{"id":4007,"date":"2017-12-07T09:00:58","date_gmt":"2017-12-07T14:00:58","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4007"},"modified":"2026-04-05T20:55:41","modified_gmt":"2026-04-06T01:55:41","slug":"s3eva2007tii_t1-edp-eliptica-problema-de-frontera","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2007tii_t1-edp-eliptica-problema-de-frontera\/","title":{"rendered":"s3Eva2007TII_T1 EDP Eliptica, problema de frontera"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2007TII_T1 EDP Eliptica, problema de frontera<\/a><\/p>\n\n\n\n

Con los datos del ejercicio se plantean de la siguiente forma en los bordes:<\/p>\n\n\n\n

\"EDP<\/figure>\n\n\n\n

La ecuaci\u00f3n a resolver es:<\/p>\n\n\n \\frac{\\delta^2u}{\\delta x^2} +\\frac{\\delta^2 u}{\\delta y^2} = 4 <\/span>\n\n\n\n

Que en su forma discreta, con diferencias divididas centradas es:<\/p>\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} = 4 <\/span>\n\n\n\n

Al agrupar constantes se convierte en:<\/p>\n\n\n u[i-1,j]-2u[i,j]+u[i+1,j] + <\/span>\n\n\n + \\frac{(\\Delta x)^2}{(\\Delta y)^2} \\Big(u[i,j-1]-2u[i,j]+u[i,j+1] \\Big)= 4(\\Delta x)^2 <\/span>\n\n\n\n

Siendo \u0394x= 1\/3 y \u0394y =2\/3, se mantendr\u00e1 la relaci\u00f3n \u03bb = (\u0394x\/\u0394y)2<\/sup><\/p>\n\n\n u[i-1,j]-2u[i,j]+u[i+1,j] + <\/span>\n\n\n + \\lambda u[i,j-1]-2\\lambda u[i,j]+\\lambda u[i,j+1] = <\/span>\n\n\n = 4(\\Delta x)^2 <\/span>\n\n\n\n

\n\n\n u[i-1,j]-2(1+\\lambda)u[i,j]+u[i+1,j] + <\/span>\n\n\n + \\lambda u[i,j-1]+\\lambda u[i,j+1] = 4(\\Delta x)^2 <\/span>\n\n\n\n
\n\n\n\n

Se pueden realizar las iteraciones para los nodos y reemplaza los valores de frontera:<\/p>\n\n\n\n

i=1 y j=1<\/p>\n\n\n u[0,1]-2(1+\\lambda)u[1,1]+u[2,1] + <\/span>\n\n\n + \\lambda u[1,0]+\\lambda u[1,2] = 4(\\Delta x)^2 <\/span>\n\n\n \\Delta y^2-2(1+\\lambda)u[1,1]+u[2,1] + <\/span>\n\n\n + \\Delta x^2+\\lambda u[1,2] = 4(\\Delta x)^2 <\/span>\n\n\n -2(1+\\lambda)u[1,1]+u[2,1] +\\lambda u[1,2] <\/span>\n\n\n = 4(\\Delta x)^2 - \\Delta y^2 - \\Delta x^2 <\/span>\n\n\n -2(1+0.25)u[1,1]+u[2,1] +0.25 u[1,2] <\/span>\n\n\n = 4(1\/3)^2 - (2\/3)^2- (1\/3)^2 <\/span>\n\n\n -2.5u[1,1]+u[2,1] +0.25 u[1,2] = -0.1111<\/span>\n\n\n\n

i=2 , j=1<\/p>\n\n\n u[1,1]-2(1+\\lambda)u[2,1]+u[3,1] + <\/span>\n\n\n + \\lambda u[2,0]+\\lambda u[2,2] = 4(\\Delta x)^2 <\/span>\n\n\n u[1,1]-2(1+\\lambda)u[2,1]+(\\delta y-1)^2 + <\/span>\n\n\n + \\Delta x^2 +\\lambda u[2,2] = 4(\\Delta x)^2 <\/span>\n\n\n u[1,1]-2(1+\\lambda)u[2,1] + \\lambda u[2,2] <\/span>\n\n\n = 4(\\Delta x)^2 - (\\Delta y-1)^2 - \\Delta x^2 <\/span>\n\n\n u[1,1]-2(1+0.25)u[2,1] + 0.25 u[2,2] <\/span>\n\n\n = 4(1\/3)^2 - (2\/3-1)^2 - (1\/3)^2 <\/span>\n\n\n u[1,1]-2.5u[2,1] + 0.25 u[2,2] = 0.2222 <\/span>\n\n\n\n

i=1 , j=2<\/p>\n\n\n u[0,2]-2(1+\\lambda)u[1,2]+u[2,2] + <\/span>\n\n\n + \\lambda u[1,1]+\\lambda u[1,3] = 4(\\Delta x)^2 <\/span>\n\n\n (2\\Delta y^2)-2(1+\\lambda)u[1,2]+u[2,2] + <\/span>\n\n\n + \\lambda u[1,1]+\\lambda (\\Delta x-1)^2 = 4(\\Delta x)^2 <\/span>\n\n\n -2(1+\\lambda)u[1,2]+u[2,2] + \\lambda u[1,1]<\/span>\n\n\n = 4(\\Delta x)^2 - (2\\Delta y^2) -\\lambda (\\Delta x-1)^2 <\/span>\n\n\n -2(1+0.25)u[1,2]+u[2,2] + 0.25 u[1,1]<\/span>\n\n\n = 4(1\/3)^2 - (2(2\/3)^2] -0.25 (1\/3-1)^2 <\/span>\n\n\n -2.5u[1,2]+u[2,2] + 0.25 u[1,1] = -0.5555 <\/span>\n\n\n\n

i=2, j=2<\/p>\n\n\n u[1,2]-2(1+\\lambda)u[2,2]+u[3,2] + <\/span>\n\n\n + \\lambda u[2,1]+\\lambda u[2,3] = 4(\\Delta x)^2 <\/span>\n\n\n u[1,2]-2(1+\\lambda)u[2,2]+(\\Delta y-1)^2 + <\/span>\n\n\n + \\lambda u[2,1]+\\lambda (\\Delta x-1)^2 = 4(\\Delta x)^2 <\/span>\n\n\n u[1,2]-2(1+\\lambda)u[2,2] + \\lambda u[2,1] <\/span>\n\n\n =4(\\Delta x)^2- (\\Delta y-1)^2 -\\lambda (\\Delta x-1)^2 <\/span>\n\n\n u[1,2]-2(1+0.25)u[2,2] + 0.25 u[2,1] <\/span>\n\n\n =4(1\/3)^2- (2\/3-1)^2 -0.25(1\/3-1)^2 <\/span>\n\n\n u[1,2]-2.5u[2,2] + 0.25 u[2,1] =-0.2222 <\/span>\n\n\n\n

Obtiene el sistema de ecuaciones a resolver:<\/p>\n\n\n -2.5u[1,1]+u[2,1] +0.25 u[1,2] = -0.1111<\/span>\n\n\n u[1,1]-2.5u[2,1] + 0.25 u[2,2] = 0.2222 <\/span>\n\n\n -2.5u[1,2]+u[2,2] + 0.25 u[1,1] = -0.5555 <\/span>\n\n\n u[1,2]-2.5u[2,2] + 0.25 u[2,1] =-0.2222 <\/span>\n\n\n\n

Se escribe en forma matricial Ax=B<\/p>\n\n\n\\begin {bmatrix} -2.5 && 1&&0.25&&0 \\\\1&&-2.5&&0&&0.25\\\\0.25&&0&&-2.5&&1\\\\0&&0.25&&1&&-2.5\\end{bmatrix} \\begin {bmatrix} u[1,1] \\\\ u[2,1]\\\\ u[1,2]\\\\u[2,2] \\end{bmatrix} = \\begin {bmatrix}-0.1111\\\\0.2222\\\\-0.5555\\\\-0.2222 \\end{bmatrix}<\/span>\n\n\n\n

que se resuelve con un algoritmo:<\/p>\n\n\n

\nimport numpy as np\n\nA = np.array([[-2.5,1,0.25,0],\n              [1,-2.5,0,0.25],\n              [0.25,0,-2.5,1],\n              [0,0.25,1,-2.5]])\nB = np.array([-0.1111,0.2222,-0.5555,-0.2222])\n\nx = np.linalg.solve(A,B)\n\nprint(x)\n\n<\/pre><\/div>\n\n\n
y los resultados son:<\/p>\n\n\n\n
[ 0.05762549 -0.04492835  0.31156835  0.20901451]<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2007TII_T1 EDP Eliptica, problema de frontera Con los datos del ejercicio se plantean de la siguiente forma en los bordes: La ecuaci\u00f3n a resolver es: Que en su forma discreta, con diferencias divididas centradas es: Al agrupar constantes se convierte en: Siendo \u0394x= 1\/3 y \u0394y =2\/3, se mantendr\u00e1 la relaci\u00f3n \u03bb = (\u0394x\/\u0394y)2 Se […]<\/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":[50],"tags":[58,54],"class_list":["post-4007","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4007","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=4007"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4007\/revisions"}],"predecessor-version":[{"id":23916,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4007\/revisions\/23916"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4007"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4007"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4007"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}