Вопрос по Latex

[Ksunitta]

Может тут найдутся, кто сможет помочь..И такой вопрос не сочтется недопустимым правилами
В общем, делаю в латехе таблицу, она получилась очень широкой, не влезала, пришлось поменять ориентирование листа на ландшафтный, то бишь сейчас документ выглядит

так

Собственно,проблема в нумерации страницы, надо, чтобы номер страницы был внизу,а не сбоку да еще и повернут, то есть

так

Пробовала через lscape, но как-то не получается..Честно гуглила - пусто.

Помогите,пожалуйста,или отправьте туда,где помогут.

@темы: Программы

Comments

[en_trope]

Как насчет показать код?

[Ksunitta]

en_trope, прошу прощения, только сейчас доползла до инета

\documentclass[twoside,12pt]{report}
 %\documentstyle[twoside,11pt]{report}
 \textwidth 14.5cm
\textheight 21cm \setlength{\evensidemargin}{1.3cm}

\usepackage{graphicx}
\usepackage{amssymb, amsmath, amsxtra}
\usepackage{epsfig,subfigure}
\usepackage{pdflscape}
\usepackage{latexsym}
\usepackage{bm}
\usepackage{wrapfig}
\usepackage{float,times}
\usepackage{fancyhdr}
\usepackage{lscape}
%\usepackage{amsmath}
                       %  \pagestyle{plain}
\usepackage{chapterbib}

\usepackage{longtable}
\usepackage{oldgerm}

\usepackage{mathrsfs}


 \usepackage{ntheorem}
           \theoremstyle{plain}
                      {\theorembodyfont{\rmfamily}
                      \theoremseparator{.}
                       \newtheorem{exa}{Example}[section]
           \newtheorem{thm}{Theorem}[section]
           \theoremstyle{plain}
           \newtheorem{cor}{Corollary}[section]
           \theoremstyle{plain} \newtheorem{lem}{Lemma}[section]
           \theoremstyle{plain}
           \newtheorem{defn}{Definition}[section]
            \newtheorem{exer}{Exercise}[section]
           \theoremstyle{plain}
           \newtheorem{rem}{Remark}[section]
            \newtheorem{prop}{Proposition}[section]}


\def\R{\mbox{I$\!$R}}
\setcounter{secnumdepth}{3}
%\addtocounter{secnumdepth}{1}
\def\thepart{\Roman{part}}
\def\thechapter{\arabic{chapter}}
\def\thesection{\arabic{section}}
\def\thesubsectio n{\arabic{section}.\arabic{subsection}}
\def\thesubsubsection{\arabic{section}.\arabic{subsection}.\arabic{subsubsection}}
% \def\theparagraph{\S~\arabic{paragraph}}

 %\def\theequation{\arabic{section}.\arabic{equation}}
\def\thefigure{\arabic{chapter}.\arabic{figure}}






\begin{document}
\large

%и так далее текст 
%=====================================================================================================================

\section{Conservation laws of the homogeneous Monge-Amp\`{e}re equation}

Let us consider the homogeneous Monge-Amp\`{e}re equation (\ref{kse}), where $u=u(x,y)$. 
It admits the following operators (see [2]):
$$
\begin{aligned}
&X_1=\frac{ \partial}{ \partial x},
\hspace{3em}X_2=\frac{ \partial}{ \partial y},
\hspace{2.1em}X_3=\frac{ \partial}{ \partial u},
\hspace{2.6em}X_4=x\frac{ \partial}{ \partial x},\\
&X_5=y\frac{ \partial}{ \partial y},
\hspace{2.6em}X_6=u\frac{ \partial}{ \partial u},
\hspace{1.5em}X_7=y\frac{ \partial}{ \partial x},
\hspace{2em}X_8=x\frac{ \partial}{ \partial y},\\
&X_9=x\frac{ \partial}{ \partial u},
\hspace{2.6em}X_{10}=y\frac{ \partial}{ \partial u},
\hspace{1.1em}X_{11}=u\frac{ \partial}{ \partial x},
\hspace{1.7em}X_{12}=u\frac{ \partial}{ \partial y},\\
&X_{13}=x^2\frac{ \partial}{ \partial x}+xy\frac{ \partial}{ \partial y}+xu\frac{ \partial}{ \partial u},
\hspace{1em}X_{14}=xy\frac{ \partial}{ \partial x}+y^2 \frac{\partial}{\partial y}+yu\frac{\partial}{\partial u},\\
&\hspace{7em}X_{15}=xu\frac{\partial}{\partial x}+yu\frac{\partial}{\partial y}+u^2\frac{\partial}{\partial u}
\end{aligned}
$$
We use algorithm from [1].
Using the formal Lagrangian 
$$
L=vF(x,u,u_{(1)},...,u_{(s)}),
$$
we conctruct the adjoint equation for system (\ref{kse})
$$
F^*=\displaystyle\frac{\delta L}{\delta u},
$$
where $\displaystyle\frac{\delta}{\delta u} = \frac{\partial}{\partial u}-D_i\frac{\partial}{\partial u_i}+D_iD_j\frac{\partial}{\partial u_{ij}}-\ldots$ - is the variational differentiations.
In our case the adjoint equation  has the form
$$
F^*=2v_{xy}u_{xy}-v_{yy}u_{xx}-v_{xx}u_{yy}.
$$
Let find a function $\varphi(x,y,u)$ such that after the substitution $v=\varphi(x,y,u)$
we obtain
\begin{equation}
\label{kseFZ}
F^* \mid_{v=\varphi(x,y,u)}=\lambda(u_{xy}^2-u_{xx}u_{yy}).
\end{equation}
Splitting this equation by variables we obtain the next system
$$
\begin{aligned}
&-2\phi_u=-\lambda,\\
& \quad2\phi_{yu}=0,\\
&\quad 2\phi_{xu}=0,\\
& \quad2\phi_{xy}=0,\\
& \quad2\phi_{yy}=0,\\
&\quad 2\phi_{xx}=0,\\
\end{aligned}
$$
whence it follows $\varphi(x,y,u)=A_1u+A_2y+A_3x+A_4$, where  $A_1,A_2,A_3,A_4- const.$

Let us consider the operator
$$
X_1=\displaystyle\frac{\partial}{\partial x}.
$$
The formal  Lagrangian  has the form 
$$
L=\displaystyle\frac{v}{2}u_{xy}^2+\frac{v}{2}u_{yx}^2-vu_{xx}u_{yy}.
$$
The operator  is written to the Lie-B\"{a}clund form: 
$$
X=\xi^i\displaystyle\frac{\partial}{\partial  x_i}+\eta \frac{\partial}{\partial  u}, i=1,...,n\quad  \rightarrow \quad X_{L_{i}}=W_i \frac{\partial}{\partial  u}, W_i=\eta-\xi^ju_j,j=1,...,n.\\
$$
In our case we obtain 
$$ 
X_{L_{1}}=-u_x\displaystyle\frac{\partial}{\partial u}, \quad W=-u_x.\\
$$
Then, the conservation laws have the form
$$
D_i(C^i)\mid_{(\ref{kse})}=0,
$$
where $C^i=W^\alpha \left[ \frac{ \partial L}{ \partial u^\alpha_i}-D_j(\frac{ \partial L}{ \partial u^\alpha_{ij}})+D_iD_k(\frac{ \partial L}{ \partial u^\alpha_{ijk}})-...\right] \\
 \phantom{\frac{ \partial L}{ \partial u^\alpha_i}-D_j(\frac{ \partial L}{ \partial u^\alpha_{ij}})+D_iD_k(\frac{ \partial L}{ \partial u^\alpha_{ijk}})} +D_j(W^\alpha) \left[   \frac{ \partial L}{ \partial u^\alpha_{ij}}-D_k( \frac{ \partial L}{ \partial u^\alpha_{ijk}})+...\right]+D_jD_k(W^\alpha)\left[  \frac{ \partial L}{ \partial u^\alpha_{ijk}}\right]$.\\
For our case the conservation vector has the form:
$$
\begin{aligned}
&C^1=A_3u_{xy}u_y+3A_1u_{xy}u_xu_y,\\
&C^2=-A_3u_{xx}u_y-3A_1u_xu_{xx}u_y,\\  
\end{aligned}
$$
and the conservation law is written  in the form
$$
D_x\left(C^1\right)+D_y\left(C^2\right)=0
$$.

Thus the all conservation laws for the operators $X_1,\dots,X_{15}$ have been derived. Result see in Table 1.
\begin{landscape}
%\parbox{10.3\textwidth}
%\rotatebox{90}{
\begin{longtable}{|l|l|}
\multicolumn{2}{l}{\hspace{40em}Table 1.}\\
\hline
Operators	&Conservation vectors\\
\hline 
$X_1$  	&$C^{\strut1}=A_3u_{xy}u_y+3A_1u_{xy}u_xu_y$,\\
	  	& $C^2=-A_3u_{xx}u_y-3A_1u_xu_{xx}u_y$\\  
\hline
$X_2$ 	&$ C^{\strut1}=-A_2u_xu_{yy}-3A_1u_yu_xu_{yy},$\\
		&$ C^2=A_2u_xu_{xy}+3A_1u_yu_xu_{xy}$\\
\hline
$X_3$		&$C^{\strut1}=-2A_1u_{xy}u_y,$\\
		&$C^2=2A_1u_yu_{xx}$\\ 
\hline
$X_4$		&$C^1=A_3xu_{xy}u_y-\displaystyle\frac{ \strut 3}{ \strut 2}A_1xu_{x}^2u_{yy}+(A_1u+A_2y+A_3x+A_4)u_xu_{yy}$,\\ 
	  	&$C^2=-A_3u_xu_y-A_3xu_{xx}u_y+\displaystyle\frac{ \strut 1}{ \strut 2}A_2u_x^2+\frac{ \strut 1}{ \strut 2}A_1u_x^2u_y+\frac{ \strut 3}{ \strut 2}A_1xu_x^2u_{xy}-(A_1u+A_2y+A_3x+A_4)u_xu_{xy}$\\	
\hline
$X_5$		&$ C^1=\displaystyle\frac{1}{2}A_3u_y^2-A_2yu_{yy}u_x+\frac{\strut3}{\strut2}A_1u_y^2u_x+\frac{\strut3}{\strut2}A_1yu_y^2u_{xy}+(A_1u+A_2y+A_3x+A_4)u_{yy}u_x$\\
		&$C^2=A_2yu_{xy}u_x-\displaystyle\frac{\strut3}{\strut2}A_1yu_y^2u_{xx}-A_3u_yu_x-A_1u_x^2u_y-(A_1u+A_2y+A_3x+A_4)u_{xy}u_x $\\
		
\hline
$X_6$		&$C^{\strut1}=-A_3u_y^2+2A_2u_yu_x-2A_1uu_{xy}u_y+2(A_1u+A_2y+A_3x+A_4)u_{xy}u_y$\\
		&$C^2=-A_2u_x^2+2A_1uu_yu_{xx}-2(A_1u+A_2y+A_3x+A_4)u_{xx}u_y $\\
\hline
$X_7$		&$C^1=A_3u_xu_y+A_3yu_{xy}u_y-\displaystyle\frac{\strut3}{2}A_1yu_x^2u_{yy},$\\
\hline
\newpage
\multicolumn{2}{l}{\hspace{35em}Table 1. Extension}\\
\hline
		&$C^2=-A_3yu_{xx}u_y+\displaystyle\frac{\strut3}{2} A_1yu_x^2u_{xy}-\frac{\strut1}{2}A_3u_x^2-\frac{\strut1}{2}A_1u_x^3$\\
%\newpage
\hline
$X_8$ 	&$C^1=-A_2xu_{yy}u_x+\displaystyle\frac{\strut3}{\strut2}A_1xu_y^2u_{xy}-\frac{\strut1}{\strut2}A_2u_y^2-\frac{\strut1}{\strut2}A_1u_y^3,$\\
		&$C^2=A_2u_yu_x+A_2xu_{xy}u_x-\displaystyle\frac{3}{2}A_1xu_y^2u_{xx}$\\
\hline
$X_9$		&$ C^{\strut1}=-2A_1xu_{xy}u_y+A_2u_y+A_1u_y^{2},$\\
		&$C^2=-A_2u_x+2A_1xu_{xx}u_y$\\
\hline
%и так далее еще куча строк в таблице
\end{longtable}
\end{landscape}

\begin{thebibliography}{99}
%------------------------------------
\bibitem{ibr1011} 
N.H.Ibragimov. Nonlinear self-adjointness in constructing conservation laws // Archivas of ALGA. ALGA Publications. 2010-2011. -- V. 7-8. -- P. 1-99.

\bibitem{bgihand} 
Handbook of Lie Group Analysis of Differential Equations // Edited by N.H.Ibragimov. CRC Press, Boca Raton, Florida, 1996. -- Vol. 1. -- 380 P. 

\end{thebibliography}

\end{document}

Ну вот как-то так. Половину ненужного убрала, но суть осталась. Здесь lscape убран, что-то не найду тот вариант.Видимо, как неполучившйся, удалила.

[en_trope]

Думаю, в пределах одного документа так не сделать. Может разбить на два?
Из первого выкинуть таблицы и перед библиографией поставить \setcounter{page}{6}
Второй:

\documentclass[twoside,12pt]{report}
 %\documentstyle[twoside,11pt]{report}


% \textwidth 14.5cm
%\textheight 21cm \setlength{\evensidemargin}{1.3cm}

\usepackage{graphicx}
\usepackage{amssymb, amsmath, amsxtra}
\usepackage{epsfig,subfigure}
\usepackage{pdflscape}
\usepackage{latexsym}
\usepackage{bm}
\usepackage{wrapfig}
\usepackage{float,times}
\usepackage{fancyhdr}
\usepackage{vmargin}
%\usepackage{amsmath}
                       %  \pagestyle{plain}
\usepackage{chapterbib}

\usepackage{longtable}
\usepackage{oldgerm}

\usepackage{mathrsfs}


 \usepackage{ntheorem}
           \theoremstyle{plain}
                      {\theorembodyfont{\rmfamily}
                      \theoremseparator{.}
                       \newtheorem{exa}{Example}[section]
           \newtheorem{thm}{Theorem}[section]
           \theoremstyle{plain}
           \newtheorem{cor}{Corollary}[section]
           \theoremstyle{plain} \newtheorem{lem}{Lemma}[section]
           \theoremstyle{plain}
           \newtheorem{defn}{Definition}[section]
            \newtheorem{exer}{Exercise}[section]
           \theoremstyle{plain}
           \newtheorem{rem}{Remark}[section]
            \newtheorem{prop}{Proposition}[section]}

\usepackage[papersize={29.7 cm,21 cm}]{geometry}
\geometry{left=4cm}
\geometry{right=4cm}
\geometry{top=4cm}
\geometry{bottom=1cm}

\def\R{\mbox{I$\!$R}}
\setcounter{secnumdepth}{3}
%\addtocounter{secnumdepth}{1}
\def\thepart{\Roman{part}}
\def\thechapter{\arabic{chapter}}
\def\thesection{\arabic{section}}
\def\thesubsectio n{\arabic{section}.\arabic{subsection}}
\def\thesubsubsection{\arabic{section}.\arabic{subsection}.\arabic{subsubsection}}
% \def\theparagraph{\S~\arabic{paragraph}}

 %\def\theequation{\arabic{section}.\arabic{equation}}
\def\thefigure{\arabic{chapter}.\arabic{figure}}
\begin{document}
\setcounter{page}{4}
%\parbox{10.3\textwidth}
%\rotatebox{90}{

\begin{longtable}{|l|l|}
\multicolumn{2}{l}{\hspace{40em}Table 1.}\\
\hline
Operators	&Conservation vectors\\
\hline 
$X_1$  	&$C^{\strut1}=A_3u_{xy}u_y+3A_1u_{xy}u_xu_y$,\\
	  	& $C^2=-A_3u_{xx}u_y-3A_1u_xu_{xx}u_y$\\  
\hline
$X_2$ 	&$ C^{\strut1}=-A_2u_xu_{yy}-3A_1u_yu_xu_{yy},$\\
		&$ C^2=A_2u_xu_{xy}+3A_1u_yu_xu_{xy}$\\
\hline
$X_3$		&$C^{\strut1}=-2A_1u_{xy}u_y,$\\
		&$C^2=2A_1u_yu_{xx}$\\ 
\hline
$X_4$		&$C^1=A_3xu_{xy}u_y-\displaystyle\frac{ \strut 3}{ \strut 2}A_1xu_{x}^2u_{yy}+(A_1u+A_2y+A_3x+A_4)u_xu_{yy}$,\\ 
	  	&$C^2=-A_3u_xu_y-A_3xu_{xx}u_y+\displaystyle\frac{ \strut 1}{ \strut 2}A_2u_x^2+\frac{ \strut 1}{ \strut 2}A_1u_x^2u_y+\frac{ \strut 3}{ \strut 2}A_1xu_x^2u_{xy}-(A_1u+A_2y+A_3x+A_4)u_xu_{xy}$\\	
\hline
$X_5$		&$ C^1=\displaystyle\frac{1}{2}A_3u_y^2-A_2yu_{yy}u_x+\frac{\strut3}{\strut2}A_1u_y^2u_x+\frac{\strut3}{\strut2}A_1yu_y^2u_{xy}+(A_1u+A_2y+A_3x+A_4)u_{yy}u_x$\\
		&$C^2=A_2yu_{xy}u_x-\displaystyle\frac{\strut3}{\strut2}A_1yu_y^2u_{xx}-A_3u_yu_x-A_1u_x^2u_y-(A_1u+A_2y+A_3x+A_4)u_{xy}u_x $\\
		
\hline
$X_6$		&$C^{\strut1}=-A_3u_y^2+2A_2u_yu_x-2A_1uu_{xy}u_y+2(A_1u+A_2y+A_3x+A_4)u_{xy}u_y$\\
		&$C^2=-A_2u_x^2+2A_1uu_yu_{xx}-2(A_1u+A_2y+A_3x+A_4)u_{xx}u_y $\\
\hline
$X_7$		&$C^1=A_3u_xu_y+A_3yu_{xy}u_y-\displaystyle\frac{\strut3}{2}A_1yu_x^2u_{yy},$\\
\hline
\newpage

\multicolumn{2}{l}{\hspace{35em}Table 1. Extension}\\
\hline
		&$C^2=-A_3yu_{xx}u_y+\displaystyle\frac{\strut3}{2} A_1yu_x^2u_{xy}-\frac{\strut1}{2}A_3u_x^2-\frac{\strut1}{2}A_1u_x^3$\\
%\newpage
\hline
$X_8$ 	&$C^1=-A_2xu_{yy}u_x+\displaystyle\frac{\strut3}{\strut2}A_1xu_y^2u_{xy}-\frac{\strut1}{\strut2}A_2u_y^2-\frac{\strut1}{\strut2}A_1u_y^3,$\\
		&$C^2=A_2u_yu_x+A_2xu_{xy}u_x-\displaystyle\frac{3}{2}A_1xu_y^2u_{xx}$\\
\hline
$X_9$		&$ C^{\strut1}=-2A_1xu_{xy}u_y+A_2u_y+A_1u_y^{2},$\\
		&$C^2=-A_2u_x+2A_1xu_{xx}u_y$\\
\hline
%и так далее еще куча строк в таблице
\end{longtable}

\end{document}

Но тут опять же не по центру... Даже не знаю, если честно, как сделать красиво.

[Ksunitta]

en_trope, ооо....По сравнению с тем,что я ктр день бьюсь - уже сдвиг.
Правда, я,честно говоря, не знаю, пойдет ли разбиение на 2 документа, т.к. для статьи в журнал пишу,у них там требований куча. Хотя, как научрук скажет,конечно.

Но все равно ОГРОМНОЕ спасибо вам, уже результат есть, над центрированием я поработаю.