mn6d Runge Kutta – Matemáticas y Métodos Numéricos

R-K es un método de solución para ecuaciones diferenciales ordinarias. Es un procedimiento iterativo para calcular la solución de la ecuación diferencial paso a paso a partir de una condición inicial. El más común es el de 4to orden.

Método de Runge – Kutta 4to. orden

K₁ =  h f(x_i,y_i) 
K₂ =  h f[x_i + ½ h, y_i + ½ K₁]
K₃ =  h f[x_i + ½ h, y_i + ½ K₂]
K₄ =  h f[x_i + h, y_i + K₃]
K =  1/6  [K₁ + 2K₂ + 2K₃ + K₄]
y_i+1 = y_i + K
xn+1 = xn + h

Ejemplo. Resolver la ecuación diferencial y’ = 2xy condición inicial y(1) = 1

n = 0 x = 1.0 y = 1.0 h = 0.1

K₁ =  (h)(2x₀y₀) =  (0.1)(2)(1)(1)  =  0.2
K₂ =  (0.1) f[x₀ + ½ (0.1), y₀ + ½ (0.2)] = 0.1 [2(1+0.05)(1+0.1)] = 0.231
K₃ =  (0.1) f[x₀ + ½ (0.1), y₀ + ½ (0.231)] = 0.1 [2(1.05)(1.1155)] = 0.234255
K₄ =  (0.1) f[x₀ + 0.1, y₀ + 0.234255] =  0.1 [2(1.1)(1.234255)] = 0.2715361
K =    [K₁ + 2K₂ + 2K₃ + K₄] = [0.2 + 2(0.231) +2(0.234255) + 0.2715361)]
K = 0.23367435

y₁ = y₀ + K   = 1 + 0.23367435 = 1.23367435

n = 1               x₁ = 1.1          y₁ = 1.23367435

K₁ =  (h)(2x₁y₁) =  (0.1)(2)(1.1)(1.23367435)  =  0.27140836
K₂ =  0.1[2(1.15)(1.36937853)] =  0.314957061555
K₃ =  0.1[2(1.15)(1.391152880778)] =   0.3199651625788
K₄ =  (0.1)[2(1.2)(1.553639512578)] =    0.3728734830187
K =  [0.27141 + 2(0.31496) +2(0.31997) + 0.37287)]   =  0.31902
y₂ = y₁ + K   =  1.23367435 +  0.31902  = 1.55269

n = 2 x₂ = 1.2 y₂ = 1.55269

K₁ =  (h)(2x₂y₂) =  (0.1)(2)(1.2)(1.55269)  =  0.3726469
K₂ =  0.1[2(1.25)(1.73901345)] =  0.43475471
K₃ =  0.1[2(1.25)(1.77006668125)] =   0.44251819
K₄ =  (0.1)[2(1.3)(1.99520819)] =    0.51875553
y₂ = y₁ + K   =  1.55269 +  0.44099  = 1.99369

Resolviendo con hoja de cálculo

x	y	K1	K2	K3	K4	K
1.0	1	0.2	0.231	0.2342	0.271	0.233
1.1	1.233	0.271	0.314	0.3199	0.372	0.319
1.2	1.552	0.372	0.434	0.4425	0.518	0.440
1.3	1.993	0.518	0.608	0.6204	0.731	0.617
1.4	2.611	0.731	0.863	0.8825	1.048	0.878
1.5	3.490	1.047	1.244	1.2748	1.524	1.268
1.6	4.758	1.522	1.821	1.8708	2.254	1.860
1.7	6.618	2.250	2.710	2.7909	3.387	2.773
1.8	9.392	3.381	4.100	4.2337	5.177	4.204
1.9	13.596	5.166	6.310	6.5333	8.0520	6.484
2.0	20.081	8.032	9.879	10.258	12.742	10.175
2.1	30.256	12.707	15.742	16.395	20.526	16.251
2.2	46.508	20.463	25.533	26.673	33.663	26.423
2.3	72.931	33.548	42.161	44.186	56.216	43.743
2.4	116.675	56.004	70.891	74.532	95.607	73.745
2.5	190.421	95.210	121.393	128.070	165.615	126.62

El error es mínimo, por lo tanto, el método de Runge-Kutta es adecuado para resolver la ecuación diferencial.

X	Y	real	% error
1.0	1	1	0
1.1	1.23367435	1.23367806	0.00030072
1.2	1.5526954	1.55270722	0.00076128
1.3	1.99368677	1.99371553	0.00144273
1.4	2.61163323	2.61169647	0.00242142
1.5	3.49021064	3.49034296	0.00379106
1.6	4.75855167	4.75882125	0.00566476
1.7	6.61882741	6.61936868	0.00817715
1.8	9.39225233	9.39333129	0.01148646
1.9	13.5969054	13.5990509	0.01577667
2.0	20.0812668	20.0855369	0.02125956
2.1	30.2567165	30.2652443	0.02817676
2.2	46.5083522	46.5254744	0.03680182
2.3	72.9318516	72.9664685	0.04744218
2.4	116.675363	116.745926	0.06044105
2.5	190.421096	190.566268	0.0761793

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