{"id":6694,"date":"2020-09-22T19:59:28","date_gmt":"2020-09-23T00:59:28","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=6694"},"modified":"2026-01-16T10:25:35","modified_gmt":"2026-01-16T15:25:35","slug":"3eva2020paoi_t3-edp-parabolica-centro-triangulo","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva20\/3eva2020paoi_t3-edp-parabolica-centro-triangulo\/","title":{"rendered":"3Eva2020PAOI_T3 EDP Parab\u00f3lica centro tri\u00e1ngulo"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">3ra Evaluaci\u00f3n 2020-2021 PAO I. 22\/Septiembre\/2020<\/h2>\n\n\n\n<p><strong>Tema 3<\/strong>. (35 puntos) Desarrolle con el m\u00e9todo impl\u00edcito para aproximar la soluci\u00f3n de la EDP Parab\u00f3lica<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{\\partial u}{\\partial t} - c^2 \\frac{\\partial ^2 u}{\\partial x^2} = g(x) <\/span>\n\n\n\n<p>con las siguientes condiciones:<\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-28f84493 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<p>u(0,t) = 0<\/p>\n\n\n\n<p>u(1,t) = 0<\/p>\n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<p>u(x,0) = f(x)<\/p>\n<\/div>\n<\/div>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> f(x) = \\begin{cases} 5x , &amp; 0 \\le x \\le 0.5 \\\\ 5(1-x) , &amp; 0.5 \\lt x \\le 1\\end{cases}<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> g(x) = 2 , 0 \\le x \\le 1<\/span>\n\n\n\n<figure class=\"wp-block-image alignright size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"640\" height=\"480\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2020\/09\/3Eva2020PAOI_T3_EDP_ParabolicaCondInicial.png\" alt=\"3Eva2020PAOI_T3 EDP Parab\u00f3lica Condici\u00f3n Inicial\" class=\"wp-image-21039\" style=\"width:350px\" \/><\/figure>\n\n\n\n<p>Considere para h=0.25, k=0.05, c=1<\/p>\n\n\n\n<p>a. Realice la gr\u00e1fica de malla,<\/p>\n\n\n\n<p>b. Escriba las ecuaciones para las derivadas en forma discreta<\/p>\n\n\n\n<p>c. Desarrolle y obtenga el modelo discreto para u(x<sub>i<\/sub>,t<sub>j<\/sub>) que sea de inter\u00e9s<\/p>\n\n\n\n<p>d. Realice al menos tres iteraciones en el eje tiempo.<\/p>\n\n\n\n<p>e. Estime el error (solo plantear) y revise convergencia<\/p>\n\n\n\n<p><strong>R\u00fabrica<\/strong>: literal a (5 puntos), literal b (5puntos), literal c (10 puntos), literal d (10 puntos), literal e (5 puntos)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p><strong>Nota<\/strong>: gr\u00e1fica de f(x) en Python<\/p>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: python; title: ; notranslate\" title=\"\">\n#3ra Evaluaci\u00f3n 2020-2021 PAO I. 22\/Septiembre\/2020\n# Grafica condici\u00f3n inicial\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\nf1 = lambda x: 5*x\nf2 = lambda x: 5*(1-x)\nfx = lambda x: np.piecewise(x,\n                            &#x5B;x&lt;0.5, x&gt;=0.5],\n                            &#x5B;f1,f2])\na = 0 # intervalo &#x5B;a,b] en eje x\nb = 1\nmuestras = 11\n\n# PROCEDIMIENTO\nxi = np.linspace(a,b, muestras)\nfi = fx(xi)\n\n# SALIDA\nplt.plot(xi,fi)\nplt.xlabel('xi')\nplt.ylabel('u(x,0)')\nplt.title('u(x,0)')\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"<p>3ra Evaluaci\u00f3n 2020-2021 PAO I. 22\/Septiembre\/2020 Tema 3. (35 puntos) Desarrolle con el m\u00e9todo impl\u00edcito para aproximar la soluci\u00f3n de la EDP Parab\u00f3lica con las siguientes condiciones: u(0,t) = 0 u(1,t) = 0 u(x,0) = f(x) Considere para h=0.25, k=0.05, c=1 a. Realice la gr\u00e1fica de malla, b. Escriba las ecuaciones para las derivadas en [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[28],"tags":[57],"class_list":["post-6694","post","type-post","status-publish","format-standard","hentry","category-mn-3eva20","tag-edp"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/6694","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=6694"}],"version-history":[{"count":8,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/6694\/revisions"}],"predecessor-version":[{"id":21042,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/6694\/revisions\/21042"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=6694"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=6694"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=6694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}