{"id":8415,"date":"2022-08-31T18:15:08","date_gmt":"2022-08-31T23:15:08","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8415"},"modified":"2026-04-05T20:46:25","modified_gmt":"2026-04-06T01:46:25","slug":"s2eva2022paoi_t3-edp-parabolica-barra-enfriada-en-centro","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2022paoi_t3-edp-parabolica-barra-enfriada-en-centro\/","title":{"rendered":"s2Eva2022PAOI_T3 EDP parab\u00f3lica barra enfriada en centro"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2022PAOI_T3 EDP parab\u00f3lica barra enfriada en centro<\/a><\/p>\n\n\n\n

Para la ecuaci\u00f3n dada con \u0394x = 1\/3, \u0394t = 0.02, en una revisi\u00f3n r\u00e1pida para cumplir la convergencia dt<dx\/10, condici\u00f3n que debe verificarse con la expresi\u00f3n obtenida para \u03bb al desarrollar el ejercicio.<\/p>\n\n\n \\frac{\\partial U}{\\partial t} - \\frac{1}{9} \\frac{\\partial ^2 U}{\\partial x^2} = 0<\/span>\n\n\n 0 \\leq x \\leq 2, t >0 <\/span>\n\n\n\n

literal a. gr\u00e1fica de malla<\/h2>\n\n\n\n
\"2Eva2022PAOI_T3<\/figure>\n\n\n\n

literal b. <\/h2>\n\n\n\n

Ecuaciones de diferencias divididas a usar<\/p>\n\n\n \\frac{\\partial U}{\\partial t} - \\frac{1}{9} \\frac{\\partial ^2 U}{\\partial x^2} = 0<\/span>\n\n\n \\frac{\\partial ^2 U}{\\partial x^2} = 9 \\frac{\\partial U}{\\partial t}<\/span>\n\n\n \\frac{u_{i+1,j}-2u_{i,j}+u_{i-1,j}}{(\\Delta x)^2} = 9 \\frac{u_{i,j+1}-u_{i,j}}{\\Delta t}<\/span>\n\n\n\n

se agrupan las constantes,<\/p>\n\n\n \\frac{\\Delta t}{9(\\Delta x)^2} \\Big(u[i-1,j]-2u[i,j]+u[i+1,j] \\Big) = u[i,j+1]-u[i,j]<\/span>\n\n\n\n

literal d Determine el valor de \u03bb<\/h2>\n\n\n \\lambda = \\frac{\\Delta t}{9(\\Delta x)^2} =\\frac{0.02}{9(1\/3)^2} = 0.02 <\/span>\n\n\n\n

valor de \u03bb que es menor que 1\/2, por lo que el m\u00e9todo converge.<\/p>\n\n\n\n

continuando luego con la ecuaci\u00f3n general,<\/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 u[i,j] + \\lambda u[i+1,j] \\Big) = u[i,j+1]-u[i,j]<\/span>\n\n\n\n

literal c. <\/h2>\n\n\n\n

Encuentre las ecuaciones considerando las condiciones dadas en el problema.<\/p>\n\n\n \\lambda u[i-1,j]+(1-2 \\lambda ) u[i,j] + \\lambda u[i+1,j] = u[i,j+1]<\/span>\n\n\n\n

el punto que no se conoce su valor es u[i,j+1] que es la ecuaci\u00f3n buscada.<\/p>\n\n\n u[i,j+1] = \\lambda u[i-1,j]+(1-2 \\lambda ) u[i,j] + \\lambda u[i+1,j] <\/span>\n\n\n\n

literal e iteraciones<\/h2>\n\n\n\n

iteraci\u00f3n  i=1, j=0<\/strong><\/p>\n\n\n u[1,1] = \\lambda u[0,0]+(1-2 \\lambda ) u[1,0] + \\lambda u[2,0] <\/span>\n\n\n u[1,1] =0.02 \\cos \\Big( \\frac{\\pi}{2}(0-3)\\Big) + (1-2(0.02) ) \\cos \\Big( \\frac{\\pi}{2}\\big(\\frac{1}{3}-3\\big)\\Big)<\/span>\n\n\n + 0.02 \\cos \\Big( \\frac{\\pi}{2}\\big( \\frac{2}{3}-3\\big) \\Big) <\/span>\n\n\n u[1,1] =0.02(0)+(0.96)(-0.5)+0.02(-0.8660)=-0.4973<\/span>\n\n\n\n

iteraci\u00f3n  i=2, j=0<\/strong><\/p>\n\n\n u[2,1] = \\lambda u[1,0]+(1-2 \\lambda ) u[2,0] + \\lambda u[3,0] <\/span>\n\n\n u[2,1] = 0.02 \\cos \\Big( \\frac{\\pi}{2}(\\frac{1}{3}-3)\\Big) + (1-2(0.02) ) \\cos \\Big( \\frac{\\pi}{2}(\\frac{2}{3}-3)\\Big)+<\/span>\n\n\n + 0.02 \\cos \\Big( \\frac{\\pi}{2}\\big(\\frac{3}{3}-3\\big)\\Big) <\/span>\n\n\n u[2,1] = 0.02 (-0.5) + (0.96 ) (-0.866025) + 0.02 (-1) =-0.8614<\/span>\n\n\n\n

iteraci\u00f3n  i=3, j=0<\/strong><\/p>\n\n\n u[3,1] = \\lambda u[2,0]+(1-2 \\lambda ) u[3,0] + \\lambda u[4,0] <\/span>\n\n\n u[3,1] = 0.02 \\cos \\Big( \\frac{\\pi}{2}\\big( \\frac{2}{3}-3\\big)\\Big)+(1-2 (0.02) ) \\cos \\Big( \\frac{\\pi}{2}(1-3)\\Big) +<\/span>\n\n\n + 0.02 \\cos \\Big( \\frac{\\pi}{2}\\big(\\frac{4}{3}-3\\big)\\Big) <\/span>\n\n\n u[3,1] = 0.02 (-0.866025)+(0.96 ) (-1) + 0.02 (-0,866025) = -0,9946 <\/span>\n\n\n\n

literal f<\/h2>\n\n\n\n

La cotas de errores de truncamiento en la ecuaci\u00f3n corresponden a segunda derivada O(hx<\/sub>2<\/sup>) y el de primera derivada O(ht<\/sub>), al reemplazar los valores ser\u00e1 la suma}<\/p>\n\n\n\n

O(hx<\/sub>2<\/sup>) + O(ht<\/sub>) = (1\/3)2<\/sup> + 0.02 = 0,1311<\/p>\n\n\n\n

literal g<\/h2>\n\n\n\n

Resultados usando el algoritmo en Python<\/p>\n\n\n\n

Tabla de resultados\n[[ 0.      0.      0.      0.      0.      0.      0.      0.      0.       0.    ]\n [-0.5    -0.4973 -0.4947 -0.492  -0.4894 -0.4867 -0.4841 -0.4815 -0.479   -0.4764]\n [-0.866  -0.8614 -0.8568 -0.8522 -0.8476 -0.8431 -0.8385 -0.8341 -0.8296  -0.8251]\n [-1.     -0.9946 -0.9893 -0.984  -0.9787 -0.9735 -0.9683 -0.9631 -0.9579  -0.9528]\n [-0.866  -0.8614 -0.8568 -0.8522 -0.8476 -0.8431 -0.8385 -0.8341 -0.8296  -0.8251]\n [-0.5    -0.4973 -0.4947 -0.492  -0.4894 -0.4867 -0.4841 -0.4815 -0.479   -0.4764]\n [ 0.      0.      0.      0.      0.      0.      0.      0.      0.       0.    ]]\n<\/code><\/pre>\n\n\n\n
\"2Eva2022PAOI_T3<\/figure>\n\n\n\n
Instrucciones en Python<\/h2>\n\n\n
\n# EDP parab\u00f3licas d2u\/dx2  = K du\/dt\n# m\u00e9todo expl\u00edcito, usando diferencias finitas\n# 2Eva_2022PAOI_T3 EDP parab\u00f3lica barra enfriada en centro\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\n# Valores de frontera\nTa = 0\nTb = 0\nT0 = lambda x: np.cos((np.pi\/2)*(x-3))\n# longitud en x\na = 0.0\nb = 2.0\n# Constante K\nK = 9\n# Tama\u00f1o de paso\ndx = 1\/3\ndt = 0.02\ntramos = int(np.round((b-a)\/dx,0))\nmuestras = tramos + 1\n# iteraciones en tiempo\nn = 10\n\n# PROCEDIMIENTO\n# iteraciones en longitud\nxi = np.linspace(a,b,muestras)\nm = len(xi)\nultimox = m-1\n\n# Resultados en tabla u[x,t]\nu = np.zeros(shape=(m,n), dtype=float)\n\n# valores iniciales de u[:,j]\nj=0\nultimot = n-1\nu[0,j]= Ta\nu[1:ultimox,j] = T0(xi[1:ultimox])\nu[ultimox,j] = Tb\n\n# factores P,Q,R\nlamb = dt\/(K*dx**2)\nP = lamb\nQ = 1 - 2*lamb\nR = lamb\n\n# Calcula U para cada tiempo + dt\nj = 0\nwhile not(j&gt;=ultimot):\n    u[0,j+1] = Ta\n    for i in range(1,ultimox,1):\n        u[i,j+1] = P*u[i-1,j] + Q*u[i,j] + R*u[i+1,j]\n    u[m-1,j+1] = Tb\n    j=j+1\n\n# SALIDA\nprint('Tabla de resultados')\nnp.set_printoptions(precision=2)\nprint(u)\n\n# Gr\u00e1fica\nsalto = int(n\/10)\nif (salto == 0):\n    salto = 1\nfor j in range(0,n,salto):\n    vector = u[:,j]\n    plt.plot(xi,vector)\n    plt.plot(xi,vector, '.r')\nplt.xlabel('x[i]')\nplt.ylabel('t[j]')\nplt.title('Soluci\u00f3n EDP parab\u00f3lica')\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2022PAOI_T3 EDP parab\u00f3lica barra enfriada en centro Para la ecuaci\u00f3n dada con \u0394x = 1\/3, \u0394t = 0.02, en una revisi\u00f3n r\u00e1pida para cumplir la convergencia dt<dx\/10, condici\u00f3n que debe verificarse con la expresi\u00f3n obtenida para \u03bb al desarrollar el ejercicio. literal a. gr\u00e1fica de malla literal b. Ecuaciones de diferencias divididas a usar […]<\/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-8415","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\/8415","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=8415"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8415\/revisions"}],"predecessor-version":[{"id":23897,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8415\/revisions\/23897"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8415"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8415"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8415"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}