{"id":3157,"date":"2019-01-31T10:15:51","date_gmt":"2019-01-31T15:15:51","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=3157"},"modified":"2026-04-05T20:12:29","modified_gmt":"2026-04-06T01:12:29","slug":"s2eva2018tii_t3-edp","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2018tii_t3-edp\/","title":{"rendered":"s2Eva2018TII_T3 EDP"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2018TII_T3 EDP<\/a><\/p>\n\n\n\n

Se indica en el enunciado que b = 0
<\/strong><\/p>\n\n\n \\frac{\\delta u}{\\delta t} = \\frac{\\delta ^2 u}{\\delta x^2} + b\\frac{\\delta u}{\\delta x}<\/span>\n\n\n\n

simplificando la ecuaci\u00f3n a:<\/p>\n\n\n \\frac{\\delta u}{\\delta t} = \\frac{\\delta ^2 u}{\\delta x^2}<\/span>\n\n\n\n

Reordenando la ecuaci\u00f3n a la forma estandarizada:<\/p>\n\n\n \\frac{\\delta ^2 u}{\\delta x^2} = \\frac{\\delta u}{\\delta t} <\/span>\n\n\n\n

Seleccione un m\u00e9todo: expl\u00edcito o impl\u00edcito.
Si el m\u00e9todo es expl\u00edcito, las diferencias finitas a usar son hacia adelante y centrada:<\/p>\n\n\n U'(x_i,t_j) = \\frac{U(x_i,t_{j+1})-U(x_i,t_j)}{\\Delta t} + O(\\Delta t) <\/span>\n\n\n U''(x_i,t_j) = \\frac{U(x_{i+1},t_j)-2U(x_{i},t_j)+U(x_{i-1},t_j)}{\\Delta x^2} + O(\\Delta x^2) <\/span>\n\n\n\n

como referencia se usa la gr\u00e1fica.<\/p>\n\n\n\n

<\/p>\n\n\n\n

Se selecciona la esquina inferior derecha como 0,  por la segunda ecuaci\u00f3n de condiciones y facilidad de c\u00e1lculo. (No hubo indicaci\u00f3n durante el examen que muestre lo contrario)<\/p>\n\n\n\n

condiciones de frontera U(0,t)=0, U(1,t)=1\ncondiciones de inicio U(x,0)=0, 0\u2264x\u22641<\/code><\/pre>\n\n\n\n
aunque lo m\u00e1s recomendable ser\u00eda cambiar la condici\u00f3n de inicio a:<\/p>\n\n\n\n
condiciones de inicio U(x,0)=0, 0<x<1<\/code><\/pre>\n\n\n\n
Siguiendo con el tema de la ecuaci\u00f3n, al reemplazar las diferencias finitas en la ecuaci\u00f3n:<\/p>\n\n\n\n
\n\n\n \\frac{U(x_{i+1},t_j)-2U(x_{i},t_j)+U(x_{i-1},t_j)}{\\Delta x^2} = <\/span>\n\n\n = \\frac{U(x_i,t_{j+1})-U(x_i,t_j)}{\\Delta t} <\/span>\n\n\n\n
\n\n\n\n
se reagrupan los t\u00e9rminos que son constantes y los t\u00e9rminos de error se acumulan:<\/p>\n\n\n \\frac{\\Delta t}{\\Delta x^2} \\Big[U(x_{i+1},t_j)-2U(x_i,t_j)+U(x_{i-1},t_j) \\Big] = U(x_i,t_{j+1})-U(x_i,t_j) <\/span>\n\n\n\n
siendo,<\/p>\n\n\n \\lambda= \\frac{\\Delta t}{\\Delta x^2}<\/span>\n\n\n error \\cong O(\\Delta t) + O(\\Delta x^2) <\/span>\n\n\n\n
continuando con la ecuaci\u00f3n, se simplifica la escritura usando s\u00f3lo los \u00edndices i,j y se reordena de izquierda a derecha como en la gr\u00e1fica<\/p>\n\n\n \\lambda \\Big[U[i-1,j]-2U[i,j]+U[i+1,j] \\Big] = U[i,j+1]-U]i,j] <\/span>\n\n\n \\lambda U[i-1,j]+(-2\\lambda+1)U[i,j]+\\lambda U[i+1,j] = U[i,j+1] <\/span>\n\n\n U[i,j+1] = \\lambda U[i-1,j]+(-2\\lambda+1)U[i,j]+\\lambda U[i+1,j] <\/span>\n\n\n U[i,j+1] = P U[i-1,j]+QU[i,j]+R U[i+1,j] <\/span>\n\n\n P=R = \\lambda <\/span>\n\n\n Q = -2\\lambda+1 <\/span>\n\n\n\n
En las iteraciones, el valor de P,Q y R se calculan a partir de \u03bb \u2264 1\/2<\/p>\n\n\n\n
iteraciones: j=0, i=1<\/p>\n\n\n\n
U[1,1] = P*0+Q*0+R*0 = 0<\/p>\n\n\n\n
j=0, i=2<\/p>\n\n\n\n
U[2,1] = P*0+Q*0+R*0=0<\/p>\n\n\n\n
j=0, i=3<\/p>\n\n\n U[3,1] = P*0+Q*0+R*1=R=\\lambda=\\frac{1}{2} <\/span>\n\n\n\n
iteraciones: j=1, i=1<\/p>\n\n\n\n
U[1,2] = P*0+Q*0+R*0 = 0<\/p>\n\n\n\n
j=1, i=2<\/p>\n\n\n U[2,2] = P*0+Q*0+R*\\lambda = \\lambda ^2 = \\frac{1}{4} <\/span>\n\n\n\n
j=1, i=3<\/p>\n\n\n U[3,2] = P*0+Q*\\frac{1}{4}+R (\\lambda) <\/span>\n\n\n U[3,2] = (-2\\lambda +1) \\frac{1}{4}+\\lambda^2 = \\Big(-2\\frac{1}{2}+1\\Big) \\frac{1}{4}+\\Big(\\frac{1}{2}\\Big)^2 <\/span>\n\n\n U[3,2] =0\\frac{1}{4} + \\frac{1}{4} = \\frac{1}{4} <\/span>\n\n\n\n
\n\n\n\n
Literal b. Para el c\u00e1lculo del error:<\/p>\n\n\n \\lambda \\leq \\frac{1}{2}<\/span>\n\n\n\\frac{\\Delta t}{\\Delta x^2} \\leq \\frac{1}{2}<\/span>\n\n\n \\Delta t \\leq \\frac{\\Delta x^2}{2}<\/span>\n\n\n\n
en el enunciado se indica h = 0.25 = \u00bc = \u0394 x<\/p>\n\n\n \\Delta t \\leq \\frac{(1\/4)^2}{2} = \\frac{1}{32}<\/span>\n\n\n error \\cong O(\\Delta t) + O(\\Delta x^2)<\/span>\n\n\n error \\cong \\frac{\\Delta x^2}{2}+ \\Delta x^2<\/span>\n\n\n error \\cong \\frac{3}{2}\\Delta x^2<\/span>\n\n\n error \\cong \\frac{3}{2}( \\frac{1}{4})^2<\/span>\n\n\n error \\cong \\frac{3}{32} = 0.09375<\/span>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2018TII_T3 EDP Se indica en el enunciado que b = 0 simplificando la ecuaci\u00f3n a: Reordenando la ecuaci\u00f3n a la forma estandarizada: Seleccione un m\u00e9todo: expl\u00edcito o impl\u00edcito.Si el m\u00e9todo es expl\u00edcito, las diferencias finitas a usar son hacia adelante y centrada: como referencia se usa la gr\u00e1fica. Se selecciona la esquina inferior derecha […]<\/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-3157","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\/3157","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=3157"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3157\/revisions"}],"predecessor-version":[{"id":23858,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3157\/revisions\/23858"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=3157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=3157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=3157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}