{"id":8763,"date":"2023-01-24T14:00:57","date_gmt":"2023-01-24T19:00:57","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8763"},"modified":"2026-07-27T08:32:15","modified_gmt":"2026-07-27T13:32:15","slug":"s2eva2022paoii_t2-edo-poblacion-protestantes-sociedad","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2022paoii_t2-edo-poblacion-protestantes-sociedad\/","title":{"rendered":"s2Eva2022PAOII_T2 EDO - poblaci\u00f3n de protestantes en una sociedad"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2022PAOII_T2 EDO - poblaci\u00f3n de protestantes en una sociedad<\/a><\/p>\n\n\n\n

El sistema de ecuaciones para el ejercicio se expresa como:<\/p>\n\n\n \\frac{\\delta}{\\delta t}x(t) = b x(t) - d (x(t))^2<\/span>\n\n\n \\frac{\\delta}{\\delta t}y(t) = b y(t) - d (y(t))^2 +r b (x(t)-y(t))<\/span>\n\n\n\n

literal a<\/h2>\n\n\n\n

Simplificando la nomenclatura, la poblaci\u00f3n total del pa\u00eds se describe con:<\/p>\n\n\n x' = b x - d x^2 <\/span>\n\n\n\n

La poblaci\u00f3n de protestantes se describe con;<\/p>\n\n\n y' = b y - d y^2 +r b (x-y)<\/span>\n\n\n\n

Sustituyendo constantes, y considerando x(0)=1 ; y(0)=0.01 ; h=0.5 valores del enunciado:<\/p>\n\n\n x' = 0.02 x - 0.015 x^2 <\/span>\n\n\n y' = 0.02 y - 0.015 y^2 +0.1(0.02) (x-y)<\/span>\n\n\n\n

El planteamiento de Runge-Kutta se hace junto a la primera iteraci\u00f3n, adem\u00e1s de encontrarse en las instrucciones con Python.<\/p>\n\n\n\n

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

Se describen 3 iteraciones usando los resultados de la tabla con el algoritmo, para mostrar la comprensi\u00f3n del algoritmo.<\/p>\n\n\n\n

i<\/th>ti<\/th>xi<\/th>yi<\/th><\/tr><\/thead>
0<\/td>0<\/td>1<\/td>0.01<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

t = 0 para iniciar las iteraciones, h=0.5<\/p>\n\n\n\n K1x = 0.5 \\left(0.02 (1) - 0.015 (1)^2 \\right) = 0.0025 <\/span>\n\n\n\n K1y = 0.5\\left(0.02 (0.01) - 0.015 (0.01)^2 +0.1(0.02) (1-0.01)\\right) = 0.001089 <\/span>\n\n\n\n

.<\/p>\n\n\n\n K2x = 0.5 \\left(0.02 (1+0.0025) - 0.015 (1+0.0025)^2 \\right) = 0.00248 <\/span>\n\n\n\n K2y = 0.5\\Big(0.02 (0.01+0.00108) - 0.015 (0.01+0.00108)^2 <\/span>\n\n\n\n + 0.1(0.02) ((1+0.0025)-(0.01+0.00108))\\Big) = 0.001101 <\/span>\n\n\n\n

La poblaci\u00f3n del pa\u00eds en la iteraci\u00f3n se actualiza a:<\/p>\n\n\n\n x_1 = 1 + \\frac{0.0025+0.00248}{2} = 1.0025 <\/span>\n\n\n\n

La poblaci\u00f3n de protestantes en la iteraci\u00f3n se actualiza a:<\/p>\n\n\n\n y_1 = 0.01 + \\frac{0.001089+0.001101}{2} = 0.01109 <\/span>\n\n\n\n

t1<\/sub> = 0 + 0.5 =0.5<\/p>\n\n\n\n

Se actualiza la tabla con los valores obtenidos:<\/p>\n\n\n\n

i<\/th>ti<\/th>xi<\/th>yi<\/th><\/tr><\/thead>
0<\/td>0<\/td>1<\/td>0.01<\/td><\/tr>
1<\/td>0.5<\/td>1.0025<\/td>0.01109<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

t=0.5<\/p>\n\n\n\n K1x = 0.5 \\left( 0.02 (1.0025) - 0.015 (1.0025)^2 \\right)= 0.002487 <\/span>\n\n\n\n K1y = 0.5\\left( 0.02 (0.01109) - 0.015 (0.01109)^2 +0.1(0.02) (1.0025-0.01109)\\right)= 0.001101 <\/span>\n\n\n\n

.<\/p>\n\n\n\n K2x = 0.5 \\left(0.02 (1.0025+ 0.002487) - 0.015 (1.0025+ 0.002487)^2 \\right) = 0.002474 <\/span>\n\n\n\n K2y = 0.5 \\Big( 0.02 (0.01109+0.001101) - 0.015(0.01109+0.001101)^2 <\/span>\n\n\n\n + 0.1(0.02)((1.0025+ 0.002487)-(0.01109+0.001101))\\Big) = 0.001113 <\/span>\n\n\n\n

.<\/p>\n\n\n\n x_2 = 1.0025 + \\frac{0.002487+0.002474}{2} = 1.0050 <\/span>\n\n\n\n y_2 = 0.01109 + \\frac{0.001101+0.001113}{2} = 0.01220<\/span>\n\n\n\n

t2<\/sub> = 0.5 + 0.5 = 1<\/p>\n\n\n\n

i<\/th>ti<\/th>xi<\/th>yi<\/th><\/tr><\/thead>
0<\/td>0<\/td>1<\/td>0.01<\/td><\/tr>
1<\/td>0.5<\/td>1.0025<\/td>0.01109<\/td><\/tr>
2<\/td>1<\/td>1.0050<\/td>0.01220<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

t=1<\/p>\n\n\n\n K1x = 0.5 \\left( 0.02 (1.0050) - 0.015 (1.0050)^2 \\right)= 0.002474 <\/span>\n\n\n\n K1y = 0.5 \\left(0.02 (0.01220) - 0.015 (0.01220)^2 +0.1(0.02) (1.0050-0.01220) \\right)= 0.001113 <\/span>\n\n\n\n

.<\/p>\n\n\n\n K2x = 0.5 \\left(0.02 (1.0050+ 0.002474) - 0.015 (1.0050+ 0.002474)^2 \\right) = 0.002462 <\/span>\n\n\n\n K2y = 0.5 \\Big(0.02 (0.01220+0.001113) - 0.015 (0.01220+0.001113)^2 <\/span>\n\n\n\n + 0.1(0.02) ((1.0050+ 0.002474)-(0.01220+0.001113))\\Big) = 0.001126 <\/span>\n\n\n\n

.<\/p>\n\n\n\n x_3 = 1.0050 + \\frac{0.002474+0.002462}{2} = 1.0074 <\/span>\n\n\n\ny_3 = 0.01220 + \\frac{0.001113+0.001126}{2} = 0.01332<\/span>\n\n\n\n

t3<\/sub> = 1 + 0.5 = 1.5<\/p>\n\n\n\n

i<\/th>ti<\/th>xi<\/th>yi<\/th><\/tr><\/thead>
0<\/td>0<\/td>1<\/td>0.01<\/td><\/tr>
1<\/td>0.5<\/td>1.0025<\/td>0.01109<\/td><\/tr>
2<\/td>1<\/td>1.0050<\/td>0.01220<\/td><\/tr>
3<\/td>1.5<\/td>1.0074<\/td>0.01332<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Resultado con el algoritmo<\/h2>\n\n\n\n

Para obtener los datos de las iteraciones, primero se ejecuta el algoritmo para pocas iteraciones.
Para la pregunta sobre 200 a\u00f1os, se incrementa las iteraciones a 2 por a\u00f1o y las condiciones iniciales, es decir 401 muestras.<\/p>\n\n\n\n

 EDO f,g con Runge-Kutta 2 Orden\ni  [ xi,  yi,  zi ]\n   [ K1y,  K1z,  K2y,  K2z ]\n0 [0.   1.   0.01]\n  [0. 0. 0. 0.]\n1 [0.5      1.002494 0.011095]\n  [0.0025   0.001089 0.002487 0.001101]\n2 [1.       1.004975 0.012203]\n  [0.002487 0.001101 0.002475 0.001114]\n3 [1.5      1.007444 0.013323]\n  [0.002475 0.001114 0.002462 0.001126]\n4 [2.       1.0099   0.014455]\n  [0.002462 0.001126 0.00245  0.001138]\n5 [2.5      1.012343 0.0156  ]\n  [0.00245  0.001138 0.002437 0.001151]\n6 [3.       1.014774 0.016757]\n  [0.002437 0.001151 0.002424 0.001163]\n7 [3.5      1.017192 0.017926]\n  [0.002424 0.001163 0.002412 0.001176]\n8 [4.       1.019597 0.019109]\n  [0.002412 0.001176 0.002399 0.001189]\n9 [4.5      1.02199  0.020304]\n  [0.002399 0.001189 0.002386 0.001202]\n10 [5.       1.02437  0.021512]\n   [0.002386 0.001202 0.002374 0.001214]\n...<\/code><\/pre>\n\n\n\n
Observaci\u00f3n<\/strong>: La poblaci\u00f3n identificada como protestante, continua creciendo, mientras que la proporci\u00f3n de \"conformistas\" se reduce seg\u00fan los par\u00e1metros indicados en el ejercicio. Los valores de natalidad y defunci\u00f3n cambian con el tiempo mucho m\u00e1s en a\u00f1os por otras variables, por lo que se deben realizar ajustes si se pretende extender el modelo.<\/p>\n\n\n\n
\"2Eva2022PAOII_T2<\/figure>\n\n\n\n
\"2Eva2022PAOII_T2<\/figure>\n\n\n\n
Algoritmo en Python<\/p>\n\n\n
\n# Modelo predador-presa de Lotka-Volterra\n# 2Eva2022PAOII_T2 EDO - poblaci\u00f3n de protestantes en una sociedad\n# Sistemas EDO con Runge Kutta de 2do Orden\nimport numpy as np\n \n# INGRESO\n# Par\u00e1metros de las ecuaciones\nb = 0.02\nd = 0.015\nr = 0.1\n\n# Ecuaciones\nf = lambda t,x,y : (b-d*x)*x\ng = lambda t,x,y : (b-d*y)*y + r*b*(x-y)\n\n# Condiciones iniciales\nt0 = 0\nx0 = 1\ny0 = 0.01\n\n# par\u00e1metros del algoritmo\nh = 0.5\nmuestras = 401\n \n# Algoritmo como funci\u00f3n\ndef rungekutta2_fg(f,g,x0,y0,z0,h,muestras,\n                   vertabla=False, precision=6):\n    ''' solucion a EDO d2y\/dx2 con Runge-Kutta 2do Orden,\n    f(x,y,z) = z #= y'\n    g(x,y,z) = expresion d2y\/dx2 con z=y'\n    tambien es solucion a sistemas edo f() y g()\n    x0,y0,z0 son valores iniciales, h es tamano de paso,\n    muestras es la cantidad de puntos a calcular.\n    '''\n    tamano = muestras + 1\n    tabla = np.zeros(shape=(tamano,3+4),dtype=float)\n    # incluye el punto [x0,y0,z0,K1y,K1z,K2y,K2z]\n    tabla[0] = [x0,y0,z0,0,0,0,0]\n      \n    xi = x0 # valores iniciales\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * f(xi,yi,zi)\n        K1z = h * g(xi,yi,zi)\n          \n        K2y = h * f(xi+h, yi + K1y, zi + K1z)\n        K2z = h * g(xi+h, yi + K1y, zi + K1z)\n  \n        yi = yi + (K1y+K2y)\/2\n        zi = zi + (K1z+K2z)\/2\n        xi = xi + h\n          \n        tabla[i] = [xi,yi,zi,K1y,K1z,K2y,K2z]\n          \n    if vertabla==True:\n        np.set_printoptions(precision)\n        print('EDO f,g con Runge-Kutta 2 Orden')\n        print('i ','[ xi,  yi,  zi',']')\n        print('   [ K1y,  K1z,  K2y,  K2z ]')\n        for i in range(0,tamano,1):  \n            txt = ' '\n            if i>=10:\n                txt = '  '\n            print(str(i),tabla[i,0:3])\n            print(txt,tabla[i,3:])\n      \n    return(tabla)\n \n# PROCEDIMIENTO\ntabla = rungekutta2_fg(f,g,t0,x0,y0,h,\n                       muestras,vertabla=True)\n# SALIDA\nprint('Sistemas EDO RK2')\n##print('i ','[ xi,  yi,  zi',']')\n##print('   [ K1y,  K1z,  K2y,  K2z ]')\n##for i in range(0,tamano,1):  \n##    txt = ' '\n##    if i>=10:\n##        txt = '  '\n##    print(str(i),tabla[i,0:3])\n##    print(txt,tabla[i,3:])\n<\/pre><\/div>\n\n\n
Instrucciones para la parte gr\u00e1fica<\/p>\n\n\n
\n# GRAFICA ---------------------\nimport matplotlib.pyplot as plt\n  \ntitulo = 'Sistemas EDO Runge-Kutta 2ord - Protestantes'\netiq_x = 'x = tiempo'\netiq_y = 'y = poblaci\u00f3n pa\u00eds'\netiq_z = 'z = protestantes'\ni = muestras # iteraci\u00f3n en gr\u00e1fica\n  \ntitulo = titulo+', i='+str(i)\nxi = tabla[:,0]\nyi = tabla[:,1]\nzi = tabla[:,2]\nK1y = tabla[:,3]\nK1z = tabla[:,4]\nK2y = tabla[:,5]\nK2z = tabla[:,6]\n  \nplt.subplot(211)\nplt.plot(xi[0],yi[0],'o',\n         color='red', label ='[t0,y0]')\nif i<20: # evita muchos puntos\n    plt.plot(xi[1:i+2],yi[1:i+2],'o',\n             color='green', label ='[t[i],y[i]]')\nplt.plot(xi[0:i+2],yi[0:i+2],\n         color='blue',label='y(t)')\nif i<muestras: # gr\u00e1fica para una iteraci\u00f3n\n    plt.plot(xi[i+1],yi[i+1],'o',color='orange',\n             label ='[x[i+1],y[i+1]]')\n    plt.plot(xi[i:i+3],yi[i:i+3],'.',color='gray')\n    plt.plot(xi[i:i+2],[yi[i],yi[i]], color='orange',\n             label='h',linestyle='dashed')\n    plt.plot([xi[i+1]-0.02*h,xi[i+1]-0.02*h],\n             [yi[i],yi[i]+K1y[i+1]],\n             color='green',label='K1y',linestyle='dashed')\n    plt.plot([xi[i+1]+0.02*h,xi[i+1]+0.02*h],\n             [yi[i],yi[i]+K2y[i+1]],\n             color='magenta',label='K2y',linestyle='dashed')\n    plt.plot([xi[i+1]-0.02*h,xi[i+1]+0.02*h],\n             [yi[i]+K1y[i+1],yi[i]+K2y[i+1]],\n             color='magenta')\nif np.min(yi[0:i+1])<0: # linea 0\n    plt.axhline(0, color='red')\nplt.ylabel(etiq_y)\nplt.title(titulo)\nplt.legend()\nplt.grid()\nplt.tight_layout()\n  \nplt.subplot(212)\nplt.plot(xi[0],zi[0],'o',\n         color='red', label ='[t0,z0]')\nif i<20: # evita muchos puntos\n    plt.plot(xi[1:i+2],zi[1:i+2],'o',\n             color='green', label ='[t[i],v[i]]')\nplt.plot(xi[0:i+2],zi[0:i+2],\n         color='green',label='z(t)')\nif i<muestras: # gr\u00e1fica para una iteraci\u00f3n\n    plt.plot(xi[i+1],zi[i+1],'o',color='orange',\n             label ='[t[i+1],z[i+1]]')\n    plt.plot(xi[i:i+3],zi[i:i+3],'.',color='gray')\n    plt.plot(xi[i:i+2],[zi[i],zi[i]], color='orange',\n             label='h',linestyle='dashed')\n    plt.plot([xi[i+1]-0.02*h,xi[i+1]-0.02*h],\n             [zi[i],zi[i]+K1z[i+1]],\n             color='green',label='K1z',linestyle='dashed')\n    plt.plot([xi[i+1]+0.02*h,xi[i+1]+0.02*h],\n             [zi[i],zi[i]+K2z[i+1]],\n             color='magenta',label='K2z',linestyle='dashed')\n    plt.plot([xi[i+1]-0.02*h,xi[i+1]+0.02*h],\n             [zi[i]+K1z[i+1],zi[i]+K2z[i+1]],\n             color='magenta')\n  \nplt.xlabel(etiq_x)\nplt.ylabel(etiq_z)\nplt.legend()\nplt.grid()\nplt.tight_layout()\n  \n#plt.show() #comentar para la siguiente gr\u00e1fica\n \n# gr\u00e1fica yi vs zi\nfig_yz, graf3 = plt.subplots()\ngraf3.plot(yi,zi)\n  \ngraf3.set_title(titulo)\ngraf3.set_xlabel(etiq_y)\ngraf3.set_ylabel(etiq_z)\ngraf3.grid()\nplt.show() #comentar para la siguiente gr\u00e1fica\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2022PAOII_T2 EDO - poblaci\u00f3n de protestantes en una sociedad El sistema de ecuaciones para el ejercicio se expresa como: literal a Simplificando la nomenclatura, la poblaci\u00f3n total del pa\u00eds se describe con: La poblaci\u00f3n de protestantes se describe con; Sustituyendo constantes, y considerando x(0)=1 ; y(0)=0.01 ; h=0.5 valores del enunciado: El planteamiento de […]<\/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-8763","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\/8763","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=8763"}],"version-history":[{"count":9,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8763\/revisions"}],"predecessor-version":[{"id":25109,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8763\/revisions\/25109"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8763"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8763"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}