{"id":4100,"date":"2019-08-29T11:32:01","date_gmt":"2019-08-29T16:32:01","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4100"},"modified":"2026-04-05T20:11:31","modified_gmt":"2026-04-06T01:11:31","slug":"s2eva2019ti_t3-edp-eliptica-placa-6x5","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2019ti_t3-edp-eliptica-placa-6x5\/","title":{"rendered":"s2Eva2019TI_T3 EDP El\u00edptica Placa 6x5"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2019TI_T3 EDP El\u00edptica Placa 6\u00d75<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n se discretiza con diferencias divididas centradas<\/p>\n\n\n\\frac{\\partial^2 u}{\\partial x^2} (x,y)+\\frac{\\partial ^2 u}{\\partial y^2 } (x,y) = -\\frac{q}{K}<\/span>\n\n\n \\frac{u_{i+1,j}-2u_{i,j}+u_{i-1,j}}{(\\Delta x)^2} + \\frac{u_{i,j+1}-2u_{i,j}+u_{i,j-1}}{(\\Delta y)^2}= -\\frac{q}{K} <\/span>\n\n\n\n

Para procesar los datos se toma como referencia la malla<\/p>\n\n\n\n

se agrupan los t\u00e9rminos conocidos de las diferencias divididas<\/p>\n\n\n \\frac{(\\Delta y)^2}{(\\Delta x)^2} \\Big( u_{i+1,j}-2u_{i,j} +u_{i-1,j} \\Big)<\/span>\n\n\n + u_{i,j+1}-2u_{i,j}+u_{i,j-1}=-\\frac{q}{K}(\\Delta y)^2 <\/span>\n\n\n\n

Considerando que hx<\/sub> =2, hy<\/sub>= 2.5 son diferentes, se mantiene el valor de \u03bb para el desarrollo.<\/p>\n\n\n \\lambda = \\frac{(\\Delta y)^2}{(\\Delta x)^2} = \\frac{(2.5)^2}{(2)^2} = 1.5625<\/span>\n\n\n \\lambda u_{i+1,j}-2\\lambda u_{i,j} +\\lambda u_{i-1,j} + u_{i,j+1}-2u_{i,j}+u_{i,j-1}=-\\frac{q}{K}(\\Delta y)^2 <\/span>\n\n\n\n

Se agrupan t\u00e9rminos de u que son iguales<\/p>\n\n\n \\lambda u_{i+1,j}-2(1+\\lambda) u_{i,j} +\\lambda u_{i-1,j} + u_{i,j+1}+u_{i,j-1}=-\\frac{q}{K}(\\Delta y)^2 <\/span>\n\n\n\n

con lo que se puede generar el sistema de ecuaciones a resolver para los nodos de la malla.<\/p>\n\n\n\n

i = 1 , j=1<\/p>\n\n\n 1.5625 u_{2,1}-2(1+ 1.5625) u_{1,1} + 1.5625 u_{0,1} + <\/span>\n\n\n +u_{1,2}+u_{1,0}=-\\frac{1.5}{1.04}(2.5)^2 <\/span>\n\n\n 1.5625 u_{2,1}-5.125u_{1,1} + 1.5625 y_1(5-y_1) +<\/span>\n\n\n+ 0+x_i(6-x_i)=-9.0144 <\/span>\n\n\n 1.5625 u_{2,1}-5.125u_{1,1} + 1.5625[ 1(5-1)]+ <\/span>\n\n\n+ 0+2(6-2)=-9.0144 <\/span>\n\n\n 1.5625 u_{2,1}-5.125 u_{1,1} =<\/span>\n\n\n =-9.0144 - 1.5625 [1(5-1)] - 0-2(6-2)<\/span>\n\n\n\n

 <\/p>\n\n\n 1.5625 u_{2,1}-5.125 u_{1,1} =-23.2644<\/span>\n\n\n\n

i=2, j=1<\/p>\n\n\n 1.5625 u_{3,1}-2(1+1.5625) u_{2,1} +1.5625 u_{1,1} + <\/span>\n\n\n+u_{2,2}+u_{2,0}=-9.0144 <\/span>\n\n\n 1.5625 (0)-5.125 u_{2,1} +1.5625 u_{1,1} + 0 +x_2(6-x_2)=-9.0144 <\/span>\n\n\n-5.125 u_{2,1} +1.5625 u_{1,1} + 0 +4(6-4)=-9.0144 <\/span>\n\n\n\n

 <\/p>\n\n\n-5.125 u_{2,1} +1.5625 u_{1,1} =-17.0144 <\/span>\n\n\n\n

las ecuaciones a resolver son:<\/p>\n\n\n 1.5625 u_{2,1}-5.125 u_{1,1} =-23.2644<\/span>\n\n\n-5.125 u_{2,1} +1.5625 u_{1,1} =-17.0144 <\/span>\n\n\n\n

cuya soluci\u00f3n es:<\/p>\n\n\n u_{2,1} = 5.1858<\/span>\n\n\n u_{1,1} =6.1204 <\/span>\n\n\n\n

Resuelto usando:<\/p>\n\n\n\n

import<\/span> numpy as<\/span> np\nA = np.array([[1.5625,-5.125],\n             [-5.125, 1.5625]])\nB = np.array([-23.2644,-17.0144])\nx = np.linalg.solve(A,B)\nprint<\/span>(x)<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2019TI_T3 EDP El\u00edptica Placa 6\u00d75 La ecuaci\u00f3n se discretiza con diferencias divididas centradas Para procesar los datos se toma como referencia la malla se agrupan los t\u00e9rminos conocidos de las diferencias divididas Considerando que hx =2, hy= 2.5 son diferentes, se mantiene el valor de \u03bb para el desarrollo. Se agrupan t\u00e9rminos de u […]<\/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-4100","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\/4100","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=4100"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4100\/revisions"}],"predecessor-version":[{"id":23855,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4100\/revisions\/23855"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}