Posted By

Bangonkali on 03/30/12

Tagged

Versions (?)

Last Edited at 03/30/12 04:04am

Statistics

Viewed 468 times

Favorited by 0 user(s)

Related snippets

Runge-Kutta (2nd Order): Heunâ€™s method

/ Published in: C++

This example solves the following ordinary differential equation: y' = ((4 * exp(0.8*x)) - (0.5*y)); graph here: http://j.mp/H3K42D using the Heunâ€™s method. Heunâ€™s method is a second order Runge-Kutta Numerical Method for solving ordinary differential equations. More indepth discussion on the Book Numerical Methods 6th Edition - Chapra Page 722.

Highlighted TextExpand

/* 
 * File:   main.cpp
 * Author: Bangonkali (bangonkali@gmail.com)
 *
 * Heunâ€™s Method
 * More indepth discussion on the Book
 * Numerical Methods 6th Edition - Chapra
 * Page 722
 * 
 * This example solves the following ordinary differential equation:
 * y' = ((4 * exp(0.8*x)) - (0.5*y)); graph here: http://j.mp/H3K42D
 * using the Heunâ€™s method. Heunâ€™s method is a second order Runge-Kutta
 * Numerical Method for solving ordinary differential equations.
 * 
 * Test Run:
 * at f(4.000000000000000083266726846886740531772375106811523437500000):
 * estd y: 75.338965077860455168967668271307047689333558082580566406250000
 * true y: 75.338962609158578537238426520161738153547048568725585937500000
 * true e: 0.000003284801148791693124426737734205834079448393936218053568
 * true_max_error: 0.000003284801148791693124426737734205834079448393936218053568
 * 
 * Created on March 29, 2012, 10:03 PM
 */

#include <cstdlib>
#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;

// the formula of y-prime
long double f(long double x, long double y) {
    return ((4 * exp(0.8*x)) - (0.5*y));
}

// the correct analytical solution
long double g(long double x) {
    return (4/1.3)*(exp(0.8*x)-exp(-0.5*x))+2*exp(-0.5*x);
}

// predictor
long double hf0(long double x1, long double y1, long double h) {
    return y1 + (f(x1, y1) * h);
}

// corrector
long double hf0_corrector(long double x1, long double y1, long double x2, long double y2, long double h) {
    return y1 + (((f(x1, y1) + f(x2, y2)) / 2)*h);
}

// get error
long double get_error(long double a, long double b) {
    return abs(((a-b)*100/a));
}

int main(int argc, char** argv) {
    cout << fixed;
    cout << setprecision (60);
    
    long double x0 = 0; // initial condition
    long double y0 = 2; // initial condition
    long double step = 1e-3;  // stepping
    long double x_max = 4;  // evaluate integral from x0 to x_max
    
    long double predictor = 0; 
    long double target_error = 2e-5; // using the error approximation 
    long double error = 100; // error place holder
    long double corrector_new = 0;
    unsigned long int i = 0;
    long double y_true; // true value of y based on g(x) above
    long double e_true = 100; // previous true error
    long double e_true_now = 100; // current true error
    long double corrector; // corrector place holder
    
    long double true_max_error = 0;
    
    for (x0 = 0; x0<=x_max; x0=x0+step) {
        target_error = 2e-5;
        error = 100;
        e_true = 100;
        e_true_now = 101;
        corrector_new = 0;
        i = 0;
        
        y_true = g(x0 + step);
        // cout << "true y: " << y_true << endl;
        
        predictor = hf0(x0, y0, step);
        // cout << "predictor = " << predictor << endl;
        
        corrector = hf0_corrector(x0, y0, (x0 + step), predictor, step);
        // cout << "corrector = " << corrector << endl;
        
        while (e_true > 1) {
            corrector_new = hf0_corrector(x0, y0, (x0 + step), corrector, step);
            error = get_error(corrector, corrector_new);
            e_true_now = get_error(corrector, y_true);
            corrector = corrector_new;
            i++;
            
            // cout << "true e: " << e_true << endl;
            if (e_true_now == e_true) break;
            
            e_true = e_true_now;
        }
        
//        cout << "iterations: " << i << endl;
//        cout << "estd y: " << corrector << endl;
//        cout << "true e: " << e_true << endl;
//        cout << endl;
        
        if (true_max_error < e_true) {
            true_max_error = e_true;
        }
        y0 = corrector;
    }
    
    cout << "at f(" << x0 << "): " << endl;
    cout << "estd y: " << corrector << endl;
    cout << "true y: " << y_true << endl;
    cout << "true e: " << e_true << endl;
    cout << "true_max_error: " << true_max_error << endl;
    
    return 0;
}

Report this snippet Tweet

Comments

Subscribe to comments

Comment:

You need to login to post a comment.