Formula
Fourth order Runge-Kutta (RK4) method for second order differential equation formula
`k_1=f(x_n,y_n,z_n)`
`l_1=g(x_n,y_n,z_n)`
`k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`k_3=f(x_n+h/2,y_n+(hk_2)/2,z_n+(hl_2)/2)`
`l_3=g(x_n+h/2,y_n+(hk_2)/2,z_n+(hl_2)/2)`
`k_4=f(x_n+h,y_n+hk_3,z_n+hl_3)`
`l_4=g(x_n+h,y_n+hk_3,z_n+hl_3)`
`y_(n+1)=y_n+h/6(k_1+2k_2+2k_3+k_4)`
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Examples
Find y(0.2) for `y''=1+2xy-x^2z`, `x_0=0, y_0=1, z_0=0`, with step length 0.1 using Runge-Kutta 4 method (second order differential equation) Solution:Given `y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, h=0.1, y(0.2)=?`
put `(dy)/(dx)=z` and differentiate w.r.t. x, we obtain `(d^2y)/(dx^2)=(dz)/(dx)`
We have system of equations
`(dy)/(dx)=z=f(x,y,z)`
`(dz)/(dx)=1+2xy-x^2z=g(x,y,z)`
Fourth order Runge-Kutta (RK4) method for second order differential equation formula
`k_1=f(x_n,y_n,z_n)`
`l_1=g(x_n,y_n,z_n)`
`k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`k_3=f(x_n+h/2,y_n+(hk_2)/2,z_n+(hl_2)/2)`
`l_3=g(x_n+h/2,y_n+(hk_2)/2,z_n+(hl_2)/2)`
`k_4=f(x_n+h,y_n+hk_3,z_n+hl_3)`
`l_4=g(x_n+h,y_n+hk_3,z_n+hl_3)`
`y_(n+1)=y_n+h/6(k_1+2k_2+2k_3+k_4)`
`z_(n+1)=z_n+h/6(l_1+2l_2+2l_3+l_4)`
for `n=0,x_0=0,y_0=1,z_0=0`
`k_1=f(x_0,y_0,z_0)`
`=f(0,1,0)`
`=0`
`l_1=g(x_0,y_0,z_0)`
`=g(0,1,0)`
`=1`
`k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)`
`=f(0.05,1,0.05)`
`=0.05`
`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)`
`=g(0.05,1,0.05)`
`=1.0999`
`k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)`
`=f(0.05,1.0025,0.055)`
`=0.055`
`l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)`
`=g(0.05,1.0025,0.055)`
`=1.1001`
`k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)`
`=f(0.1,1.0055,0.11)`
`=0.11`
`l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)`
`=g(0.1,1.0055,0.11)`
`=1.2`
Now,
`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`
`=1+0.1/6[0+2(0.05)+2(0.055)+(0.11)]`
`=1.0053`
`z_1=z_0+h/6(l_1+2l_2+2l_3+l_4)`
`=0+0.1/6[1+2(1.0999)+2(1.1001)+(1.2)]`
`=0.11`
`x_1=x_0+h=0+0.1=0.1`
for `n=1,x_1=0.1,y_1=1.0053,z_1=0.11`
`k_1=f(x_1,y_1,z_1)`
`=f(0.1,1.0053,0.11)`
`=0.11`
`l_1=g(x_1,y_1,z_1)`
`=g(0.1,1.0053,0.11)`
`=1.2`
`k_2=f(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`
`=f(0.15,1.0108,0.17)`
`=0.17`
`l_2=g(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`
`=g(0.15,1.0108,0.17)`
`=1.2994`
`k_3=f(x_1+h/2,y_1+(hk_2)/2,z_1+(hl_2)/2)`
`=f(0.15,1.0138,0.175)`
`=0.175`
`l_3=g(x_1+h/2,y_1+(hk_2)/2,z_1+(hl_2)/2)`
`=g(0.15,1.0138,0.175)`
`=1.3002`
`k_4=f(x_1+h,y_1+hk_3,z_1+hl_3)`
`=f(0.2,1.0228,0.24)`
`=0.24`
`l_4=g(x_1+h,y_1+hk_3,z_1+hl_3)`
`=g(0.2,1.0228,0.24)`
`=1.3995`
Now,
`y_2=y_1+h/6(k_1+2k_2+2k_3+k_4)`
`=1.0053+0.1/6[0.11+2(0.17)+2(0.175)+(0.24)]`
`=1.0227`
`x_2=x_1+h=0.1+0.1=0.2`
`:.y(0.2)=1.0227`
| `n` | `x_n` | `y_n` | `z_n` | `k_1` | `l_1` | `k_2` | `l_2` | `k_3` | `l_3` | `k_4` | `l_4` | `x_(n+1)` | `y_(n+1)` | `z_(n+1)` |
| 0 | 0 | 1 | 0 | 0 | 1 | 0.05 | 1.0999 | 0.055 | 1.1001 | 0.11 | 1.2 | 0.1 | 1.0053 | 0.11 |
| 1 | 0.1 | 1.0053 | 0.11 | 0.11 | 1.2 | 0.17 | 1.2994 | 0.175 | 1.3002 | 0.24 | 1.3995 | 0.2 | 1.0227 | |
This material is intended as a summary. Use your textbook for detail explanation.
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