Runge-Kutta method: Difference between revisions
Go solution
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Line 136:
y(9.0) = 451.56245928 Error: 4.07232e-05
y(10.0) = 675.99994902 Error: 5.09833e-05</pre>
=={{header|Go}}==
{{works with|Go1}}
<lang go>package main
import (
"fmt"
"math"
)
type ypFunc func(t, y float64) float64
type ypStepFunc func(t, y, dt float64) float64
// newRKStep takes a function representing a differential equation
// and returns a function that performs a single step of the forth-order
// Runge-Kutta method.
func newRK4Step(yp ypFunc) ypStepFunc {
return func(t, y, dt float64) float64 {
dy1 := dt * yp(t, y)
dy2 := dt * yp(t+dt/2, y+dy1/2)
dy3 := dt * yp(t+dt/2, y+dy2/2)
dy4 := dt * yp(t+dt, y+dy3)
return y + (dy1+2*(dy2+dy3)+dy4)/6
}
}
// example differential equation
func yprime(t, y float64) float64 {
return t * math.Sqrt(y)
}
// exact solution of example
func actual(t float64) float64 {
t = t*t + 4
return t * t / 16
}
func main() {
t0, tFinal := 0, 10 // task specifies times as integers,
dtPrint := 1 // and to print at whole numbers.
y0 := 1. // initial y.
dtStep := .1 // step value.
t, y := float64(t0), y0
ypStep := newRK4Step(yprime)
for t1 := t0 + dtPrint; t1 <= tFinal; t1 += dtPrint {
printErr(t, y) // print intermediate result
for steps := int(float64(dtPrint)/dtStep + .5); steps > 1; steps-- {
y = ypStep(t, y, dtStep)
t += dtStep
}
y = ypStep(t, y, float64(t1)-t) // adjust step to integer time
t = float64(t1)
}
printErr(t, y) // print final result
}
func printErr(t, y float64) {
fmt.Printf("y(%.1f) = %f Error: %e\n", t, y, math.Abs(actual(t)-y))
}</lang>
{{out}}
<pre>
y(0.0) = 1.000000 Error: 0.000000e+00
y(1.0) = 1.562500 Error: 1.457219e-07
y(2.0) = 3.999999 Error: 9.194792e-07
y(3.0) = 10.562497 Error: 2.909562e-06
y(4.0) = 24.999994 Error: 6.234909e-06
y(5.0) = 52.562489 Error: 1.081970e-05
y(6.0) = 99.999983 Error: 1.659460e-05
y(7.0) = 175.562476 Error: 2.351773e-05
y(8.0) = 288.999968 Error: 3.156520e-05
y(9.0) = 451.562459 Error: 4.072316e-05
y(10.0) = 675.999949 Error: 5.098329e-05
</pre>
=={{header|Tcl}}==
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