C program to solve equation in runge kutta method?

Answer 1

PROGRAM :-

/* Runge Kutta for a set of first order differential equations */

#include

#include

#define N 2 /* number of first order equations */

#define dist 0.1 /* stepsize in t*/

#define MAX 30.0 /* max for t */

FILE *output; /* internal filename */

void runge4(double x, double y[], double step); /* Runge-Kutta function */

double f(double x, double y[], int i); /* function for derivatives */

void main()

{

double t, y[N];

int j;

output=fopen("osc.dat", "w"); /* external filename */

y[0]=1.0; /* initial position */

y[1]=0.0; /* initial velocity */

fprintf(output, "0\t%f\n", y[0]);

for (j=1; j*dist<=MAX ;j++) /* time loop */

{

t=j*dist;

runge4(t, y, dist);

fprintf(output, "%f\t%f\n", t, y[0]);

}

fclose(output);

}

void runge4(double x, double y[], double step)

{

double h=step/2.0, /* the midpoint */

t1[N], t2[N], t3[N], /* temporary storage arrays */

k1[N], k2[N], k3[N],k4[N]; /* for Runge-Kutta */

int i;

for (i=0;i

{

t1[i]=y[i]+0.5*(k1[i]=step*f(x,y,i));

}

for (i=0;i

{

t2[i]=y[i]+0.5*(k2[i]=step*f(x+h, t1, i));

}

for (i=0;i

{

t3[i]=y[i]+ (k3[i]=step*f(x+h, t2, i));

}

for (i=0;i

{

k4[i]= step*f(x+step, t3, i);

}

for (i=0;i

{

y[i]+=(k1[i]+2*k2[i]+2*k3[i]+k4[i])/6.0;

}

}

double f(double x, double y[], int i)

{

if (i==0)

x=y[1]; /* derivative of first equation */

if (i==1)

x= -0.2*y[1]-y[0]; /* derivative of second equation */

return x;

}

Answer 2

/* Runge Kutta for a set of first order differential equations */

#include <stdio.h>

#include <math.h>

#define N 2 /* number of first order equations */

#define dist 0.1 /* stepsize in t*/

#define MAX 30.0 /* max for t */

FILE *output; /* internal filename */

main()

{

double t, y[N];

int j;

void runge4(double x, double y[], double step); /* Runge-Kutta function */

double f(double x, double y[], int i); /* function for derivatives */

output=fopen("osc.dat", "w"); /* external filename */

y[0]=1.0; /* initial position */

y[1]=0.0; /* initial velocity */

fprintf(output, "0\t%f\n", y[0]);

for (j=1; j*dist<=MAX ;j++) /* time loop */

{

t=j*dist;

runge4(t, y, dist);

fprintf(output, "%f\t%f\n", t, y[0]);

}

fclose(output);

}

void runge4(double x, double y[], double step)

{

double h=step/2.0, /* the midpoint */

t1[N], t2[N], t3[N], /* temporary storage arrays */

k1[N], k2[N], k3[N],k4[N]; /* for Runge-Kutta */

int i;

for (i=0;i<N;i++) t1[i]=y[i]+0.5*(k1[i]=step*f(x, y, i));

for (i=0;i<N;i++) t2[i]=y[i]+0.5*(k2[i]=step*f(x+h, t1, i));

for (i=0;i<N;i++) t3[i]=y[i]+ (k3[i]=step*f(x+h, t2, i));

for (i=0;i<N;i++) k4[i]= step*f(x+step, t3, i);

for (i=0;i<N;i++) y[i]+=(k1[i]+2*k2[i]+2*k3[i]+k4[i])/6.0;

}

double f(double x, double y[], int i)

{

if (i==0) return(y[1]); /* derivative of first equation */

if (i==1) return(-0.2*y[1]-y[0]); /* derivative of second equation */

}

Answer 3

public class RungeKutta { public static final int STEPS = 100; public static double deriv(double x, double y) { return x * Math.sqrt(1 + y*y); } public static void main(String[] args) { // `h' is the size of each step. double h = 1.0 / STEPS; double k1, k2, k3, k4; double x, y; int i; y = 0; for (i=0; i<STEPS; i++) { x = i * h; y += h * deriv(x, y); } System.out.println("Using the Euler method " + "The value at x=1 is:"); System.out.println(y); y = 0; for (i=0; i<STEPS; i++) { x = i * h; k1 = h * deriv(x, y); k2 = h * deriv(x + h/2, y + k1/2); k3 = h * deriv(x + h/2, y + k2/2); k4 = h * deriv(x + h, y + k3); y += k1/6 + k2/3+ k3/3 + k4/6; } System.out.println(); System.out.println("Using 4th order Runge-Kutta " + "The value at x=1 is:"); System.out.println(y); System.out.println(); System.out.println("The value really is:"); y = (Math.exp(0.5) - Math.exp(-0.5)) / 2; System.out.println(y); } }

Answer 4

using c language to implement Runge-Kutta method in numerical analysis