\documentclass[12pt,a4paper]{article}
\usepackage[utf8x]{inputenc}
\usepackage[brazil]{babel}
\usepackage{ucs}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{makeidx}
\usepackage[pdftex]{hyperref}
\usepackage{graphicx}
\usepackage{color} % Pacote de cores do Latex
\usepackage[usenames,dvipsnames]{xcolor} %Pacote com 68 variaçoes de cores do Latex
% Definiçao de Cores
% \definecolor{''nome''}{''model (RGB ou rgb)''}{''codigo do espectro da cor''}
\definecolor{orange}{rgb}{1,0.5,0}
\definecolor{light-green-blue}{RGB}{220,255,255}
\begin{document}
\pagecolor{light-green-blue}
\colorbox{white}{Dada a \textcolor{red}{EDO:}}
\[
\dfrac{d y}{d x} = f(x,y)
\]
\colorbox{white}{RK de {\color{Blue} ordem m:}}
\[
y_{i+1} = y _{i} + h \phi (x_{i},y_{i},h) + O(h^{m+1})
\]
\section{ \colorbox{white}{Controle do {\color{orange} Passo RK}} }
Erro local de truncamento:
\[
{\color{Bittersweet} e_{a} = \frac{h^{m+1}}{(m+1)!} f^{m}(x_{i},y^{i}) = k *h^{m+1}}
\]
Em que $e_{a}$ e o valor aproximado
\[
{\color{Fuchsia} e_{t} = kh^{m+1} + O(h^{m+2})}
\]
\[
{\color{Brown}y^{*}_{n+1}\textit{ : Soluçao Exata em }[x_{n},x_{n+1}]}
\]
\begin{equation}
{\color {OrangeRed} y^{*}_{n+1} - y _{n+1,1} \cong Kh_{1}^{m+1}*\frac{(x_{n+1}-x_{n})}{h_{1}} }
\label{eq:1}
\end{equation}
\begin{equation}
{\color {Red} y^{*}_{n+1} - y_{n+1,2} \cong k h_{2}^{m+1} \frac{x_{n+1}-x_{n}}{h_{2}} }
\label{eq:2}
\end{equation}
Dividindo a equaçao \ref{eq:1} pela equaçao \ref{eq:2}:
\[
{\color{Maroon} \frac{y^{*}_{n+1} - y _{n+1,1}}{y^{*}_{n+1} - y_{n+1,2}} = \frac{h_{1}^{m+1}}{h_{2}^{m+1}}*\frac{h_{2}}{h_{1}} }
\]
\[
{\color{Magenta} y^{*}_{n+1} \cong \frac{y_{n+1} - y _{n+1,2}(h_{1}/h_{2})^{m}}{1-(h_{1}/h_{2})^{m}}}
\]
Escolhendo $h_{2} = h_{1}/2$
\begin{equation}
{\color{MidnightBlue} y^{*}_{n+1} \cong \frac{y_{n+1,1}-2^{m}y_{n+1,2}}{1-2^{m}} }
\label{eq:3}
\end{equation}
\[
x_{n+1} - x_{n} = h_{1}
\]
Substituindo a equaçao \ref{eq:3} na equaçao 1:
\begin{equation}
{\color {RawSienna} e_{a} = k h_{1}^{m+1} = \frac{2^{m}(y_{n+1,2}-y_{n+1,1})}{2^{m}-1} }
\label{eq:4}
\end{equation}
Acho K (equaçao \ref{eq:4}), $y*_{n+1}$ (\ref{eq:3})
\[
E^{\%} = \frac{k h^{m}}{y^{*}_{n+1}}
\]
RK (m =4)
\[
e_{a} = kh_{1}^{5} = \frac{16}{15}(y_{n+1,2}-y_{n+1,1})
\]
\section{Exemplo} Resolver o sistema de EDO's
\[
\dfrac{d ^{2} y}{d t^{2}} = \frac{1}{2}x + y + \sin(\frac{1}{2}xt)-2 \frac{d y}{d t}
\]
\[
\dfrac{d^{2} x}{d t^{2}} = 2 y^{2} - x + \frac{d y}{d t} - 3 \frac{d x}{d t}
\]
Condiçoes iniciais:
\[
x(0)=1,y(0)=2, x'(0)=0, y'(0)= \frac{1}{2}
\]
Utilizar passo k = 0,1 e calcular x e y para t = 0,1 (1o Passo)
a) M. de Euler
b) M de R.K de 4a ordem
\end{document}
Não há comentários.
Comentar