{"id":4802,"date":"2020-02-11T20:45:27","date_gmt":"2020-02-12T01:45:27","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4802"},"modified":"2026-04-05T20:58:51","modified_gmt":"2026-04-06T01:58:51","slug":"s3eva2019tii_t2-diferenciacion-valor-en-frontera","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva20\/s3eva2019tii_t2-diferenciacion-valor-en-frontera\/","title":{"rendered":"s3Eva2019TII_T2 Diferenciaci\u00f3n, valor en frontera"},"content":{"rendered":"\n

Ejercicio<\/strong>: 3Eva2019TII_T2 Diferenciaci\u00f3n, valor en frontera<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n de problema de valor de frontera, con h = 1\/4 = 0.25:<\/p>\n\n\n y'' = -(x+1)y' + 2y + (1-x^2) e^{-x} <\/span>\n\n\n\n

0 \u2264 x \u2264 1
y(0) = -1
y(1) = 0<\/p>\n\n\n\n

Se interpreta en la tabla como los valores que faltan por encontrar:<\/p>\n\n\n\n

i<\/td>0<\/td>1<\/td>2<\/td>3<\/td>4<\/td><\/tr>
xi<\/td>0<\/td>1\/4<\/td>1\/2<\/td>3\/4<\/td>1<\/td><\/tr>
yi<\/td>-1<\/td> <\/td> <\/td> <\/td>0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Por ejemplo, se usan las derivadas en diferencias finitas divididas centradas para segunda derivada y hacia adelante para primera derivada.<\/p>\n\n\n\n

Semejante al procedimiento aplicado para m\u00e9todos con EDO.<\/p>\n\n\n f''(x_i) = \\frac{f(x_{i+1})-2f(x_{i})+f(x_{i-1})}{h^2} + O(h^2) <\/span>\n\n\n f'(x_i) = \\frac{f(x_{i+1})-f(x_i)}{h} + O(h) <\/span>\n\n\n\n

Se formula entonces la ecuaci\u00f3n en forma discreta, usando solo los \u00edndices para los puntos yi:<\/p>\n\n\n \\frac{y[i+1]-2y[i]+y[i-1]}{h^2} = -(x_i +1) \\frac{y[i+1]-y[i]}{h} <\/span>\n\n\n + 2y[i] + (1-x_i ^2) e^{-x_i} <\/span>\n\n\n\n

\n\n\n\n

se multiplica todo por h2<\/sup><\/p>\n\n\n y[i+1]-2y[i]+y[i-1] = -h(x_i +1)(y[i+1]-y[i]) +<\/span>\n\n\n + 2 h^2 y[i] + h^2 (1-x_i ^2) e^{-x_i} <\/span>\n\n\n\n

\n\n\n y[i+1]-2y[i]+y[i-1] = -h(x_i +1)y[i+1] + <\/span>\n\n\n+ h(x_i +1)y[i] + 2 h^2 y[i] + h^2 (1-x_i ^2) e^{-x_i} <\/span>\n\n\n\n
\n\n\n y[i+1](1 +h(x_i +1)) + y[i](-2 - h(x_i +1)-2 h^2)+y[i-1] = <\/span>\n\n\n = h^2 (1-x_i ^2) e^{-x_i} <\/span>\n\n\n\n

Ecuaci\u00f3n que se aplica en cada uno de los puntos desconocidos con i =1,2,3<\/p>\n\n\n\n

\n\n\n\n

i = 1<\/p>\n\n\n y[2](1 +h(x_1 +1)) + y[1](-2 - h(x_1 +1)-2 h^2)+y[0] = <\/span>\n\n\n = h^2 (1-x_1 ^2) e^{-x_1} <\/span>\n\n\n y[2]\\Big(1 +\\frac{1}{4}\\Big(\\frac{1}{4} +1\\Big)\\Big) + y[1]\\Big(-2 - \\frac{1}{4}\\Big(\\frac{1}{4} +1\\Big)-2 \\Big(\\frac{1}{4}\\Big)^2\\Big) -1 = <\/span>\n\n\n = \\Big(\\frac{1}{4}\\Big)^2 \\Big(1-\\Big(\\frac{1}{4}\\Big) ^2\\Big) e^{-1\/4} <\/span>\n\n\n y[2]\\Big(1 +\\frac{5}{16} \\Big) + y[1]\\Big(-2 - \\frac{5}{16}- \\frac{2}{16}\\Big) = <\/span>\n\n\n = 1+ \\frac{1}{16} \\Big(1-\\frac{1}{16}\\Big) e^{-1\/4} <\/span>\n\n\n \\frac{21}{16} y[2]- \\frac{39}{16}y[1] = 1+ \\frac{15}{16^2} e^{-1\/4} <\/span>\n\n\n - \\frac{39}{16}y[1] + \\frac{21}{16} y[2] = 1+ \\frac{15}{16^2} e^{-1\/4} <\/span>\n\n\n\n

multiplicando ambos lados de la ecuacion por 16 y reordenando<\/p>\n\n\n - 39 y[1] + 21 y[2] = 16+ \\frac{15}{16} e^{-1\/4} <\/span>\n\n\n\n

\n\n\n\n

i = 2<\/p>\n\n\n y[3](1 +h(x_2 +1)) + y[2](-2 - h(x_2 +1)-2 h^2)+y[1] = <\/span>\n\n\n = h^2 (1-x_2 ^2) e^{-x_2} <\/span>\n\n\n y[3]\\Big(1 +\\frac{1}{4}\\Big(\\frac{1}{2} +1\\Big)\\Big) + y[2]\\Big(-2 - \\frac{1}{4}\\Big(\\frac{1}{2} +1\\Big)-2 \\Big(\\frac{1}{4}\\Big)^2\\Big)+y[1] = <\/span>\n\n\n = \\Big(\\frac{1}{4}\\Big)^2 \\Big(1-\\Big(\\frac{1}{2}\\Big) ^2\\Big) e^{-\\frac{1}{2}} <\/span>\n\n\n y[3]\\Big(1 +\\frac{3}{8}\\Big) + y[2]\\Big(-2 - \\frac{1}{4}\\Big(\\frac{3}{2}\\Big)- \\frac{1}{8}\\Big)+y[1] = <\/span>\n\n\n = \\Big(\\frac{1}{16}\\Big) \\Big(1-\\frac{1}{4}\\Big) e^{-\\frac{1}{2}} <\/span>\n\n\n \\frac{11}{8} y[3] - \\frac{21}{8} y[2]+y[1] = \\frac{1}{16} \\Big(\\frac{3}{4}\\Big) e^{-\\frac{1}{2}} <\/span>\n\n\n\n

multiplicando ambos lados por 8 y reordenando,<\/p>\n\n\n 8y[1] - 21 y[2] + 11 y[3] = \\frac{3}{8} e^{-\\frac{1}{2}} <\/span>\n\n\n\n

\n\n\n\n

i = 3<\/p>\n\n\n y[4](1 +h(x_3 +1)) + y[3](-2 - h(x_3 +1)-2 h^2)+y[2] = <\/span>\n\n\n = h^2 (1-x_3 ^2) e^{-x_3} <\/span>\n\n\n (0) \\Big(1 +h\\Big(x_3 +1\\Big)\\Big) + y[3]\\Big(-2 - \\frac{1}{4}\\Big(\\frac{3}{4} +1\\Big)-2 \\frac{1}{16}\\Big)+y[2] = <\/span>\n\n\n = \\frac{1}{16}\\Big (1-\\Big(\\frac{3}{4}\\Big) ^2 \\Big) e^{-3\/4} <\/span>\n\n\n y[3]\\Big(-2 - \\frac{7}{16} - \\frac{2}{16}\\Big)+y[2] = \\frac{1}{16}\\Big (1-\\frac{9}{16}\\Big) e^{-3\/4} <\/span>\n\n\n\n

multiplicando todo por 16 y reordenando:<\/p>\n\n\n 16 y[2] - 41 y[3] = \\frac{7}{16} e^{-3\/4} <\/span>\n\n\n\n

\n\n\n\n

Con lo que se puede crear la forma matricial del sistema de ecuaciones Ax=B<\/p>\n\n\n \\begin{bmatrix} -39 && 21 && 0\\\\8 && 21 && 11 \\\\ 0 && 16 && 41 \\end{bmatrix}\\begin{bmatrix} y[1]\\\\ y[2] \\\\y[3] \\end{bmatrix} =\\begin{bmatrix} 16+ \\frac{15}{16} e^{-1\/4} \\\\ \\frac{3}{8} e^{-\\frac{1}{2}} \\\\ \\frac{7}{16} e^{-3\/4} \\end{bmatrix}<\/span>\n\n\n\n

con lo que resolviendo la matriz se obtienen los valroes para y[1], y[2], y[3]<\/p>\n\n\n\n

soluci\u00f3n de matriz: \n[-0.59029143 -0.29958287 -0.12195088]<\/code><\/pre>\n\n\n\n
con lo que se completan los puntos de la tabla,<\/p>\n\n\n\n
Soluci\u00f3n de ecuaci\u00f3n\nx[i] =  [ 0.   0.25        0.5         0.75        1.  ]\ny[i] =  [-1.  -0.59029143 -0.29958287 -0.12195088  0.  ]<\/code><\/pre>\n\n\n\n
con la siguiente gr\u00e1fica:<\/p>\n\n\n\n
\"3e2019tii_t2<\/figure>\n\n\n\n
\n\n\n\n
Algoritmo para soluci\u00f3n de matriz con Python<\/h2>\n\n\n\n
tarea<\/strong><\/em>: modificar para cambiar el valor del tama\u00f1o de paso.<\/p>\n\n\n
\n# Problema de Frontera\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\n\nh = 1\/4\ny0 = -1\ny1 = 0\n\nxi = np.arange(0,1+h,h)\nn = len(xi)\nyi = np.zeros(n,dtype=float)\nyi[0] = y0\nyi[n-1] = y1\n    \nA = np.array([[-39.0, 21,  0],\n              [  8.0, -21, 11],\n              [  0.0, 16, -41]])\nB = np.array([16+(15\/16)*np.exp(-1\/4),\n              (3\/8)*np.exp(-1\/2),\n              (7\/16)*np.exp(-3\/4)])\n\n# PROCEDIMIENTO\nx = np.linalg.solve(A,B)\n\nfor i in range(1,n-1,1):\n    yi[i] = x[i-1]\n\n# SALIDA\nprint('soluci\u00f3n de matriz: ')\nprint(x)\nprint('Soluci\u00f3n de ecuaci\u00f3n')\nprint('x[i] = ',xi)\nprint('y[i] = ',yi)\n\n# Grafica\nplt.plot(xi,yi,'ro')\nplt.plot(xi,yi)\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2019TII_T2 Diferenciaci\u00f3n, valor en frontera La ecuaci\u00f3n de problema de valor de frontera, con h = 1\/4 = 0.25: 0 \u2264 x \u2264 1y(0) = -1y(1) = 0 Se interpreta en la tabla como los valores que faltan por encontrar: i 0 1 2 3 4 xi 0 1\/4 1\/2 3\/4 1 yi -1 […]<\/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":[51],"tags":[58,54],"class_list":["post-4802","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4802","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=4802"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4802\/revisions"}],"predecessor-version":[{"id":23923,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4802\/revisions\/23923"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4802"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4802"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4802"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}