
\documentclass{article}
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%TCIDATA{Created=Thu Oct 17 15:29:05 2013}
%TCIDATA{LastRevised=Thu Oct 17 18:12:38 2013}
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%$R\left( A,x\right) =-EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}}{H^{2}}-f_{0}$
%$f_{0}=1$
%$E=1$
%$H=1$
%$p\left( x\right) =f_{0}\frac{x}{2E}\left( x-H\right) $
%$p_{s}\left( x\right) =-\frac{1}{2}f_{0}\frac{H^{2}}{\pi E}\sin \left( \frac{\pi }{H}x\right) $
%$p_{2}(x)=-f_{0}\frac{H^{2}}{E}\frac{\sqrt{2}}{\pi ^{2}}\sin \left( \frac{\pi }{H}x\right) $
%$p_{5}\left( x\right) =-\frac{1}{2\pi }\sin \left( \frac{\pi }{H}x\right) $
%$p_{6}(x)=-\frac{3}{4\sqrt{\left( 2+\sqrt{2}\right) }\pi }\sin \left( \frac{\pi }{H}x\right) $
%$p_{3}(x)=-\frac{2}{3}\frac{\sqrt{3}}{\pi ^{2}}\sin \left( \frac{\pi }{H}x\right) $
%$p_{4}(x)=-\frac{2}{\sqrt{\left( 2-\sqrt{2}\right) }\pi ^{2}}\sin \left( \frac{\pi }{H}x\right) $
%$p_{c}\left( x\right) =-f_{0}\frac{H^{2}}{E\pi ^{2}}\sin \left( \frac{\pi }{H}x\right) $
%}


\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\section{Collocation Method}

Now, taking $p\left( x\right) =A\sin \left( \frac{\pi }{H}x\right) \in V$
and substitute it in $(S2),$

\begin{equation*}
E\frac{d^{2}}{dx^{2}}\left( A\sin \left( \frac{\pi }{H}x\right) \right)
-f_{0},
\end{equation*}

the residual equation is obtained:

\begin{equation*}
R\left( A,x\right) =-\frac{EA\pi ^{2}}{H^{2}}\left( \sin \frac{\pi }{H}%
x\right) -f_{0}.
\end{equation*}

The collocation method requires that the residual equation of the aproximate
solution be zero at as many points as there are undetermined parameters.
This example has only one parameter $A,$ the point $x=\frac{H}{2}$ is
selected, then $R\left( A,\frac{H}{2}\right) =-\frac{EA\pi ^{2}}{H^{2}}%
\left( \sin \frac{\pi }{H}\frac{H}{2}\right) -f_{0}=0$. Therefore the
solution is : $A=-\frac{H^{2}f_{0}}{E\pi ^{2}}.$

Desarrolo en clase$\bigskip $

$E\frac{d^{2}}{dx^{2}}\left( A\sin \left( \frac{\pi }{H}x\right) \right)
-f_{0}=\allowbreak -EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}}{%
H^{2}}-f_{0}$

$R\left( A,x\right) =-EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}}{%
H^{2}}-f_{0}$\FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{\special{language
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\TEXUX{$A$};valid_file "T";tempfilename 'MUU15S0E.wmf';tempfile-properties
"XPR";}} Grafica del residuo

$R\left( A,\frac{H}{2}\right) =\allowbreak -EA\frac{\pi ^{2}}{H^{2}}-f_{0}=0$

$-EA\frac{\pi ^{2}}{H^{2}}-f_{0}=0$, Solution is : $\left\{ A=-f_{0}\frac{%
H^{2}}{E\pi ^{2}}\right\} $

\bigskip Soluci\'{o}n exacta:

$p\left( x\right) =f_{0}\frac{x}{2E}\left( x-H\right) $

$\frac{x}{2}\left( x-1\right) $

$p_{c}\left( x\right) =-f_{0}\frac{H^{2}}{E\pi ^{2}}\sin \left( \frac{\pi }{H%
}x\right) $

Comparaci\'{o}n entre la soluci\'{o}n exacta y la obtenida por colocaci\'{o}%
n.

$-\frac{1}{\pi ^{2}}\sin \left( \pi x\right) $\FRAME{itbpF}{3in}{2.0003in}{%
0in}{}{}{}{\special{language "Scientific Word";type "MAPLEPLOT";width
3in;height 2.0003in;depth 0in;display "USEDEF";plot_snapshots TRUE;function
\TEXUX{$-\frac{1}{\pi ^{2}}\sin \left( \pi x\right) $};linecolor
"black";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$\frac{x}{2}\left( x-1\right) $};linecolor "red";linestyle
1;linethickness 1;pointstyle "point";xmin "0";xmax "1";xviewmin
"0.000000";xviewmax "1.000000";yviewmin "-0.127500";yviewmax
"0.002550";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle
"normal";xis \TEXUX{x};var1name \TEXUX{$x$};valid_file "T";tempfilename
'MUU15S0B.wmf';tempfile-properties "XPR";}}

\bigskip

$E\frac{d^{2}}{dx^{2}}\left( -f_{0}\frac{H^{2}}{E\pi ^{2}}\sin \left( \frac{%
\pi }{H}x\right) \right) =\allowbreak f_{0}\sin \frac{\pi }{H}x$

$f_{0}\sin \frac{\pi }{H}\frac{H}{2}-f_{0}=\allowbreak 0.$

\bigskip 

Otro punto

$x=\frac{H}{4}$

$R\left( A,\frac{H}{4}\right) =\allowbreak -\frac{1}{2}EA\sqrt{2}\frac{\pi
^{2}}{H^{2}}-f_{0}$

$-\frac{1}{2}EA\sqrt{2}\frac{\pi ^{2}}{H^{2}}-f_{0}=0$, Solution is : $%
\left\{ A=-f_{0}\frac{H^{2}}{E}\frac{\sqrt{2}}{\pi ^{2}}\right\} $

\bigskip $f_{0}=1$

$E=1$

$H=1$

$p_{2}(x)=-f_{0}\frac{H^{2}}{E}\frac{\sqrt{2}}{\pi ^{2}}\sin \left( \frac{%
\pi }{H}x\right) $\FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{\special{language
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"point";function \TEXUX{$p(x)$};linecolor "red";linestyle 1;linethickness
1;pointstyle "point";xmin "0";xmax "1";xviewmin "0.000000";xviewmax
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45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};valid_file "T";tempfilename 'MUU15S0C.wmf';tempfile-properties
"XPR";}} Grafica de la soluci\'{o}n exacta y la del punto $\frac{H}{4}.$

\bigskip 

Otro punto.

$R\left( A,\frac{H}{3}\right) =\allowbreak -\frac{1}{2}A\sqrt{3}\pi ^{2}-1$

\bigskip

$-\frac{1}{2}A\sqrt{3}\pi ^{2}-1=0$, Solution is : $\left\{ A=-\frac{2}{3}%
\frac{\sqrt{3}}{\pi ^{2}}\right\} $

\bigskip

$p_{3}(x)=-\frac{2}{3}\frac{\sqrt{3}}{\pi ^{2}}\sin \left( \frac{\pi }{H}%
x\right) $\FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{\special{language
"Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth
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\TEXUX{$-\frac{2}{3}\frac{\sqrt{3}}{\pi ^{2}}\sin \left( \frac{\pi
}{H}x\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle
"point";function \TEXUX{$p(x)$};linecolor "red";linestyle 1;linethickness
1;pointstyle "point";xmin "0";xmax "1";xviewmin "0.000000";xviewmax
"1.000000";yviewmin "-0.127500";yviewmax "0.002550";rangeset"X";phi 45;theta
45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};valid_file "T";tempfilename 'MUU1BW0H.wmf';tempfile-properties
"XPR";}}Grafica de la soluci\'{o}n exacta y la del punto $\frac{H}{3}.$

\bigskip 

\bigskip

$\bigskip $

$p(x),p_{c}(x),p_{2}(x),p_{3}(x)$\FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{%
\special{language "Scientific Word";type "MAPLEPLOT";width 3in;height
2.0003in;depth 0in;display "USEDEF";plot_snapshots TRUE;function
\TEXUX{$p_{c}(x),p_{2}(x),p_{3}(x)$};linecolor "black";linestyle
1;linethickness 1;pointstyle "point";function \TEXUX{$p(x)$};linecolor
"red";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$p_{4}(x)$};linecolor "blue";linestyle 1;linethickness 1;pointstyle
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4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};valid_file "T";tempfilename 'MUU5JM00.wmf';tempfile-properties
"XPR";}}

\bigskip

$R\left( A,\frac{H}{8}\right) =\allowbreak -\frac{1}{2}A\sqrt{\left( 2-\sqrt{%
2}\right) }\pi ^{2}-1$

\bigskip

$-\frac{1}{2}A\sqrt{\left( 2-\sqrt{2}\right) }\pi ^{2}-1=0$, Solution is : $%
\left\{ A=-\frac{2}{\sqrt{\left( 2-\sqrt{2}\right) }\pi ^{2}}\right\} $

\bigskip

$p_{4}(x)=-\frac{2}{\sqrt{\left( 2-\sqrt{2}\right) }\pi ^{2}}\sin \left( 
\frac{\pi }{H}x\right) $

\bigskip

\section{Subdomain Method}

In this method the integral of the residual equation is aproximate to zero
as many subintervals as thera are undetermined parameters. In this example
there are only one parameter, therefore the complete interval is used:

\begin{equation*}
\int_{0}^{H}R\left( x\right) dx=\int_{0}^{H}\left( -\frac{EA\pi ^{2}}{H^{2}}%
\left( \sin \frac{\pi }{H}x\right) -f_{0}\right) dx=0.
\end{equation*}
Metodo de subdominios.

para el problema de juego

$E\frac{d^{2}}{dx^{2}}\left( p\left( x\right) \right) -f_{0}=0,0<x<H,$

$p\left( 0\right) =p\left( H\right) =0.$

\bigskip

se tiene el residuo

$R\left( A,x\right) =\allowbreak -EA\left( \sin \frac{\pi }{H}x\right) \frac{%
\pi ^{2}}{H^{2}}-f_{0}$

$\int_{0}^{H}\left( -EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}}{%
H^{2}}-f_{0}\right) dx=\allowbreak -\frac{2\pi EA+f_{0}H^{2}}{H}=\allowbreak
-2\pi A-1$

$-\frac{2\pi EA+f_{0}H^{2}}{H}=0$, Solution is : $\left\{ A=-\frac{1}{2}f_{0}%
\frac{H^{2}}{\pi E}\right\} $

$p_{s}\left( x\right) =-\frac{1}{2}f_{0}\frac{H^{2}}{\pi E}\sin \left( \frac{%
\pi }{H}x\right) $

$p\left( x\right) $\FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{\special{language
"Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth
0in;display "USEDEF";plot_snapshots TRUE;function \TEXUX{$p\left( x\right)
$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$p_{s}\left( x\right) $};linecolor "maroon";linestyle 2;linethickness
2;pointstyle "point";function \TEXUX{$p_{2}\left( x\right) $};linecolor
"blue";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$p_{5}(x)$};linecolor "green";linestyle 1;linethickness 1;pointstyle
"point";function \TEXUX{$p_{6}(x)$};linecolor "yellow";linestyle
4;linethickness 1;pointstyle "point";xmin "0";xmax "1";xviewmin "0";xviewmax
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45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};valid_file "T";tempfilename 'MUU2UC00.wmf';tempfile-properties
"XPR";}} Note que se enciman las soluciones $p_{s}$ y $p_{5}.$

$\int_{0}^{H}\left( p\left( x\right) -p_{s}\left( x\right) \right)
dx=\allowbreak 1.\,\allowbreak 798\,8\times 10^{-2}$

$\int_{0}^{H}\left( p\left( x\right) -p_{2}\left( x\right) \right)
dx=\allowbreak 7.\,\allowbreak 887\,8\times 10^{-3}$

\bigskip

Otro subdominio

\bigskip

$\int_{0}^{H/2}\left( -EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}}{%
H^{2}}-f_{0}\right) dx=\allowbreak -\frac{1}{2}-\pi A$

\bigskip

$-\frac{1}{2}-\pi A=0$, Solution is : $\left\{ A=-\frac{1}{2\pi }\right\} $

\bigskip

$p_{5}\left( x\right) =-\frac{1}{2\pi }\sin \left( \frac{\pi }{H}x\right) $

\bigskip

Otro subdominio

\bigskip

$\int_{H/8}^{7H/8}\left( -EA\left( \sin \frac{\pi }{H}x\right) \frac{\pi ^{2}%
}{H^{2}}-f_{0}\right) dx=\allowbreak -\sqrt{\left( 2+\sqrt{2}\right) }\pi A-%
\frac{3}{4}$

\bigskip

$-\sqrt{\left( 2+\sqrt{2}\right) }\pi A-\frac{3}{4}=0$, Solution is : $%
\left\{ A=-\frac{3}{4\sqrt{\left( 2+\sqrt{2}\right) }\pi }\right\} $

\bigskip

$p_{6}(x)=-\frac{3}{4\sqrt{\left( 2+\sqrt{2}\right) }\pi }\sin \left( \frac{%
\pi }{H}x\right) $

\bigskip

$\int_{0}^{H}\left( p\left( x\right) -p_{6}\left( x\right) \right)
dx=\allowbreak -1.\,\allowbreak 081\,4\times 10^{-3}$ y esta es la mejor
soluci\'{o}n hasta el momento.

\end{document}
