{"id":1077,"date":"2018-02-10T13:21:20","date_gmt":"2018-02-10T18:21:20","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=1077"},"modified":"2026-04-05T20:15:13","modified_gmt":"2026-04-06T01:15:13","slug":"s2eva2017tii_t4-problema-con-valor-de-frontera","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2017tii_t4-problema-con-valor-de-frontera\/","title":{"rendered":"s2Eva2017TII_T4 Problema con valor de frontera"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2017tii_t4-problema-con-valor-de-frontera\/\" data-type=\"post\" data-id=\"1077\">2Eva2017TII_T4 Problema con valor de frontera<\/a><\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{d^2T}{dx^2} + \\frac{1}{x}\\frac{dT}{dx} +S =0 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> 0 \\leq x \\leq 1 <\/span>\n\n\n\n<p>Las diferencias finitas a usar son:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{dT}{dx} =\\frac{T_{i+1} - T_{i-1}}{2h}+O(h^2) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{d^2T}{dx^2}=\\frac{T_{i+1} - 2T_i + T_{i-1}}{h^2}+O(h^2) <\/span>\n\n\n\n<p>que al reemplazar el la ecuaci\u00f3n:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{T_{i+1} - 2T_i + T_{i-1}}{h^2} + \\frac{1}{x_i}\\frac{T_{i+1} -T_{i-1}}{2h}+S=0 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> 2x_i (T_{i+1} - 4h x_i T_i + T_{i-1}) + h (T_{i+1} - T_{i-1}) = -2h^2 S x_i <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> T_{i+1}(2x_i + h) - 4x_i T_i + T_{i-1}(2x_i - h) = -2h^2 S x_i <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> T_{i-1}(2x_i - h) - 4x_i T_i + T_{i+1}(2x_i + h)= -2h^2 S x_i <\/span>\n\n\n\n<p>con lo que se puede crear un sistema de ecuaciones para cada valor xi con T<sub>0<\/sub>=2 y T<sub>4<\/sub> =1 que son parte del enunciado del problema.<\/p>\n\n\n\n<p>Siendo h = 0.25 = 1\/4,&nbsp; y se indica al final que S=1, se crea un sistema de ecuaciones a resolver,<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>x = &#091;0, 1\/4, 1\/4, 3\/4, 1]<\/code><\/pre>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> T_{i-1}\\Big[2x_i - \\frac{1}{4} \\Big] - 4x_i T_i + T_{i+1}\\Big[2x_i + \\frac{1}{4} \\Big] = -2\\Big(\\frac{1}{4}\\Big)^2 (1) x_i <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> T_{i-1}\\Big[2x_i -\\frac{1}{4}\\Big] - 4x_i T_i + T_{i+1}\\Big[2x_i + \\frac{1}{4}\\Big] = -\\frac{1}{8} x_i <\/span>\n\n\n\n<p>se sustituye con los valores conocidos para cada i:<\/p>\n\n\n\n<pre class=\"wp-block-code alignwide\"><code>i=1: \nT<sub>0<\/sub>&#091;2(1\/4) - 1\/4] - 4(1\/4)T<sub>1<\/sub> + T<sub>2<\/sub>&#091;2(1\/4) + 1\/4] = -(1\/8)(1\/4)\n\n     - T<sub>1<\/sub> + (3\/4)T<sub>2<\/sub> = -1\/32 - (1\/4)(2)\n     - T<sub>1<\/sub> + (3\/4)T<sub>2<\/sub> = -17\/32\n\ni=2: \nT<sub>1<\/sub>&#091;2(1\/2) - 1\/4] - 4(1\/2)T<sub>2<\/sub> + T<sub>3<\/sub>&#091;2(1\/2) + 1\/4] = -(1\/8)(1\/2)\n\n     (3\/4)T<sub>1<\/sub> - 2T<sub>2<\/sub> + (5\/4)T<sub>3<\/sub> = -1\/16\n\ni=3: \nT<sub>2<\/sub>&#091;2(3\/4) - 1\/4] - 4(3\/4)T<sub>3<\/sub> + T<sub>4<\/sub>&#091;2(3\/4) + 1\/4] = -(1\/8)(3\/4)\n\n     (5\/4)T<sub>2<\/sub> - 3T<sub>3<\/sub> = -3\/32 - (7\/4)(1)\n     (5\/4)T<sub>2<\/sub> - 3T<sub>3<\/sub> = -59\/32\n<\/code><\/pre>\n\n\n\n<p>se ponen las ecuaciones en matrices para resolver con alg\u00fan m\u00e9todo num\u00e9rico:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A = &#091;&#091; -1, 3\/4,   0],\n     &#091;3\/4,  -2, 5\/4],\n     &#091;  0, 5\/4,  -3]]\nB = &#091;-17\/32, -1\/16, -59\/32]\nnp.linalg.solve(A,B)\narray(&#091; 1.54,  1.34,  1.17])<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 2Eva2017TII_T4 Problema con valor de frontera Las diferencias finitas a usar son: que al reemplazar el la ecuaci\u00f3n: con lo que se puede crear un sistema de ecuaciones para cada valor xi con T0=2 y T4 =1 que son parte del enunciado del problema. Siendo h = 0.25 = 1\/4,&nbsp; y se indica al [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[58,54],"class_list":["post-1077","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1077","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=1077"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1077\/revisions"}],"predecessor-version":[{"id":23866,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1077\/revisions\/23866"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1077"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1077"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1077"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}