{"id":7678,"date":"2021-08-31T20:01:19","date_gmt":"2021-09-01T01:01:19","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=7678"},"modified":"2026-04-05T20:49:58","modified_gmt":"2026-04-06T01:49:58","slug":"s2eva2021paoi_t3-edp-eliptica-valores-frontera-fxgy","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2021paoi_t3-edp-eliptica-valores-frontera-fxgy\/","title":{"rendered":"s2Eva2021PAOI_T3 EDP El\u00edptica con valores en la frontera f(x) g(y)"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2021PAOI_T3 EDP El\u00edptica con valores en la frontera f(x) g(y)<\/a><\/p>\n\n\n\n

Dada la EDP el\u00edptica<\/p>\n\n\n \\frac{\\partial ^2 u}{\\partial x^2} +\\frac{\\partial^2 u}{\\partial y^2} = 0 <\/span>\n\n\n 0 \\lt x \\lt \\frac{1}{2}, 0 \\lt y\\lt \\frac{1}{2} <\/span>\n\n\n\n

Se convierte a la versi\u00f3n discreta usando diferencias divididas centradas:<\/p>\n\n\n\n

\"2E2020PAOIT3<\/figure>\n\n\n\n
\n\n\n \\frac{u[i-1,j]-2u[i,j]+u[i+1,j]}{\\Delta x^2} +<\/span>\n\n\n + \\frac{u[i,j-1]-2u[i,j]+u[i,j+1]}{\\Delta y^2} = 0 <\/span>\n\n\n\n
\n\n\n\n

Se agrupan los t\u00e9rminos \u0394x, \u0394y semejante a formar un \u03bb al multiplicar todo por \u0394y2<\/sup><\/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 + \\frac{\\Delta y^2}{\\Delta y^2}\\Big(u[i,j-1]-2u[i,j]+u[i,j+1]\\Big) = 0<\/span>\n\n\n\n

\n\n\n\n

los tama\u00f1os de paso en ambos ejes son de igual valor, se simplifica la ecuaci\u00f3n<\/p>\n\n\n \\lambda= \\frac{\\Delta y^2}{\\Delta x^2} = 1<\/span>\n\n\n\n

\n\n\n u[i-1,j]-2u[i,j]+u[i+1,j] +<\/span>\n\n\n + u[i,j-1]-2u[i,j]+u[i,j+1] = 0<\/span>\n\n\n\n
\n\n\n u[i-1,j]-4u[i,j]+u[i+1,j] +<\/span>\n\n\n + u[i,j-1]+u[i,j+1] = 0<\/span>\n\n\n\n
\n\n\n\n

que permite plantear las ecuaciones para cada punto en posici\u00f3n [i,j]<\/p>\n\n\n\n

En cada iteraci\u00f3n se requiere el uso de los valores en la frontera<\/p>\n\n\n u(x,0)=0, 0 \\leq x \\leq \\frac{1}{2} <\/span>\n\n\nu(0,y)=0 , 0\\leq y \\leq \\frac{1}{2} <\/span>\n\n\n u\\Big(x,\\frac{1}{2} \\Big) = 200 x , 0 \\leq x \\leq \\frac{1}{2} <\/span>\n\n\n u\\Big(\\frac{1}{2} ,y \\Big) = 200 y , 0 \\leq y \\leq \\frac{1}{2} <\/span>\n\n\n\n

Iteraciones<\/strong><\/p>\n\n\n\n

i=1, j=1<\/p>\n\n\n u[0,1]-4u[1,1]+u[2,1] +<\/span>\n\n\n + u[1,0]+u[1,2] = 0<\/span>\n\n\n\n

usando los valores en la frontera,<\/p>\n\n\n 0-4u[1,1]+u[2,1] + 0+u[1,2] = 0<\/span>\n\n\n -4u[1,1]+u[2,1] + u[1,2] = 0<\/span>\n\n\n\n

\n\n\n\n

i=2, j=1<\/p>\n\n\n u[1,1]-4u[2,1]+u[3,1] +<\/span>\n\n\n + u[2,0]+u[2,2] = 0<\/span>\n\n\n\n

usando los valores en la frontera,<\/p>\n\n\n u[1,1]-4u[2,1]+200(1\/6) + 0+u[2,2] = 0<\/span>\n\n\n u[1,1]-4u[2,1] + u[2,2] = -200(1\/6)<\/span>\n\n\n\n

\n\n\n\n

i=1, j=2<\/p>\n\n\n u[0,2]-4u[1,2]+u[2,2] +<\/span>\n\n\n + u[1,1]+u[1,3] = 0<\/span>\n\n\n 0 - 4u[1,2]+u[2,2] + u[1,1]+200\\frac{1}{6} = 0<\/span>\n\n\n - 4u[1,2] + u[2,2]+u[1,1] = -200\\frac{1}{6}<\/span>\n\n\n\n

\n\n\n\n

i=2, j=2<\/p>\n\n\n u[1,2]-4u[2,2]+u[3,2] +<\/span>\n\n\n + u[2,1]+u[2,3] = 0<\/span>\n\n\n u[1,2]-4u[2,2]+200\\frac{2}{6} + u[2,1]+200\\frac{2}{6} = 0<\/span>\n\n\n u[1,2]-4u[2,2]+ u[2,1] = -(2)200\\frac{2}{6} <\/span>\n\n\n\n

Sistema de ecuaciones a resolver:<\/p>\n\n\n \\begin{bmatrix} -4 & 1 & 1 & 0\\\\1 & -4 0 & 1\\\\1& 0 & -4 & 1 \\\\ 0 & 1 & 1 & -4\\end{bmatrix} \\begin{bmatrix} u[1,1]\\\\u[2,1] \\\\u[1,2]\\\\u[2,2] \\end{bmatrix} = \\begin{bmatrix} 0\\\\-200(1\/6)\\\\-200(1\/6)\\\\-200(4\/6) \\end{bmatrix} <\/span>\n\n\n\n

Resolviendo el sistema se tiene:<\/p>\n\n\n\n

[11.11111111 22.22222222 22.22222222 44.44444444]<\/code><\/pre>\n\n\n\n
Instrucciones Python<\/h2>\n\n\n
\nimport numpy as np\n\nA = np.array([[-4, 1, 1, 0],\n              [ 1,-4, 0, 1],\n              [ 1, 0,-4, 1],\n              [ 0, 1, 1,-4]])\n\nB = np.array([0,-200*(1\/6),-200*(1\/6),-200*(4\/6)])\n\nx= np.linalg.solve(A,B)\n\nprint(x)\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2021PAOI_T3 EDP El\u00edptica con valores en la frontera f(x) g(y) Dada la EDP el\u00edptica Se convierte a la versi\u00f3n discreta usando diferencias divididas centradas: Se agrupan los t\u00e9rminos \u0394x, \u0394y semejante a formar un \u03bb al multiplicar todo por \u0394y2 los tama\u00f1os de paso en ambos ejes son de igual valor, se simplifica la ecuaci\u00f3n […]<\/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":[49],"tags":[58,54],"class_list":["post-7678","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7678","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=7678"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7678\/revisions"}],"predecessor-version":[{"id":23906,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7678\/revisions\/23906"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=7678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=7678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=7678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}