How to resolve the algorithm Runge-Kutta method step by step in the Nim programming language

Published on 12 May 2024 09:40 PM
#Nim

How to resolve the algorithm Runge-Kutta method step by step in the Nim programming language

Table of Contents

Problem Statement

Given the example Differential equation: With initial condition: This equation has an exact solution:

Demonstrate the commonly used explicit   fourth-order Runge–Kutta method   to solve the above differential equation.

Starting with a given

y

n

{\displaystyle y_{n}}

and

t

n

{\displaystyle t_{n}}

calculate: then:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Runge-Kutta method step by step in the Nim programming language

Source code in the nim programming language

import math

proc fn(t, y: float): float =
    result = t * math.sqrt(y)

proc solution(t: float): float =
    result = (t^2 + 4)^2 / 16

proc rk(start, stop, step: float) =
    let nsteps = int(round((stop - start) / step)) + 1
    let delta = (stop - start) / float(nsteps - 1)
    var cur_y = 1.0
    for i in 0..(nsteps - 1):
        let cur_t = start + delta * float(i)

        if abs(cur_t - math.round(cur_t)) < 1e-5:
            echo "y(", cur_t, ") = ", cur_y, ", error = ", solution(cur_t) - cur_y
        
        let dy1 = step * fn(cur_t, cur_y)
        let dy2 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy1)
        let dy3 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy2)
        let dy4 = step * fn(cur_t + step, cur_y + dy3)
import math, strformat

proc fn(t, y: float): float =
    result = t * math.sqrt(y)

proc solution(t: float): float =
    result = (t^2 + 4)^2 / 16

proc rk(start, stop, step: float) =
    let nsteps = int(round((stop - start) / step)) + 1
    let delta = (stop - start) / float(nsteps - 1)
    var cur_y = 1.0
    for i in 0..<nsteps:
        let cur_t = start + delta * float(i)

        if abs(cur_t - math.round(cur_t)) < 1e-5:
            echo &"y({cur_t}) = {cur_y}, error = {solution(cur_t) - cur_y}"

        let dy1 = step * fn(cur_t, cur_y)
        let dy2 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy1)
        let dy3 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy2)
        let dy4 = step * fn(cur_t + step, cur_y + dy3)

        cur_y += (dy1 + 2 * (dy2 + dy3) + dy4) / 6

rk(start = 0, stop = 10, step = 0.1)
        cur_y += (dy1 + 2.0 * (dy2 + dy3) + dy4)

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