac{\partial \psi}{\partial y}, v=\frac{\partial \psi}{\partial x}$$.  \selectlanguage{russian}Применим к \selectlanguage{english}$x$ \selectlanguage{russian}уравнению операцию \selectlanguage{english}$\partial / \partial y$, \selectlanguage{russian}а к \selectlanguage{english}$y$ \selectlanguage{russian}уравнению - \selectlanguage{english}$\partial / \partial y$, \selectlanguage{russian}затем результаты вычтем один из другого. Это будет аналогично взятию \selectlanguage{english}$z$-\selectlanguage{russian}компоненты ротора. Потому есть подозрение, что данная операция эквивалентна применению приближения мелкой воды и гидростатики к уравнению движения в форме Громеки-Лэмба  \selectlanguage{english}  \begin{align*}  (1)\;  \frac{\partial}{\partial y}: \;  \frac{\partial u}{\partial t}   \to  \frac{\partial}{\partial y} \left( \frac{\partial u}{\partial t}\right)   &= \frac{\partial}{\partial y}\left[ \frac{\partial}{\partial t} \left( -\frac{\partial \psi}{\partial y} \right) \right]   = - \frac{\partial^2 \psi}{\partial y^2} \\  %newline  \frac{\partial}{\partial x}: \;  \frac{\partial v}{\partial t}  \to   \frac{\partial}{\partial x} \left( \frac{\partial v}{\partial t}\right)   &= \frac{\partial}{\partial x}\left[ \frac{\partial}{\partial t} \left( \frac{\partial \psi}{\partial x} \right) \right]   = \frac{\partial^2 \psi}{\partial x^2}  \end{align*}  \begin{align*}  (2)\;  % d/dy  \frac{\partial}{\partial y}: \;  % u*u_x  u \frac{\partial u}{\partial x}   \to  \frac{\partial}{\partial y} \left( u \frac{\partial u}{\partial x}\right)   = \frac{\partial u}{\partial y} \cdot \frac{\partial u}{\partial x} + u \frac{\partial^2 u}{\partial x \partial y}   &= \frac{\partial}{\partial y}\left[ \left( - \frac{\partial \psi}{\partial y} \right) \cdot \frac{\partial}{\partial x} \left( -\frac{\partial \psi}{\partial y} \right) \right] = \\  %newline   &= \frac{\partial^2 \psi}{\partial y^2} \cdot \frac{\partial^2 \psi}{\partial x \partial y }   + \frac{\partial \psi}{\partial y } \cdot \frac{\partial^3 \psi}{\partial x \partial y^2}\\  %newline  % d/dx  \frac{\partial}{\partial x}: \;  % u*v_x  u \frac{\partial v}{\partial x}   \to  \frac{\partial}{\partial x} \left( u \frac{\partial v}{\partial x}\right)   = \frac{\partial u}{\partial x} \cdot \frac{\partial v}{\partial x} + u \frac{\partial^2 v}{\partial x^2}   &= \frac{\partial}{\partial x}\left[ \left(- \frac{\partial \psi}{\partial y} \right) \cdot \frac{\partial}{\partial x} \left( \frac{\partial \psi}{\partial x} \right) \right] = \\  %newline   &= - \frac{\partial^2 \psi}{\partial x \partial y} \cdot \frac{\partial^2 \psi}{\partial x^2 }   - \frac{\partial \psi}{\partial y } \cdot \frac{\partial^3 \psi}{\partial x^3 }\\  \end{align*}  \begin{align*}  (3)\;  % d/dy  \frac{\partial}{\partial y}: \;  % v*u_y  v \frac{\partial u}{\partial y}   \to  \frac{\partial}{\partial y} \left( v \frac{\partial u}{\partial y}\right)   = \frac{\partial v}{\partial y} \cdot \frac{\partial u}{\partial y} + v \frac{\partial^2 u}{\partial y^2}   &= \frac{\partial}{\partial y} \left[ \frac{\partial \psi}{\partial x} \cdot \frac{\partial}{\partial y} \left( -\frac{\partial \psi}{\partial y} \right) \right] = \\  %newline   &=-\frac{\partial^2 \psi}{\partial x \partial y } \cdot \frac{\partial^2 \psi}{\partial y^2 }   + -\frac{\partial \psi}{\partial x } \cdot \frac{\partial^3 \psi}{\partial y^3 }\\  %newline  % d/dx  \frac{\partial}{\partial x}: \;  % v*v_y  v \frac{\partial v}{\partial y}   \to  \frac{\partial}{\partial x} \left( v \frac{\partial v}{\partial y}\right)   = \frac{\partial v}{\partial x} \cdot \frac{\partial v}{\partial y} + v \frac{\partial^2 v}{\partial x \partial y}   &= \frac{\partial}{\partial x}\left[ \frac{\partial \psi}{\partial x} \cdot \frac{\partial}{\partial y} \left( \frac{\partial \psi}{\partial x} \right) \right] = \\  %newline   &= \frac{\partial^2 \psi}{\partial x^2 } \cdot \frac{\partial^2 \psi}{\partial x \partial y }   - \frac{\partial \psi}{\partial x } \cdot \frac{\partial^3 \psi}{\partial x^2 \partial y }\\  \end{align*}  \end{document}