tag:blogger.com,1999:blog-38595409451779444002024-03-05T10:43:03.985-08:00Tópicos de Física ComputacionalAplicações da Linguagem de programação C/C++, e de software livre, tipo: gnuplot, scilab, lazarus e outros, em problemas da Física e da Matemática.Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.comBlogger25125tag:blogger.com,1999:blog-3859540945177944400.post-1564641828095626922019-01-01T11:52:00.001-08:002019-01-01T11:54:27.809-08:00Dinâmica - Máquina de AtwoodCaros leitores, uma pequena animação da Máquina de Atwood, assunto estudado em Dinâmica, é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, circle, polygon, arrow, e label) são implementados facilmente.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-31297558343766620842019-01-01T11:49:00.001-08:002019-01-01T11:49:32.184-08:00Calorimetria - Transição de Fases da águaCaros leitores, uma pequena animação do fenômeno de Transição de Fases da água, estudado em Calorimetria, é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, circle, polygon, e label) são implementados facilmente.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-38108391814995957592019-01-01T11:46:00.001-08:002019-01-01T11:46:22.623-08:00Conservação de Energia Caros leitores, uma pequena animação do princípio da Conservação da Energia é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, circle, polygon, e label) são implementados facilmente.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-62451654256457594612017-05-13T14:16:00.001-07:002019-01-01T11:44:18.256-08:00Espectrômetro de Massa<span style="background-color: white;"><b>Espectrômetro de Massa</b></span><br />
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Caros
leitores, uma pequena animação do Espectrômetro de Massa é aqui construída segundo os comandos
básicos do aplicativo gnuplot. Os objetos presentes na animação (rect,
circle, polygon, arrow, e label) são implementados facilmente.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-8709047407020913762017-05-13T14:08:00.001-07:002019-01-01T11:42:19.542-08:00Selecionador de Velocidades - Aplicação da Força de Lorentz<b>Selecionador de Velocidades</b><br />
<br />
Caros
leitores, uma pequena animação do Selecionador de Velocidades como aplicação da Força de Lorentz é aqui construída segundo os comandos
básicos do aplicativo gnuplot. Os objetos presentes na animação (rect,
polygon, circle, arrow, e label) são implementados facilmente.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-53202919192537235052016-05-09T23:19:00.000-07:002019-01-01T11:37:34.961-08:00Movimento Harmônico Simples - Sistema Massa - Mola<span style="background-color: white;"><b>Movimento Harmônico Simples: Sistema massa - mola</b></span><br />
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Caros leitores, uma pequena animação do sistema massa-mola e os correspondentes gráficos das funções horárias da posição, velocidade, aceleração e das energias potencial e cinética é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, circle, arrow, e label) são implementados facilmente. As equações físicas envolvidas na animação são,<br />
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1. A função horária do movimento: $$x(t)=x_m cos(wt + \phi_o)$$<br />
onde $w$ é a frequência angular, obtida por: $$w=\sqrt{\frac{k}{m}}$$<br />
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2. O período $T$, determinado por $$T=\frac{2\pi}{w}=2\pi\sqrt{\frac{m}{k}}$$<br />
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3. A força restauradora, segundo a Lei de Hooke, definida por: $$F(x)=-k.x$$<br />
onde $k$ é a constante elástica da mola, medida em $N/m$, e por último,<br />
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4. As expressões das energias mecânica, potencial e cinética, respectivamente, determinadas por:<br />
$$E_{T}=\frac{1}{2}kx^{2}_{m}, \quad \quad U(x)=\frac{1}{2}kx^2 \quad \quad \mbox{e} \quad \quad K(x)=E_{T}-U(x)$$.<br />
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O objeto mola foi construído segundo as equações paramétricas da hélice circular reta, a saber, $$x_{1}(t)=a\cos(wt), \quad \quad y_{1}(t)=a\sin(wt) \quad \quad \mbox{e} \quad \quad h(t)=bt $$<br />
onde $a$ e $b$ são parâmetros que determinam o raio e o espaçamento da hélice.<br />
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No aplicativo do gnuplot é possível alterar todos os parâmetros de entrada facilmente, até os ângulos de visão do sistema massa-mola (como é mostrado em algumas imagens), ou seja, o aplicativo é bem interativo. </div>
Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-86163807936838985912016-04-05T10:20:00.002-07:002016-04-05T10:20:36.378-07:00Animações em Física com gnuplot: Satélite em ÓrbitaMOVIMENTO DE UM SATÉLITE EM ÓRBITA<br />
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Caros leitores, uma pequena animação do movimento de um satélite em órbita em torno da Terra (não estacionário) é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (circle, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-72321831771244491952016-04-05T09:48:00.000-07:002016-04-05T09:49:34.764-07:00Animações em Física com gnuplot: Movimento de Queda Livre<span style="color: blue;"><b>MOVIMENTO M.R.U.V - QUEDA LIVRE</b></span><br />
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Caros leitores, uma ilustração simples do movimento retilíneo uniformemente variado ao longo do eixo -y, denominado de Queda Livre e os correspondentes gráficos deste movimento M.R.U.V. é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (polygon, circle, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-86103749778661281462016-04-05T09:44:00.001-07:002019-01-01T11:34:05.040-08:00Animações em Física com gnuplot: Movimento Retilíneo Uniformemente Variado<span style="color: blue;"><b>MOVIMENTO RETILÍNEO UNIFORMEMENTE VARIADO (M.R.U.V)</b></span><br />
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<span style="color: blue;"><b><br /></b></span>Caros leitores, uma ilustração simples do movimento retilíneo uniformemente variado e os correspondentes gráficos do M.R.U.V. é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (polygon, circle, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.</div>
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-51716706858154928262016-03-12T12:04:00.000-08:002019-01-01T11:42:43.419-08:00Animações em Física com gnuplot: Lançamento de um foguete inicialmente aceleradoCaros leitores, uma pequena animação da cinemática do lançamento de um foguete inicialmente acelerado e os correspondentes gráficos das velocidades Vx(t), Vy(t) e V(t) é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rectangle, arrow, circle e label) são implementados facilmente. As equações físicas envolvidas na animação são,<br />
<br />
1. Primeiro Estágio: Acelerado por um intervalo de tempo, $\Delta t = 30\, s$, com aceleração de $a = 46\, m/s^{2}$, sob o ângulo de inclinação $\theta = 70^{o}$:<br />
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1.1 As funções horárias do movimento: $$X_{1}(t)=x_{o}+v_{ox}t + \frac{1}{2} a_{x} t^{2} \qquad \mbox{e} \qquad Y_{1}(t)=y_{o}+v_{oy}t + \frac{1}{2} a_{y} t^{2}$$<br />
onde $v_{ox}=v_{o}\cos{\theta}$, $v_{oy}=v_{o}\sin{\theta}$, $a_{x}=a \cos{\theta}$ e $a_{y}=a \sin{\theta}$.<br />
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1.2. As funções da velocidade: $$V_{x}(t)=v_{ox} + a_{x} t \qquad \mbox{e} \qquad V_{y}(t)=v_{oy} + a_{y} t. $$ O módulo de $V(t)$ para qualquer instante $t$ é determinado por $$V(t) = \sqrt{(V_{x}(t))^{2}+(V_{y}(t))^{2}}.$$<br />
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2. Segundo Estágio: Para os instantes de tempo $t > T = 30 \,s$, agora, sob a ação da gravidade apenas, temos:<br />
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2.1 As funções horárias do movimento: $$X_{2}(t)=X_{1}(T)+v_{1x}(t-T) \qquad \mbox{e} \qquad Y_{2}(t)=Y_{1}(T)+v_{1y}(t-T) + \frac{1}{2} g (t-T)^{2}$$<br />
onde $v_{1x}=V(T) \cos{\theta}$, e $v_{1y}=V(T) \sin{\theta}$.<br />
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2.2. As funções da velocidade: $$V_{2x}(t)=v_{1x} \qquad \mbox{e} \qquad V_{2y}(t)=v_{1y} - g (t-T)$$.<br />
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No aplicativo do gnuplot é possível alterar todos os parâmetros de entrada facilmente, ou seja, é bem geral.</div>
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-82156331676859306742016-03-01T22:31:00.001-08:002016-03-01T22:31:33.869-08:00Animações em Física com gnuplot: Sistema massa-molaCaros leitores, uma pequena animação do sistema massa-mola e os correspondentes gráficos das energias potencial e cinética é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, circle, arrow, e label) são implementados facilmente. As equações físicas envolvidas na animação são,<br />
<br />
1. A função horária do movimento: $$x(t)=x_m cos(wt + \phi_o)$$<br />
onde $w$ é a frequência angular, obtida por: $$w=\sqrt{\frac{k}{m}}$$<br />
2. O período $T$, determinado por $$T=\frac{2\pi}{w}=2\pi\sqrt{\frac{m}{k}}$$<br />
<br />
3. A força restauradora, segundo a Lei de Hooke, definida por: $$F(x)=-k.x$$<br />
onde $k$ é a constante elástica da mola, medida em $N/m$, e por último,<br />
<br />
4. As expressões das energias mecânica, potencial e cinética, respectivamente, determinadas por:<br />
$$E_{T}=\frac{1}{2}kx^{2}_{m}, \quad \quad U(x)=\frac{1}{2}kx^2 \quad \quad \mbox{e} \quad \quad K(x)=E_{T}-U(x)$$.<br />
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O objeto mola foi construído segundo as equações paramétricas da hélice circular reta, a saber, $$x_{1}(t)=a\cos(wt), \quad \quad y_{1}(t)=a\sin(wt) \quad \quad \mbox{e} \quad \quad h(t)=bt $$<br />
onde $a$ e $b$ são parâmetros que determinam o raio e o espaçamento da hélice.<br />
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No aplicativo do gnuplot é possível alterar todos os parâmetros de entrada facilmente, até os ângulos de visão do sistema massa-mola, ou seja, é bem geral.<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-10337462892762123612016-02-25T21:54:00.001-08:002019-01-01T11:37:09.943-08:00Relatividade: Contração do Comprimento<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;">Caros leitores, uma pequena animação do tópico referente a contração do comprimento da teoria da relatividade geral é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, polygon, circle, arrow, e label) são implementados facilmente. A equação física envolvida na animação é a equação de contração do comprimento da relatividade, a saber, </span><br />
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><br /></span>
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;">$$L(v)=Lo \sqrt{1-(v/c)^2}$$</span><br />
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><br /></span>
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;">onde $c$ é a velocidade da luz, $c=300000\, km/s$. </span><br />
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><br /></span>
<span style="color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif;"><span style="background-color: white; font-size: 13px; line-height: 18.2px;">A animação abaixo simula o comportamento de um ônibus espacial se movendo desde a velocidade inicial $v_i=10000\,km/s$ até a velocidade final $v_f=280000\,km/s$ (um valor considerado próximo à velocidade da luz) com variação na velocidade de $\Delta V = 5000\,km/s$. </span></span><br />
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<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><br /></span>Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-54374368773760674992016-02-20T14:00:00.000-08:002019-01-01T11:43:57.554-08:00Animações em Física com gnuplot: Máquina Térmica e o Ciclo de CarnotCaros leitores, uma pequena animação da operação de uma máquina térmica e o correspondente ciclo de Carnot é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (rect, polygon, circle, arrow, e label) são implementados facilmente. As equações físicas envolvidas na animação são as equações de estado do gás ideal.<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-49613970452292493382016-02-10T13:35:00.003-08:002016-02-12T15:05:48.335-08:00Animações com gnuplot: Pêndulo de NewtonCaros leitores, uma pequena animação do pêndulo de Newton foi construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (label, circle e arrow) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física. O aplicativo é bem geral, aceita qualquer número de pêndulos e qualquer valor para o ângulo inicial.<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-8946501190211949942016-02-07T11:32:00.003-08:002016-02-10T13:36:18.827-08:00Lançamento Oblíquo - Foguete: Animação com gnuplot Caros leitores, uma pequena animação de um foguete é aqui construída segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (polygon, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-87194582409857475442015-12-14T19:58:00.001-08:002019-01-01T11:31:40.206-08:00Animação com gnuplot: Pêndulo Cônico<span style="background-color: white; color: blue; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><b>Pêndulo Cônico</b></span><br />
<span style="background-color: white; color: blue; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><b><br /></b></span><span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"></span>
<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;">Caros leitores, o problema do pêndulo cônico é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes (circle, arrow e label) na animação são implementados facilmente.</span><br />
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<span style="background-color: white; color: #666666; font-family: "trebuchet ms" , "trebuchet" , sans-serif; font-size: 13px; line-height: 18.2px;"><br /></span>Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-64515078033661603302015-12-14T19:47:00.001-08:002019-01-01T12:04:26.211-08:00Animação com gnuplot: Movimento Uniforme<span style="color: blue;"><b>MOVIMENTO RETILÍNEO UNIFORME (M.R.U)</b></span><br />
<span style="color: blue;"><b><br /></b></span>Caros leitores, uma ilustração simples do movimento retilíneo uniforme e os correspondentes gráficos do M.R.U. é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (polygon, circle, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.<br />
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-59588513633073341862015-04-25T22:26:00.000-07:002015-04-25T22:26:22.450-07:00Sistemas de Equações Diferenciais Ordinárias: Métodos de Runge-Kutta de Ordem p<b><br /></b>
<b>1.0 Introdução</b><br />
<br />
<br />
<div style="text-align: justify;">
Em duas postagens anteriores, apresentamos os métodos numéricos de Runge-Kutta com o intuito de resolver equações diferenciais de primeira ordem, mais especificamente em problemas de valor inicial (P.V.I) de primeira ordem. Entretanto, a maioria das equações diferenciais com importância prática, são de ordem maior que $1$ ou então são sistemas de equações diferenciais. Nesta postagem, nosso intuito é o de apresentar o procedimento de como resolver um sistema de equações diferenciais de primeira ordem e, também, o de resolver numericamente uma equação diferencial de ordem elevada, usando os métodos de Runge-Kutta.</div>
<br />
<b>1.1 - Sistemas de Equações Diferenciais Ordinárias</b><br />
<br />
<div style="text-align: justify;">
Consideremos um sistema de $n$ equações diferenciais de primeira ordem:</div>
<br />
\begin{equation}<br />
\begin{array}{llll}<br />
y'_1=f_{1}(x, y_1, y_2, \cdots, y_n)\\<br />
y'_2=f_{2}(x, y_1, y_2, \cdots, y_n)\\<br />
\vdots\\<br />
y'_n=f_{n}(x, y_1, y_2, \cdots, y_n)<br />
\end{array}<br />
\end{equation}<br />
<div style="text-align: justify;">
o qual pode ser escrito, como: $$\mathbf{y' = f(x, y)} , $$ onde $\mathbf{y}$, $\mathbf{y'}$ e $\mathbf{f}$ são vetores com componentes $\mathbf{y_i}$, $\mathbf{y'_i}$ e $\mathbf{f_{i}(i=1, 2, \cdots, n)}$, respectivamente. Para que esse sistema possua uma única solução é necessário impormos uma condição adicional sobre $\mathbf{y}$. Esta condição é usualmente da forma: $$\mathbf{y(x_{0}) = y_{0}} , $$ para um dado número $\mathbf{x_0}$ e um vetor $\mathbf{y_0}$</div>
<br />
<div style="text-align: justify;">
Agora descreveremos como os métodos apresentados nas postagens anteriores para a solução de equações diferenciais de primeira ordem podem ser aplicados para resolver sistemas de equações diferenciais de $1^a$ ordem. Para efeito de simplicidade, e sem perda de generalidade, consideramos apenas o caso em que $n=2$, isto é, o sistema possui apenas duas equações, e para maior clareza usaremos a notação:</div>
\begin{equation} <br />
\left\{<br />
\begin{array}{llll}<br />
y' = f(x, y, z)\\ <br />
z' = g(x, y, z)\\ <br />
y(x_{0})=y_{0}\\ <br />
z(x_{0})=z_{0} \ \ \ x \in \left[x_{0}, b \right]<br />
\end{array}<br />
\right. <br />
\end{equation}<br />
<br />
<div style="text-align: justify;">
Assim, se desejamos resolver o sistema $(2)$ pelo método de Euler (Runge-Kutta de ordem 1), teremos:</div>
<br />
\begin{equation}<br />
\begin{array}{ll}<br />
y_{i+1}=y_i + hf(x_i, y_i, z_i)\\<br />
z_{i+1}=z_i + hg(x_i, y_i, z_i)\\<br />
\end{array}<br />
\end{equation}<br />
que será aplicado passo a passo.<br />
<br />
<div style="text-align: justify;">
No caso de escolhermos o método de Runge-Kutta de $4^a$ ordem para resolvermos o sistema $(2)$, teremos as seguintes expressões:</div>
<br />
\begin{equation}<br />
\begin{array}{ll}<br />
y_{i+1}=y_i + \frac{h}{6}(k_1 + 2k_2 + 2k_3 +k_4)\\<br />
z_{i+1}=z_i + \frac{h}{6}(l_1 + 2l_2 + 2l_3 + l_4)\\<br />
\end{array}<br />
\end{equation}<br />
onde<br />
<ul>
<li>$k_1 = f(x_i, y_i, z_i)$</li>
<li>$l_1 = g(x_i, y_i, z_i)$</li>
<li>$k_2 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{1}, z_i + \frac{h}{2}l_1)$ </li>
<li>$l_2 = g(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{1}, z_i + \frac{h}{2}l_1)$ </li>
<li>$k_3 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{2}, z_i + \frac{h}{2}l_2)$</li>
<li>$l_3 = g(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{2}, z_i + \frac{h}{2}l_2)$ </li>
<li>$k_4 = f(x_{i}+h, y_{i} + hk_{3}, z_i + l_3)$ </li>
<li>$l_4 = g(x_{i}+h, y_{i} + hk_{3}, z_i + l_3)$ </li>
</ul>
<br />
<br />
<div style="text-align: justify;">
Nota: A aplicação de um método numérico a um sistema de equações ordinárias de primeira ordem se processa como no caso de uma única equação, só que aqui devemos aplicar o método numérico a cada uma das componentes do vetor.</div>
<div style="text-align: justify;">
<br /></div>
<b>1.2 - Equações Diferenciais de Ordem Elevada</b><br />
<br />
<div style="text-align: justify;">
Finalmente, mostraremos como equações de ordem superiores a 1 podem ser escritas e portanto resolvidas como um sistema de equações de primeira ordem. Consideremos a equação diferencial de ordem $n$: $$y^{(n)} = f(x, y, y', \cdots, y^{(n-1)}).$$</div>
<div style="text-align: justify;">
com as condições iniciais: $$y(x_0) = y_0, \ \ \ y'(x_0) = y'_0, \ \ \cdots, \ \ y^{(n-1)}(x_0)=y^{(n-1)}_0.$$</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Novamente, para simplicidade, mas sem perda de generalidade, consideremos a equação diferencial de segunda ordem:</div>
\begin{equation} <br />
\left\{<br />
\begin{array}{lll}<br />
\frac{d^2\,y}{dx^2}+q(x)\frac{dy}{dx} = r(x,y)\\ <br />
<br />
y(x_{0})=y_{0}\\ <br />
\frac{d\,y}{dx}|_{x_0}=y'_{0}<br />
\end{array}<br />
\right. <br />
\end{equation}<br />
<br />
<div style="text-align: justify;">
Podemos resolver qualquer equação diferencial de ordem elevada reduzindo-a a um sistema de equações diferenciais de primeira ordem. Para tanto basta fazer uma mudança adequada de variável, para a equação de segunda ordem, (5). Fazendo a seguinte mudança de variável: $$z(x, y) = \frac{d\,y}{dx}, $$ </div>
a equação (5) é escrita como um conjunto de duas equações diferenciais de $1^a$ ordem, ou seja, o P.V.I, definido por (5), se reduz a:<br />
\begin{equation} <br />
\left\{<br />
\begin{array}{llll}<br />
\frac{d\,y}{dx} = z(x, y)\\ <br />
\frac{d\,z}{dx}= r(x, y) - q(x)z(x, y)\\ <br />
y(x_{0})=y_{0}\\<br />
z(x_{0})=z_{0} <br />
\end{array}<br />
\right. <br />
\end{equation}<br />
<div style="text-align: justify;">
Observe que o conjunto de equações em (6) pode ser escrito na forma:</div>
\begin{equation} <br />
\left\{<br />
\begin{array}{llll}<br />
y' = f(x, y, z)\\ <br />
z' = g(x, y, z)\\ <br />
y(x_{0})=y_{0}\\<br />
z(x_{0})=z_{0} <br />
\end{array}<br />
\right. <br />
\end{equation}<br />
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Nota: Ao reduzir o P.V.I. definido pela equação $(5)$, a um sistema de E.D.O. de primeira ordem, o problema passa a ser análogo ao P.V.I definido pelo sistema (2).</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Portanto, para a solução da equação diferencial ordinária de $2^a$ ordem $(5)$, é necessário que as equações diferenciais ordinárias de $1^a$ ordem, apresentadas em $(7)$ sejam resolvidas. A seguir, apresenta-se a solução destas equações, usando o método de Euler e o Método de Runge-Kutta de $4^a$ ordem.</div>
<br />
<b>1.3 - Exemplo usando o Método de Euler e o de Runge-Kutta de $4^a$ Ordem</b><br />
<br />
A
equação diferencial ordinária de $2^{a}$ ordem escolhida será:<br />
<br />
\begin{equation}<br />
\frac{d^{2}y}{dx^{2}}-y=e^{x}<br />
\end{equation}<br />
com $x$ variando de $0$ a $1$, com as seguintes condições iniciais:<br />
\begin{equation}<br />
\left\{<br />
\begin{array}{ll}<br />
y(0)=1\\<br />
\frac{dy}{dx}|_{x_0}=0<br />
\end{array}<br />
\right. <br />
\end{equation}<br />
<br />
<div style="text-align: justify;">
Primeiro passo: é reduzir esta equação diferencial de $2^a$ ordem para um conjunto de $2$ equações diferenciais ordinárias de $1^a$ ordem:</div>
\begin{equation}<br />
\left\{<br />
\begin{array}{ll}<br />
\frac{dy}{dx} = z \\<br />
\frac{dz}{dx} =y+e^{x}<br />
\end{array}<br />
\right.<br />
\end{equation}
<br />
<div style="text-align: justify;">
De acordo com as condições iniciais em $(9)$, tem-se $x_0 = 0$ e $y_0 = 1$. </div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
A solução da equação diferencial ordinária de $2^a$ ordem, $(8)$, com a condição inicial $(9)$ é:</div>
\begin{equation}<br />
y(x)=\frac{1}{4}[e^{x}(1+2x)+3e^{-x}].<br />
\end{equation} <br />
<div style="text-align: justify;">
A
solução numérica é encontrada com a avaliação das equações do Método de
Euler (Runge-Kutta de $1^a$ ordem), e do Método de Runge-Kutta de $4^a$ ordem, apresentadas respectivamente pelas equações (3) e (4), com
a condição inicial, $x_0=0$ e $y_0=0$.</div>
<br />
<div style="text-align: justify;">
Nota: vale lembrar que, a solução aproximada da equação diferencial de segunda ordem encontra-se na primeira componente do vetor, isto é, apenas nos interessa o valor de $y_i$, apesar de termos de calcular, a cada passo, todas as componentes do vetor.</div>
<br />
Para efeitos de
comparação, usaremos três valores para o passo h=0.1 (Figura 1), h=0.01 (Figura 2) e h=0.001 (Figura 3), juntamente com a solução algébrica (8).<br />
<br />
<b>1.3.1 - Resultados obtidos pelos Métodos de Euler e de Runge-Kutta de $4^a$ ordem</b><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNxf4m0h9TQzdU1hvBrdOGoU_7dcGVCWjzO0ZfYqKN9t1Fmp7lEkuQIs6VLqQ3uXF7GDuUiXiYcnp_5rdg9cBzrxeiNZ4NKtyf4-Rlm7Yc66L1I5ptmB7kv6ZK2kl1Ma3Uk6YlcQgPAF1I/s1600/Captura+de+tela+de+2015-04-26+01:32:11.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNxf4m0h9TQzdU1hvBrdOGoU_7dcGVCWjzO0ZfYqKN9t1Fmp7lEkuQIs6VLqQ3uXF7GDuUiXiYcnp_5rdg9cBzrxeiNZ4NKtyf4-Rlm7Yc66L1I5ptmB7kv6ZK2kl1Ma3Uk6YlcQgPAF1I/s1600/Captura+de+tela+de+2015-04-26+01:32:11.png" height="388" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 1: Solução da E.D.O. de $2^a$ ordem com passo $h=0.1$</td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_WFnBlVdM0FG6HIQchKEDXO8gGjaC6Mw4Uv8wDM6qiqgy6QQpCkgwH_PBhS65e5_LvNmuoQtRzxX7e_67HzZMsH3g9YvsqjdQnsLMeknySsUBgecHyTNr0kT5TbdZea4eiGXZVEf1wkJ4/s1600/Captura+de+tela+de+2015-04-26+01:32:59.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_WFnBlVdM0FG6HIQchKEDXO8gGjaC6Mw4Uv8wDM6qiqgy6QQpCkgwH_PBhS65e5_LvNmuoQtRzxX7e_67HzZMsH3g9YvsqjdQnsLMeknySsUBgecHyTNr0kT5TbdZea4eiGXZVEf1wkJ4/s1600/Captura+de+tela+de+2015-04-26+01:32:59.png" height="379" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 2: Solução da E.D.O. de $2^a$ ordem com passo $h=0.01$</td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiW7RyZtV1_TQlbCje6X90B49FSunC2QhIRs-Khfu8hKZX5oUlDdNUkh4xTJi_6dNMHR1-zS1KE3pkWXL7SFqYXl0LyOPsViVHBbYnbDK5_3k3Gt_ftkYM59bpYrIOcsYD-omaPZo9TcvxS/s1600/Captura+de+tela+de+2015-04-26+01:34:00.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiW7RyZtV1_TQlbCje6X90B49FSunC2QhIRs-Khfu8hKZX5oUlDdNUkh4xTJi_6dNMHR1-zS1KE3pkWXL7SFqYXl0LyOPsViVHBbYnbDK5_3k3Gt_ftkYM59bpYrIOcsYD-omaPZo9TcvxS/s1600/Captura+de+tela+de+2015-04-26+01:34:00.png" height="384" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 3: Solução da E.D.O. de $2^a$ ordem com passo $h=0.001$</td></tr>
</tbody></table>
<br />
<b>1.4 - Métodos de Runge-Kutta implementado em linguagem de programação. </b><br />
<br />
Um
programa escrito em C/C++ é apresentado a seguir. Uma parte do
código-fonte pode ser retirada sem problemas, pois trata-se de rotinas
voltadas para mostrar a interatividade entre o aplicativo gnuplot (para
plotar gráficos) e a linguagem programação C/C++. Os dados obtidos são
impressos em um arquivo de nome "runge_kutta.txt", podendo ser plotado
usando o Origin, qtiplot, MatLab, SciLab, dentre outros.<br />
<br />
<br />
/*========================*/ <br />
<span style="font-size: x-small;">/* Métodos de Runge-Kutta - E.D.O. de 2ª ordem */</span><br />
<span style="font-size: x-small;">/*=============================*/ <br /> </span><br />
<span style="font-size: x-small;">#include <stdio.h><br />#include <stdlib.h><br />#include <cmath><br /><br />FILE *fp;<br />static char nome[]="runge_kutta.txt";<br /><br /><span style="color: red;">void scrip_gnuplot(char str[25])<br />{<br /> fp=fopen("rk4_2_ordem.plt", "w");<br /> </span></span><span style="font-size: x-small;"><span style="color: red;">fprintf(fp, "reset\n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set title \"Métodos de Runge Kutta\" \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set xlabel \"Xn\" \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set ylabel \"Yn\" \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set key bottom \n" );</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set term jpeg \n" );</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set output \"Runge_Kutta.jpg\" \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "plot \'%s\' u 1:2 t \" Euler \" w lp ls 7 lc 1 lw 1, \'%s\' u 1:3 t \" Runge-Kutta 4a ordem \" w lp ls 7 lc 2 lw 1\n", str, str);</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "replot \'%s\' u 1:4 t \" Analitica \" w l lc -1\n", str);</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "set output; set term wxt \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "replot \n");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fprintf(fp, "pause -1 \"Continuar?\" ");</span></span><br /><span style="font-size: x-small;"><span style="color: red;"> fclose(fp);</span></span><br /><span style="font-size: x-small;"><span style="color: red;"></span></span><span style="font-size: x-small;"><span style="color: red;">}</span><br /><br />double f(double x, double y, double z)<br />{<br /> return z; <span style="color: red;">// equação diferencial: dy/dx=f(x,y,z) </span><br />}<br /><br />double g(double x, double y, double z)<br />{ <br /> return (y+exp(x)); <span style="color: red;">// equação diferencial: dz/dx=g(x,y,z)</span><br />}<br /><br />double y(double x) <span style="color: red;">// solução analitica da equação diferencial</span><br />{<br /> return (0.25*(exp(x)*(1.0+2.0*x)+3.0*exp(-x))); <br />}<br /><br />double euler(double x, double y, double *z, double h)<br />{<br /> y = y + h*f(x,y,*z); <span style="color: red;"> // método de euler</span><br /> *z = *z + h*g(x,y,*z);<br />return y; <br />}<br /><br />double runge_kutta_4(double x, double y, double *z, double h)<br />{<br />double k1, k2, k3, k4, l1, l2, l3, l4;<br /><br /> k1 = h*f(x, y, *z); <span style="color: red;">// método de runge-kutta de ordem 4</span><br /> l1 = h*g(x, y, *z);<br /> k2 = h*f(x+0.5*h, y+0.5*k1, *z+0.5*l1);<br /> l2 = h*g(x+0.5*h, y+0.5*k1, *z+0.5*l1);<br /> k3 = h*f(x+0.5*h, y+0.5*k2, *z+0.5*l2);<br /> l3 = h*g(x+0.5*h, y+0.5*k2, *z+0.5*l2);<br /> k4 = h*f(x+h, y+k3, *z+l3);<br /> l4 = h*g(x+h, y+k3, *z+l3);<br /> y = y + (k1+2*k2+2*k3+k4)/6.0;<br /> *z = *z + (l1+2*l2+2*l3+l4)/6.0;<br />return y; <br />}<br /><br />int main(void)<br />{<br />double x0, y0, z0, xmax, h; // variaveis de entrada<br /><br />double xn, yn_e, yn_rk4, *zn_e, *zn_rk4; // variaveis incrementadas<br />double z0_e, z0_rk4;<br /><br />// Dados de entrada:<br />x0=0.0;<br />y0=1.0;<br />z0=0.0;<br />xmax=1.0;<br />h=0.01;<br />// -----------------<br /> <br /> fp=fopen(nome,"w");<br /><br /> xn = x0;<br /> yn_e = yn_rk4 = y0;<br /> <br /> z0_e = z0_rk4 = z0;<br /> <br /> zn_e = &z0_e; <br /> zn_rk4 = &z0_rk4;<br /> while (xn<xmax)<br /> {<br /> fprintf(fp, "%f \t %f \t %f \t %f\n", xn, yn_e, yn_rk4, y(xn));<br /> <br /> yn_e = euler(xn, yn_e, zn_e, h);<br /> yn_rk4 = runge_kutta_4(xn, yn_rk4, zn_rk4, h);<br /> xn = xn+h;<br /> }<br /> fclose(fp);<br /> <span style="color: red;">scrip_gnuplot(nome);<br /> system("gnuplot rk4_2_ordem.plt");</span><br /> return 0;<br />}</span><br />
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</div>
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<b>1.5 - Conclusão</b></div>
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<br /></div>
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<b> </b>A
resolução numérica pode ter grau crescente de aproximação, de acordo
com o método numérico escolhido e da escolha do valor do passo de
integração. Comparando os resultados da solução numérica usando o método de Euler com os resultados da solução usando o método de Runge-Kutta de $4^a$ ordem, observa-se que neste segundo método a precisão é sempre maior, mesmo com o uso de um passo $10$ vezes maior. </div>
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Assim,
diante dos exemplos estudados, concluímos que o Método de Runge-Kutta
de $4^a$ ordem é bastante eficaz para resolução de sistemas de EDOs de primeira ordem, o que explica o porquê de ser o método de passo simples mais
utilizado por engenheiros e cientistas na resolução numérica de equações
diferenciais ordinárias.<br />
<br /></div>
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<b>1.6 - Referências</b></div>
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<br /></div>
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[1]
W. E. Boyce, C. R. Diprima, "Equações Diferenciais Elementares e
Problemas de Valor de Contorno", LTC, Rio de Janeiro, 2011.</div>
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[2] M. A. G. Ruggiero, V. L. R. Lopes, "Cálculo Numérico: Aspectos Teóricos Computacionais", Makron Books, São Paulo, 1996.</div>
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[3] G. Baratto, "Solução de Equações Diferenciais Ordinárias Usando Métodos
Numéricos", Departamento de Eletrônica e Computação - Estudo de Casos em
Engenharia Elétrica - UFSM, 2007.<br />
[4] L. N. de Andrade, "Cálculo Numérico: Introdução à Matemática Computacional", Departamento de Matemática - UFPB, 2013.</div>
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<br /></div>
Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-88385760310577714582015-04-24T23:13:00.001-07:002015-04-25T22:00:08.120-07:00Solução Numérica de Equações Diferenciais Ordinárias usando o Método de Runge-Kutta<br />
<b>1.0 INTRODUÇÃO</b><br />
<br />
<br />
<div style="text-align: justify;">
A resolução de equações diferenciais é um problema importantíssimo porque possui aplicações a diversas áreas do conhecimento tais como Matemática Aplicada, Física, Engenharia, Biologia, Economia e Computação Gráfica.</div>
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<br /></div>
<div style="text-align: justify;">
Devido à impossibilidade de se determinar a solução exata na maioria dos casos, desenvolveram-se técnicas de determinação de solução numérica aproximadas da equação. </div>
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<br /></div>
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Aqui, iremos nos concentrar em problemas de valor inicial
(P.V.I.) em sua forma mais simples que são as equações diferenciais
ordinárias de primeira ordem:</div>
<span style="font-style: normal;">\begin{equation} <br />\left\{<br />\begin{array}{ll}<br />\frac{dy}{dx}=f(x,y)\\<br />y(x_0)=y_0<br />\end{array}<br />\right.<br />\end{equation} </span><br />
<div style="text-align: justify;">
Uma solução exata de um P.V.I. do tipo (1) é uma função derivável cujo gráfico passa pelo ponto $(x_0, y_0)$. Uma solução aproximada é uma tabela de valores que inicia com $(x_0, y_0)$, próximos do gráfico da função que seria a solução da equação.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEmBeYqgJhmfKrsT5Iu092W0DPYGThg5c_xQGxQl7JByIZ4VLtPSL6fvv1r1L5_n3dTc1_rVQlED0e6AJYDwLcEJOGXGZQdPMSusZizCifHKdVIP97hSGzfsxJplgpVuM8eq0SY0TXNYmg/s1600/Captura+de+tela+de+2015-04-20+03:00:28.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEmBeYqgJhmfKrsT5Iu092W0DPYGThg5c_xQGxQl7JByIZ4VLtPSL6fvv1r1L5_n3dTc1_rVQlED0e6AJYDwLcEJOGXGZQdPMSusZizCifHKdVIP97hSGzfsxJplgpVuM8eq0SY0TXNYmg/s1600/Captura+de+tela+de+2015-04-20+03:00:28.png" height="146" width="400" /></a></div>
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Nesta postagem, nosso intuito é o de apresentar as idéias básicas e as principais propriedades dos métodos de Runge-Kutta. Uma demonstração breve do método de Runge-Kutta de $2^a$ ordem é apresentada. E, em seguida, apresentamos as regras de implementação dos métodos de Runge-Kutta de $3^a$ e $4^a$ ordem, concluindo com exemplos práticos de sua aplicação.</div>
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<br />
<b>1.1 - OS MÉTODOS DE RUNGE-KUTTA</b><br />
<br />
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Os métodos de Runge-Kutta são uma família de métodos
iterativos para aproximar numericamente a solução de uma equação
diferencial ordinária. Os matemáticos alemães Carl David Runge (1856-1927)
e Martin Wilhelm Kutta (1867-1944), construíram os métodos de tal modo que sejam
equivalentes a aproximar a solução exata de uma equação diferencial
pelos primeiros $n$ termos de uma expansão em série de Taylor.
</div>
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<br />
As
características de eficiência e simplicidade de implementação, conferem
aos métodos de Runge-Kutta grande popularidade, sendo um dos mais
utilizados por engenheiros e cientistas.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxKFmuQ8uTR-P1yxnMYtZRn6K4rihSYiE9ezt4KCR2XRi51AcVeRfzDXSr9s5rUHSTXS-TJWCk0YJnDHXT3ZxMOh26Ym37mxoDRIw8iH7guejQW-G7RgrhN_aShsbwRK2eULLmRst5P-Nm/s1600/Captura+de+tela+de+2015-04-21+19:01:28.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxKFmuQ8uTR-P1yxnMYtZRn6K4rihSYiE9ezt4KCR2XRi51AcVeRfzDXSr9s5rUHSTXS-TJWCk0YJnDHXT3ZxMOh26Ym37mxoDRIw8iH7guejQW-G7RgrhN_aShsbwRK2eULLmRst5P-Nm/s1600/Captura+de+tela+de+2015-04-21+19:01:28.png" height="246" width="400" /></a></div>
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<b>1.1.1 - Propriedades dos Métodos de Runge-Kutta</b><br />
<br />
<b> </b>A idéia básica destes métodos é aproveitar as qualidades dos métodos de série de Taylor (ordem elevada) e ao mesmo tempo eliminar sua maior dificuldade que é o cálculo de derivadas de $f(x,y)$ que torna os métodos de série de Taylor computacionalmente inaceitáveis.<br />
<br />
Características dos métodos de Runge-Kutta de ordem $p$: <br />
<ul>
<li>são de passo simples (auto-iniciantes);</li>
<li>não exigem o cálculo de derivadas parciais de $f(x,y)$;</li>
<li>necessitam apenas do cálculo de $f(x,y)$ em determinados pontos do intervalo $[x_{i}, x_{i+1}]$ (os quais dependem da ordem dos métodos);</li>
<li>expandindo-se $f(x,y)$ por Taylor em torno de $(x_i , y_i)$ e agrupando-se os termos em relação às potências de $h$, a expressão do método de Runge-Kutta coincide com a do método de Taylor de mesma ordem. </li>
</ul>
<br />
Na postagem "Solução Numérica de Equações Diferenciais Ordinárias usando o Método de Euler", vimos que o método de Euler para resolução do P.V.I, $y'=f(x,y)$, $y(x_0)=y_0$, consiste na aplicação das fórmulas:<br />
\begin{equation}<br />
x_{i+1}=x_i+h<br />
\end{equation}
e
\begin{equation}<br />
y_{i+1}=y_i+k_1, \ \ \mbox{para} \ \ i=0, 1, 2, \cdots, <br />
\end{equation} <br />
onde $k_1=hf(x_i, y_i)$ e $h$ chamado passo ou incremento de integração é próximo de $0$.<br />
<br />
O método de Runge-Kutta é um aperfeiçoamento do método de Euler e consiste em somar ao $y_i$ não apenas um valor de $k_1$, mas uma média de vários valores de $k_1, \,k_2, \,k_3, \cdots $. <br />
<br />
<b>1.1.2 - Síntese do Método de Runge-Kutta de Ordem $p$</b><br />
<b><br /></b>
<br />
<ul>
<li>Fórmula geral do Método de Runge-Kutta: $$y_{i+1}=y_i + h \phi(x_i, y_i; h)$$, </li>
<li>$\phi(x_i, y_i; h)$ é chamada <i>função incremento, </i>e pode ser interpretada como a inclinação no intervalo considerado. </li>
<li>Fórmula geral da função incremento de ordem $p$: $$\phi(x, y; h)=b_1k_1+b_2k_2+ \cdots + b_pk_p , $$ </li>
<ul>
<li> $k_1=f(x, y)$</li>
<li> $k_2=f(x+c_2h, y+ha_{21}k_1)$, </li>
<li> $k_3=f(x+c_3h, y+h(a_{31}k_1+a_{32}k_2))$, </li>
<li> $\cdots$ </li>
<li> $k_p=f(x+c_ph, y+h(a_{p1}k_1+ \cdots + a_{p, p-1}k_{p-1}))$. </li>
</ul>
</ul>
<ul>
<li>$b_i$, $c_i$ e $a_{ij}$: constantes obtidas igualando-se a fórmula geral de Runge-Kutta com os termos da expansão em série de Taylor de mesma ordem.</li>
<li>$k_i$: relações de recorrência (cálculo computacional eficiente).</li>
</ul>
<br />
Uma demonstração completa do método de Runge-Kutta pode ser encontrada em livros como a referência bibliográfica [2]. No que segue, limitar-nos-emos a apresentar as fórmulas necessárias.<br />
<br />
<b>1.2 - Métodos de Runge-Kutta de $2^a$ Ordem</b><br />
<br />
<b> </b>O método de Runge-Kutta de $2^a$ ordem concorda com a precisão do método de série de Taylor até os termos de $2^a$ ordem em h: $$y_{i+1}=y_i+hf(x_i, y_i) + \frac{h^2}{2}\left[\frac{\partial f(x_i, y_i)}{\partial x} + f(x_i, y_i)\frac{\partial f(x_i, y_i)}{\partial y} \right]$$<br />
Usando a notação simplificada, $f_i=f(x_i, y_i)$, temos:<br />
\begin{equation}<br />
y_{i+1}=y_i+hf_i + \frac{h^2}{2}\left[\frac{\partial f_i}{\partial x} + f_i \frac{\partial f_i}{\partial y} \right] <br />
\end{equation}<br />
Considerando, agora, a definição do método de Runge-Kutta de $2^a$ ordem:<br />
\begin{equation} <br />
y_{i+1}=y_i+h(b_{1}f(x_i, y_i) + b_{2}f(x_{i}+c_{2}h, y_{i}+ha_{21}k_1)<br />
\end{equation} <br />
Expandindo $f(x, y)$, em série de Taylor, em torno de $(x_i, y_i)$ (retendo somente os termos de derivada primeira), temos:<br />
$$f(x_{i}+c_{2}h, y_{i}+ha_{21}k_1) \approx f_i + c_{2}h \frac{\partial f_i}{\partial x}+a_{21}h f_i \frac{\partial f_i}{\partial y}$$<br />
Substituindo esse resultado na equação anterior, e rearranjando, teremos:<br />
\begin{equation}<br />
y_{i+1}=y_i+h(b_1 + b_2)f_i + h^2 \left(b_{2}c_{2}\frac{\partial f_i}{\partial x} + b_{2}a_{21}f_i \frac{\partial f_i}{\partial y} \right) <br />
\end{equation}<br />
Assim, para haver concordância com o método de série de Taylor, resultado (4), é preciso que: $$b_1+b_2 = 1 \ \ \ \mbox{e} \ \ \ b_{2}c_{2}=\frac{1}{2} \ \ \ \mbox{e} \ \ \ b_{2}a_{21}=\frac{1}{2}.$$<br />
<ul>
<li>Resultado: Sistema não linear com 3 equações e 4 incógnitas, o que corresponde a uma variedade de métodos de segunda ordem. </li>
</ul>
<br />
<ul>
</ul>
Apresentaremos aqui os dois mais conhecidos métodos de Runge-Kutta de $2^a$ ordem.</div>
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<b>1.2.1 - Métodos de Euler Modificado ($b_2=1$, $b_1=0$, $c_2=a_{21}=1/2$)</b><br />
<b><br /></b>
Dado um P.V.I, $y' = f(x, y)$, $y(x_0)=x_0$. A implementação deste método é dada pelas expressões:<br />
<br />
<ul>
<li>$y_{i+1} = y_i + h.k_2$</li>
<li>$k_1 = f(x_i, y_i)$</li>
<li>$k_2 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{1})$ </li>
<li>com $x_{i+1} = x_i + h$, para $h>0$ próximo de $0$, $i=0, 1, 2, \cdots $ .</li>
</ul>
<br />
Para cada valor inteiro de $i$, a partir de $i=0$, calculam-se: $$x_{i+1} \ \ \rightarrow \ \ k_1 \ \ \rightarrow \ \ k_2 \ \ \rightarrow \ \ y_{i+1}$$<br />
Repete-se essa sequência de cálculos recursivamente, até chegar no valor de $y_i$ desejado.<br />
<br />
<b>1.2.2 - Métodos de Euler Melhorado ou de Heun ($b_2=1/2$, $b_1=1/2$, $c_2=a_{21}=1$)</b><br />
<b><br /></b>
Dado um P.V.I, $y' = f(x, y)$, $y(x_0)=x_0$. A implementação deste método é dada pelas expressões:<br />
<br />
<ul>
<li>$y_{i+1} = y_i + \frac{h}{2}\left(k_1 + k_2 \right)$</li>
<li>$k_1 = f(x_i, y_i)$</li>
<li>$k_2 = f(x_{i}+h, y_{i}+hk_{1})$ </li>
<li>com $x_{i+1} = x_i + h$, para $h>0$ próximo de $0$, $i=0, 1, 2, \cdots $ .</li>
</ul>
<br />
Para cada valor inteiro de $i$, a partir de $i=0$, calculam-se: $$x_{i+1} \
\ \rightarrow \ \ k_1 \ \ \rightarrow \ \ k_2 \ \ \rightarrow \
\ y_{i+1}$$
<br />
Repete-se essa sequência de cálculos recursivamente, até chegar no valor de $y_i$ desejado.<br />
<br />
<br />
<b><b>1.3 - Métodos de Runge-Kutta de Ordem</b> Superiores</b><br />
<br />
<b> </b>De modo semelhante, podem ser deduzidas as fórmulas de Runge-Kutta de ordens superiores.<b> </b><br />
<b> </b>Em cada ordem, também haverá uma infinidade de versões. A seguir serão fornecidas apenas as fórmulas mais conhecidas para os métodos de Runge-Kutta de $3^a$ e $4^a$ ordens:<br />
<br />
<b>1.3.1 - Métodos Runge-Kutta de $3^a$ Ordem</b><br />
<b><br /></b>
Dado um P.V.I, $y' = f(x, y)$, $y(x_0)=x_0$. A implementação deste método é dada pelas expressões:<br />
<ul>
<li>$y_{i+1} = y_i + \frac{h}{6}\left(k_1 + 4k_2 + k_3 \right)$</li>
<li>$k_1 = f(x_i, y_i)$</li>
<li>$k_2 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{1})$ </li>
<li>$k_3 = f(x_{i}+h, y_{i} - hk_{1} + 2hk_{2})$ </li>
<li>com $x_{i+1} = x_i + h$, para $h>0$ próximo de $0$, $i=0, 1, 2, \cdots $.</li>
</ul>
<br />
Para cada valor inteiro de $i$, a partir de $i=0$, calculam-se:
$$x_{i+1} \ \ \rightarrow \ \ k_1 \ \ \rightarrow \ \ k_2 \ \
\rightarrow \ \ k_3 \ \
\rightarrow \ \ y_{i+1}$$<br />
Repete-se essa sequência de cálculos recursivamente, até chegar no valor de $y_i$ desejado.<br />
<br />
<br />
<b>1.3.2 - Métodos Runge-Kutta de $4^a$ Ordem</b><br />
<b><br /></b>
Dado um P.V.I, $y' = f(x, y)$, $y(x_0)=x_0$. A implementação deste método é dada pelas expressões:<br />
<ul>
<li>$y_{i+1} = y_i + \frac{h}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)$</li>
<li>$k_1 = f(x_i, y_i)$</li>
<li>$k_2 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{1})$ </li>
<li>$k_3 = f(x_{i}+\frac{h}{2}, y_{i}+\frac{h}{2}k_{2})$</li>
<li>$k_4 = f(x_{i}+h, y_{i} + hk_{3})$ </li>
<li>com $x_{i+1} = x_i + h$, para $h>0$ próximo de $0$, $i=0, 1, 2, \cdots $.</li>
</ul>
<br />
Para cada valor inteiro de $i$, a partir de $i=0$, calculam-se:
$$x_{i+1} \ \ \rightarrow \ \ k_1 \ \ \rightarrow \ \ k_2 \ \
\rightarrow \ \ k_3 \ \
\rightarrow \ \ k_4 \ \
\rightarrow \ \ y_{i+1}$$<br />
Repete-se essa sequência de cálculos recursivamente, até chegar no valor de $y_i$ desejado.<br />
<br />
Nota: O método de Runge-Kutta de $4^a$ ordem é o mais usado na solução numérica de problemas com equações diferenciais ordinárias.<br />
<br />
<br />
<b>1.4 - Exemplo usando o Método de Runge-Kutta de $2^a$ e $4^a$ Ordem</b><br />
<br />
<b> </b>A
equação diferencial ordinária de $1^{a}$ ordem escolhida será (Boyce,
W. E., Diprima, R. C. Equações Diferenciais Elementares e Problemas de
Valor de Contorno. Ed. 7, pp. 420):<span style="font-style: normal;"><span style="font-style: normal;"> </span></span><br />
<span style="font-style: normal;"><span style="font-style: normal;">\begin{equation} <br />\left\{<br />\begin{array}{ll}<br />\frac{dy}{dx}=1-x+4.y\\<br />y(0)=1<br />\end{array}<br />\right.<br />\end{equation} </span></span><br />
Esta equação será resolvida de $x=0$ a $x=2$. A solução da equação diferencial (7) com a condição inicial (P.V.I.) é conhecida:<br />
\begin{equation}<br />
y(x)=\frac{1}{4}.x-\frac{3}{16}+\frac{19}{16}e^{4.x}<br />
\end{equation} <br />
A solução numérica é encontrada com a avaliação das equações do Método de Euler (Runge-Kutta de $1^a$ ordem), Euler modificado (Runge-Kutta de $2^a$ ordem) e do Método de Runge-Kutta de $4^a$ ordem, apresentadas acima, com
a condição inicial, $x_0=0$ e $y_0=1$.<br />
<br />
Para efeitos de
comparação, usaremos dois valores para o passo h=0.1 (Figura 2) e
h=0.01 (Figura 3), juntamente com a solução algébrica (8).<br />
<br />
<br />
<b>1.4.1 - Resultados obtidos pelos Métodos de Euler, Euler modificado e de Runge-Kutta de $4^a$ ordem</b><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjofnWrj0BLQoL_AotwJEQ12zhiQl6p63xxi7cRAV53Gijm_7dL-ah0fOZaBxSiYg7Tjc41OHoICcPZe3TUiFDf1tErQubAwkJkLU9L_ZQjhiQ7H0nHlDEbO_rJRc45FcPxd60XxSGIhSr9/s1600/Captura+de+tela+de+2015-04-25+03:05:46.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjofnWrj0BLQoL_AotwJEQ12zhiQl6p63xxi7cRAV53Gijm_7dL-ah0fOZaBxSiYg7Tjc41OHoICcPZe3TUiFDf1tErQubAwkJkLU9L_ZQjhiQ7H0nHlDEbO_rJRc45FcPxd60XxSGIhSr9/s1600/Captura+de+tela+de+2015-04-25+03:05:46.png" height="384" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 2: Métodos de Runge-Kutta com passo de integração $h=0.1$.</td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiutcuKKfgGtsffJIQPo-yGUIB6_AZzqQl21_FecGlY7Svm8wLHfIlV0PV53APjSmSdpS9PZnDWJT8fi31W8Tpj6qWpVE7tvsJpeZjnDl0YZWp22ULE5ZpCnxCKX0znRyesVcFCBHJfY9Ky/s1600/Captura+de+tela+de+2015-04-25+03:06:22.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiutcuKKfgGtsffJIQPo-yGUIB6_AZzqQl21_FecGlY7Svm8wLHfIlV0PV53APjSmSdpS9PZnDWJT8fi31W8Tpj6qWpVE7tvsJpeZjnDl0YZWp22ULE5ZpCnxCKX0znRyesVcFCBHJfY9Ky/s1600/Captura+de+tela+de+2015-04-25+03:06:22.png" height="380" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 3: Métodos de Runge-Kutta com passo de integração $h=0.01$.</td></tr>
</tbody></table>
<br />
<br />
Por torna-se um processo demasiadamente repetitivo pelo número de iterações,
é comum utilizar o método com o auxílio de uma linguagem de
programação. <br />
<br />
<b><b> </b> </b><br />
<b>1.5 - Métodos de Runge-Kutta implementado em linguagem de programação. </b><br />
<br />
Um
programa escrito em C/C++ é apresentado a seguir. Uma parte do
código-fonte pode ser retirada sem problemas, pois trata-se de rotinas
voltadas para mostrar a interatividade entre o aplicativo gnuplot (para
plotar gráficos) e a linguagem programação C/C++. Os dados obtidos são
impressos em um arquivo de nome "runge_kutta.txt", podendo ser plotado
usando o Origin, qtiplot, MatLab, SciLab, dentre outros. <br />
<br />
<br />
<span style="font-size: x-small;">/*=================== */</span><br />
<span style="font-size: x-small;"><span style="font-size: x-small;">/* Métodos de Runge-Kutta */</span></span><br />
<span style="font-size: x-small;"><span style="font-size: x-small;"><span style="font-size: x-small;">/*=================== */</span></span></span><br />
<span style="font-size: x-small;"><span style="font-size: x-small;"></span></span><br />
<span style="font-size: x-small;">#include <stdio.h><br />#include <stdlib.h><br />#include <cmath><br /><br />FILE *fp;<br /><br />static char nome[]="runge_kutta.txt";<br /><br /><span style="color: red;">void scrip_gnuplot(char str[25])<br />{<br /> fp=fopen("rk_2_4.plt", "w");<br /> <br /> fprintf(fp, "reset\n");<br /> fprintf(fp, "set title \"Metodos de Runge Kutta\" \n");<br /> fprintf(fp, "set xlabel \"Xn\" \n");<br /> fprintf(fp, "set ylabel \"Yn\" \n");<br /> fprintf(fp, "set key left \n");<br /> fprintf(fp, "set term jpeg \n" );<br /> fprintf(fp, "set output \"RK_2_4.jpeg\" \n");<br /> fprintf(fp, "plot \'%s\' u 1:2 t \" Euler \" w lp ls 7 lc 1 lw 1, \'%s\' u 1:3 t \" Euler Modificado \" w lp ls 7 lc 2 lw 1\n", str, str);<br /> fprintf(fp, "replot \'%s\' u 1:4 t \" Runge-Kutta 4a ordem \" w lp ls 7 lc 3 lw 1, \'%s\' u 1:5 t \" Analitica \" w l lc -1\n", str, str);<br /> fprintf(fp, "set output; set term wxt \n");<br /> fprintf(fp, "replot \n");<br /> fprintf(fp, "pause -1 \"Continuar?\" ");<br /> fclose(fp);<br />}</span><br />// Definição da equação diferencial: dy/dx=f(x,y)<br />double f(double x, double y)<br />{<br /> return (1-x+4*y);<br />}<br />// Definiçao da solução analitica da equação diferencial<br />double y(double x)<br />{<br /> return ((1.0/4.0)*x - (3.0/16.0) + (19.0/16.0)*exp(4.0*x));<br />}<br />// Definiçao do método de euler<br />double euler(double x, double y, double h)<br />{<br />double k1;<br /><br /> k1 = f(x,y);<br /> y = y + h*k1; <br /> <br />return y; <br />}<br />// Definiçao do método de euler modificado<br />double euler_modificado(double x, double y, double h)<br />{<br />double k1, k2;<br /><br /> k1 = f(x,y);<br /> k2 = f(x+0.5*h, y+0.5*h*k1);<br /> y = y + h*k2; <br /> <br />return y; <br />}<br />// Definiçao do método de runge-kutta de 4a ordem<br />double runge_kutta_4(double x, double y, double h)<br />{<br />double k1, k2, k3, k4;<br /><br /> k1 = f(x,y);<br /> k2 = f(x+0.5*h, y+0.5*h*k1);<br /> k3 = f(x+0.5*h, y+0.5*h*k2);<br /> k4 = f(x+h, y+h*k3);<br /> y = y + h*(k1+2*k2+2*k3+k4)/6.0;<br /><br />return y; <br />}<br /><br />int main(void)<br />{<br />double x0, y0, xmax, h; // variaveis de entrada<br />double xn, yn_e, yn_em, yn_rk4;<br />double y_e, y_em, y_rk4;<br /><br />// Dados de entrada:<br />x0=0.0;<br />xmax=2.0;<br />y0=1.0;<br />h=0.01;<br />// -----------------<br /> fp=fopen(nome,"w");<br /> <br /> xn=x0;<br /> yn_e=y0; <br /> yn_em=y0; <br /> yn_rk4=y0;<br /> while (xn<xmax)<br /> {<br /> fprintf(fp, "%f \t %f \t %f \t %f \t %f\n", xn, yn_e, yn_em, yn_rk4, y(xn));<br /> <br /> y_e = euler(xn, yn_e, h);<br /> y_em =euler_modificado(xn, yn_em, h);<br /> y_rk4=runge_kutta_4(xn, yn_rk4, h);<br /> xn=xn+h;<br /> yn_e = y_e;<br /> yn_em =y_em;<br /> yn_rk4=y_rk4;<br /> }<br /> fclose(fp);<br /><br /> <span style="color: red;">scrip_gnuplot(nome);<br /> system("gnuplot rk_2_4.plt");</span><br /> return 0;<br />}</span><br />
<br />
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<br /></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<b>1.6 - Conclusão</b></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<br /></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<b> </b>A resolução numérica pode ter grau crescente de aproximação, de acordo com o método numérico escolhido e da escolha do valor do passo de integração. Para os valores do passo de integração analisados, observamos que o método de Euler não se mostra uma boa escolha, pois seus resultados não são nada precisos, quando comparado aos dois outros métodos. Já no primeiro caso, obtivemos uma
boa aproximação por parte do método de Runge-Kutta de $4^a$ ordem. E, no segundo caso, já se verifica uma coincidência entre os métodos de Euler modificado e de Runge-Kutta de $4^a$ ordem com solução exata conhecida até onde se observou com passo de integração
$h=0.01$, ou seja, o "erro" ou diferença entre a aproximação e a
solução exata é praticamente nula para todos os pontos. </div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
Assim,
diante dos exemplos estudados, concluímos que o Método de Runge-Kutta de $4^a$ ordem é bastante eficaz para resolução de EDOs de primeira ordem, o que explica o porquê de ser o método de passo simples mais utilizado por engenheiros e cientistas na resolução numérica de equações diferenciais ordinárias.<br />
<br /></div>
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<b>1.7 - Referências</b></div>
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<br /></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
[1]
W. E. Boyce, C. R. Diprima, "Equações Diferenciais Elementares e
Problemas de Valor de Contorno", LTC, Rio de Janeiro, 2011.</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
[2] M. A. G. Ruggiero, V. L. R. Lopes, "Cálculo Numérico: Aspectos Teóricos Computacionais", Makron Books, São Paulo, 1996.</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
[3]
G. Baratto, "Solução de Equações Diferenciais Ordinárias Usando Métodos
Numéricos", Departamento de Eletrônica e Computação - Estudo de Casos em
Engenharia Elétrica - UFSM, 2007.<br />
[4] L. N. de Andrade, "Cálculo Numérico: Introdução à Matemática Computacional", Departamento de Matemática - UFPB, 2013.</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<br />
<br />
Nesta
postagem, o nosso objetivo, além de apresentar este importante tema
da Matemática, é o de fornecer exemplos da interatividade entre as
linguagem de programação C/C++ e o aplicativo gnuplot, que juntos
minimizam as tarefas de uma pesquisa. </div>
</div>
Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-8766066026603423802015-04-18T11:48:00.001-07:002015-04-18T12:09:07.383-07:00Solução Numérica de Equações Diferenciais Ordinárias usando o Método de Euler<style type="text/css">P { margin-bottom: 0.21cm; }P.western { }A:link { }</style><b>1.0 INTRODUÇÃO</b><br />
<br />
<br />
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
Uma
<i>Equação Diferencial</i> é uma equação que envolve derivadas
de uma ou mais funções. Elas servem para descrever o comportamento
de sistemas dinâmicos e possuem enorme aplicação em diversas
áreas, como em engenharia (no estudo do comportamento de um circuito
elétrico ou do movimento oscilatório de estruturas), Física
(lançamento de foquetes, oscilações forçadas, ...), Biologia
(crescimento de populações de bactérias) ou economia (aplicações
financeiras). Elas são classificadas de acordo com o seu tipo,
ordem e grau. Se uma equação diferencial envolve derivadas de uma
função de uma única variável independente, ela é dita ser
<i>Equação Diferencial Ordinária. </i><span style="font-style: normal;">Caso
a equação diferencial envolva as derivadas parciais de uma função
de duas ou mais variáveis independentes, é uma </span><i>Equação
Diferencial Parcial.</i><br />
<br />
<i><span style="font-style: normal;">Uma
equação diferencial ordinária (ou E.D.O.) de ordem </span><i>n</i><span style="font-style: normal;"> pode ser expressa na seguinte forma:</span></i><span style="font-style: normal;"> </span><br />
$$\frac{d^n\,y}{dx^n}=F \left(x, y,\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}, \cdots, \frac{d^{n-1}y}{dx^{n-1}} \right)$$<br />
<span style="font-style: normal;">onde $x$ é a variável independente, $y$ é uma função desta variável independente e $\frac{d^k\,y}{dx^k}$, com $k=1,2,3, \cdots, n$ são as derivadas de $y$ em relação a $x$.<br />
<br />
Há vários métodos que resolvem analiticamente uma E.D.O., entretanto nem sempre é possível obter uma solução analítica. Neste caso, os métodos numéricos são uma saída para se encontrar uma solução aproximada.<br />
<br />
Nesta postagem, vamos nos concentrar em problemas de valor inicial (P.V.I.) em sua forma mais simples que são as equações diferenciais ordinárias de primeira ordem:<br />
<br />
<span style="font-style: normal;">\begin{equation} <br />\left\{<br />\begin{array}{ll}<br />\frac{dy}{dx}=f(x,y)\\<br />y(x_0)=y_0<br />\end{array}<br />\right.<br />\end{equation} </span></span></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<br />
onde $y_0$ é um número dado. Os problemas de valor inicial (P.V.I) de ordem superior podem ser reduzidos a sistemas de primeira ordem através de variáveis auxiliares, o que permite a utilização do método numérico aqui apresentado.<br />
<br />
Para resolver numericamente uma E.D.O. com P.V.I. (equação 1), supõem-se que ela satisfaz às condições de existência e unicidade. Esta solução numérica será encontrada para um conjunto finito de pontos (um intervalo fechado $[a,b]$) no eixo das abscissas.<br />
<br />
Tomando-se $m$ subintervalos deste intervalo $[a, b]$, sendo $m \geq 1$, é possível determinar $m+1$ pontos onde as soluções numéricas devem ser calculados. Estes pontos $x_j \in [a, b]$ são igualmente espaçados entre si por um fator $h$, onde $x_j=x_0+j.h$, sendo que $x_0=a, \ \ x_m=b, \ \ h=(b-a)/m \ \ \mbox{e} \ \ j=0, 1, 2, \cdots, m$. O conjunto $\left\{x_0, x_1, x_2, \cdots, x_m \right\}$ obtido denomina-se rede ou malha de $[a, b]$.<br />
<br />
Para facilitar a interpretação dos métodos, convenciona-se a seguinte notação:<br />
<ul>
<li>$y(x_j)$ é a solução exata do P.V.I., obtida analiticamente;</li>
<li>$y_j$ é a solução numérica.</li>
</ul>
Observação: Na solução numérica não se determina a expressão literal da função $y(x)$, mas sim uma solução aproximada do P.V.I. num conjunto discreto de pontos. <br />
<br />
<br />
<b>1.1 - MÉTODO DE EULER</b><br />
<br />
O método de Euler, também conhecido como método da reta tangente, é um dos métodos mais antigos que se conhece para solução de equações diferenciais ordinárias. O método é muito atraente por sua simplicidade, mas não muito usado em problemas práticos, dentre muitos outros métodos, pois apesar de simples, para conseguir "boas" aproximações é necessário um número maior de cálculos. <br />
<br />
<b>1.1.1 - Derivação da Fórmula de Euler</b><br />
<br />
<b> </b>Seja uma E.D.O. com P.V.I. dada pela equação 1. O que se deseja é encontrar as aproximações $y_1, y_2, \cdots, y_m$ para as soluções exatas $y(x_1), y(x_2), \cdots, y(x_m)$.<br />
<br />
Sendo que o ponto inicial $(x_0, y_0)$ é fornecido pelo problema, o primeiro passo então é a busca de $y_1$. Para isto, aproximando-se a solução $y(x)$ por uma Série de Taylor no ponto $x=x_0$ e truncando no segundo termo, tem-se:<br />
<br />
\begin{equation}<br />
y(x)=y(x_0)+y'(x_0).(x-x_0) <br />
\end{equation} <br />
<br />
Para $x=x_1$ tem-se:<br />
<br />
\begin{equation}<br />
y(x_1)=y(x_0)+(x_1-x_0).y'(x_0)<br />
\end{equation} <br />
Lembrando que, como os valores exatos $y(x_n)$ são desconhecidos, são utilizados os valores aproximados $y_n$ e que $x_1-x_0=h$ e $y'(x_0)=f(x_0,y_0)$, onde $h$ é a distância entre os pontos consecutivos $x_n$ e $x_{n+1}$, então:<br />
\begin{equation}<br />
y_1=y_0+h.f(x_0, y_0)<br />
\end{equation} Para encontrar $y_2$ na abscissa $x=x_2$ adota-se o mesmo procedimento (tomando-se ($x_1, y_1$) como ponto de partida). Assim, a solução aproximada é:<br />
\begin{equation}<br />
y_2=y_1+h.f(x_1, y_1)<br />
\end{equation} </div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
Se utilizarmos esse processo recursivas vezes chegaremos a uma fórmula geral, e esse procedimento é chamado método de Euler:<br />
\begin{equation}<br />
y_{n+1}=y_n+h.f(x_n, y_n)<br />
\end{equation} </div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
onde $x_n=x_{n-1}+h=x_0+n.h$, com $n=0, 1, 2, \cdots, m-1$.</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<br />
Assim a solução numérica fornecida pelo método de Euler é a poligonal com vértices $(x_0, y_0), (x_1, y_1), (x_2, y_2), \cdots, (x_{n+1}, y_{n+1})$, onde os segmentos de reta no intervalo $(x_k, x_{k+1})$, $k=0, 1, 2, \cdots, n$ têm como equação $L_k(x)=y_k+f(x_k,y_k)(x-x_k)$.<br />
<br />
</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
</div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
Na figura 1, ilustramos a idéia do método de Euler. Veja que quanto menor o valor da diferença entre $x_{n+1}$ e $x_n$ menor o erro da estimativa para $y_{n+1}$. <br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a 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" 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" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 1: Ilustração do método de Euler</td></tr>
</tbody></table>
<b>1.1.2 - Exemplo usando o Método de Euler</b><br />
<br />
<b> </b>A equação diferencial ordinária de $1^{a}$ ordem escolhida será (Boyce, W. E., Diprima, R. C. Equações Diferenciais Elementares e Problemas de Valor de Contorno. Ed. 7, pp. 420):<br />
<span style="font-style: normal;"><span style="font-style: normal;">\begin{equation} <br />\left\{<br />\begin{array}{ll}<br />\frac{dy}{dx}=1-x+4.y\\<br />y(0)=1<br />\end{array}<br />\right.<br />\end{equation} </span></span><br />
Esta equação será resolvida de $x=0$ a $x=2$. A solução da equação diferencial (7) com a condição inicial (P.V.I.) é conhecida:<br />
\begin{equation}<br />
y(x)=\frac{1}{4}.x-\frac{3}{16}+\frac{19}{16}e^{4.x}<br />
\end{equation} <br />
A solução numérica é encontrada com a avaliação das equações (6):<br />
\begin{equation}<br />
x_{n+1}=x_n+h<br />
\end{equation} </div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
e<br />
\begin{equation}<br />
y_{n+1}=y_n+h.f(x_n, y_n)=y_n+h.(1-x_n+4.y_n)<br />
\end{equation} <br />
Com a condição inicial, $x_0=0$ e $y_0=1$, os próximos valores são calculados com o uso recursivo das equações (9) e (10). Para efeitos de comparação, usaremos dois valores para o passo h=0.01 (Figura 2) e h=0.001 (Figura 3), juntamente com a solução algébrica (8).<br />
<br />
<b>1.1.3 - Resultados</b><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibgZ8HBl3EwLVkiCCtjKSnsFskyrGtGFaMszettL-S3uEGkez1NtrW2SBhyz3vjzM9KqEo4eaui36oTb576Wut-szk5Ys1YzIzzso2Iz8nU-cQalBlEEKyCLVHhXotRZ86PuII5kBxzh0g/s1600/Metodo_Euler.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibgZ8HBl3EwLVkiCCtjKSnsFskyrGtGFaMszettL-S3uEGkez1NtrW2SBhyz3vjzM9KqEo4eaui36oTb576Wut-szk5Ys1YzIzzso2Iz8nU-cQalBlEEKyCLVHhXotRZ86PuII5kBxzh0g/s1600/Metodo_Euler.jpg" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 2: Gráfico apresentando a solução numérica com h=0.01 (método de Euler) e análitica (eq. 8)</td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiL89GH6c_JOgfxlHPRpY9t3ZM7KIe5x7S8MZpfc26co1FRe-M9ovqgc3ZqyhPCyKjVTCzjmsXbeO1ofov8Zt0xj07Yyx7Is18CoJSFH8m9wThQg3cEn_B8m0W9TooHRKdY7xhwTyYuKoD0/s1600/Metodo_Euler.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiL89GH6c_JOgfxlHPRpY9t3ZM7KIe5x7S8MZpfc26co1FRe-M9ovqgc3ZqyhPCyKjVTCzjmsXbeO1ofov8Zt0xj07Yyx7Is18CoJSFH8m9wThQg3cEn_B8m0W9TooHRKdY7xhwTyYuKoD0/s1600/Metodo_Euler.jpg" height="300" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Figura 3: Gráfico apresentando a solução numérica com h=0.001 (método de Euler) e análitica (eq. 8)</td></tr>
</tbody></table>
Por torna-se um processo demasiadamente repetitivo pelo número de iterações, é comum utilizar o método com o auxílio de uma linguagem de programação. <br />
<br />
<b>1.2 - Método de Euler implementado em linguagem de programação. </b><br />
<br />
Um programa escrito em C/C++ é apresentado a seguir. Uma parte do código-fonte pode ser retirada sem problemas, pois trata-se de rotinas voltadas para mostrar a interatividade entre o aplicativo gnuplot (para plotar gráficos) e a linguagem programação C/C++. Os dados obtidos são impressos em um arquivo de nome "metodo_euler.txt", podendo ser plotado usando o Origin, qtiplot, MatLab, SciLab, dentre outros. <br />
<br />
<br />
<span style="font-size: x-small;">/*==========================*/ </span></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<span style="font-size: x-small;">/* Método de Euler */</span></div>
<div align="JUSTIFY" class="western" style="margin-bottom: 0cm;">
<span style="font-size: x-small;">/*==========================*/ <br /><br />#include <stdio.h><br />#include <stdlib.h><br />#include <cmath><br /><br />FILE *fp;<br /><br />static char nome[]="metodo_euler.txt";<br /><br /><span style="color: red;">// Função para plotar o gráfico via o aplicativo gnuplot (opcional no programa)</span><br /><span style="color: red;">void scrip_gnuplot(char str[25])<br />{<br /> fp=fopen("metodo_euler.plt", "w");<br /> <br /> fprintf(fp, "reset\n");<br /> fprintf(fp, "set title \"Metodo de Euler\" \n");<br /> fprintf(fp, "set xlabel \"Xn\" \n");<br /> fprintf(fp, "set ylabel \"Yn\" \n");<br /> fprintf(fp, "set term jpeg \n" );<br /> fprintf(fp, "set output \"Metodo_Euler.jpg\" \n");<br /> fprintf(fp, "plot \'%s\' u 1:2 t \" Numerica \" w lp ls 7 lc 1 lw 1, \'%s\' u 1:3 t \" Analitica \" w l lc -1\n", str, str);<br /> fprintf(fp, "set output; set term wxt \n");<br /> fprintf(fp, "replot \n");<br /> fprintf(fp, "pause -1 \"Continuar?\" ");<br /> fclose(fp);<br />}</span><br /><br />// Definição da equação diferencial: dy/dx=f(x,y)<br />double f(double x, double y)<br />{<br /> return (1-x+4*y);<br />}<br />// Definiçao da solução analitica da equação diferencial<br />double y(double x)<br />{<br /> return ((1.0/4.0)*x - (3.0/16.0) + (19.0/16.0)*exp(4.0*x));<br />}<br /><br />int main(void)<br />{<br />double x0, y0, xmax, h; // variaveis de entrada<br />double xn, yn, xn1, yn1;<br /><br />// Dados de entrada:<br />x0=0.0;<br />xmax=2.0;<br />y0=1.0;<br />h=0.001; // passo h<br />// -----------------<br /> fp=fopen(nome,"w");<br /> <br /> xn=x0;<br /> yn=y0; <br /> while (xn<xmax)<br /> {<br /> fprintf(fp, "%f \t %f \t %f\n", xn, yn, y(xn));<br /> yn1=yn+h*f(xn,yn);<br /> xn1=xn+h;<br /> <br /> xn=xn1;<br /> yn=yn1;<br /> }<br /> fclose(fp);<br /> <span style="color: red;">scrip_gnuplot(nome); // chamada opcional no programa<br /> system("gnuplot metodo_euler.plt"); // chamada opcional no programa</span><br /> return 0;<br />}</span></div>
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<br /></div>
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<b>1.3 - Conclusão</b></div>
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<br /></div>
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<b> </b>Através do Método Numérico de Euler, obtemos uma boa aproximação para a solução $y(x)$ de uma E.D.O. de primeira ordem. No primeiro caso, obtivemos uma aproximação considerada "boa" em boa parte do intervalo considerado, $[0, 2]$. Já no segundo caso, obtivemos uma aproximação coincidente com a solução exata conhecida até onde se observou com passo de integração $h=0.001$, ou seja, o "erro" ou diferença entre a aproximação e a solução exata é praticamente nula para todos os pontos. </div>
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Assim, diante dos exemplos estudados, concluímos que o Método Numérico de Euler é eficaz para resolução de EDOs de primeira ordem, apresentando em alguns casos erros pequenos que podem ser melhores ajustados através da escolha do passo de integração. </div>
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<br /></div>
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<b>1.4 - Referências</b></div>
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<br /></div>
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[1] W. E. Boyce, C. R. Diprima, "Equações Diferenciais Elementares e Problemas de Valor de Contorno", LTC, Rio de Janeiro, 2011.</div>
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[2] M. A. G. Ruggiero, V. L. R. Lopes, "Cálculo Numérico: Aspectos Teóricos Computacionais", Makron Books, São Paulo, 1996.</div>
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[3] G. Baratto, "Solução de Equações Diferenciais Ordinárias Usando Métodos Numéricos", Depto de Eletrônica e Computação - Estudo de Casos em Engenharia Elétrica, 2007.</div>
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<br />
<br />
Nesta
postagem, o nosso objetivo, além de apresentar este importante tema
da Matemática, é o de fornecer exemplos da interatividade entre as
linguagem de programação C/C++ e o aplicativo gnuplot, que juntos
minimizam as tarefas de uma pesquisa. </div>
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<b></b></div>
Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-45563108723651017292015-02-26T11:24:00.001-08:002015-02-26T12:07:47.547-08:00Caos em Sistemas dinâmicos: Diagrama de Bifurgação do Mapa Logístico<style type="text/css">P { margin-bottom: 0.21cm; }P.western { }A:link { }</style>
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Edward
Lorenz, em 1963, estudando a previsão do tempo por um modelo
dinâmico definido por poucas e simples equações diferenciais, fez
uma descoberta que surpreendeu o mundo. Seu modelo seguia um padrão,
que não se encaixava nas categorias possíveis de sistemas
dinâmicos da época, exibindo um comportamento bastante complexo. Sistemas
como o de Lorenz são denominados “caótico-determinísticos”
ou seja, embora apresentem um comportamento aperiódico e
imprevisível, a sua dinâmica é governada por equações
diferenciais ou equações diferenças determinísticas simples.</div>
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<br /></div>
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Um bom
exemplo de um sistema que pode gerar padrões complexos apesar de ter
uma representação simples é o chamado mapa logístico, considerado
como sistema dinâmico exemplar, por sua importância histórica e
por sua natureza didática, e que não é mais do que um sistema
unidimensional definido por,</div>
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<br /></div>
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x(n+1)=<span style="font-family: Liberation Serif, serif;">μ</span>.x(n).[1-x(n)],</div>
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<br /></div>
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onde à
medida que se vai mudando o valor da constante <span style="font-family: Liberation Serif, serif;">μ</span>,
também se vai mudando o comportamento do sistema. A seguir vamos
analisar alguns casos, para diferentes valores de <span style="font-family: Liberation Serif, serif;">μ</span>,
dada a condição inicial Xo=0.55, sendo os espaços de fase
apresentados, aqueles que fazem corresponder x(n+1) a x(n).
</div>
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<br /></div>
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Nesta
postagem, o nosso objetivo, além de apresentar este importante tema
da Física, é o de fornecer exemplos da interatividade entre as
linguagem de programação C/C++ e o aplicativo gnuplot, que juntos
minimizam as tarefas de uma pesquisa. </div>
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<br />
<br /></div>
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<b><span style="color: blue;">Informações sobre o Código Fonte em C/C++ para gerar o mapa logístico</span></b><br />
<br />
O código fonte em C/C++ gera o arquivo de dados de nome "mapa_logistico.txt" e o
arquivo-script do gnuplot de nome "mapa_logistico.plt", e em seguida chama o
script do gnuplot através da função system( ), e o resultado é o gráfico
gerado em formato .jpg no seu diretório corrente.</div>
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<br /></div>
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/* MAPA LOGISTICO */<br />
<br />
#include <stdio.h><br />
#include <stdlib.h><br />
<br />
FILE *fp;<br />
<br />
static char nome[]="mapa_logistico.txt";<br />
<br />
void scrip_gnuplot(char str[25], float m)<br />
{<br />
fp=fopen("mapa_logistico.plt", "w");<br />
<br />
fprintf(fp, "reset\n");<br />
fprintf(fp, "set title \"Mapa Logístico para mi=%.2f\" \n", m);<br />
fprintf(fp, "set xlabel \"Numero de Iteracoes (N)\" \n");<br />
fprintf(fp, "set ylabel \"Valor das Iteracoes (X)\" \n");<br />
fprintf(fp, "set term jpeg \n" );<br />
fprintf(fp, "set output \"Mapa_Logistico.jpg\" \n");<br />
fprintf(fp, "plot \'%s\' t \"mapa x versus n\" w lp ls 7 lc 3 lw 2\n", str);<br />
fprintf(fp, "set output; set term wxt\" \n");<br />
fprintf(fp, "replot \n");<br />
fprintf(fp, "pause -1 \"Continuar?\" ");<br />
fclose(fp);<br />
}<br />
<br />
int main(void)<br />
{<br />
int k;<br />
float n, mi;<br />
double x;<br />
<br />
fp=fopen(nome,"w");<br />
// definindo os valores iniciais de mi, x e n<br />
mi = 4.0; <br />
x = 0.8; <br />
n = 1.0;<br />
for(k=1; k<=100; k++)<br />
{<br />
n=n+1.0;<br />
x = mi*x*(1.0-x);<br />
fprintf(fp,"%.1f \t %lf\n", n, x);<br />
}<br />
fclose(fp);<br />
<br />
scrip_gnuplot(nome, mi);<br />
<br />
system("gnuplot mapa_logistico.plt");<br />
return 0;<br />
}</div>
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<span style="color: blue;"><b>RESULTADOS OBTIDOS</b></span></div>
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<br /></div>
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<span style="color: blue;"><span style="color: black;"><span style="font-family: Comic Sans MS, Arial, Times New Roman; font-size: small;">Parte (I): Iteração
do mapa logístico com o código fonte acima<a href="http://www.geocities.ws/projeto_caos_ufg/minicurso/downloads/index.html" target="_blank"></a>
no intuito de observar o atrator para diferentes valores de
<i>μ</i>. </span></span><b><br /></b></span></div>
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<br /></div>
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</div>
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</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjF80r3JidBI88pbe8vceH9IU26qiN3ZyNj4axccJh2K5q_GYLrGVJVTokdR3RmM2dKiiVsYT0zgA0dnycQ66pHlMTGY-kXmy469kJm86S_wlmBiBDmSnvyMu_Z6NYPHlf7cTkaHNHkoBrs/s1600/Mapa_Henon_2.9.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjF80r3JidBI88pbe8vceH9IU26qiN3ZyNj4axccJh2K5q_GYLrGVJVTokdR3RmM2dKiiVsYT0zgA0dnycQ66pHlMTGY-kXmy469kJm86S_wlmBiBDmSnvyMu_Z6NYPHlf7cTkaHNHkoBrs/s1600/Mapa_Henon_2.9.jpg" height="300" width="400" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigbB7cK8LTYaZH4xdwKc9sLsrnFfOt0AVDezCN4HXLTLmLqR1VGysF1Kwr_9pfl9_BVirWDW4r-iUd6fIYJ8qk7LLc9Su2dBEjLKuGIMbLx0SkMwKkmu8ag3yv2hSe8bTidLfYGRpdQCU_/s1600/Mapa_Henon_3.0.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigbB7cK8LTYaZH4xdwKc9sLsrnFfOt0AVDezCN4HXLTLmLqR1VGysF1Kwr_9pfl9_BVirWDW4r-iUd6fIYJ8qk7LLc9Su2dBEjLKuGIMbLx0SkMwKkmu8ag3yv2hSe8bTidLfYGRpdQCU_/s1600/Mapa_Henon_3.0.jpg" height="300" width="400" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8UkMMqV0QwL6t_4OFntdaxx0zPj74YGBBEs5GcCPNjtpo-sdCWGY73WWWPQ89F9cdORcR6eJANbJMwwf0RaTyHRbZEyaumLadoKWUXxpjWKAEu4rl9A58GSCrKfwEcoO3-U0fcPM3mO44/s1600/Mapa_Henon_3.5.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8UkMMqV0QwL6t_4OFntdaxx0zPj74YGBBEs5GcCPNjtpo-sdCWGY73WWWPQ89F9cdORcR6eJANbJMwwf0RaTyHRbZEyaumLadoKWUXxpjWKAEu4rl9A58GSCrKfwEcoO3-U0fcPM3mO44/s1600/Mapa_Henon_3.5.jpg" height="300" width="400" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNF_wiViT4WNltoTEopJlc9Cvcm_xvUXOANCdhFsEfyt-Ou98SZtgTQXE1QPjjTH-Lj5kGYwbL1ApO6znSFQfHoRxVvXpMhBElsZQtDURHPPNmoSw0sw4g9878FO2Gy6n6z0ol4w8iFZT9/s1600/Mapa_Henon_3.7.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNF_wiViT4WNltoTEopJlc9Cvcm_xvUXOANCdhFsEfyt-Ou98SZtgTQXE1QPjjTH-Lj5kGYwbL1ApO6znSFQfHoRxVvXpMhBElsZQtDURHPPNmoSw0sw4g9878FO2Gy6n6z0ol4w8iFZT9/s1600/Mapa_Henon_3.7.jpg" height="300" width="400" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2-NmvkUQN1wRCgR4tG70x7TAPa_fWhjj6BwG0NcwqnH5_Jx4qWRbzfNHOE8RUhsJLeMPKKtQhD3KH0FFdPgdJMKPk9oEUjvZuzP1R1LItSe_XHN8hrAlIlv3HBrcfj4jvOApcXpzf1H6S/s1600/Mapa_Henon_4.0.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2-NmvkUQN1wRCgR4tG70x7TAPa_fWhjj6BwG0NcwqnH5_Jx4qWRbzfNHOE8RUhsJLeMPKKtQhD3KH0FFdPgdJMKPk9oEUjvZuzP1R1LItSe_XHN8hrAlIlv3HBrcfj4jvOApcXpzf1H6S/s1600/Mapa_Henon_4.0.jpg" height="300" width="400" /></a></div>
<br />
<b><span style="color: blue;">Informações do Código Fonte em C/C++ para gerar o diagrama de bifurgação para o mapa logístico</span></b><br />
<br />
<div style="text-align: justify;">
Este código fonte em C/C++ gera o arquivo de dados de nome "diagrama.txt" e o
arquivo-script do gnuplot de nome "diagrama.plt", e em seguida chama o
script do gnuplot através da função system( ), e o resultado é o gráfico
gerado em formato .jpg no seu diretório corrente.</div>
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</div>
<b><span style="color: blue;"></span></b><br />
<br />
/* DIAGRAMA DE BIFURGAÇÃO DO MAPA LOGISTICO */<br />
<br />
#include <stdio.h><br />
#include <stdlib.h><br />
<br />
FILE *fp;<br />
static char nome[]="diagrama.txt";<br />
void scrip_gnuplot(char str[25])<br />
{<br />
fp=fopen("diagrama.plt", "w");<br />
<br />
fprintf(fp, "reset\n");<br />
fprintf(fp, "set samples 500\n");<br />
fprintf(fp, "set xr[1.9:4.1]\n");<br />
fprintf(fp, "set yr[0:1.1]\n");<br />
fprintf(fp, "set title \"Diagrama do Mapa Logistico\" \n");<br />
fprintf(fp, "set xlabel \"Coeficiente mi\" \n");<br />
fprintf(fp, "set ylabel \"Atrator X\" \n");<br />
fprintf(fp, "set term jpeg \n" );<br />
fprintf(fp, "set output \"Diagrama.jpg\" \n");<br />
fprintf(fp, "plot \'%s\' notitle w p ls 1 lw 0\n", str);<br />
fprintf(fp, "set output; set term wxt\" \n");<br />
fprintf(fp, "replot \n");<br />
fprintf(fp, "pause -1 \"Continuar?\" ");<br />
fclose(fp);<br />
}<br />
<br />
int main(void)<br />
{<br />
int k, n;<br />
float mi;<br />
double x;<br />
<br />
fp=fopen(nome,"w");<br />
// definindo o valoar inicial de mi<br />
mi = 2.0;<br />
for(k=1; k<=500; k++)<br />
{<br />
mi = mi + 0.004;<br />
x = 0.8;<br />
for(n=1; n<=1000; n++)<br />
{<br />
x = mi*x*(1-x);<br />
if (n>500)<br />
{<br />
fprintf(fp,"%f \t %lf\n", mi, x);<br />
}<br />
}<br />
} <br />
fclose(fp);<br />
scrip_gnuplot(nome);<br />
system("gnuplot diagrama.plt");<br />
return 0;<br />
}<b><span style="color: blue;"><br /></span></b><span style="color: blue;"><span style="color: black;"><span style="font-family: Comic Sans MS, Arial, Times New Roman; font-size: small;">Parte (II): </span></span></span><span style="font-family: Comic Sans MS, Arial, Times New Roman; font-size: small;">Construção
do diagrama do mapa logístico a partir do código acima
para observar a rota de duplicação de período. </span><br />
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</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEju1wLMT8DinJXm-T16CQAjfOgPq6UIJnqTdl2SG6WIO_ZOeI5kZWE6Fq83wJDHEKkQXWGfydNGzzUCtqVqZznsTOUPyYBI6xM21cAAu7_rxaAntM2vOhoS4Gow2OFeAfMiF1NXyuh445zJ/s1600/Diagrama.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEju1wLMT8DinJXm-T16CQAjfOgPq6UIJnqTdl2SG6WIO_ZOeI5kZWE6Fq83wJDHEKkQXWGfydNGzzUCtqVqZznsTOUPyYBI6xM21cAAu7_rxaAntM2vOhoS4Gow2OFeAfMiF1NXyuh445zJ/s1600/Diagrama.jpg" height="300" width="400" /></a></div>
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<div style="text-align: justify;">
<span style="font-family: Comic Sans MS, Arial, Times New Roman; font-size: small;">Neste diagrama estão representados apenas os pontos
referentes aos atratores dos mapas logísticos, para
diferentes valores de μ. Não estão representados
os pontos do <b>transiente</b>! <br />
Vemos que o atrator foi ficando cada vez mais complicado:
para os valores de μ entre 2,0 e aproximadamente 2,9
é do tipo ponto fixo, na primeira <b>bifurcação</b>,
acima de 3,0, é duplo ciclo até aproximadamente
3,4, onde já passam a ser 4 pontos de repetição,
depois 8 e assim por diante... A cada bifurcação
ocorre uma <b>duplicação de período</b>
até o sistema entrar em regime caótico. Por
isso essa <b>rota</b> para o Caos ficou conhecida
como <b>rota de duplicação de período</b>. </span></div>
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Para
maiores informações sobre a teoria dos caos, o link a seguir é
uma ótima fonte de pesquisa:
<a href="http://www.geocities.ws/projeto_caos_ufg/">http://www.geocities.ws/projeto_caos_ufg/</a></div>
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com1tag:blogger.com,1999:blog-3859540945177944400.post-85180243403291051132015-02-25T10:49:00.002-08:002015-02-25T10:56:48.657-08:00Interação entre a Linguagem C e o Aplicativo gnuplot: O Mapa de Hénon<div style="text-align: justify;">
O <b>mapa de Hénon</b>, proposto em 1976 pelo astrônomo francês Michel Hénon como um modelo para descrever a seção de Poincaré do sistema de Lorentz a tempo contínuo, é governado por três equações diferenciais ordinárias não lineares de primeira ordem. </div>
<div style="text-align: justify;">
<br /></div>
O mapa de Hénon é definido pelo seguinte conjunto de equações:<br />
<br />
<br />
<br />
x(t+1) = a - x(t)² + b.y(t) e y(t+1)=x(t).<br />
<br />
<div style="text-align: justify;">
onde <b><i>a</i></b> é o parâmetro de não-linearidade, <b><i>b</i></b> o parâmetro de dissipação do sistema, <b><i>x(t)</i></b> e <b><i>y(t)</i></b> são variáveis dinâmicas, e <b><i>t</i></b> = 0, 1, 2, ...., é o tempo discreto. É um dos exemplos mais estudados de sistemas dinâmicos pois, apesar de ser descrito por um conjunto de equações bastante simples, o mapa de Hénon apresenta uma dinâmica extremamente rica. Por exemplo, para os valores dos parâmetros <i><b>a=1.4</b></i> e <i><b>b=0.3</b></i>, que foram os valores utilizados por M. Hénon em seu trabalho, o mapa apresenta um comportamento caótico. No entanto, para outros conjuntos de valores de parâmetros, o mapa de Hénon pode convergir para uma órbita periódica, caótica ou mesmo divergir. </div>
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Neste post, temos dois objetivos, o primeiro é o de apresentar o mapa de Hénon para três conjuntos de valores para os parâmetros <i><b>a</b></i> e <i><b>b</b></i>, conforme as figuras abaixo, e o segundo é o de apresentar a interação entre a linguagem de programação C/C++ com o aplicativo gnuplot, que juntos otimizam bastante as tarefas do pesquisador. </div>
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O código fonte em C gera o arquivo de dados de nome "mapa.txt" e o arquivo-script do gnuplot de nome "mapa.plt", e em seguida chama o script do gnuplot através da função system( ), e o resultado é o gráfico gerado em formato .jpeg no seu diretório corrente automáticamente. </div>
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<b><span style="color: blue;">Código fonte em C</span></b></div>
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#include <stdio.h><br />
#include <stdlib.h><br />
#include <math.h><br />
<br />
#define a 1.4 // Parâmetros do Mapa de Hénon<br />
#define b 0.3 // (a, b) = (1.2,0.2); (1.3,0.3) e (1.4,0.3)<br />
#define T 1000 // Tempo total discreto<br />
<br />
double x[T], y[T];<br />
FILE *fp;<br />
<br />
static char nome[]="mapa.txt";<br />
<br />
void scrip_gnuplot(char str[20])<br />
{<br />
fp=fopen("mapa.plt", "w");<br />
<br />
fprintf(fp, "reset\n");<br />
fprintf(fp, "set title \"Mapa de Hénon para a=%.2f, b=%.2f \" \n", a, b);<br />
fprintf(fp, "set xl \"Xt+1\" \n");<br />
fprintf(fp, "set yl \"Yt+1\" \n");<br />
fprintf(fp, "set term jpeg \n" );<br />
fprintf(fp, "set output \"Mapa_Henon.jpg\" \n");<br />
fprintf(fp, "plot \'%s\' u ($1):($2) t \'%s\' w p ls 7 lc 3 \n", str,str);<br />
fprintf(fp, "set output; set term wxt\" \n");<br />
fprintf(fp, "replot \n");<br />
fprintf(fp, "pause -1 \"Continuar?\" ");<br />
fclose(fp);<br />
}<br />
<br />
int main()<br />
{<br />
int t; <br />
<br />
fp=fopen(nome,"w");<br />
<br />
x[0]=1;<br />
y[0]=1;<br />
for(t=0;t<=T;t++)<br />
{<br />
x[t+1]=a-pow(x[t],2)+b*y[t];<br />
y[t+1]=x[t]; <br />
fprintf(fp,"%lf %15lf \n",x[t+1],y[t+1]);<br />
}<br />
fclose(fp); <br />
<br />
scrip_gnuplot(nome);<br />
system("gnuplot mapa.plt");<br />
return 0;<br />
}<br />
// --------------------------------------------------------------------</div>
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<b><span style="color: blue;"><br /></span></b></div>
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<b><span style="color: blue;">Mapa de Hénon</span></b></div>
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Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-16044638845554635222015-02-20T17:05:00.006-08:002019-01-01T12:10:03.462-08:00Animação com gnuplot: Satélite em órbita<span style="color: blue;"><b>Satélite em órbita em torno da Terra</b></span><br />
<span style="color: blue;"><b><br /></b></span>
Caros leitores, o problema do satélite em órbita é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes na animação (circle, arrow, e label) são implementados facilmente. As equações físicas utilizadas são as equações trabalhadas nas aulas de Física.<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-5313866374459696472015-02-20T16:48:00.001-08:002019-01-01T11:30:32.439-08:00Animação com gnuplot: Pêndulo Simples<span style="color: blue;"><b>Pêndulo Simples</b></span><br />
<span style="color: blue;"><b><br /></b></span>
<br />
Caros leitores, o problema do pêndulo simples é aqui construído segundo os comandos básicos do aplicativo gnuplot. Os objetos presentes (circle, arrow e label) na animação são implementados facilmente.<br />
<br />
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<br />Anonymoushttp://www.blogger.com/profile/09054344204176446253noreply@blogger.com0tag:blogger.com,1999:blog-3859540945177944400.post-13308546956511207912015-02-20T16:35:00.001-08:002019-01-01T12:20:18.669-08:00Animação com gnuplot: Problema do Plano Inclinado<div>
<b><span style="color: blue;">Plano Inclinado</span></b></div>
<div>
<b><span style="color: blue;"><br /></span></b></div>
Caros leitores, o problema do plano inclinado é aqui construído segundo os comandos básicos do aplicativo gnuplot. A dificuldade maior está na construção do polígono que desce o plano inclinado. Os demais objetos (arrow, polygon, circle, label) são bastante simples.<br />
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