s2Eva_2024PAOII_T2 Mayoría entre grupos Azules y Rojos

Ejercicio: 2Eva_2024PAOII_T2 Mayoría entre grupos Azules y Rojos

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La población anual del país se describe con x(t), con tasas de natalidad a = 0.018 y mortalidad b = 0.012,

f(t,x,,y)=δxδt=0.018x0.012x2 f(t,x,,y) = \frac{\delta x}{\delta t} = 0.018 x - 0.012 x^2

La población de Rojos es minoría y se describe con y(t). Los valore iniciales son: x(0)=2 , y(0)=0.5 y tamaño de paso h=0.5

g(t,x,y)=δyδt=0.026x0.017y2+0.19(0.012)(xy) g(t,x,y) = \frac{\delta y}{\delta t} = 0.026x - 0.017 y^2 +0.19 (0.012) (x-y)

el algoritmo de Runge-Kutta para sistemas de ecuaciones aplicado al ejercicio:

K1x=hf(t,x,y)=h(0.018x0.012x2) K1x = h f(t,x,y) = h (0.018 x - 0.012 x^2) K1y=hg(t,x,y)=h(0.026x0.017y2+0.19(0.012)(xy)) K1y = h g(t,x,y) = h \Big(0.026x - 0.017 y^2 +0.19 (0.012) (x-y)\Big) K2x=hf(t+h,x+K1x,y+K1y) K2x = h f(t+h,x+K1x,y+K1y) =h(0.018(x+K1x)0.012(x+K1x)2) = h (0.018 (x+K1x) - 0.012 (x+K1x)^2) K2y=hg(t+h,x+K1x,y+K1y) K2y = h g(t+h,x+K1x,y+K1y) =h(0.026(x+K1x)0.017(y+K1y)2 = h \Big(0.026(x+K1x) - 0.017 (y + K1y)^2 +0.19(0.012)((x+K1x)(y+K1y)))+0.19 (0.012) ((x+K1x)-(y + K1y))\Big) x[i+1]=x[i]+K1x+K2x2 x[i+1] = x[i] + \frac{K1x+K2x}{2} y[i+1]=y[i]+K1y+K2y2 y[i+1] = y[i] + \frac{K1y+K2y}{2} t[i+1]=t[i]+h t[i+1] = t[i] + h

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Desarrolle tres iteraciones con expresiones completas para x(t), y(t) con tamaño de paso h=0.5.

itera = 0

K1x=0.5(0.018(2)0.012(2)2)=0.006 K1x = 0.5 (0.018 (2) - 0.012 (2)^2) = -0.006 K1y=0.5(0.026(2)0.017(0.5)2+0.19(0.012)((2)(0.5))) K1y = 0.5 \Big(0.026(2) - 0.017 (0.5)^2 +0.19 (0.012) ((2)-(0.5))\Big) =0.006085= 0.006085 K2x=0.5(0.018(20.006)0.012(20.006)2)=0.00591 K2x = 0.5 (0.018 (2-0.006) - 0.012 (2-0.006)^2) = -0.00591 K2y=0.5(0.026(20.006)0.017(0.5+0.006085)2 K2y = 0.5 \Big(0.026(2-0.006) - 0.017 (0.5 + 0.006085)^2 +0.19(0.012)((20.006)(0.5+0.006085)))=0.006098+0.19 (0.012) ((2-0.006)-(0.5 + 0.006085))\Big) = 0.006098 x[1]=2+0.006+0.005912=1.9940 x[1] = 2 + \frac{-0.006+-0.00591}{2} = 1.9940 y[1]=0.5+0.006085+0.0060982=0.5060 y[1] = 0.5 + \frac{0.006085+0.006098}{2} = 0.5060 t[1]=0+0.5=0.5 t[1] = 0 + 0.5 =0.5

itera = 1

K1x=0.5(0.018(1.9940)0.012(1.9940)2) K1x = 0.5 (0.018 (1.9940) - 0.012 (1.9940)^2) =0.005911=-0.005911 K1y=0.5(0.0261.99400.017(0.5060)2+0.19(0.012)(1.9940y)) K1y = 0.5 \Big(0.0261.9940 - 0.017 (0.5060)^2 +0.19 (0.012) (1.9940-y)\Big) =0.006098= 0.006098 K2x=0.5(0.018(1.99400.005911)0.012(1.99400.005911)2 K2x = 0.5 (0.018 (1.9940-0.005911) - 0.012 (1.9940-0.005911)^2 =0.005823= -0.005823 K2y=0.5(0.026(1.99400.005911)0.017(0.5060+0.006098)2 K2y = 0.5 \Big(0.026(1.9940-0.005911) - 0.017 (0.5060 + 0.006098)^2 +0.19(0.012)((1.99400.005911)(0.5060+0.006098))) +0.19 (0.012) ((1.9940-0.005911)-(0.5060 + 0.006098))\Big) =0.006111 = 0.006111 x[2]=1.9940+0.0059110.0058232=1.988178 x[2] = 1.9940 + \frac{-0.005911-0.005823}{2} = 1.988178 y[2]=0.5060+0.006098+0.0061112=0.512196 y[2] = 0.5060 + \frac{0.006098+0.006111}{2} = 0.512196 t[2]=0.5+0.5=1 t[2] = 0.5 + 0.5 = 1

itera = 2

K1x=0.5(0.018(1.988178)0.012(1.988178)2)=0.005824 K1x = 0.5 (0.018 (1.988178) - 0.012 (1.988178)^2) =-0.005824 K1y=0.5(0.026(0.512196)0.017(0.512196)2 K1y = 0.5 \Big(0.026(0.512196) - 0.017 (0.512196)^2 +0.19(0.012)(1.9881780.512196))=0.006111+0.19 (0.012) (1.988178-0.512196)\Big) = 0.006111 K2x=0.5(0.018(1.9881780.005824)0.012(1.9881780.005824)2) K2x = 0.5 (0.018 (1.988178-0.005824) - 0.012 (1.988178-0.005824)^2) =0.005737=-0.005737 K2y=0.5(0.026(0.512196+0.006111)0.017(0.512196+0.006111)2 K2y = 0.5 \Big(0.026(0.512196+0.006111) - 0.017 (0.512196 + 0.006111 )^2 +0.19(0.012)((1.9881780.005911)(0.512196+0.006111)) +0.19 (0.012) ((1.988178-0.005911)-(0.512196 + 0.006111)\Big) =0.006124=0.006124 x[3]=(1.988178)+0.0058240.0057372=1.982398 x[3] = (1.988178) + \frac{-0.005824-0.005737}{2} = 1.982398 y[3]=0.512196+0.006111+0.0061242=0.518314 y[3] = 0.512196 + \frac{0.006111+0.006124}{2} = 0.518314 t[3]=1+0.5=1.5 t[3] = 1 + 0.5 = 1.5

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Los resultados para x(t) de las primeras iteraciones indican que la población total del país disminuye en el intervalo observado. Se puede comprobar al usar el algoritmo para mas iteraciones como se muestra en la gráfica.

Mayoría Azules Rojos

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Los resultados para y(t) muestran que la población clasificada como Rojo aumenta en el intervalo observado. Sin embargo la mayoría sigue siendo Azul para las tres iteraciones realizadas.

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La población y(t) alcanza la mitad de la población cuanto t cambia de 30.5 a 31. Tiempo en el que de realizar elecciones, ganarían los Rojos.

Usando el algoritmo, se añade una columna con la condición que MayoriaY = yi>xi/2, que se convierte a 1 o verdadero cuando se cumple la condición. Con lo que se encuentra el cambio de mayoría a Rojo.

Runge-Kutta Segundo Orden
i  [ ti,  xi,  yi ]
   [ K1x,  K1y,  K2x,  K2y , MayoriaY]
0 [0.  2.  0.5]
   [0. 0. 0. 0. 0.]
1 [0.5      1.994045 0.506092]
  [-0.006     0.006085 -0.00591   0.006098  0.      ]
2 [1.       1.988178 0.512196]
  [-0.005911  0.006098 -0.005823  0.006111  0.      ]
3 [1.5      1.982398 0.518314]
  [-0.005824  0.006111 -0.005737  0.006124  0.      ]
4 [2.       1.976702 0.524443]
…
   [-0.002761  0.005927 -0.002727  0.005907  0.      ]
60 [30.        1.755801  0.869617]
   [-0.002728  0.005907 -0.002695  0.005887  0.      ]
61 [30.5       1.753123  0.875494]
   [-0.002695  0.005887 -0.002662  0.005867  0.      ]
62 [31.        1.750476  0.88135 ]
   [-0.002663  0.005867 -0.002631  0.005846  1.      ]
63 [31.5       1.747861  0.887185]
   [-0.002631  0.005846 -0.002599  0.005824  1.      ]
64 [32.        1.745277  0.892998]
   [-0.002599  0.005824 -0.002568  0.005802  1.      ]
65 [32.5       1.742724  0.898789]
   [-0.002568  0.005802 -0.002538  0.00578   1.      ]
66 [33.        1.740201  0.904558]
   [-0.002538  0.00578  -0.002508  0.005757  1.      ]
67 [33.5       1.737708  0.910303]
   [-0.002508  0.005757 -0.002478  0.005734  1.      ]